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For the function

$$ f: = \cot(\sqrt x) + \frac{1}{\sqrt{x}}$$ with the initial approximate $$x_0 = \pi^2\left(n-\frac{1}{2}\right)^2$$ Show that after one iteration $$ x_{n+1} = \pi^2\left(n-\frac{1}{2}\right)^2\left(1+ \frac{1}{1+ \pi^2\left(n-\frac{1}{2}\right)^2}\right)^2$$

So we have $$ f' = \frac{-x\csc^2(\sqrt{x})-1}{2x^{3/2}}$$

Then

$$ x_{n+1} = \pi^2\left(n-\frac{1}{2}\right)^2 - \frac{\cot(\pi\left(n-\frac{1}{2}\right))+\frac{1}{\pi\left(n-\frac{1}{2}\right)}}{\frac{-\pi^2\left(n-\frac{1}{2}\right)^2\csc^2\left(\pi\left(n-\frac{1}{2}\right)\right)-1}{2\pi^3\left(n-\frac{1}{2}\right)^3}}$$

I think that

$$ \cot\left(\pi\left(n-\frac{1}{2}\right)\right) = 0, \forall n \in \mathbb{N}$$ and $$ \csc^2\left(\pi\left(n-\frac{1}{2}\right)\right) = 1, \forall n \in \mathbb{N}$$

So,

$$ x_{n+1} = \pi^2\left(n-\frac{1}{2}\right)^2 - \frac{\frac{1}{\pi\left(n-\frac{1}{2}\right)}}{\frac{-\pi^2\left(n-\frac{1}{2}\right)^2-1}{2\pi^3\left(n-\frac{1}{2}\right)^3}}$$

$$ = \pi^2\left(n-\frac{1}{2}\right)^2+ \frac{2\pi^2\left(n-\frac{1}{2}\right)^2}{\pi^2\left(n-\frac{1}{2}\right)^2-1}$$

$$ = \pi^2\left(n-\frac{1}{2}\right)^2\left(1 + \frac{2}{1+ \pi^2\left(n-\frac{1}{2}\right)^2}\right)$$

Which is not the correct answer! Can anyone spot a flaw?

  • 0
    are you sure that this is the correct answer? I followed your calculations and get the same results, and your results implies that the answer given by the - what seems to be the solutions manual of your text - solution is not the correct one. – 2017-02-22
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    (+1)Because I'm also interested in the answer and for the question that I think is a very helpful one. – 2017-02-22
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    Yes - solving f in wolfram gives solutions approximately equal to 4 and then 24. Subbing n=1 and n=2 into the 'show that' solution also gives these values – 2017-02-22

1 Answers 1

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From your resulto of $x_1$, I followed the calculations and get the same result and a usefull thing should be this.

If we use that, calling $\Delta := \pi(n-1/2)$

$$\left(1 + \frac{1}{1+\Delta^2}\right)^2 = 1 + \frac{2}{1 + \Delta^2} + \frac{1}{(1 + \Delta^2)^2}$$

From your result we have that

$$x_1 = \Delta^2\left(1 + \frac{1}{1+\Delta^2}\right)^2 - \frac{\Delta^2}{(1 + \Delta^2)^2} = R - C$$

Where $R$ is the answer and $C$ is a positive real number.