Overview video. See also lecture on Thursday October 22 and examples from week 42.
The Universal Approximation Theorem states that a neural network can approximate any function at a single hidden layer along with one input and output layer to any given precision.
An ordinary differential equation (ODE) is an equation involving functions having one variable.
In general, an ordinary differential equation looks like $$ \begin{equation} \label{ode} f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) = 0 \end{equation} $$
where \( g(x) \) is the function to find, and \( g^{(n)}(x) \) is the \( n \)-th derivative of \( g(x) \).
The \( f\left(x, g(x), g'(x), g''(x), \, \dots \, , g^{(n)}(x)\right) \) is just a way to write that there is an expression involving \( x \) and \( g(x), \ g'(x), \ g''(x), \, \dots \, , \text{ and } g^{(n)}(x) \) on the left side of the equality sign in \eqref{ode}. The highest order of derivative, that is the value of \( n \), determines to the order of the equation. The equation is referred to as a \( n \)-th order ODE. Along with \eqref{ode}, some additional conditions of the function \( g(x) \) are typically given for the solution to be unique.
Let the trial solution \( g_t(x) \) be $$ \begin{equation} g_t(x) = h_1(x) + h_2(x,N(x,P)) \label{_auto1} \end{equation} $$
where \( h_1(x) \) is a function that makes \( g_t(x) \) satisfy a given set of conditions, \( N(x,P) \) a neural network with weights and biases described by \( P \) and \( h_2(x, N(x,P)) \) some expression involving the neural network. The role of the function \( h_2(x, N(x,P)) \), is to ensure that the output from \( N(x,P) \) is zero when \( g_t(x) \) is evaluated at the values of \( x \) where the given conditions must be satisfied. The function \( h_1(x) \) should alone make \( g_t(x) \) satisfy the conditions.
But what about the network \( N(x,P) \)?
As described previously, an optimization method could be used to minimize the parameters of a neural network, that being its weights and biases, through backward propagation.
For the minimization to be defined, we need to have a cost function at hand to minimize.
It is given that \( f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right) \) should be equal to zero in \eqref{ode}. We can choose to consider the mean squared error as the cost function for an input \( x \). Since we are looking at one input, the cost function is just \( f \) squared. The cost function \( c\left(x, P \right) \) can therefore be expressed as $$ C\left(x, P\right) = \big(f\left(x, \, g(x), \, g'(x), \, g''(x), \, \dots \, , \, g^{(n)}(x)\right)\big)^2 $$
If \( N \) inputs are given as a vector \( \boldsymbol{x} \) with elements \( x_i \) for \( i = 1,\dots,N \), the cost function becomes $$ \begin{equation} \label{cost} C\left(\boldsymbol{x}, P\right) = \frac{1}{N} \sum_{i=1}^N \big(f\left(x_i, \, g(x_i), \, g'(x_i), \, g''(x_i), \, \dots \, , \, g^{(n)}(x_i)\right)\big)^2 \end{equation} $$
The neural net should then find the parameters \( P \) that minimizes the cost function in \eqref{cost} for a set of \( N \) training samples \( x_i \).
To perform the minimization using gradient descent, the gradient of \( C\left(\boldsymbol{x}, P\right) \) is needed. It might happen so that finding an analytical expression of the gradient of \( C(\boldsymbol{x}, P) \) from \eqref{cost} gets too messy, depending on which cost function one desires to use.
Luckily, there exists libraries that makes the job for us through automatic differentiation. Automatic differentiation is a method of finding the derivatives numerically with very high precision.
An exponential decay of a quantity \( g(x) \) is described by the equation $$ \begin{equation} \label{solve_expdec} g'(x) = -\gamma g(x) \end{equation} $$
with \( g(0) = g_0 \) for some chosen initial value \( g_0 \).
The analytical solution of \eqref{solve_expdec} is $$ \begin{equation} g(x) = g_0 \exp\left(-\gamma x\right) \label{_auto2} \end{equation} $$
Having an analytical solution at hand, it is possible to use it to compare how well a neural network finds a solution of \eqref{solve_expdec}.
The program will use a neural network to solve $$ \begin{equation} \label{solveode} g'(x) = -\gamma g(x) \end{equation} $$
where \( g(0) = g_0 \) with \( \gamma \) and \( g_0 \) being some chosen values.
In this example, \( \gamma = 2 \) and \( g_0 = 10 \).
with \( h_1(x) \) ensuring that \( g_t(x) \) satisfies some conditions and \( h_2(x,N(x, P)) \) an expression involving \( x \) and the output from the neural network \( N(x,P) \) with \( P \) being the collection of the weights and biases for each layer. For now, it is assumed that the network consists of one input layer, one hidden layer, and one output layer.
In this network, there are no weights and bias at the input layer, so \( P = \{ P_{\text{hidden}}, P_{\text{output}} \} \). If there are \( N_{\text{hidden} } \) neurons in the hidden layer, then \( P_{\text{hidden}} \) is a \( N_{\text{hidden} } \times (1 + N_{\text{input}}) \) matrix, given that there are \( N_{\text{input}} \) neurons in the input layer.
The first column in \( P_{\text{hidden} } \) represents the bias for each neuron in the hidden layer and the second column represents the weights for each neuron in the hidden layer from the input layer. If there are \( N_{\text{output} } \) neurons in the output layer, then \( P_{\text{output}} \) is a \( N_{\text{output} } \times (1 + N_{\text{hidden} }) \) matrix.
Its first column represents the bias of each neuron and the remaining columns represents the weights to each neuron.
It is given that \( g(0) = g_0 \). The trial solution must fulfill this condition to be a proper solution of \eqref{solveode}. A possible way to ensure that \( g_t(0, P) = g_0 \), is to let \( F(N(x,P)) = x \cdot N(x,P) \) and \( A(x) = g_0 \). This gives the following trial solution: $$ \begin{equation} \label{trial} g_t(x, P) = g_0 + x \cdot N(x, P) \end{equation} $$
We wish that our neural network manages to minimize a given cost function.
A reformulation of out equation, \eqref{solveode}, must therefore be done, such that it describes the problem a neural network can solve for.
The neural network must find the set of weights and biases \( P \) such that the trial solution in \eqref{trial} satisfies \eqref{solveode}.
The trial solution $$ g_t(x, P) = g_0 + x \cdot N(x, P) $$
has been chosen such that it already solves the condition \( g(0) = g_0 \). What remains, is to find \( P \) such that $$ \begin{equation} \label{nnmin} g_t'(x, P) = - \gamma g_t(x, P) \end{equation} $$
is fulfilled as best as possible.
The left hand side and right hand side of \eqref{nnmin} must be computed separately, and then the neural network must choose weights and biases, contained in \( P \), such that the sides are equal as best as possible. This means that the absolute or squared difference between the sides must be as close to zero, ideally equal to zero. In this case, the difference squared shows to be an appropriate measurement of how erroneous the trial solution is with respect to \( P \) of the neural network.
This gives the following cost function our neural network must solve for: $$ \min_{P}\Big\{ \big(g_t'(x, P) - ( -\gamma g_t(x, P) \big)^2 \Big\} $$
(the notation \( \min_{P}\{ f(x, P) \} \) means that we desire to find \( P \) that yields the minimum of \( f(x, P) \))
or, in terms of weights and biases for the hidden and output layer in our network: $$ \min_{P_{\text{hidden} }, \ P_{\text{output} }}\Big\{ \big(g_t'(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) - ( -\gamma g_t(x, \{ P_{\text{hidden} }, P_{\text{output} }\}) \big)^2 \Big\} $$
for an input value \( x \).
If the neural network evaluates \( g_t(x, P) \) at more values for \( x \), say \( N \) values \( x_i \) for \( i = 1, \dots, N \), then the total error to minimize becomes $$ \begin{equation} \label{min} \min_{P}\Big\{\frac{1}{N} \sum_{i=1}^N \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \Big\} \end{equation} $$
Letting \( \boldsymbol{x} \) be a vector with elements \( x_i \) and \( C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 \) denote the cost function, the minimization problem that our network must solve, becomes $$ \min_{P} C(\boldsymbol{x}, P) $$
In terms of \( P_{\text{hidden} } \) and \( P_{\text{output} } \), this could also be expressed as $$ \min_{P_{\text{hidden} }, \ P_{\text{output} }} C(\boldsymbol{x}, \{P_{\text{hidden} }, P_{\text{output} }\}) $$
For simplicity, it is assumed that the input is an array \( \boldsymbol{x} = (x_1, \dots, x_N) \) with \( N \) elements. It is at these points the neural network should find \( P \) such that it fulfills \eqref{min}.
First, the neural network must feed forward the inputs. This means that \( \boldsymbol{x}s \) must be passed through an input layer, a hidden layer and a output layer. The input layer in this case, does not need to process the data any further. The input layer will consist of \( N_{\text{input} } \) neurons, passing its element to each neuron in the hidden layer. The number of neurons in the hidden layer will be \( N_{\text{hidden} } \).
For the \( i \)-th in the hidden layer with weight \( w_i^{\text{hidden} } \) and bias \( b_i^{\text{hidden} } \), the weighting from the \( j \)-th neuron at the input layer is: $$ \begin{aligned} z_{i,j}^{\text{hidden}} &= b_i^{\text{hidden}} + w_i^{\text{hidden}}x_j \\ &= \begin{pmatrix} b_i^{\text{hidden}} & w_i^{\text{hidden}} \end{pmatrix} \begin{pmatrix} 1 \\ x_j \end{pmatrix} \end{aligned} $$
The result after weighting the inputs at the \( i \)-th hidden neuron can be written as a vector: $$ \begin{aligned} \boldsymbol{z}_{i}^{\text{hidden}} &= \Big( b_i^{\text{hidden}} + w_i^{\text{hidden}}x_1 , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_2, \ \dots \, , \ b_i^{\text{hidden}} + w_i^{\text{hidden}} x_N\Big) \\ &= \begin{pmatrix} b_i^{\text{hidden}} & w_i^{\text{hidden}} \end{pmatrix} \begin{pmatrix} 1 & 1 & \dots & 1 \\ x_1 & x_2 & \dots & x_N \end{pmatrix} \\ &= \boldsymbol{p}_{i, \text{hidden}}^T X \end{aligned} $$
The vector \( \boldsymbol{p}_{i, \text{hidden}}^T \) constitutes each row in \( P_{\text{hidden} } \), which contains the weights for the neural network to minimize according to \eqref{min}.
After having found \( \boldsymbol{z}_{i}^{\text{hidden}} \) for every \( i \)-th neuron within the hidden layer, the vector will be sent to an activation function \( a_i(\boldsymbol{z}) \).
In this example, the sigmoid function has been chosen to be the activation function for each hidden neuron: $$ f(z) = \frac{1}{1 + \exp{(-z)}} $$
It is possible to use other activations functions for the hidden layer also.
