Prove by fixed point theorem that f(x) =exp(-x/2) has a unique fixed point in interval [0, 1]

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Introduction : In this solution, we will apply fixed point theorem to prove that f(x)=exp(-x/2) has a unique fixed point in interval [0, 1].

Question : Prove by fixed point theorem that there exists a unique fixed point of

\[ f(x) =e^{-x/2}\]

in interval [0, 1].

Proof :

Existence of fixed point

Define

\[f(x)=x-g(x)\]

\[ f(x)=x-e^{-x/2} \]

Since \(f(0)=-1<0, f(1)=0.393>0\) and f(x) is continuous. By mean value theorem there exist a point p in [0, 1]. Such that

\[f(p)=0 \]

\[p-g(p)=0\]

\[ g(p)=p \]

So p is fixed point of g.

Uniquness of fixed point

\[f(x)=e^{-x/2}\]

Since \(0≤f(x)≤1\) for all x∈[0,1]

Let x, y be any points in [0, 1], then

\[\left|f\left(x\right)-f\left(y\right)\right|=\left|e^{-\frac{x}{2}}-e^{-\frac{y}{2}}\right|\]

\[=\left|\left(1-\frac{x}{2}+\frac{1}{2!}\left(\frac{x}{2}\right)^{2}-\frac{1}{3!}\left(\frac{x}{2}\right)^{3}+…\right)-\left(1-\frac{y}{2}+\frac{1}{2!}\left(\frac{y}{2}\right)^{2}-\frac{1}{3!}\left(\frac{y}{2}\right)^{3}+…\right)\right|\]

\[=\left|\frac{1}{2}\left(x-y\right)+higher order terms\right|\]

\[\le \left|\frac{1}{2}\left(x-y\right)\right|\]

Since \(K =\frac{1}{2}<1\)

Hence f(x) is a contraction map.

So by fixed point theorem f(x) has a unique fixed point in [0, 1].


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