n-th root of x

[Imnot Arobot No.8306866]

>see iteration for sqrt(x)
>generalize to n-th root
amirite?
[math] \displaystyle \lim_{x \rightarrow \infty}S_i=S_{ \infty}= \sqrt[n]{x}[/math]

[Imnot Arobot No.8306868]

>tested a few values
converges quickly to correct result

[Imnot Arobot No.8306876]

oops, should be
[math] \displaystyle \lim_{i \rightarrow \infty}S_i=S_{ \infty}= \sqrt[n]{x}[/math]

[Anonymous No.8307176]

To solve an equation

[math] f(x) = 0 [/math]

you may want to try and rewrite it as an equation

[math] F(x) = x [/math]

and, if F fulfills some convexity conditions, the solution then is

[math] x = F(...F(F(F(FF(F(i))))))= [/math]

where i is some initial value.
https://en.wikipedia.org/wiki/Fixed-point_theorem

[math] x - s^n = 0 [/math]

[math] x \, s^{n-1} - s = 0 [/math]

(we may no intruduce an auxilliary k)

[math] x \, s^{n-1} - s + k\, s = k\,s [/math]

[math] \dfrac{1}{k} \left( x \, s^{n-1} - s + k\, s \right) = s [/math]

[math] \dfrac{1}{k} \left( x \, s^{n-1} + (k-1)\, s \right) = s [/math]

Now, for fixed x,

[math] F(s) := \dfrac{1}{k} \left( x \, s^{n-1} + (k-1)\, s \right) [/math]

applied to some start value i and iterated a lot will go to x solving

[math] x - s^n = 0 [/math]

i.e. the n'th square root.
It's even a more general expression than your, as you had k=n. But yoou can make it even more variable and if F fulfills some conditions the limit will still be the solution.

[Anonymous No.8307182]

To solve an equation

[math] f(x) = 0 [/math]

you may want to try and rewrite it as an equation

[math] F(x) = x [/math]

and, if F fulfills some convexity conditions, the solution then is

[math] x = F(...F(F(F(FF(F(i))))))= [/math]

where i is some initial value.
https://en.wikipedia.org/wiki/Fixed-point_theorem

[math] x - s^n = 0 [/math]

[math] x \, s^{1-n} - s = 0 [/math]

(we may no introduce an auxiliary k)

[math] x \, s^{1-n} - s + k\, s = k\,s [/math]

[math] \dfrac{1}{k} \left( x \, s^{1-n} - s + k\, s \right) = s [/math]

[math] \dfrac{1}{k} \left( x \, s^{1-n} + (k-1)\, s \right) = s [/math]

Now, for fixed x,

[math] F(s) := \dfrac{1}{k} \left( x \, s^{1-n} + (k-1)\, s \right) [/math]

applied to some start value i and iterated a lot will go to x solving

[math] x - s^n = 0 [/math]

i.e. the n'th square root.
It's even a more general expression than your, as you had k=n. But you can make it even more variable and if F fulfills some conditions the limit will still be the solution.

[Anonymous No.8307190]

will go to s solving*

[Imnot Arobot No.8308162]

forgot to mention
[math]S_1 =1[/math]

[Imnot Arobot No.8308163]

>>8307182
wat

[Anonymous No.8308191]

>>8306866
>CS major detected

If Si-1 is the nth root, what does an extra iteration do?

[Anonymous No.8308203]

>>8307182
Just use newton's method

[Anonymous No.8308257]

>>8308203
this is the easier way to do it.
additionally this proves quadratic convergence of the iteration for suitable initial values.

[Anonymous No.8308593]

>>8308163
wat