Taylor Series and the radius of convergence

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...

• Do you mean $f'(x)=3(x-1)^2$? – delt3 Mar 22 '17 at 12:02
• The $(-1)$ in the first derivative is wrong. – Emilio Novati Mar 22 '17 at 12:04
• Hint: $f(x)=-1+3x-3x^2+x^3$. Oops... question fully solved. – 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 $a=0$.
• 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? – CS14 Mar 22 '17 at 17:34
• 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. – Emilio Novati Mar 22 '17 at 17:57
• $f(x) = -1 +3x -6x^2+6x^3$ Does this seem okay for the series? – CS14 Mar 22 '17 at 22:47