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, not sure how to put together the Taylor Series if the above is correct...

  • $\begingroup$ Do you mean $f'(x)=3(x-1)^2$? $\endgroup$ – delt3 Mar 22 '17 at 12:02
  • $\begingroup$ The $(-1)$ in the first derivative is wrong. $\endgroup$ – Emilio Novati Mar 22 '17 at 12:04
  • 1
    $\begingroup$ Hint: $f(x)=-1+3x-3x^2+x^3$. Oops... question fully solved. $\endgroup$ – Did Mar 22 '17 at 12:24


After correcting the first derivative ( as suggested in the comments), use exactly the formula of the series with $a=0$.

Since the forth derivative is null only the first four terms of the series are not null.

  • $\begingroup$ Would it then be, $$ f(x)= (x-1)^3 + 3(x-1)^2 + 6(x-1) + 6 $$ as the Taylor Series? Would the radius of convergence be infinity then? $\endgroup$ – CS14 Mar 22 '17 at 17:34
  • $\begingroup$ Your series is wrong. Use the formula in : en.wikipedia.org/wiki/Taylor_series#Definition , with $a=0$.Obviously the series is the development of the cube $(x-1)^3$ and has radius of convergence infinite. $\endgroup$ – Emilio Novati Mar 22 '17 at 17:57
  • $\begingroup$ $ f(x) = -1 +3x -6x^2+6x^3 $ Does this seem okay for the series? $\endgroup$ – CS14 Mar 22 '17 at 22:47

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