The output \( \boldsymbol{x}_i^{\text{hidden}} \) from each \( i \)-th hidden neuron is: $$ \boldsymbol{x}_i^{\text{hidden} } = f\big( \boldsymbol{z}_{i}^{\text{hidden}} \big) $$
The outputs \( \boldsymbol{x}_i^{\text{hidden} } \) are then sent to the output layer.
The output layer consists of one neuron in this case, and combines the output from each of the neurons in the hidden layers. The output layer combines the results from the hidden layer using some weights \( w_i^{\text{output}} \) and biases \( b_i^{\text{output}} \). In this case, it is assumes that the number of neurons in the output layer is one.
The procedure of weighting the output neuron \( j \) in the hidden layer to the \( i \)-th neuron in the output layer is similar as for the hidden layer described previously. $$ \begin{aligned} z_{1,j}^{\text{output}} & = \begin{pmatrix} b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} \end{pmatrix} \begin{pmatrix} 1 \\ \boldsymbol{x}_j^{\text{hidden}} \end{pmatrix} \end{aligned} $$
Expressing \( z_{1,j}^{\text{output}} \) as a vector gives the following way of weighting the inputs from the hidden layer: $$ \boldsymbol{z}_{1}^{\text{output}} = \begin{pmatrix} b_1^{\text{output}} & \boldsymbol{w}_1^{\text{output}} \end{pmatrix} \begin{pmatrix} 1 & 1 & \dots & 1 \\ \boldsymbol{x}_1^{\text{hidden}} & \boldsymbol{x}_2^{\text{hidden}} & \dots & \boldsymbol{x}_N^{\text{hidden}} \end{pmatrix} $$
In this case we seek a continuous range of values since we are approximating a function. This means that after computing \( \boldsymbol{z}_{1}^{\text{output}} \) the neural network has finished its feed forward step, and \( \boldsymbol{z}_{1}^{\text{output}} \) is the final output of the network.
The next step is to decide how the parameters should be changed such that they minimize the cost function.
The chosen cost function for this problem is $$ C(\boldsymbol{x}, P) = \frac{1}{N} \sum_i \big(g_t'(x_i, P) - ( -\gamma g_t(x_i, P) \big)^2 $$
In order to minimize the cost function, an optimization method must be chosen.
Here, gradient descent with a constant step size has been chosen.
The idea of the gradient descent algorithm is to update parameters in a direction where the cost function decreases goes to a minimum.
In general, the update of some parameters \( \boldsymbol{\omega} \) given a cost function defined by some weights \( \boldsymbol{\omega} \), \( C(\boldsymbol{x}, \boldsymbol{\omega}) \), goes as follows: $$ \boldsymbol{\omega}_{\text{new} } = \boldsymbol{\omega} - \lambda \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega}) $$
for a number of iterations or until $ \big|\big| \boldsymbol{\omega}_{\text{new} } - \boldsymbol{\omega} \big|\big|$ becomes smaller than some given tolerance.
The value of \( \lambda \) decides how large steps the algorithm must take in the direction of $ \nabla_{\boldsymbol{\omega}} C(\boldsymbol{x}, \boldsymbol{\omega})$. The notation \( \nabla_{\boldsymbol{\omega}} \) express the gradient with respect to the elements in \( \boldsymbol{\omega} \).
In our case, we have to minimize the cost function \( C(\boldsymbol{x}, P) \) with respect to the two sets of weights and biases, that is for the hidden layer \( P_{\text{hidden} } \) and for the output layer \( P_{\text{output} } \) .
This means that \( P_{\text{hidden} } \) and \( P_{\text{output} } \) is updated by $$ \begin{aligned} P_{\text{hidden},\text{new}} &= P_{\text{hidden}} - \lambda \nabla_{P_{\text{hidden}}} C(\boldsymbol{x}, P) \\ P_{\text{output},\text{new}} &= P_{\text{output}} - \lambda \nabla_{P_{\text{output}}} C(\boldsymbol{x}, P) \end{aligned} $$
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
# Assuming one input, hidden, and output layer
def neural_network(params, x):
# Find the weights (including and biases) for the hidden and output layer.
# Assume that params is a list of parameters for each layer.
# The biases are the first element for each array in params,
# and the weights are the remaning elements in each array in params.
w_hidden = params[0]
w_output = params[1]
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
## Hidden layer:
# Add a row of ones to include bias
x_input = np.concatenate((np.ones((1,num_values)), x_input ), axis = 0)
z_hidden = np.matmul(w_hidden, x_input)
x_hidden = sigmoid(z_hidden)
## Output layer:
# Include bias:
x_hidden = np.concatenate((np.ones((1,num_values)), x_hidden ), axis = 0)
z_output = np.matmul(w_output, x_hidden)
x_output = z_output
return x_output
# The trial solution using the deep neural network:
def g_trial(x,params, g0 = 10):
return g0 + x*neural_network(params,x)
# The right side of the ODE:
def g(x, g_trial, gamma = 2):
return -gamma*g_trial
# The cost function:
def cost_function(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial(x,P)
# Find the derivative w.r.t x of the neural network
d_net_out = elementwise_grad(neural_network,1)(P,x)
# Find the derivative w.r.t x of the trial function
d_g_t = elementwise_grad(g_trial,0)(x,P)
# The right side of the ODE
func = g(x, g_t)
err_sqr = (d_g_t - func)**2
cost_sum = np.sum(err_sqr)
return cost_sum / np.size(err_sqr)
# Solve the exponential decay ODE using neural network with one input, hidden, and output layer
def solve_ode_neural_network(x, num_neurons_hidden, num_iter, lmb):
## Set up initial weights and biases
# For the hidden layer
p0 = npr.randn(num_neurons_hidden, 2 )
# For the output layer
p1 = npr.randn(1, num_neurons_hidden + 1 ) # +1 since bias is included
P = [p0, p1]
print('Initial cost: %g'%cost_function(P, x))
## Start finding the optimal weights using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_grad = grad(cost_function,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of two arrays;
# one for the gradient w.r.t P_hidden and
# one for the gradient w.r.t P_output
cost_grad = cost_function_grad(P, x)
P[0] = P[0] - lmb * cost_grad[0]
P[1] = P[1] - lmb * cost_grad[1]
print('Final cost: %g'%cost_function(P, x))
return P
def g_analytic(x, gamma = 2, g0 = 10):
return g0*np.exp(-gamma*x)
# Solve the given problem
if __name__ == '__main__':
# Set seed such that the weight are initialized
# with same weights and biases for every run.
npr.seed(15)
## Decide the vales of arguments to the function to solve
N = 10
x = np.linspace(0, 1, N)
## Set up the initial parameters
num_hidden_neurons = 10
num_iter = 10000
lmb = 0.001
# Use the network
P = solve_ode_neural_network(x, num_hidden_neurons, num_iter, lmb)
# Print the deviation from the trial solution and true solution
res = g_trial(x,P)
res_analytical = g_analytic(x)
print('Max absolute difference: %g'%np.max(np.abs(res - res_analytical)))
# Plot the results
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(x, res_analytical)
plt.plot(x, res[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('x')
plt.ylabel('g(x)')
plt.show()
It is also possible to extend the construction of our network into a more general one, allowing the network to contain more than one hidden layers.
The number of neurons within each hidden layer are given as a list of integers in the program below.
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
# The neural network with one input layer and one output layer,
# but with number of hidden layers specified by the user.
def deep_neural_network(deep_params, x):
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consists of
# parameters to all the hidden
# layers AND the output layer.
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
# Due to multiple hidden layers, define a variable referencing to the
# output of the previous layer:
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output
# The trial solution using the deep neural network:
def g_trial_deep(x,params, g0 = 10):
return g0 + x*deep_neural_network(params, x)
# The right side of the ODE:
def g(x, g_trial, gamma = 2):
return -gamma*g_trial
# The same cost function as before, but calls deep_neural_network instead.
def cost_function_deep(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial_deep(x,P)
# Find the derivative w.r.t x of the neural network
d_net_out = elementwise_grad(deep_neural_network,1)(P,x)
# Find the derivative w.r.t x of the trial function
d_g_t = elementwise_grad(g_trial_deep,0)(x,P)
# The right side of the ODE
func = g(x, g_t)
err_sqr = (d_g_t - func)**2
cost_sum = np.sum(err_sqr)
return cost_sum / np.size(err_sqr)
# Solve the exponential decay ODE using neural network with one input and one output layer,
# but with specified number of hidden layers from the user.
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
# num_hidden_neurons is now a list of number of neurons within each hidden layer
# The number of elements in the list num_hidden_neurons thus represents
# the number of hidden layers.
# Find the number of hidden layers:
N_hidden = np.size(num_neurons)
## Set up initial weights and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 )
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: %g'%cost_function_deep(P, x))
## Start finding the optimal weights using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_deep_grad = grad(cost_function_deep,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
# in the hidden layers and output layers evaluated at x.
cost_deep_grad = cost_function_deep_grad(P, x)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_deep_grad[l]
print('Final cost: %g'%cost_function_deep(P, x))
return P
def g_analytic(x, gamma = 2, g0 = 10):
return g0*np.exp(-gamma*x)
# Solve the given problem
if __name__ == '__main__':
npr.seed(15)
## Decide the vales of arguments to the function to solve
N = 10
x = np.linspace(0, 1, N)
## Set up the initial parameters
num_hidden_neurons = np.array([10,10])
num_iter = 10000
lmb = 0.001
P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
res = g_trial_deep(x,P)
res_analytical = g_analytic(x)
plt.figure(figsize=(10,10))
plt.title('Performance of a deep neural network solving an ODE compared to the analytical solution')
plt.plot(x, res_analytical)
plt.plot(x, res[0,:])
plt.legend(['analytical','dnn'])
plt.ylabel('g(x)')
plt.show()
A logistic model of population growth assumes that a population converges toward an equilibrium. The population growth can be modeled by $$ \begin{equation} \label{log} g'(t) = \alpha g(t)(A - g(t)) \end{equation} $$
where \( g(t) \) is the population density at time \( t \), \( \alpha > 0 \) the growth rate and \( A > 0 \) is the maximum population number in the environment. Also, at \( t = 0 \) the population has the size \( g(0) = g_0 \), where \( g_0 \) is some chosen constant.
In this example, similar network as for the exponential decay using Autograd has been used to solve the equation. However, as the implementation might suffer from e.g numerical instability and high execution time (this might be more apparent in the examples solving PDEs), using a library like TensorFlow is recommended. Here, we stay with a more simple approach and implement for comparison, the simple forward Euler method.
Here, we will model a population \( g(t) \) in an environment having carrying capacity \( A \). The population follows the model $$ \begin{equation} \label{solveode_population} g'(t) = \alpha g(t)(A - g(t)) \end{equation} $$
where \( g(0) = g_0 \).
