function f is given by an equation: $$f(x)=\frac{1}{3+x^3}$$ Find the taylor expansion in a point $x_0=0$ and calculate radiu of the convergence. Could you explain how to find taylor expansion of such series? and thus how to find the radius of convrgence?

  • $\begingroup$ For the radius of convergence, you do not need the Taylor series here. Just search for the closest pole in the complex plane. $\endgroup$ Mar 5, 2015 at 19:15
  • $\begingroup$ but i want taylor as well $\endgroup$
    – kurkowski
    Mar 5, 2015 at 19:15

2 Answers 2


Hint. Use the (famous) expansion $$ \frac{1}{1-x} = \sum_{n \in \mathbb{N}}x^n $$ To do this, factor by 3 and consider $x' = -(x^3)/3$.


We know that

$\frac{1}{1-y} = \sum_{n=1}^{\infty} y^n$ for $|y| < 1$

Rewrite your function as $f(x) = \frac{1}{3(1-\frac{-x^3}{3})}$.

Finally, $y = -x^3/3$


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