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Consider the function

$$K(u) = \frac 1 {\sqrt {2\pi}} \left( \Bbb e ^{-\frac 1 2 \left( \frac {u-5.3} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-1.6} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-2.1} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-1.7} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-1.9} h \right)^2 } \right) .$$

I know how to compute the derivative of each term in order to find its extrema, but how should I proceed in order to find the extrema of the whole sum?

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  • $\begingroup$ You may just differentiate each term. But you will end up with numerical values only. $\endgroup$ – Olivier Oloa Dec 22 '16 at 21:39
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There is no hope of explicitly finding an expression for a root, so I'm going to get an expression that will be easier to evaluate.

Your sum, ignoring the constant factor, is a case of

$\begin{array}\\ f(x) &=\sum_{k=1}^n e^{-\frac12\left(\frac{x-a_k}{h}\right)^2}\\ &=\sum_{k=1}^n e^{-\frac12\left(\frac{x^2-2xa_k+a_k^2}{h^2}\right)}\\ &=e^{-\frac{x^2}{2h^2}}\sum_{k=1}^n e^{-\frac12\left(\frac{-2xa_k+a_k^2}{h^2}\right)}\\ \end{array} $

Differentiating,

$\begin{array}\\ f'(x) &=-\sum_{k=1}^n \frac12\left(\frac{2x-2a_k}{h^2}\right) e^{-\frac12\left(\frac{x^2-2xa_k+a_k^2}{h^2}\right)}\\ &=e^{-\frac12\left(\frac{x^2}{h^2}\right)}\sum_{k=1}^n -\left(\frac{x-a_k}{h^2}\right)e^{-\frac12\left(\frac{-2xa_k+a_k^2}{h^2}\right)}\\ &=e^{-\frac{x^2}{2h^2}}\sum_{k=1}^n -\left(\frac{x-a_k}{h^2}\right)e^{-\left(\frac{-2xa_k+a_k^2}{2h^2}\right)}\\ &=-e^{-\frac{x^2}{2h^2}}\sum_{k=1}^n \left(\frac{x-a_k}{h^2}\right)e^{-\left(\frac{a_k^2}{2h^2}\right)}e^{\left(\frac{xa_k}{h^2}\right)}\\ &=-e^{-\frac{x^2}{2h^2}}\sum_{k=1}^n \left(\frac{x-a_k}{h^2}\right)e^{-\left(\frac{a_k^2}{2h^2}\right)}\left(e^{\left(\frac{a_k}{h^2}\right)}\right)^x\\ &=-e^{-\frac{x^2}{2h^2}}\sum_{k=1}^n \left(\frac{x-a_k}{h^2}\right)e^{-b_k}c_k^x \qquad\text{where }b_k = \frac{a_k^2}{2h^2}, c_k=e^{\left(\frac{a_k}{h^2}\right)}\\ \end{array} $

You can split the sum into $\sum_{k=1}^n \left(\frac{x-a_k}{h^2}\right)e^{-b_k}c_k^x =\frac{x}{h^2}\sum_{k=1}^n e^{-b_k}c_k^x-\sum_{k=1}^n \frac{a_k}{h^2}e^{-b_k}c_k^x $ which might help a little.

I don't see anything further.

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A desmos graph of the function and its derivative is somewhat instructive.

https://www.desmos.com/calculator/h4y1mlc4fl

The value of $h$ is controlled by a slider.

One can see that it has a global maximum but not a global minimum.

For certain values of $h$ it does have a local minimum and two local maxima. For values of $h$ near $0$ it has multiple local maxima.

I suspect that an analytical solution will be somewhat difficult, but one can obtain numerical values from the intersection of the $x$-axis with the graph of the derivative by clicking on the $x$-axis or on the graph of the derivative (in red).

I am including a snapshot of both the graph and its derivative for $h=1.$

enter image description here

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