In this example, we let \( \alpha = 2 \), \( A = 1 \), and \( g_0 = 1.2 \).
We will get a slightly different trial solution, as the boundary conditions are different compared to the case for exponential decay.
A possible trial solution satisfying the condition \( g(0) = g_0 \) could be $$ h_1(t) = g_0 + t \cdot N(t,P) $$
with \( N(t,P) \) being the output from the neural network with weights and biases for each layer collected in the set \( P \).
The analytical solution is $$ g(t) = \frac{Ag_0}{g_0 + (A - g_0)\exp(-\alpha A t)} $$
The network will be the similar as for the exponential decay example, but with some small modifications for our problem.
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
# Function to get the parameters.
# Done such that one can easily change the paramaters after one's liking.
def get_parameters():
alpha = 2
A = 1
g0 = 1.2
return alpha, A, g0
def deep_neural_network(P, x):
# N_hidden is the number of hidden layers
N_hidden = np.size(P) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
# Due to multiple hidden layers, define a variable referencing to the
# output of the previous layer:
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = P[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = P[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output
def cost_function_deep(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial_deep(x,P)
# Find the derivative w.r.t x of the trial function
d_g_t = elementwise_grad(g_trial_deep,0)(x,P)
# The right side of the ODE
func = f(x, g_t)
err_sqr = (d_g_t - func)**2
cost_sum = np.sum(err_sqr)
return cost_sum / np.size(err_sqr)
# The right side of the ODE:
def f(x, g_trial):
alpha,A, g0 = get_parameters()
return alpha*g_trial*(A - g_trial)
# The trial solution using the deep neural network:
def g_trial_deep(x, params):
alpha,A, g0 = get_parameters()
return g0 + x*deep_neural_network(params,x)
# The analytical solution:
def g_analytic(t):
alpha,A, g0 = get_parameters()
return A*g0/(g0 + (A - g0)*np.exp(-alpha*A*t))
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
# num_hidden_neurons is now a list of number of neurons within each hidden layer
# Find the number of hidden layers:
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 )
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: %g'%cost_function_deep(P, x))
## Start finding the optimal weigths using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_deep_grad = grad(cost_function_deep,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
# in the hidden layers and output layers evaluated at x.
cost_deep_grad = cost_function_deep_grad(P, x)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_deep_grad[l]
print('Final cost: %g'%cost_function_deep(P, x))
return P
if __name__ == '__main__':
npr.seed(4155)
## Decide the vales of arguments to the function to solve
Nt = 10
T = 1
t = np.linspace(0,T, Nt)
## Set up the initial parameters
num_hidden_neurons = [100, 50, 25]
num_iter = 1000
lmb = 1e-3
P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)
g_dnn_ag = g_trial_deep(t,P)
g_analytical = g_analytic(t)
# Find the maximum absolute difference between the solutons:
diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
print("The max absolute difference between the solutions is: %g"%diff_ag)
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(t, g_analytical)
plt.plot(t, g_dnn_ag[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('t')
plt.ylabel('g(t)')
plt.show()
A straightforward way of solving an ODE numerically, is to use Euler's method.
Euler's method uses Taylor series to approximate the value at a function \( f \) at a step \( \Delta x \) from \( x \): $$ f(x + \Delta x) \approx f(x) + \Delta x f'(x) $$
In our case, using Euler's method to approximate the value of \( g \) at a step \( \Delta t \) from \( t \) yields $$ \begin{aligned} g(t + \Delta t) &\approx g(t) + \Delta t g'(t) \\ &= g(t) + \Delta t \big(\alpha g(t)(A - g(t))\big) \end{aligned} $$ along with the condition that \( g(0) = g_0 \).
Let \( t_i = i \cdot \Delta t \) where \( \Delta t = \frac{T}{N_t-1} \) where \( T \) is the final time our solver must solve for and \( N_t \) the number of values for \( t \in [0, T] \) for \( i = 0, \dots, N_t-1 \).
For \( i \geq 1 \), we have that $$ \begin{aligned} t_i &= i\Delta t \\ &= (i - 1)\Delta t + \Delta t \\ &= t_{i-1} + \Delta t \end{aligned} $$
Now, if \( g_i = g(t_i) \) then $$ \begin{equation} \begin{aligned} g_i &= g(t_i) \\ &= g(t_{i-1} + \Delta t) \\ &\approx g(t_{i-1}) + \Delta t \big(\alpha g(t_{i-1})(A - g(t_{i-1}))\big) \\ &= g_{i-1} + \Delta t \big(\alpha g_{i-1}(A - g_{i-1})\big) \end{aligned} \end{equation} \label{odenum} $$ for \( i \geq 1 \) and \( g_0 = g(t_0) = g(0) = g_0 \).
Equation \eqref{odenum} could be implemented in the following way, extending the program that uses the network using Autograd:
# Assume that all function definitions from the example program using Autograd
# are located here.
if __name__ == '__main__':
npr.seed(4155)
## Decide the vales of arguments to the function to solve
Nt = 10
T = 1
t = np.linspace(0,T, Nt)
## Set up the initial parameters
num_hidden_neurons = [100,50,25]
num_iter = 1000
lmb = 1e-3
P = solve_ode_deep_neural_network(t, num_hidden_neurons, num_iter, lmb)
g_dnn_ag = g_trial_deep(t,P)
g_analytical = g_analytic(t)
# Find the maximum absolute difference between the solutons:
diff_ag = np.max(np.abs(g_dnn_ag - g_analytical))
print("The max absolute difference between the solutions is: %g"%diff_ag)
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(t, g_analytical)
plt.plot(t, g_dnn_ag[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('t')
plt.ylabel('g(t)')
## Find an approximation to the funtion using forward Euler
alpha, A, g0 = get_parameters()
dt = T/(Nt - 1)
# Perform forward Euler to solve the ODE
g_euler = np.zeros(Nt)
g_euler[0] = g0
for i in range(1,Nt):
g_euler[i] = g_euler[i-1] + dt*(alpha*g_euler[i-1]*(A - g_euler[i-1]))
# Print the errors done by each method
diff1 = np.max(np.abs(g_euler - g_analytical))
diff2 = np.max(np.abs(g_dnn_ag[0,:] - g_analytical))
print('Max absolute difference between Euler method and analytical: %g'%diff1)
print('Max absolute difference between deep neural network and analytical: %g'%diff2)
# Plot results
plt.figure(figsize=(10,10))
plt.plot(t,g_euler)
plt.plot(t,g_analytical)
plt.plot(t,g_dnn_ag[0,:])
plt.legend(['euler','analytical','dnn'])
plt.xlabel('Time t')
plt.ylabel('g(t)')
plt.show()
The Poisson equation for \( g(x) \) in one dimension is $$ \begin{equation} \label{poisson} -g''(x) = f(x) \end{equation} $$
where \( f(x) \) is a given function for \( x \in (0,1) \).
The conditions that \( g(x) \) is chosen to fulfill, are $$ \begin{align*} g(0) &= 0 \\ g(1) &= 0 \end{align*} $$
This equation can be solved numerically using programs where e.g Autograd and TensorFlow are used. The results from the networks can then be compared to the analytical solution. In addition, it could be interesting to see how a typical method for numerically solving second order ODEs compares to the neural networks.
Here, the function \( g(x) \) to solve for follows the equation $$ -g''(x) = f(x),\qquad x \in (0,1) $$
where \( f(x) \) is a given function, along with the chosen conditions $$ \begin{aligned} g(0) = g(1) = 0 \end{aligned}\label{cond} $$
In this example, we consider the case when \( f(x) = (3x + x^2)\exp(x) \).
For this case, a possible trial solution satisfying the conditions could be $$ g_t(x) = x \cdot (1-x) \cdot N(P,x) $$
The analytical solution for this problem is $$ g(x) = x(1 - x)\exp(x) $$
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
# Due to multiple hidden layers, define a variable referencing to the
# output of the previous layer:
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
# num_hidden_neurons is now a list of number of neurons within each hidden layer
# Find the number of hidden layers:
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 )
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: %g'%cost_function_deep(P, x))
## Start finding the optimal weigths using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_deep_grad = grad(cost_function_deep,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
# in the hidden layers and output layers evaluated at x.
cost_deep_grad = cost_function_deep_grad(P, x)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_deep_grad[l]
print('Final cost: %g'%cost_function_deep(P, x))
return P
## Set up the cost function specified for this Poisson equation:
# The right side of the ODE
def f(x):
return (3*x + x**2)*np.exp(x)
def cost_function_deep(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial_deep(x,P)
# Find the derivative w.r.t x of the trial function
d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
right_side = f(x)
err_sqr = (-d2_g_t - right_side)**2
cost_sum = np.sum(err_sqr)
return cost_sum/np.size(err_sqr)
# The trial solution:
def g_trial_deep(x,P):
return x*(1-x)*deep_neural_network(P,x)
# The analytic solution;
def g_analytic(x):
return x*(1-x)*np.exp(x)
if __name__ == '__main__':
npr.seed(4155)
## Decide the vales of arguments to the function to solve
Nx = 10
x = np.linspace(0,1, Nx)
## Set up the initial parameters
num_hidden_neurons = [200,100]
num_iter = 1000
lmb = 1e-3
P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
g_dnn_ag = g_trial_deep(x,P)
g_analytical = g_analytic(x)
# Find the maximum absolute difference between the solutons:
max_diff = np.max(np.abs(g_dnn_ag - g_analytical))
print("The max absolute difference between the solutions is: %g"%max_diff)
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(x, g_analytical)
plt.plot(x, g_dnn_ag[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('x')
plt.ylabel('g(x)')
plt.show()
The Poisson equation is possible to solve using Taylor series to approximate the second derivative.
Using Taylor series, the second derivative can be expressed as $$ g''(x) = \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} + E_{\Delta x}(x) $$
where \( \Delta x \) is a small step size and \( E_{\Delta x}(x) \) being the error term.
Looking away from the error terms gives an approximation to the second derivative: $$ \begin{equation} \label{approx} g''(x) \approx \frac{g(x + \Delta x) - 2g(x) + g(x-\Delta x)}{\Delta x^2} \end{equation} $$
If \( x_i = i \Delta x = x_{i-1} + \Delta x \) and \( g_i = g(x_i) \) for \( i = 1,\dots N_x - 2 \) with \( N_x \) being the number of values for \( x \), \eqref{approx} becomes $$ \begin{aligned} g''(x_i) &\approx \frac{g(x_i + \Delta x) - 2g(x_i) + g(x_i -\Delta x)}{\Delta x^2} \\ &= \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \end{aligned} $$
Since we know from our problem that $$ \begin{aligned} -g''(x) &= f(x) \\ &= (3x + x^2)\exp(x) \end{aligned} $$
along with the conditions \( g(0) = g(1) = 0 \), the following scheme can be used to find an approximate solution for \( g(x) \) numerically: $$ \begin{equation} \begin{aligned} -\Big( \frac{g_{i+1} - 2g_i + g_{i-1}}{\Delta x^2} \Big) &= f(x_i) \\ -g_{i+1} + 2g_i - g_{i-1} &= \Delta x^2 f(x_i) \end{aligned} \end{equation} \label{odesys} $$
for \( i = 1, \dots, N_x - 2 \) where \( g_0 = g_{N_x - 1} = 0 \) and \( f(x_i) = (3x_i + x_i^2)\exp(x_i) \), which is given for our specific problem.
