Xrfgtjtk

I have a function $f(x)=cos(x/4)e^{-frac{x^2}{4}}$ that has a fixed point in between [0,1] and I know that a point is fixed if $f(x)=x$. Solving it for $x$ is a manually is messy, through a computer program i get $x=0.825..$, though I think I am only suppose to show that there is one (and only one) fixed point in that interval.

1. To show that there is a fixed point of $f(x)=cos(frac x4)e^{-x^2/4}$ in $[0,1]$

We may show that there is a fixed point of $f(x)$ if there is an $xin[0,1]$ such that $f(x)=x$, i.e. that there is a root of $g(x)=f(x)-x$. We may also show that $g(0)=cos(0)e^{0}=1$ but $g(1)=frac{cos(1/4)}{e^{1/4}}-1approx-0.245$. Then since $g(0)geq0geq g(1)$ and since $f(x)$ has no discontinuities in the interval, it must have at least one fixed point in $[0,1]$ by a change of sign.

2. To numerically evaluate such a fixed point

From the same reasoning in part 1, we may look for the root of $g(x)$. Using the Newton-Raphson method, we may construct a sequence that rapidly converges to the root, with an appropriate initial guess.

$$begin{align}g'(x)&=-frac14e^{-x^2/4}left(sinfrac x4+2xcosfrac x4right)-1quadtext{by the product rule}\
x_{n+1}&=x_n-frac{g(x_n)}{g'(x_n)}quadtext{by the Newton-Raphson method}\
&=x_n+4frac{cosfrac {x_n}4+xe^{x^2/4}}{sinfrac {x_n}4+2x_ncosfrac {x_n}4+e^{x^2/4}}end{align}$$

We may write this as a python3 code as follows and set $x_0=0.8$.

import math as m
def dg(x):
return -0.25*m.exp(-x**2/4)*(m.sin(x/4)+2*x*m.cos(x/4))-1
def g(x):
return m.cos(x/4)*m.exp(-x**2/4)-x
def x(n):
if n==0:return 0.8
else: return x(n-1)-g(x(n-1))/dg(x(n-1))
for n in range(0,6):print(n,x(n))

The output of this program shows us that there is a fixed point at $x=0.82547ldots$ which may again be validated by a change of sign.

3. To show that there is only one such fixed point

We can show that there is only one fixed point of $f(x)$ by noting that for all $xin[0,1]$, all three of $e^{-x^2/4}$, $cosfrac x4$ and $sinfrac x4$ are positive. So $e^{-x^2/4}left(sinfrac x4+2xcosfrac x4right)geq0$ and $g'(x)$ is strictly negative. So there are no turning points of $g(x)$ in $[0,1]$ and it has only one root, meaning that $f(x)$ has only one fixed point.

4. [Partial answer] To analytically evaluate the fixed point

As we can express $cos(x)$ in terms of complex exponentials, we can show that the fixed point, $alpha$, is the solution to the following equation.
$$frac{e^{ix/4}-e^{-ix/4}}{2}e^{-x^2/4}=x$$

This may be simplified further with $y=x/4$.
$$8ye^{4y^2+iy}-e^{2iy}-1=0$$

I think it's unlikely that there will be a simple closed form for $alpha$ and it's likely that it's transcendental too.

If $g:[0,1] to [0,1]$ is continuous, Then $g$ has a fixed point. Try to prove this !

Your function $f$ therefore has a fixed point, this means that $h(x):=f(x)-x$ has a zero in [0,1].

Observe that $h'(x)<0$ for all $ x in [0,1]$. Thus $h$ has only one zero in [0,1]. Hence $f$ has only one fixed point in [0,1]