week 48
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@@ -266,9 +266,9 @@ MathJax.Hub.Config({
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<h2 id="___sec5" class="anchor">The equations </h2>
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<p>
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Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables). We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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Suppose we define a polynomial transformation of degree two only. We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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$$
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\boldsymbol{z}_i = \phi(\boldsymbol{x}_i)^T =\left(1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right).
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\boldsymbol{z}_i^T = \phi(\boldsymbol{x}_i)^T =\left[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right].
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$$
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<p>
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@@ -290,7 +290,7 @@ $$
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For the above example, the kernel reads
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$$
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]^T\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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$$
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<p>
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@@ -294,10 +294,10 @@ plt.show()
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<h2 id="___sec5">The equations </h2>
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<p>
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Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables). We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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Suppose we define a polynomial transformation of degree two only. We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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<p> <br>
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$$
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\boldsymbol{z}_i = \phi(\boldsymbol{x}_i)^T =\left(1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right).
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\boldsymbol{z}_i^T = \phi(\boldsymbol{x}_i)^T =\left[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right].
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$$
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<p> <br>
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@@ -327,7 +327,7 @@ $$
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For the above example, the kernel reads
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<p> <br>
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$$
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]^T\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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$$
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<p> <br>
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@@ -346,9 +346,9 @@ plt.show()
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<h2 id="___sec5">The equations </h2>
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<p>
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Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables). We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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Suppose we define a polynomial transformation of degree two only. We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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$$
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\boldsymbol{z}_i = \phi(\boldsymbol{x}_i)^T =\left(1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right).
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\boldsymbol{z}_i^T = \phi(\boldsymbol{x}_i)^T =\left[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right].
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$$
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<p>
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@@ -370,7 +370,7 @@ $$
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For the above example, the kernel reads
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$$
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]^T\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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$$
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<p>
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@@ -351,9 +351,9 @@ plt<span style="color: #666666">.</span>show()
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<h2 id="___sec5">The equations </h2>
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<p>
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Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with \( x_i \) and \( y_i \) as variables). We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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Suppose we define a polynomial transformation of degree two only. We define a vector \( \boldsymbol{x}_i=[x_i,y_i] \) and have
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$$
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\boldsymbol{z}_i = \phi(\boldsymbol{x}_i)^T =\left(1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right).
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\boldsymbol{z}_i^T = \phi(\boldsymbol{x}_i)^T =\left[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right].
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$$
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<p>
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@@ -375,7 +375,7 @@ $$
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For the above example, the kernel reads
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$$
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]^T\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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K(\boldsymbol{x}_i,\boldsymbol{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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$$
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<p>
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Binary file not shown.
@@ -138,7 +138,7 @@
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"source": [
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"## The equations\n",
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"\n",
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"Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with $x_i$ and $y_i$ as variables). We define a vector $\\boldsymbol{x}_i=[x_i,y_i]$ and have"
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"Suppose we define a polynomial transformation of degree two only. We define a vector $\\boldsymbol{x}_i=[x_i,y_i]$ and have"
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]
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},
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{
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@@ -146,7 +146,7 @@
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"metadata": {},
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"source": [
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"$$\n",
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"\\boldsymbol{z}_i = \\phi(\\boldsymbol{x}_i)^T =\\left(1, \\sqrt{2}x_i, \\sqrt{2}y_i, x_i^2, y_i^2, \\sqrt{2}x_iy_i\\right).\n",
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"\\boldsymbol{z}_i^T = \\phi(\\boldsymbol{x}_i)^T =\\left[1, \\sqrt{2}x_i, \\sqrt{2}y_i, x_i^2, y_i^2, \\sqrt{2}x_iy_i\\right].\n",
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"$$"
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]
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},
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@@ -211,7 +211,7 @@
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"metadata": {},
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"source": [
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"$$\n",
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"K(\\boldsymbol{x}_i,\\boldsymbol{x}_j)=[1, \\sqrt{2}x_i, \\sqrt{2}y_i, x_i^2, y_i^2, \\sqrt{2}x_iy_i]^T\\begin{bmatrix} 1\\\\ \\sqrt{2}x_j \\\\ \\sqrt{2}y_j \\\\ x_j^2\\\\ y_i^2\\\\ \\sqrt{2}x_jy_j \\end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.\n",
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"K(\\boldsymbol{x}_i,\\boldsymbol{x}_j)=[1, \\sqrt{2}x_i, \\sqrt{2}y_i, x_i^2, y_i^2, \\sqrt{2}x_iy_i]\\begin{bmatrix} 1\\\\ \\sqrt{2}x_j \\\\ \\sqrt{2}y_j \\\\ x_j^2\\\\ y_i^2\\\\ \\sqrt{2}x_jy_j \\end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.\n",
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"$$"
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]
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},
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@@ -119,10 +119,10 @@ plt.show()
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!split
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===== The equations =====
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Suppose we define a polynomial transformation of degree two only (we continue to live in a plane with $x_i$ and $y_i$ as variables). We define a vector $\bm{x}_i=[x_i,y_i]$ and have
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Suppose we define a polynomial transformation of degree two only. We define a vector $\bm{x}_i=[x_i,y_i]$ and have
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!bt
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\[
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\bm{z}_i = \phi(\bm{x}_i)^T =\left(1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right).
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\bm{z}_i^T = \phi(\bm{x}_i)^T =\left[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i\right].
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\]
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!et
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@@ -148,7 +148,7 @@ K(\bm{x}_i,\bm{x}_j)=\bm{z}_i^T\bm{z}_j= \phi(\bm{x}_i)^T\phi(\bm{x}_j).
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For the above example, the kernel reads
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!bt
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\[
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K(\bm{x}_i,\bm{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]^T\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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K(\bm{x}_i,\bm{x}_j)=[1, \sqrt{2}x_i, \sqrt{2}y_i, x_i^2, y_i^2, \sqrt{2}x_iy_i]\begin{bmatrix} 1\\ \sqrt{2}x_j \\ \sqrt{2}y_j \\ x_j^2\\ y_i^2\\ \sqrt{2}x_jy_j \end{bmatrix}=1+2x_ix_j+2y_iy_j+(x_ix_j)^2+(y_iy_j)^2+2x_ix_jy_iy_j.
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\]
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!et
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