Banach contraction theorem solution verification

Question

I have to find the limit of sequence

$ x_0= 2019 $

$ x_n= \frac{3}{5}x_{n-1} + 2 \sqrt{x_{n-1}} +3$

for $n=1,2,3,...$

using Banach theorem.

I assumed that $f(x)=\frac{3}{5}x+2 \sqrt x+3$

First I want to prove that f(x) is a contraction mapping:

EDIT: As spotted in the comment my proof for the inequality was incorect so know I am stuck at:

$|f(x)-f(y)|=|\frac{3}{5}x+2 \sqrt x+3 - \frac{3}{5}y-2 \sqrt y-3|=|\frac{3}{5}(x-y)+2(\sqrt x -\sqrt y)|$

I thought about using the inequality $|\sqrt x- \sqrt y|< \sqrt{|x-y|}$ but I have no idea how to proceed from here.

The next step would be that if f(x) is a contraction mapping then the limit would be a positive solution to

$\frac{3}{5}x+2 \sqrt x+3=x$

After some calculation I get that $x=\frac{5}{2}+\frac{5 \sqrt55}{2}$ and that is our limit. Is this a right way to do solve this problem?

Answer 1

So first we have

$$x_{n+1}=f(x_n)=\frac{3}{5}x_n+2\sqrt{x_n}+3 \ \ \ (0)$$

We have $$ f'(x)=\frac{3}{5}-\dfrac{1}{\sqrt{x}} $$

So $$f(x) \geq \dfrac{48}{5} \ \ \text{when} \ x\geq (\dfrac{3}{5})^2 \ \ \ (2)$$

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Further more we study $f(x)-x$ in order to know the monotony.

$$ f'(x)-1=\dfrac{-2}{5}-\dfrac{1}{\sqrt{x}} \leq 0$$ for $x >0 \ \ \ (2)$

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Now by iteration we show :

• $\forall n \geq 0, x_n>0 $ using (0)
• $\forall n \geq 0, x_{n+1}-x_n\leq 0$ using (3)

So because $x_n$ is decreasing and lower-bounded using (2) it converges.

You know that limits is a fix point of $f$ .

Hence your result.