The equation can be rewritten into a matrix equation: $$ \begin{aligned} \begin{pmatrix} 2 & -1 & 0 & \dots & 0 \\ -1 & 2 & -1 & \dots & 0 \\ \vdots & & \ddots & & \vdots \\ 0 & \dots & -1 & 2 & -1 \\ 0 & \dots & 0 & -1 & 2\\ \end{pmatrix} \begin{pmatrix} g_1 \\ g_2 \\ \vdots \\ g_{N_x - 3} \\ g_{N_x - 2} \end{pmatrix} &= \Delta x^2 \begin{pmatrix} f(x_1) \\ f(x_2) \\ \vdots \\ f(x_{N_x - 3}) \\ f(x_{N_x - 2}) \end{pmatrix} \\ \boldsymbol{A}\boldsymbol{g} &= \boldsymbol{f}, \end{aligned} $$
which makes it possible to solve for the vector \( \boldsymbol{g} \).
We can then compare the result from this numerical scheme with the output from our network using Autograd:
import autograd.numpy as np
from autograd import grad, elementwise_grad
import autograd.numpy.random as npr
from matplotlib import pyplot as plt
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assumes input x being an one-dimensional array
num_values = np.size(x)
x = x.reshape(-1, num_values)
# Assume that the input layer does nothing to the input x
x_input = x
# Due to multiple hidden layers, define a variable referencing to the
# output of the previous layer:
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_values)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_values)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output
def solve_ode_deep_neural_network(x, num_neurons, num_iter, lmb):
# num_hidden_neurons is now a list of number of neurons within each hidden layer
# Find the number of hidden layers:
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 )
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: %g'%cost_function_deep(P, x))
## Start finding the optimal weigths using gradient descent
# Find the Python function that represents the gradient of the cost function
# w.r.t the 0-th input argument -- that is the weights and biases in the hidden and output layer
cost_function_deep_grad = grad(cost_function_deep,0)
# Let the update be done num_iter times
for i in range(num_iter):
# Evaluate the gradient at the current weights and biases in P.
# The cost_grad consist now of N_hidden + 1 arrays; the gradient w.r.t the weights and biases
# in the hidden layers and output layers evaluated at x.
cost_deep_grad = cost_function_deep_grad(P, x)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_deep_grad[l]
print('Final cost: %g'%cost_function_deep(P, x))
return P
## Set up the cost function specified for this Poisson equation:
# The right side of the ODE
def f(x):
return (3*x + x**2)*np.exp(x)
def cost_function_deep(P, x):
# Evaluate the trial function with the current parameters P
g_t = g_trial_deep(x,P)
# Find the derivative w.r.t x of the trial function
d2_g_t = elementwise_grad(elementwise_grad(g_trial_deep,0))(x,P)
right_side = f(x)
err_sqr = (-d2_g_t - right_side)**2
cost_sum = np.sum(err_sqr)
return cost_sum/np.size(err_sqr)
# The trial solution:
def g_trial_deep(x,P):
return x*(1-x)*deep_neural_network(P,x)
# The analytic solution;
def g_analytic(x):
return x*(1-x)*np.exp(x)
if __name__ == '__main__':
npr.seed(4155)
## Decide the vales of arguments to the function to solve
Nx = 10
x = np.linspace(0,1, Nx)
## Set up the initial parameters
num_hidden_neurons = [200,100]
num_iter = 1000
lmb = 1e-3
P = solve_ode_deep_neural_network(x, num_hidden_neurons, num_iter, lmb)
g_dnn_ag = g_trial_deep(x,P)
g_analytical = g_analytic(x)
# Find the maximum absolute difference between the solutons:
plt.figure(figsize=(10,10))
plt.title('Performance of neural network solving an ODE compared to the analytical solution')
plt.plot(x, g_analytical)
plt.plot(x, g_dnn_ag[0,:])
plt.legend(['analytical','nn'])
plt.xlabel('x')
plt.ylabel('g(x)')
## Perform the computation using the numerical scheme
dx = 1/(Nx - 1)
# Set up the matrix A
A = np.zeros((Nx-2,Nx-2))
A[0,0] = 2
A[0,1] = -1
for i in range(1,Nx-3):
A[i,i-1] = -1
A[i,i] = 2
A[i,i+1] = -1
A[Nx - 3, Nx - 4] = -1
A[Nx - 3, Nx - 3] = 2
# Set up the vector f
f_vec = dx**2 * f(x[1:-1])
# Solve the equation
g_res = np.linalg.solve(A,f_vec)
g_vec = np.zeros(Nx)
g_vec[1:-1] = g_res
# Print the differences between each method
max_diff1 = np.max(np.abs(g_dnn_ag - g_analytical))
max_diff2 = np.max(np.abs(g_vec - g_analytical))
print("The max absolute difference between the analytical solution and DNN Autograd: %g"%max_diff1)
print("The max absolute difference between the analytical solution and numerical scheme: %g"%max_diff2)
# Plot the results
plt.figure(figsize=(10,10))
plt.plot(x,g_vec)
plt.plot(x,g_analytical)
plt.plot(x,g_dnn_ag[0,:])
plt.legend(['numerical scheme','analytical','dnn'])
plt.show()
A partial differential equation (PDE) has a solution here the function is defined by multiple variables. The equation may involve all kinds of combinations of which variables the function is differentiated with respect to.
In general, a partial differential equation for a function \( g(x_1,\dots,x_N) \) with \( N \) variables may be expressed as $$ \begin{equation} \label{PDE} f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) = 0 \end{equation} $$
where \( f \) is an expression involving all kinds of possible mixed derivatives of \( g(x_1,\dots,x_N) \) up to an order \( n \). In order for the solution to be unique, some additional conditions must also be given.
The problem our network must solve for, is similar to the ODE case. We must have a trial solution \( g_t \) at hand.
For instance, the trial solution could be expressed as $$ \begin{align*} g_t(x_1,\dots,x_N) = h_1(x_1,\dots,x_N) + h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) \end{align*} $$ where \( h_1(x_1,\dots,x_N) \) is a function that ensures \( g_t(x_1,\dots,x_N) \) satisfies some given conditions. The neural network \( N(x_1,\dots,x_N,P) \) has weights and biases described by \( P \) and \( h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) \) is an expression using the output from the neural network in some way.
The role of the function \( h_2(x_1,\dots,x_N,N(x_1,\dots,x_N,P)) \), is to ensure that the output of \( N(x_1,\dots,x_N,P) \) is zero when \( g_t(x_1,\dots,x_N) \) is evaluated at the values of \( x_1,\dots,x_N \) where the given conditions must be satisfied. The function \( h_1(x_1,\dots,x_N) \) should alone make \( g_t(x_1,\dots,x_N) \) satisfy the conditions.
The network tries then the minimize the cost function following the same ideas as described for the ODE case, but now with more than one variables to consider. The concept still remains the same; find a set of parameters \( P \) such that the expression \( f \) in \eqref{PDE} is as close to zero as possible.
As for the ODE case, the cost function is the mean squared error that the network must try to minimize. The cost function for the network to minimize is $$ \begin{equation*} C\left(x_1, \dots, x_N, P\right) = \left( f\left(x_1, \, \dots \, , x_N, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1}, \dots , \frac{\partial g(x_1,\dots,x_N) }{\partial x_N}, \frac{\partial g(x_1,\dots,x_N) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(x_1,\dots,x_N) }{\partial x_N^n} \right) \right)^2 \end{equation*} $$
If we let \( \boldsymbol{x} = \big( x_1, \dots, x_N \big) \) be an array containing the values for \( x_1, \dots, x_N \) respectively, the cost function can be reformulated into the following: $$ C\left(\boldsymbol{x}, P\right) = f\left( \left( \boldsymbol{x}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}) }{\partial x_N^n} \right) \right)^2 $$
If we also have \( M \) different sets of values for \( x_1, \dots, x_N \), that is \( \boldsymbol{x}_i = \big(x_1^{(i)}, \dots, x_N^{(i)}\big) \) for \( i = 1,\dots,M \) being the rows in matrix \( X \), the cost function can be generalized into $$ \begin{equation*} C\left(X, P \right) = \sum_{i=1}^M f\left( \left( \boldsymbol{x}_i, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1}, \dots , \frac{\partial g(\boldsymbol{x}_i) }{\partial x_N}, \frac{\partial g(\boldsymbol{x}_i) }{\partial x_1\partial x_2}, \, \dots \, , \frac{\partial^n g(\boldsymbol{x}_i) }{\partial x_N^n} \right) \right)^2. \end{equation*} $$
In one spatial dimension, the equation reads $$ \begin{equation*} \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} \end{equation*} $$
where a possible choice of conditions are $$ \begin{align*} g(0,t) &= 0 ,\qquad t \geq 0 \\ g(1,t) &= 0, \qquad t \geq 0 \\ g(x,0) &= u(x),\qquad x\in [0,1] \end{align*} $$ with \( u(x) \) being some given function.
For this case, we want to find \( g(x,t) \) such that $$ \begin{equation} \frac{\partial g(x,t)}{\partial t} = \frac{\partial^2 g(x,t)}{\partial x^2} \end{equation} \label{diffonedim} $$
and $$ \begin{align*} g(0,t) &= 0 ,\qquad t \geq 0 \\ g(1,t) &= 0, \qquad t \geq 0 \\ g(x,0) &= u(x),\qquad x\in [0,1] \end{align*} $$ with \( u(x) = \sin(\pi x) \).
First, let us set up the deep neural network. The deep neural network will follow the same structure as discussed in the examples solving the ODEs. First, we will look into how Autograd could be used in a network tailored to solve for bivariate functions.
The only change to do here, is to extend our network such that functions of multiple parameters are correctly handled. In this case we have two variables in our function to solve for, that is time \( t \) and position \( x \). The variables will be represented by a one-dimensional array in the program. The program will evaluate the network at each possible pair \( (x,t) \), given an array for the desired \( x \)-values and \( t \)-values to approximate the solution at.
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# x is now a point and a 1D numpy array; make it a column vector
num_coordinates = np.size(x,0)
x = x.reshape(num_coordinates,-1)
num_points = np.size(x,1)
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assume that the input layer does nothing to the input x
x_input = x
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output[0][0]
The cost function must then iterate through the given arrays containing values for \( x \) and \( t \), defines a point \( (x,t) \) the deep neural network and the trial solution is evaluated at, and then finds the Jacobian of the trial solution.
