- #1
ecastro
- 254
- 8
I tried calculating the partial derivative of
##\varphi\left(x, y\right) = \sum_\lambda\left\{H\left(\lambda\right) \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2\right\}##
with respect to ##a_n## and equating it to zero to minimise the function (please check my computation).
\begin{eqnarray*}
\frac{\partial \varphi\left(x, y\right)}{\partial a_n} &=& \frac{\partial}{\partial a_n} \sum_\lambda\left\{H\left(\lambda\right) \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2\right\} \\
&=& \sum_\lambda \frac{\partial}{\partial a_n} \left\{H\left(\lambda\right) \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2\right\} \\
&=& \sum_\lambda H\left(\lambda\right) \frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 + \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 \frac{\partial}{\partial a_n} H\left(\lambda\right)
\end{eqnarray*}
Since ##H\left(\lambda\right)## is not a function of ##a_n##, its partial derivative is zero, and thus the second term vanishes. For the partial differentiation in the first term
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 &=& 2\left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] \times \frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right].
\end{eqnarray*}
The second term of the previous equation is
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] &=& \frac{\partial}{\partial a_n} C_E\left(\lambda; x, y\right) + \frac{\partial}{\partial a_n} \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)
\end{eqnarray*}
The first term is not a function of ##a_n##, thus it also vanishes, while the second term is
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \sum_n a_n\left(x, y\right) e_n\left(\lambda\right) &=& \sum_n \left[\frac{\partial}{\partial a_n} a_n\left(x, y\right) e_n\left(\lambda\right)\right] \\
&=& \sum_n \left[a_n\left(x, y\right)\frac{\partial}{\partial a_n} e_n\left(\lambda\right) + e_n\left(\lambda\right) \frac{\partial}{\partial a_n} a_n\left(x, y\right)\right] \\
&=& \sum_n e_n\left(\lambda\right)
\end{eqnarray*}
Plugging-in everything back,
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] &=& \sum_n e_n\left(\lambda\right) \\
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 &=& 2\left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] \times \sum_n e_n\left(\lambda\right) \\
\frac{\partial \varphi\left(x, y\right)}{\partial a_n} &=& \sum_\lambda H\left(\lambda\right) \times 2\left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] \times \sum_m e_m\left(\lambda\right) \\
&=& \sum_\lambda \left[2 H\left(\lambda\right) C_E\left(\lambda; x, y\right) \sum_m e_m\left(\lambda\right) + 2 H\left(\lambda\right) \sum_n a_n\left(x, y\right) e_n\left(\lambda\right) \sum_m e_m\left(\lambda\right)\right]
\end{eqnarray*}
Note that I changed the last summation from ##n## to ##m## to avoid confusion with the other summation.
From here, I need to equate it to zero to minimise the function and solve for ##a_n##, but I don't know how.
##\varphi\left(x, y\right) = \sum_\lambda\left\{H\left(\lambda\right) \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2\right\}##
with respect to ##a_n## and equating it to zero to minimise the function (please check my computation).
\begin{eqnarray*}
\frac{\partial \varphi\left(x, y\right)}{\partial a_n} &=& \frac{\partial}{\partial a_n} \sum_\lambda\left\{H\left(\lambda\right) \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2\right\} \\
&=& \sum_\lambda \frac{\partial}{\partial a_n} \left\{H\left(\lambda\right) \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2\right\} \\
&=& \sum_\lambda H\left(\lambda\right) \frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 + \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 \frac{\partial}{\partial a_n} H\left(\lambda\right)
\end{eqnarray*}
Since ##H\left(\lambda\right)## is not a function of ##a_n##, its partial derivative is zero, and thus the second term vanishes. For the partial differentiation in the first term
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 &=& 2\left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] \times \frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right].
\end{eqnarray*}
The second term of the previous equation is
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] &=& \frac{\partial}{\partial a_n} C_E\left(\lambda; x, y\right) + \frac{\partial}{\partial a_n} \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)
\end{eqnarray*}
The first term is not a function of ##a_n##, thus it also vanishes, while the second term is
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \sum_n a_n\left(x, y\right) e_n\left(\lambda\right) &=& \sum_n \left[\frac{\partial}{\partial a_n} a_n\left(x, y\right) e_n\left(\lambda\right)\right] \\
&=& \sum_n \left[a_n\left(x, y\right)\frac{\partial}{\partial a_n} e_n\left(\lambda\right) + e_n\left(\lambda\right) \frac{\partial}{\partial a_n} a_n\left(x, y\right)\right] \\
&=& \sum_n e_n\left(\lambda\right)
\end{eqnarray*}
Plugging-in everything back,
\begin{eqnarray*}
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] &=& \sum_n e_n\left(\lambda\right) \\
\frac{\partial}{\partial a_n} \left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right]^2 &=& 2\left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] \times \sum_n e_n\left(\lambda\right) \\
\frac{\partial \varphi\left(x, y\right)}{\partial a_n} &=& \sum_\lambda H\left(\lambda\right) \times 2\left[C_E\left(\lambda; x, y\right) + \sum_n a_n\left(x, y\right) e_n\left(\lambda\right)\right] \times \sum_m e_m\left(\lambda\right) \\
&=& \sum_\lambda \left[2 H\left(\lambda\right) C_E\left(\lambda; x, y\right) \sum_m e_m\left(\lambda\right) + 2 H\left(\lambda\right) \sum_n a_n\left(x, y\right) e_n\left(\lambda\right) \sum_m e_m\left(\lambda\right)\right]
\end{eqnarray*}
Note that I changed the last summation from ##n## to ##m## to avoid confusion with the other summation.
From here, I need to equate it to zero to minimise the function and solve for ##a_n##, but I don't know how.