What is the Taylor Series for $f(x)=(x-1)^3$ centered at $x=0$? What is the radius of convergence?

\begin{align}f'(x)&= (3)(x-1)^2\\ f''(x)&=6(x-1)\\ f^{(3)}(x)&=6\\ f^{(4)}(x)&=0\\ &\vdots\end{align}

Using the definition of Taylor Series, $$ f(x)=−1+3x−2x^2+x^3 $$

Does this seem okay for the Taylor series? With radius of convergence equal infinity?

  • $\begingroup$ Looks like you forgot to divide $f^{(n)}(0)$ by $n!$. $\endgroup$ – Daniel Schepler Mar 23 '17 at 0:44
  • $\begingroup$ @DanielSchepler Would this be correct now? $\endgroup$ – CS14 Mar 23 '17 at 0:48
  • $\begingroup$ Shouldn't it be: $$f(x)=-1+3x\color{red}{-3}x^2+x^3$$ Since: $$\frac{-6}{2!}x^2=-3x^2$$ The Taylor Series expansion will thus be consistent with the binomial expansion of $(x-1)^3$ as we expect. $\endgroup$ – projectilemotion Mar 23 '17 at 0:59
  • $\begingroup$ @projectilemotion My mistake! $\endgroup$ – CS14 Mar 23 '17 at 1:06
  • $\begingroup$ Using the error term, En(x), how would I should that it converges to all x? $\endgroup$ – CS14 Mar 23 '17 at 1:24

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