A possible trial solution for this PDE is $$ g_t(x,t) = h_1(x,t) + x(1-x)tN(x,t,P) $$
with \( A(x,t) \) being a function ensuring that \( g_t(x,t) \) satisfies our given conditions, and \( N(x,t,P) \) being the output from the deep neural network using weights and biases for each layer from \( P \).
To fulfill the conditions, \( A(x,t) \) could be: $$ h_1(x,t) = (1-t)\Big(u(x) - \big((1-x)u(0) + x u(1)\big)\Big) = (1-t)u(x) = (1-t)\sin(\pi x) $$ since \( (0) = u(1) = 0 \) and \( u(x) = \sin(\pi x) \).
The Jacobian is used because the program must find the derivative of the trial solution with respect to \( x \) and \( t \).
This gives the necessity of computing the Jacobian matrix, as we want to evaluate the gradient with respect to \( x \) and \( t \) (note that the Jacobian of a scalar-valued multivariate function is simply its gradient).
In Autograd, the differentiation is by default done with respect to the first input argument of your Python function. Since the points is an array representing \( x \) and \( t \), the Jacobian is calculated using the values of \( x \) and \( t \).
To find the second derivative with respect to \( x \) and \( t \), the Jacobian can be found for the second time. The result is a Hessian matrix, which is the matrix containing all the possible second order mixed derivatives of \( g(x,t) \).
# Set up the trial function:
def u(x):
return np.sin(np.pi*x)
def g_trial(point,P):
x,t = point
return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)
# The right side of the ODE:
def f(point):
return 0.
# The cost function:
def cost_function(P, x, t):
cost_sum = 0
g_t_jacobian_func = jacobian(g_trial)
g_t_hessian_func = hessian(g_trial)
for x_ in x:
for t_ in t:
point = np.array([x_,t_])
g_t = g_trial(point,P)
g_t_jacobian = g_t_jacobian_func(point,P)
g_t_hessian = g_t_hessian_func(point,P)
g_t_dt = g_t_jacobian[1]
g_t_d2x = g_t_hessian[0][0]
func = f(point)
err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
cost_sum += err_sqr
return cost_sum
Having set up the network, along with the trial solution and cost function, we can now see how the deep neural network performs by comparing the results to the analytical solution.
The analytical solution of our problem is $$ g(x,t) = \exp(-\pi^2 t)\sin(\pi x) $$
A possible way to implement a neural network solving the PDE, is given below. Be aware, though, that it is fairly slow for the parameters used. A better result is possible, but requires more iterations, and thus longer time to complete.
Indeed, the program below is not optimal in its implementation, but rather serves as an example on how to implement and use a neural network to solve a PDE. Using TensorFlow results in a much better execution time. Try it!
import autograd.numpy as np
from autograd import jacobian,hessian,grad
import autograd.numpy.random as npr
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import axes3d
## Set up the network
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# x is now a point and a 1D numpy array; make it a column vector
num_coordinates = np.size(x,0)
x = x.reshape(num_coordinates,-1)
num_points = np.size(x,1)
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assume that the input layer does nothing to the input x
x_input = x
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output[0][0]
## Define the trial solution and cost function
def u(x):
return np.sin(np.pi*x)
def g_trial(point,P):
x,t = point
return (1-t)*u(x) + x*(1-x)*t*deep_neural_network(P,point)
# The right side of the ODE:
def f(point):
return 0.
# The cost function:
def cost_function(P, x, t):
cost_sum = 0
g_t_jacobian_func = jacobian(g_trial)
g_t_hessian_func = hessian(g_trial)
for x_ in x:
for t_ in t:
point = np.array([x_,t_])
g_t = g_trial(point,P)
g_t_jacobian = g_t_jacobian_func(point,P)
g_t_hessian = g_t_hessian_func(point,P)
g_t_dt = g_t_jacobian[1]
g_t_d2x = g_t_hessian[0][0]
func = f(point)
err_sqr = ( (g_t_dt - g_t_d2x) - func)**2
cost_sum += err_sqr
return cost_sum /( np.size(x)*np.size(t) )
## For comparison, define the analytical solution
def g_analytic(point):
x,t = point
return np.exp(-np.pi**2*t)*np.sin(np.pi*x)
## Set up a function for training the network to solve for the equation
def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
## Set up initial weigths and biases
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: ',cost_function(P, x, t))
cost_function_grad = grad(cost_function,0)
# Let the update be done num_iter times
for i in range(num_iter):
cost_grad = cost_function_grad(P, x , t)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_grad[l]
print('Final cost: ',cost_function(P, x, t))
return P
if __name__ == '__main__':
### Use the neural network:
npr.seed(15)
## Decide the vales of arguments to the function to solve
Nx = 10; Nt = 10
x = np.linspace(0, 1, Nx)
t = np.linspace(0,1,Nt)
## Set up the parameters for the network
num_hidden_neurons = [100, 25]
num_iter = 250
lmb = 0.01
P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
## Store the results
g_dnn_ag = np.zeros((Nx, Nt))
G_analytical = np.zeros((Nx, Nt))
for i,x_ in enumerate(x):
for j, t_ in enumerate(t):
point = np.array([x_, t_])
g_dnn_ag[i,j] = g_trial(point,P)
G_analytical[i,j] = g_analytic(point)
# Find the map difference between the analytical and the computed solution
diff_ag = np.abs(g_dnn_ag - G_analytical)
print('Max absolute difference between the analytical solution and the network: %g'%np.max(diff_ag))
## Plot the solutions in two dimensions, that being in position and time
T,X = np.meshgrid(t,x)
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
s = ax.plot_surface(T,X,g_dnn_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Analytical solution')
s = ax.plot_surface(T,X,G_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Difference')
s = ax.plot_surface(T,X,diff_ag,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
## Take some slices of the 3D plots just to see the solutions at particular times
indx1 = 0
indx2 = int(Nt/2)
indx3 = Nt-1
t1 = t[indx1]
t2 = t[indx2]
t3 = t[indx3]
# Slice the results from the DNN
res1 = g_dnn_ag[:,indx1]
res2 = g_dnn_ag[:,indx2]
res3 = g_dnn_ag[:,indx3]
# Slice the analytical results
res_analytical1 = G_analytical[:,indx1]
res_analytical2 = G_analytical[:,indx2]
res_analytical3 = G_analytical[:,indx3]
# Plot the slices
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t1)
plt.plot(x, res1)
plt.plot(x,res_analytical1)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t2)
plt.plot(x, res2)
plt.plot(x,res_analytical2)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t3)
plt.plot(x, res3)
plt.plot(x,res_analytical3)
plt.legend(['dnn','analytical'])
plt.show()
The wave equation is $$ \begin{equation*} \frac{\partial^2 g(x,t)}{\partial t^2} = c^2\frac{\partial^2 g(x,t)}{\partial x^2} \end{equation*} $$
with \( c \) being the specified wave speed.
Here, the chosen conditions are $$ \begin{align*} g(0,t) &= 0 \\ g(1,t) &= 0 \\ g(x,0) &= u(x) \\ \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} &= v(x) \end{align*} $$ where \( \frac{\partial g(x,t)}{\partial t} \Big |_{t = 0} \) means the derivative of \( g(x,t) \) with respect to \( t \) is evaluated at \( t = 0 \), and \( u(x) \) and \( v(x) \) being given functions.
The wave equation to solve for, is $$ \begin{equation} \label{wave} \frac{\partial^2 g(x,t)}{\partial t^2} = c^2 \frac{\partial^2 g(x,t)}{\partial x^2} \end{equation} $$
where \( c \) is the given wave speed. The chosen conditions for this equation are $$ \begin{aligned} g(0,t) &= 0, &t \geq 0 \\ g(1,t) &= 0, &t \geq 0 \\ g(x,0) &= u(x), &x\in[0,1] \\ \frac{\partial g(x,t)}{\partial t}\Big |_{t = 0} &= v(x), &x \in [0,1] \end{aligned} \label{condwave} $$
In this example, let \( c = 1 \) and \( u(x) = \sin(\pi x) \) and \( v(x) = -\pi\sin(\pi x) \).
The trial solution becomes slightly different since we have other conditions than in the example of solving the diffusion equation. Here, a possible trial solution \( g_t(x,t) \) is $$ g_t(x,t) = h_1(x,t) + x(1-x)t^2N(x,t,P) $$
where $$ h_1(x,t) = (1-t^2)u(x) + tv(x) $$
Note that this trial solution satisfies the conditions only if \( u(0) = v(0) = u(1) = v(1) = 0 \), which is the case in this example.
