# Newton's Method, Analytical Formula

Currently I am learning about Netwon's Method. Given the function f: $\frac{1}{5} x^5 - \frac{2}{3}x^3 + x$ and $x^{(0)} = \sqrt{\frac{ 25+2\sqrt{55} }{27} }$, I want to analytically determine the sequence $x^{(n+1)} = \Psi(x^{(n)})$ with $\Psi(x) := x - \frac{f(x)}{f'(x)}$. Can someone show me how this is done in this particular case?

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Interessting starting point, why did you choose that one ? –  Dominic Michaelis Mar 5 '13 at 11:00
butterfly effect. –  TestGuest Mar 5 '13 at 11:42

## 1 Answer

You already have the sequence, given by a recursion. For arbitrary sequence it is not possible to find any formula (even not a recursive one).

As this is a numerical method, why do you want to do it analytical? If you want the analytical solutions use $$\frac{1}{5} x^5 - \frac{2}{3} x^3 + x= x( \frac{1}{5} x^4 - \frac{2}{3} x^2 +1)$$ so $0$ is a solution and the others you get by using pq formal with $z=x^2$

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Thanks. Why is it for arbitrary sequences not possible to find a recursion? Isn't the recursion given by Newton's Method? I.e. can I not construct a recursion via this method? –  TestGuest Mar 5 '13 at 11:03
Oh i mean with arbitrary sequences really arbitrary sequences. Newton methods always give you the recursion, but there are sequences which don't come from an newton iteration –  Dominic Michaelis Mar 5 '13 at 11:05
Thank you, understood –  TestGuest Mar 5 '13 at 11:08
In case you are interested, I figured that the function is always oscillating and hence the values are always x_0,-x_0,x_0 etc. I wonder if there was a way to see this just by inspecting the formula? –  TestGuest Mar 5 '13 at 18:11
the values are something like $0.233$ than $-0.0184$ –  Dominic Michaelis Mar 5 '13 at 18:25