I am trying to find the taylor series for $f(x)=(x-1)^3$ centered at $x=0$

Calculating derivative and evaluating them for $x=0$ we see that any derivative greater than and including the 5th derivative is 0.

So would the taylor series for this be,


And then since the taylor series is $0$ when calculating the radius of convergence would it just be $\infty$ since when you use the ratio test for $\sum\frac{0x^2}{n!}=0$ you would get $lim=0$ therefore for all $x$ the series converges and the radius is $\infty$

  • $\begingroup$ Taylor series just has 5 terms, not infinite in this case. $\endgroup$ – offret Apr 15 '17 at 18:33
  • $\begingroup$ the ratio test cannot be used if some coefficient is zero frequently. $\endgroup$ – Masacroso Apr 15 '17 at 18:40
  • $\begingroup$ Well, even the fourth derivative is 0. So no $x^4$ term or any higher one. The whole series is not 0 though, some of the lower powers are non-zero. Calculate those lower derivatives and you can write down the full power series. Now multiply out $(x - 1)^3$ and compare them. $\endgroup$ – badjohn Apr 15 '17 at 18:40
  • $\begingroup$ @badjohn but don't you have to consider $f^n(0)=0$? so then how would the series not be 0 $\endgroup$ – fr14 Apr 15 '17 at 18:52
  • $\begingroup$ Let's look at the first term in the Taylor series: $f(0)$ - that's $-1$ so the series starts with $-1$. I calculate $f'(0)$ as 3 so the next term is $3x$. So, although all terms of the series are zero after a while, the first few are not: $-1 + 3x + ?x^2 + ?x^3$. Try to figure out the ?s. $\endgroup$ – badjohn Apr 15 '17 at 19:04

$f'(0)=3(x-1)^2\bigg\vert_{x=0}=3$, $\;f''(0)=6(x-1)\bigg\vert_{x=0}=-6$, $\;f'''(x)=6=f'''(0)$, $\;f^{(4)}(x)=0$. So the Taylor's series of $(x-1)^3\;$ centred at $0$ is simply $$(x-1)^3=-1+3x-3x^2+x^3,$$ which you might have guessed by the binomial formula.

More generally, a polynomial is its own Taylor's series centred at $0$, and its radius of convergence is trivially infinite.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.