The analytical solution for our specific problem, is $$ g(x,t) = \sin(\pi x)\cos(\pi t) - \sin(\pi x)\sin(\pi t) $$
import autograd.numpy as np
from autograd import hessian,grad
import autograd.numpy.random as npr
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import axes3d
## Set up the trial function:
def u(x):
return np.sin(np.pi*x)
def v(x):
return -np.pi*np.sin(np.pi*x)
def h1(point):
x,t = point
return (1 - t**2)*u(x) + t*v(x)
def g_trial(point,P):
x,t = point
return h1(point) + x*(1-x)*t**2*deep_neural_network(P,point)
## Define the cost function
def cost_function(P, x, t):
cost_sum = 0
g_t_hessian_func = hessian(g_trial)
for x_ in x:
for t_ in t:
point = np.array([x_,t_])
g_t_hessian = g_t_hessian_func(point,P)
g_t_d2x = g_t_hessian[0][0]
g_t_d2t = g_t_hessian[1][1]
err_sqr = ( (g_t_d2t - g_t_d2x) )**2
cost_sum += err_sqr
return cost_sum / (np.size(t) * np.size(x))
## The neural network
def sigmoid(z):
return 1/(1 + np.exp(-z))
def deep_neural_network(deep_params, x):
# x is now a point and a 1D numpy array; make it a column vector
num_coordinates = np.size(x,0)
x = x.reshape(num_coordinates,-1)
num_points = np.size(x,1)
# N_hidden is the number of hidden layers
N_hidden = np.size(deep_params) - 1 # -1 since params consist of parameters to all the hidden layers AND the output layer
# Assume that the input layer does nothing to the input x
x_input = x
x_prev = x_input
## Hidden layers:
for l in range(N_hidden):
# From the list of parameters P; find the correct weigths and bias for this layer
w_hidden = deep_params[l]
# Add a row of ones to include bias
x_prev = np.concatenate((np.ones((1,num_points)), x_prev ), axis = 0)
z_hidden = np.matmul(w_hidden, x_prev)
x_hidden = sigmoid(z_hidden)
# Update x_prev such that next layer can use the output from this layer
x_prev = x_hidden
## Output layer:
# Get the weights and bias for this layer
w_output = deep_params[-1]
# Include bias:
x_prev = np.concatenate((np.ones((1,num_points)), x_prev), axis = 0)
z_output = np.matmul(w_output, x_prev)
x_output = z_output
return x_output[0][0]
## The analytical solution
def g_analytic(point):
x,t = point
return np.sin(np.pi*x)*np.cos(np.pi*t) - np.sin(np.pi*x)*np.sin(np.pi*t)
def solve_pde_deep_neural_network(x,t, num_neurons, num_iter, lmb):
## Set up initial weigths and biases
N_hidden = np.size(num_neurons)
## Set up initial weigths and biases
# Initialize the list of parameters:
P = [None]*(N_hidden + 1) # + 1 to include the output layer
P[0] = npr.randn(num_neurons[0], 2 + 1 ) # 2 since we have two points, +1 to include bias
for l in range(1,N_hidden):
P[l] = npr.randn(num_neurons[l], num_neurons[l-1] + 1) # +1 to include bias
# For the output layer
P[-1] = npr.randn(1, num_neurons[-1] + 1 ) # +1 since bias is included
print('Initial cost: ',cost_function(P, x, t))
cost_function_grad = grad(cost_function,0)
# Let the update be done num_iter times
for i in range(num_iter):
cost_grad = cost_function_grad(P, x , t)
for l in range(N_hidden+1):
P[l] = P[l] - lmb * cost_grad[l]
print('Final cost: ',cost_function(P, x, t))
return P
if __name__ == '__main__':
### Use the neural network:
npr.seed(15)
## Decide the vales of arguments to the function to solve
Nx = 10; Nt = 10
x = np.linspace(0, 1, Nx)
t = np.linspace(0,1,Nt)
## Set up the parameters for the network
num_hidden_neurons = [50,20]
num_iter = 1000
lmb = 0.01
P = solve_pde_deep_neural_network(x,t, num_hidden_neurons, num_iter, lmb)
## Store the results
res = np.zeros((Nx, Nt))
res_analytical = np.zeros((Nx, Nt))
for i,x_ in enumerate(x):
for j, t_ in enumerate(t):
point = np.array([x_, t_])
res[i,j] = g_trial(point,P)
res_analytical[i,j] = g_analytic(point)
diff = np.abs(res - res_analytical)
print("Max difference between analytical and solution from nn: %g"%np.max(diff))
## Plot the solutions in two dimensions, that being in position and time
T,X = np.meshgrid(t,x)
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Solution from the deep neural network w/ %d layer'%len(num_hidden_neurons))
s = ax.plot_surface(T,X,res,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Analytical solution')
s = ax.plot_surface(T,X,res_analytical,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
fig = plt.figure(figsize=(10,10))
ax = fig.gca(projection='3d')
ax.set_title('Difference')
s = ax.plot_surface(T,X,diff,linewidth=0,antialiased=False,cmap=cm.viridis)
ax.set_xlabel('Time $t$')
ax.set_ylabel('Position $x$');
## Take some slices of the 3D plots just to see the solutions at particular times
indx1 = 0
indx2 = int(Nt/2)
indx3 = Nt-1
t1 = t[indx1]
t2 = t[indx2]
t3 = t[indx3]
# Slice the results from the DNN
res1 = res[:,indx1]
res2 = res[:,indx2]
res3 = res[:,indx3]
# Slice the analytical results
res_analytical1 = res_analytical[:,indx1]
res_analytical2 = res_analytical[:,indx2]
res_analytical3 = res_analytical[:,indx3]
# Plot the slices
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t1)
plt.plot(x, res1)
plt.plot(x,res_analytical1)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t2)
plt.plot(x, res2)
plt.plot(x,res_analytical2)
plt.legend(['dnn','analytical'])
plt.figure(figsize=(10,10))
plt.title("Computed solutions at time = %g"%t3)
plt.plot(x, res3)
plt.plot(x,res_analytical3)
plt.legend(['dnn','analytical'])
plt.show()
The principal component analysis deals with the problem of fitting a low-dimensional affine subspace \( S \) of dimension \( d \) much smaller than the total dimension \( D \) of the problem at hand (our data set). Mathematically it can be formulated as a statistical problem or a geometric problem. In our discussion of the theorem for the classical PCA, we will stay with a statistical approach. Historically, the PCA was first formulated in a statistical setting in order to estimate the principal component of a multivariate random variable.
We have a data set defined by a design/feature matrix \( \boldsymbol{X} \) (see below for its definition)
Before we discuss the PCA theorem, we need to remind ourselves about the definition of the covariance and the correlation function. These are quantities
Suppose we have defined two vectors \( \hat{x} \) and \( \hat{y} \) with \( n \) elements each. The covariance matrix \( \boldsymbol{C} \) is defined as $$ \boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{cov}[\boldsymbol{x},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\ \mathrm{cov}[\boldsymbol{y},\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{y},\boldsymbol{y}] \\ \end{bmatrix}, $$ where for example $$ \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). $$ With this definition and recalling that the variance is defined as $$ \mathrm{var}[\boldsymbol{x}]=\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})^2, $$ we can rewrite the covariance matrix as $$ \boldsymbol{C}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} \mathrm{var}[\boldsymbol{x}] & \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] \\ \mathrm{cov}[\boldsymbol{x},\boldsymbol{y}] & \mathrm{var}[\boldsymbol{y}] \\ \end{bmatrix}. $$
The covariance takes values between zero and infinity and may thus lead to problems with loss of numerical precision for particularly large values. It is common to scale the covariance matrix by introducing instead the correlation matrix defined via the so-called correlation function $$ \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}]=\frac{\mathrm{cov}[\boldsymbol{x},\boldsymbol{y}]}{\sqrt{\mathrm{var}[\boldsymbol{x}] \mathrm{var}[\boldsymbol{y}]}}. $$
The correlation function is then given by values \( \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \in [-1,1] \). This avoids eventual problems with too large values. We can then define the correlation matrix for the two vectors \( \boldsymbol{x} \) and \( \boldsymbol{y} \) as $$ \boldsymbol{K}[\boldsymbol{x},\boldsymbol{y}] = \begin{bmatrix} 1 & \mathrm{corr}[\boldsymbol{x},\boldsymbol{y}] \\ \mathrm{corr}[\boldsymbol{y},\boldsymbol{x}] & 1 \\ \end{bmatrix}, $$
In the above example this is the function we constructed using pandas.
In our derivation of the various regression algorithms like Ordinary Least Squares or Ridge regression we defined the design/feature matrix \( \boldsymbol{X} \) as $$ \boldsymbol{X}=\begin{bmatrix} x_{0,0} & x_{0,1} & x_{0,2}& \dots & \dots x_{0,p-1}\\ x_{1,0} & x_{1,1} & x_{1,2}& \dots & \dots x_{1,p-1}\\ x_{2,0} & x_{2,1} & x_{2,2}& \dots & \dots x_{2,p-1}\\ \dots & \dots & \dots & \dots \dots & \dots \\ x_{n-2,0} & x_{n-2,1} & x_{n-2,2}& \dots & \dots x_{n-2,p-1}\\ x_{n-1,0} & x_{n-1,1} & x_{n-1,2}& \dots & \dots x_{n-1,p-1}\\ \end{bmatrix}, $$ with \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors/features \( p \) refering to the column numbers and the entries \( n \) being the row elements. We can rewrite the design/feature matrix in terms of its column vectors as $$ \boldsymbol{X}=\begin{bmatrix} \boldsymbol{x}_0 & \boldsymbol{x}_1 & \boldsymbol{x}_2 & \dots & \dots & \boldsymbol{x}_{p-1}\end{bmatrix}, $$ with a given vector $$ \boldsymbol{x}_i^T = \begin{bmatrix}x_{0,i} & x_{1,i} & x_{2,i}& \dots & \dots x_{n-1,i}\end{bmatrix}. $$
With these definitions, we can now rewrite our \( 2\times 2 \) correaltion/covariance matrix in terms of a moe general design/feature matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \). This leads to a \( p\times p \) covariance matrix for the vectors \( \boldsymbol{x}_i \) with \( i=0,1,\dots,p-1 \) $$ \boldsymbol{C}[\boldsymbol{x}] = \begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\ \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\ \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_1] & \mathrm{var}[\boldsymbol{x}_2] & \dots & \dots & \mathrm{cov}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\ \dots & \dots & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots & \dots \\ \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{cov}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & \mathrm{var}[\boldsymbol{x}_{p-1}]\\ \end{bmatrix}, $$ and the correlation matrix $$ \boldsymbol{K}[\boldsymbol{x}] = \begin{bmatrix} 1 & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_0,\boldsymbol{x}_{p-1}]\\ \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_0] & 1 & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_2] & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_1,\boldsymbol{x}_{p-1}]\\ \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_1] & 1 & \dots & \dots & \mathrm{corr}[\boldsymbol{x}_2,\boldsymbol{x}_{p-1}]\\ \dots & \dots & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots & \dots \\ \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_0] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_1] & \mathrm{corr}[\boldsymbol{x}_{p-1},\boldsymbol{x}_{2}] & \dots & \dots & 1\\ \end{bmatrix}, $$
The Numpy function np.cov calculates the covariance elements using the factor \( 1/(n-1) \) instead of \( 1/n \) since it assumes we do not have the exact mean values. The following simple function uses the np.vstack function which takes each vector of dimension \( 1\times n \) and produces a \( 2\times n \) matrix \( \boldsymbol{W} \) $$ \boldsymbol{W} = \begin{bmatrix} x_0 & y_0 \\ x_1 & y_1 \\ x_2 & y_2\\ \dots & \dots \\ x_{n-2} & y_{n-2}\\ x_{n-1} & y_{n-1} & \end{bmatrix}, $$
which in turn is converted into into the \( 2\times 2 \) covariance matrix \( \boldsymbol{C} \) via the Numpy function np.cov(). We note that we can also calculate the mean value of each set of samples \( \boldsymbol{x} \) etc using the Numpy function np.mean(x). We can also extract the eigenvalues of the covariance matrix through the np.linalg.eig() function.
# Importing various packages
import numpy as np
n = 100
x = np.random.normal(size=n)
print(np.mean(x))
y = 4+3*x+np.random.normal(size=n)
print(np.mean(y))
W = np.vstack((x, y))
C = np.cov(W)
print(C)
The previous example can be converted into the correlation matrix by simply scaling the matrix elements with the variances. We should also subtract the mean values for each column. This leads to the following code which sets up the correlations matrix for the previous example in a more brute force way. Here we scale the mean values for each column of the design matrix, calculate the relevant mean values and variances and then finally set up the \( 2\times 2 \) correlation matrix (since we have only two vectors).
import numpy as np
n = 100
# define two vectors
x = np.random.random(size=n)
y = 4+3*x+np.random.normal(size=n)
#scaling the x and y vectors
x = x - np.mean(x)
y = y - np.mean(y)
variance_x = np.sum(x@x)/n
variance_y = np.sum(y@y)/n
print(variance_x)
print(variance_y)
cov_xy = np.sum(x@y)/n
cov_xx = np.sum(x@x)/n
cov_yy = np.sum(y@y)/n
C = np.zeros((2,2))
C[0,0]= cov_xx/variance_x
C[1,1]= cov_yy/variance_y
C[0,1]= cov_xy/np.sqrt(variance_y*variance_x)
C[1,0]= C[0,1]
print(C)
We see that the matrix elements along the diagonal are one as they should be and that the matrix is symmetric. Furthermore, diagonalizing this matrix we easily see that it is a positive definite matrix.
