calculusnumerical methods
Derive the Newton-Raphson iteration formula for $a^{\frac{1}{5}}$ where $a$ is a real positive number and then find $3^{\frac{1}{5}}$ correct to $3$ decimal places.
My attempt:
$f(a)=a^{\frac{1}{5}}$
$f'(a)=\frac{1}{5}a^{-\frac{4}{5}}$
The Newton-Raphson iteration formula:
$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}=a_n-\frac{a^{\frac{1}{5}}}{\frac{1}{5}a^{-\frac{4}{5}}}=-4a_n$$
So each guess is $-4$ times the previous guess. Then clearly the guesses diverge and the Newton-Raphson method fails.
To visualize geometrically what's going on, I will code an interactive diagram with GNU Dr. Geo (free software of mine) from where I can drag the initial value (the red dot) and see how the method converge or not.
For example $x \rightarrow cos x + x$, comes with some mines, but not $x \rightarrow cos x +1.1x$.
When you get close to a flat area, the tangent sends you far away, even further than your initial value.
Your best option is to get close to the root in an area without nul value of the derivative. I guess it is the tough part as it depends on the function. Visualizing can help.
This article explains how to code with Pharo+DrGeo this interactive diagram and the link with the Hero method from the classical period.
Yes, the initial seed matters quite a lot. If you're "close" to one of the roots, you'll converge to that root. Exactly how close is required is complicated, though.
The image below shows the regions of attraction for the cosine function in a neighborhood of the origin in the complex plane. Initial seeds chosen from green region on the left converge to $-\pi/2$ while initial seeds chosen from yellow region on the right converge to $+\pi/2$. As you move closer to the origin, you can converge to points farther away.
Best Answer
Hint:
What you need is a formula to get the $5^{th}$ root of $3$.
So, we have:
$$x^5 = a \rightarrow f(x) = x^5 - a$$
Can you repeat the process for the Newton solution again with $a = 3$?
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