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<a class="navbar-brand" href="week46-bs.html">Week 46: Support Vector Machines and Project 3</a>
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<h2 id="a-quick-reminder-on-lagrangian-multipliers" class="anchor">A quick Reminder on Lagrangian Multipliers </h2>
<p>Consider a function of three independent variables \( f(x,y,z) \) . For the function \( f \) to be an
extreme we have
</p>
$$
df=0.
$$
<p>A necessary and sufficient condition is</p>
$$
\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0,
$$
<p>due to</p>
$$
df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz.
$$
<p>In many problems the variables \( x,y,z \) are often subject to constraints (such as those above for the margin)
so that they are no longer all independent. It is possible at least in principle to use each
constraint to eliminate one variable
and to proceed with a new and smaller set of independent varables.
</p>
<p>The use of so-called Lagrangian multipliers is an alternative technique when the elimination
of variables is incovenient or undesirable. Assume that we have an equation of constraint on
the variables \( x,y,z \)
</p>
$$
\phi(x,y,z) = 0,
$$
<p> resulting in</p>
$$
d\phi = \frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz =0.
$$
<p>Now we cannot set anymore</p>
$$
\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0,
$$
<p>if \( df=0 \) is wanted
because there are now only two independent variables! Assume \( x \) and \( y \) are the independent
variables.
Then \( dz \) is no longer arbitrary.
</p>
<p>
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