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i'm trying to solve this question.

Find the sum of the Serias using Fixed-Point iteration: Q I should use Fixed-point iteration in order to find it. $P\ge1$.

Part B - Prove that the order of convergence R - of this serias - is unique(singularity). Proof by contradiction

I've seen this solution for Part A:

Lets call S - the sum, and try to find it. $$g(S)=S=\sqrt{p-(1-p)*S} \Rightarrow S^2 = p-(1-p)*S \Rightarrow S^2 + (1-p)*S-p=0$$ By so solving this quadratic equation we achieve $S=-1$ or $S=p$

This solution cancels S=-1, and says that S=p is the answer.

Couple of questions regarding this solution:

  1. Why is it ok to use Fixed-point iteration in order to find sum of a serias? Till now i've used Fixed-point iteration to find ROOTS of an equation.

  2. Why is it ok to assign random variable S g(S)=S and then say that solving this equation says that S is the sum?

  3. Why canceling the solution of $S=-1$?

  4. Part B - how is it possible to prove that order of convergence is unique? By definition, it's the first derivative that is not 0 in the root (probarely root=p).

Thanks.

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There is no series, thus no sum of a series.

This is an expression with nested square roots, the sequence of finite, truncated expressions has (probably) a limit.

The range of the square root is the non-negative numbers, automatically disqualifying $-1$ from being the limit.

You need to prove the contractivity of $g$. Using elementary transformations and estimates you should be able to get $$ |g(S_1)-g(S_2)|\le\frac{p-1}{2\sqrt{2p-1}}|S_1-S_2| $$ You will need better estimates of the values that the iteration can reach to get contractivity for general $p>1$.

Convergence or divergence close to the fixed point is determined by if $|g'(p)|<1$. Now $g'(p)=\frac{p-1}{\sqrt{p+(p-1)p}}=1-\frac1p$

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