Quantum Chemistry/Finding maxima

The Process of finding a maximum of a function

The maximum of a function is found by first computing the derivative of the function $f^{\prime }(x)$, as it describes the rate of change of the original function $f(x)$. At a maximum, the slope of the function changes from positive where the function is increasing to negative where the function is decreasing. Therefore, the slope is equal to zero at a maximum: $f^{\prime }(x)=0$. The points at which $f^{\prime }(x)=0$ are denoted as critical points. It is important to note that not all critical points are maximums since there are other instances where the slope is equal to zero, such as a minimum or inflection point of the function. To determine whether a critical point is a maximum, the second derivative of the function $f^{″}(x)$, must be computed. It describes the concavity of the function and how its slope is changing. If $f^{″}(x)<0$, the function is concave down and the critical point is a maximum.

Example Problem

Problem

Find the maximum of the function

$r^2\exp (-r^2)$

Solution

Determining the critical points

Compute the derivative $f^{\prime }(r)$ of the function.

Since $r^2\exp (-r^2)$ takes the form $f(r)\cdot g(r)$, it can be differentiated using the product rule.

Product Rule $\frac{d}{dr}[f(r)\cdot g(r)]=f^{\prime }(r)\cdot g(r)+g^{\prime }(r)\cdot f(r)$

$\frac{d}{dr}[r^2\exp (-r^2)]=\frac{d}{dr}(r^2)\exp (-r^2)+(r^2)\frac{d}{dr}\exp (-r^2)$

$f^{\prime }(r)=2r\exp (-r^2)-2r^3\exp (-r^2)$

$f^{\prime }(r)=2r\exp (-r^2)(1-r^2)$

Since a maximum occurs when the slope is zero,

$2r\exp (-r^2)(1-r^2)=0$

The points at which the function equal zero are the critical points. Since $\exp (-r^2)$ will never equal zero,

$2r=0$ when $r=0$

$1-r^2=0$ when $r=\pm 1$

Therefore, the critical points are $0,-1,1$

Evaluating the critical points

Compute the second derivative $f^{″}(x)$ of the function.

Since $f^{\prime }(r)=2r\exp (-r^2)(1-r^2)$ again takes the form $f(r)\cdot g(r)$,

$\frac{d}{dr}[2r\exp (-r^2)(1-r^2)]=\frac{d}{dr}(2r\exp (-r^2))(1-r^2)+(2r\exp (-r^2))\frac{d}{dr}(1-r^2)$

$f^{″}(r)=(2\exp (-r^2)-4r^2\exp (-r^2))(1-r^2)+(2r\exp (-r^2))(-2r)$

Evaluating $f^{″}(r)$ at the critical points determines the concavity of the function.

When $f^{″}(r)<0$, the critical point is a maximum.

At $r=0$, $f^{″}(0)=2$.

At $r=1$, $f^{″}(1)=-4\exp (-1)$

At $r=-1$, $f^{″}(-1)=-4\exp (-1)$

Since $-4\exp (-1)<0$, the points $r=1,-1$ are maximums.