The above procedure with numpy can be made more compact if we use pandas.
We whow here how we can set up the correlation matrix using pandas, as done in this simple code
import numpy as np
import pandas as pd
n = 10
x = np.random.normal(size=n)
x = x - np.mean(x)
y = 4+3*x+np.random.normal(size=n)
y = y - np.mean(y)
X = (np.vstack((x, y))).T
print(X)
Xpd = pd.DataFrame(X)
print(Xpd)
correlation_matrix = Xpd.corr()
print(correlation_matrix)
We expand this model to the Franke function discussed above.
# Common imports
import numpy as np
import pandas as pd
def FrankeFunction(x,y):
term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2))
term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1))
term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2))
term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2)
return term1 + term2 + term3 + term4
def create_X(x, y, n ):
if len(x.shape) > 1:
x = np.ravel(x)
y = np.ravel(y)
N = len(x)
l = int((n+1)*(n+2)/2) # Number of elements in beta
X = np.ones((N,l))
for i in range(1,n+1):
q = int((i)*(i+1)/2)
for k in range(i+1):
X[:,q+k] = (x**(i-k))*(y**k)
return X
# Making meshgrid of datapoints and compute Franke's function
n = 4
N = 100
x = np.sort(np.random.uniform(0, 1, N))
y = np.sort(np.random.uniform(0, 1, N))
z = FrankeFunction(x, y)
X = create_X(x, y, n=n)
Xpd = pd.DataFrame(X)
# subtract the mean values and set up the covariance matrix
Xpd = Xpd - Xpd.mean()
covariance_matrix = Xpd.cov()
print(covariance_matrix)
We note here that the covariance is zero for the first rows and columns since all matrix elements in the design matrix were set to one (we are fitting the function in terms of a polynomial of degree \( n \)). We would however not include the intercept and wee can simply drop these elements and construct a correlation matrix without them.
We can rewrite the covariance matrix in a more compact form in terms of the design/feature matrix \( \boldsymbol{X} \) as $$ \boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}= \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}]. $$
To see this let us simply look at a design matrix \( \boldsymbol{X}\in {\mathbb{R}}^{2\times 2} \) $$ \boldsymbol{X}=\begin{bmatrix} x_{00} & x_{01}\\ x_{10} & x_{11}\\ \end{bmatrix}=\begin{bmatrix} \boldsymbol{x}_{0} & \boldsymbol{x}_{1}\\ \end{bmatrix}. $$
If we then compute the expectation value $$ \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}=\begin{bmatrix} x_{00}^2+x_{01}^2 & x_{00}x_{10}+x_{01}x_{11}\\ x_{10}x_{00}+x_{11}x_{01} & x_{10}^2+x_{11}^2\\ \end{bmatrix}, $$ which is just $$ \boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]=\begin{bmatrix} \mathrm{var}[\boldsymbol{x}_0] & \mathrm{cov}[\boldsymbol{x}_0,\boldsymbol{x}_1] \\ \mathrm{cov}[\boldsymbol{x}_1,\boldsymbol{x}_0] & \mathrm{var}[\boldsymbol{x}_1] \\ \end{bmatrix}, $$ where we wrote $$\boldsymbol{C}[\boldsymbol{x}_0,\boldsymbol{x}_1] = \boldsymbol{C}[\boldsymbol{x}]$$ to indicate that this the covariance of the vectors \( \boldsymbol{x} \) of the design/feature matrix \( \boldsymbol{X} \).
It is easy to generalize this to a matrix \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \).
We have that the covariance matrix (the correlation matrix involves a simple rescaling) is given as $$ \boldsymbol{C}[\boldsymbol{x}] = \frac{1}{n}\boldsymbol{X}^T\boldsymbol{X}= \mathbb{E}[\boldsymbol{X}^T\boldsymbol{X}]. $$ Let us now assume that we can perform a series of orthogonal transformations where we employ some orthogonal matrices \( \boldsymbol{S} \). These matrices are defined as \( \boldsymbol{S}\in {\mathbb{R}}^{p\times p} \) and obey the orthogonality requirements \( \boldsymbol{S}\boldsymbol{S}^T=\boldsymbol{S}^T\boldsymbol{S}=\boldsymbol{I} \). The matrix can be written out in terms of the column vectors \( \boldsymbol{s}_i \) as \( \boldsymbol{S}=[\boldsymbol{s}_0,\boldsymbol{s}_1,\dots,\boldsymbol{s}_{p-1}] \) and \( \boldsymbol{s}_i \in {\mathbb{R}}^{p} \).
Assume also that there is a transformation \( \boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}=\boldsymbol{C}[\boldsymbol{y}] \) such that the new matrix \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal with elements \( [\lambda_0,\lambda_1,\lambda_2,\dots,\lambda_{p-1}] \).
That is we have $$ \boldsymbol{C}[\boldsymbol{y}] = \mathbb{E}[\boldsymbol{S}^T\boldsymbol{X}^T\boldsymbol{X}T\boldsymbol{S}]=\boldsymbol{S}^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}, $$ since the matrix \( \boldsymbol{S} \) is not a data dependent matrix. Multiplying with \( \boldsymbol{S} \) from the left we have $$ \boldsymbol{S}\boldsymbol{C}[\boldsymbol{y}] = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}, $$ and since \( \boldsymbol{C}[\boldsymbol{y}] \) is diagonal we have for a given eigenvalue \( i \) of the covariance matrix that $$ \boldsymbol{S}_i\lambda_i = \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{S}_i. $$
In the derivation of the PCA theorem we will assume that the eigenvalues are ordered in descending order, that is \( \lambda_0 > \lambda_1 > \dots > \lambda_{p-1} \).
The eigenvalues tell us then how much we need to stretch the corresponding eigenvectors. Dimensions with large eigenvalues have thus large variations (large variance) and define therefore useful dimensions. The data points are more spread out in the direction of these eigenvectors. Smaller eigenvalues mean on the other hand that the corresponding eigenvectors are shrunk accordingly and the data points are tightly bunched together and there is not much variation in these specific directions. Hopefully then we could leave it out dimensions where the eigenvalues are very small. If \( p \) is very large, we could then aim at reducing \( p \) to \( l < < p \) and handle only \( l \) features/predictors.
Here's how we would proceed in setting up the algorithm for the PCA, see also discussion below here.
We will use a simple example first with two-dimensional data drawn from a multivariate normal distribution with the following mean and covariance matrix (we have fixed these quantities but will play around with them below): $$ \mu = (-1,2) \qquad \Sigma = \begin{bmatrix} 4 & 2 \\ 2 & 2 \end{bmatrix} $$ Note that the mean refers to each column of data. We will generate \( n = 10000 \) points \( X = \{ x_1, \ldots, x_N \} \) from this distribution, and store them in the \( 1000 \times 2 \) matrix \( \boldsymbol{X} \). This is our design matrix where we have forced the covariance and mean values to take specific values.
The following Python code aids in setting up the data and writing out the design matrix. Note that the function multivariate returns also the covariance discussed above and that it is defined by dividing by \( n-1 \) instead of \( n \).
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from IPython.display import display
n = 10000
mean = (-1, 2)
cov = [[4, 2], [2, 2]]
X = np.random.multivariate_normal(mean, cov, n)
Now we are going to implement the PCA algorithm. We will break it down into various substeps.
The first step of PCA is to compute the sample mean of the data and use it to center the data. Recall that the sample mean is $$ \mu_n = \frac{1}{n} \sum_{i=1}^n x_i $$ and the mean-centered data \( \bar{X} = \{ \bar{x}_1, \ldots, \bar{x}_n \} \) takes the form $$ \bar{x}_i = x_i - \mu_n. $$ When you are done with these steps, print out \( \mu_n \) to verify it is close to \( \mu \) and plot your mean centered data to verify it is centered at the origin! The following code elements perform these operations using pandas or using our own functionality for doing so. The latter, using numpy is rather simple through the mean() function.
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
# we center it ourselves
X_centered = X - X.mean(axis=0)
Alternatively, we could use the functions we discussed earlier for scaling the data set. That is, we could have used the StandardScaler function in Scikit-Learn, a function which ensures that for each feature/predictor we study the mean value is zero and the variance is one (every column in the design/feature matrix). You would then not get the same results, since we divide by the variance. The diagonal covariance matrix elements will then be one, while the non-diagonal ones need to be divided by \( 2\sqrt{2} \) for our specific case.
Now we are going to use the mean centered data to compute the sample covariance of the data by using the following equation $$ \begin{equation*} \Sigma_n = \frac{1}{n-1} \sum_{i=1}^n \bar{x}_i^T \bar{x}_i = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_n)^T (x_i - \mu_n) \end{equation*} $$ where the data points \( x_i \in \mathbb{R}^p \) (here in this example \( p = 2 \)) are column vectors and \( x^T \) is the transpose of \( x \). We can write our own code or simply use either the functionaly of numpy or that of pandas, as follows
print(df.cov())
print(np.cov(X_centered.T))
Note that the way we define the covariance matrix here has a factor \( n-1 \) instead of \( n \). This is included in the cov() function by numpy and pandas. Our own code here is not very elegant and asks for obvious improvements. It is tailored to this specific \( 2\times 2 \) covariance matrix.
# extract the relevant columns from the centered design matrix of dim n x 2
x = X_centered[:,0]
y = X_centered[:,1]
Cov = np.zeros((2,2))
Cov[0,1] = np.sum(x.T@y)/(n-1.0)
Cov[0,0] = np.sum(x.T@x)/(n-1.0)
Cov[1,1] = np.sum(y.T@y)/(n-1.0)
Cov[1,0]= Cov[0,1]
print("Centered covariance using own code")
print(Cov)
plt.plot(x, y, 'x')
plt.axis('equal')
plt.show()
Depending on the number of points \( n \), we will get results that are close to the covariance values defined above. The plot shows how the data are clustered around a line with slope close to one. Is this expected? Try to change the covariance and the mean values. For example, try to make the variance of the first element much larger than that of the second diagonal element. Try also to shrink the covariance (the non-diagonal elements) and see how the data points are distributed.
