Taylor series for $\sqrt[3]{x+1}$ I'm trying to find the Taylor series for $\sqrt[3]{x}$, but since the n-th derivative of the function $$f(x)=\sqrt[3]{x}$$
not definded at $x=0$, I've swithched to $f(x)=\sqrt[3]{x+1}$, This is my Attempt:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)x^n}{n!}$$
$$\sqrt[3]{x+1}=1+\frac{x}{3}-\frac{2x^2}{9\cdot2!}+\frac{10x^3}{27\cdot 3!}-\frac{80x^4}{81\cdot4!}+...$$
but when i plug $x=125 \iff \sqrt[3]{x+1}=\sqrt[3]{126}$ i get something way bigger than the acttual value which is about $5.01...$, you can check it here.
But I can't find the mistake in the equation.
For $x\gt 1$, expand in terms of $\frac1x&lt;1$ to ensure convergence
$$\sqrt[3]{x+1}= \sqrt[3]{x} \sqrt[3]{1+\frac1x}
= \sqrt[3]{x}  (1+\frac{1}{3x}-\frac{2}{9\cdot2!}\frac1{x^2}+\frac{10}{27\cdot 3!}\frac1{x^3}-\frac{80}{81\cdot4!}\frac1{x^4}+...)
$$
Then, set $ x=125$ to obtain the approximation
$$\sqrt[3]{126} =5(1+\frac1{3\cdot 125}-\cdots )=5.0133$$
The binomial series for $(a+x)^\alpha$ is only valid if the absolute value of x is smaller than one, unless $\alpha$ is a non-negative integer. Your value is x=125&gt;&gt;1 and $\alpha$ is not integer, thus the divergence. To get a good approximation you would have to expand around $x_0=125$, which of course requires knowledge of the third root of 126.