Now we are ready to solve for the principal components! To do so we diagonalize the sample covariance matrix \( \Sigma \). We can use the function np.linalg.eig to do so. It will return the eigenvalues and eigenvectors of \( \Sigma \). Once we have these we can perform the following tasks:
Collecting all these steps we can write our own PCA function and compare this with the functionality included in Scikit-Learn.
The code here outlines some of the elements we could include in the analysis. Feel free to extend upon this in order to address the above questions.
# diagonalize and obtain eigenvalues, not necessarily sorted
EigValues, EigVectors = np.linalg.eig(Cov)
# sort eigenvectors and eigenvalues
#permute = EigValues.argsort()
#EigValues = EigValues[permute]
#EigVectors = EigVectors[:,permute]
print("Eigenvalues of Covariance matrix")
for i in range(2):
print(EigValues[i])
FirstEigvector = EigVectors[:,0]
SecondEigvector = EigVectors[:,1]
print("First eigenvector")
print(FirstEigvector)
print("Second eigenvector")
print(SecondEigvector)
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2Dsl = pca.fit_transform(X)
print("Eigenvector of largest eigenvalue")
print(pca.components_.T[:, 0])
This code does not contain all the above elements, but it shows how we can use Scikit-Learn to extract the eigenvector which corresponds to the largest eigenvalue. Try to address the questions we pose before the above code. Try also to change the values of the covariance matrix by making one of the diagonal elements much larger than the other. What do you observe then?
We assume now that we have a design matrix \( \boldsymbol{X} \) which has been centered as discussed above. For the sake of simplicity we skip the overline symbol. The matrix is defined in terms of the various column vectors \( [\boldsymbol{x}_0,\boldsymbol{x}_1,\dots, \boldsymbol{x}_{p-1}] \) each with dimension \( \boldsymbol{x}\in {\mathbb{R}}^{n} \).
The PCA theorem states that minimizing the above reconstruction error corresponds to setting \( \boldsymbol{W}=\boldsymbol{S} \), the orthogonal matrix which diagonalizes the empirical covariance(correlation) matrix. The optimal low-dimensional encoding of the data is then given by a set of vectors \( \boldsymbol{z}_i \) with at most \( l \) vectors, with \( l < < p \), defined by the orthogonal projection of the data onto the columns spanned by the eigenvectors of the covariance(correlations matrix).
To show the PCA theorem let us start with the assumption that there is one vector \( \boldsymbol{s}_0 \) which corresponds to a solution which minimized the reconstruction error \( J \). This is an orthogonal vector. It means that we now approximate the reconstruction error in terms of \( \boldsymbol{w}_0 \) and \( \boldsymbol{z}_0 \) as
We are almost there, we have obtained a relation between minimizing the reconstruction error and the variance and the covariance matrix. Minimizing the error is equivalent to maximizing the variance of the projected data.
We could trivially maximize the variance of the projection (and thereby minimize the error in the reconstruction function) by letting the norm-2 of \( \boldsymbol{w}_0 \) go to infinity. However, this norm since we want the matrix \( \boldsymbol{W} \) to be an orthogonal matrix, is constrained by \( \vert\vert \boldsymbol{w}_0 \vert\vert_2^2=1 \). Imposing this condition via a Lagrange multiplier we can then in turn maximize $$ J(\boldsymbol{w}_0)= \boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0+\lambda_0(1-\boldsymbol{w}_0^T\boldsymbol{w}_0). $$ Taking the derivative with respect to \( \boldsymbol{w}_0 \) we obtain $$ \frac{\partial J(\boldsymbol{w}_0)}{\partial \boldsymbol{w}_0}= 2\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0-2\lambda_0\boldsymbol{w}_0=0, $$ meaning that $$ \boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0\boldsymbol{w}_0. $$ The direction that maximizes the variance (or minimizes the construction error) is an eigenvector of the covariance matrix! If we left multiply with \( \boldsymbol{w}_0^T \) we have the variance of the projected data is $$ \boldsymbol{w}_0^T\boldsymbol{C}[\boldsymbol{x}]\boldsymbol{w}_0=\lambda_0. $$
If we want to maximize the variance (minimize the construction error) we simply pick the eigenvector of the covariance matrix with the largest eigenvalue. This establishes the link between the minimization of the reconstruction function \( J \) in terms of an orthogonal matrix and the maximization of the variance and thereby the covariance of our observations encoded in the design/feature matrix \( \boldsymbol{X} \).
The proof for the other eigenvectors \( \boldsymbol{w}_1,\boldsymbol{w}_2,\dots \) can be established by applying the above arguments and using the fact that our basis of eigenvectors is orthogonal, see Murphy chapter 12.2. The discussion in chapter 12.2 of Murphy's text has also a nice link with the Singular Value Decomposition theorem. For categorical data, see chapter 12.4 and discussion therein.
For more details, see for example Vidal, Ma and Sastry, chapter 2.
For a detailed demonstration of the geometric interpretation, see Vidal, Ma and Sastry, section 2.1.2.
Principal Component Analysis (PCA) is by far the most popular dimensionality reduction algorithm. First it identifies the hyperplane that lies closest to the data, and then it projects the data onto it.
The following Python code uses NumPy’s svd() function to obtain all the principal components of the training set, then extracts the first two principal components. First we center the data using either pandas or our own code
import numpy as np
import pandas as pd
from IPython.display import display
np.random.seed(100)
# setting up a 10 x 5 vanilla matrix
rows = 10
cols = 5
X = np.random.randn(rows,cols)
df = pd.DataFrame(X)
# Pandas does the centering for us
df = df -df.mean()
display(df)
# we center it ourselves
X_centered = X - X.mean(axis=0)
# Then check the difference between pandas and our own set up
print(X_centered-df)
#Now we do an SVD
U, s, V = np.linalg.svd(X_centered)
c1 = V.T[:, 0]
c2 = V.T[:, 1]
W2 = V.T[:, :2]
X2D = X_centered.dot(W2)
print(X2D)
PCA assumes that the dataset is centered around the origin. Scikit-Learn’s PCA classes take care of centering the data for you. However, if you implement PCA yourself (as in the preceding example), or if you use other libraries, don’t forget to center the data first.
Once you have identified all the principal components, you can reduce the dimensionality of the dataset down to \( d \) dimensions by projecting it onto the hyperplane defined by the first \( d \) principal components. Selecting this hyperplane ensures that the projection will preserve as much variance as possible.
W2 = V.T[:, :2]
X2D = X_centered.dot(W2)
Scikit-Learn’s PCA class implements PCA using SVD decomposition just like we did before. The following code applies PCA to reduce the dimensionality of the dataset down to two dimensions (note that it automatically takes care of centering the data):
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2D = pca.fit_transform(X)
print(X2D)
After fitting the PCA transformer to the dataset, you can access the principal components using the components variable (note that it contains the PCs as horizontal vectors, so, for example, the first principal component is equal to
pca.components_.T[:, 0]
Another very useful piece of information is the explained variance ratio of each principal component, available via the \( explained\_variance\_ratio \) variable. It indicates the proportion of the dataset’s variance that lies along the axis of each principal component.
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
cancer = load_breast_cancer()
X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
logreg = LogisticRegression()
logreg.fit(X_train, y_train)
print("Train set accuracy from Logistic Regression: {:.2f}".format(logreg.score(X_train,y_train)))
# We scale the data
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
# Then perform again a log reg fit
logreg.fit(X_train_scaled, y_train)
print("Train set accuracy scaled data: {:.2f}".format(logreg.score(X_train_scaled,y_train)))
#thereafter we do a PCA with Scikit-learn
from sklearn.decomposition import PCA
pca = PCA(n_components = 2)
X2D_train = pca.fit_transform(X_train_scaled)
# and finally compute the log reg fit and the score on the training data
logreg.fit(X2D_train,y_train)
print("Train set accuracy scaled and PCA data: {:.2f}".format(logreg.score(X2D_train,y_train)))
We see that our training data after the PCA decomposition has a performance similar to the non-scaled data.
Instead of arbitrarily choosing the number of dimensions to reduce down to, it is generally preferable to choose the number of dimensions that add up to a sufficiently large portion of the variance (e.g., 95%). Unless, of course, you are reducing dimensionality for data visualization — in that case you will generally want to reduce the dimensionality down to 2 or 3. The following code computes PCA without reducing dimensionality, then computes the minimum number of dimensions required to preserve 95% of the training set’s variance:
pca = PCA()
pca.fit(X)
cumsum = np.cumsum(pca.explained_variance_ratio_)
d = np.argmax(cumsum >= 0.95) + 1
You could then set \( n\_components=d \) and run PCA again. However, there is a much better option: instead of specifying the number of principal components you want to preserve, you can set \( n\_components \) to be a float between 0.0 and 1.0, indicating the ratio of variance you wish to preserve:
pca = PCA(n_components=0.95)
X_reduced = pca.fit_transform(X)
One problem with the preceding implementation of PCA is that it requires the whole training set to fit in memory in order for the SVD algorithm to run. Fortunately, Incremental PCA (IPCA) algorithms have been developed: you can split the training set into mini-batches and feed an IPCA algorithm one minibatch at a time. This is useful for large training sets, and also to apply PCA online (i.e., on the fly, as new instances arrive).
Scikit-Learn offers yet another option to perform PCA, called Randomized PCA. This is a stochastic algorithm that quickly finds an approximation of the first d principal components. Its computational complexity is \( O(m \times d^2)+O(d^3) \), instead of \( O(m \times n^2) + O(n^3) \), so it is dramatically faster than the previous algorithms when \( d \) is much smaller than \( n \).
The kernel trick is a mathematical technique that implicitly maps instances into a very high-dimensional space (called the feature space), enabling nonlinear classification and regression with Support Vector Machines. Recall that a linear decision boundary in the high-dimensional feature space corresponds to a complex nonlinear decision boundary in the original space. It turns out that the same trick can be applied to PCA, making it possible to perform complex nonlinear projections for dimensionality reduction. This is called Kernel PCA (kPCA). It is often good at preserving clusters of instances after projection, or sometimes even unrolling datasets that lie close to a twisted manifold. For example, the following code uses Scikit-Learn’s KernelPCA class to perform kPCA with an
from sklearn.decomposition import KernelPCA
rbf_pca = KernelPCA(n_components = 2, kernel="rbf", gamma=0.04)
X_reduced = rbf_pca.fit_transform(X)
Locally Linear Embedding (LLE) is another very powerful nonlinear dimensionality reduction (NLDR) technique. It is a Manifold Learning technique that does not rely on projections like the previous algorithms. In a nutshell, LLE works by first measuring how each training instance linearly relates to its closest neighbors (c.n.), and then looking for a low-dimensional representation of the training set where these local relationships are best preserved (more details shortly).
There are many other dimensionality reduction techniques, several of which are available in Scikit-Learn.
Here are some of the most popular: