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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,

$$\sum\frac{0x^2}{n!}=0$$

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 $\infty$

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  • $\begingroup$ 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
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$f'(0)=3(x-1)^2\bigg\vert_{x=0}=3$, $\;f''(0)=6(x-1)\bigg\vert_{x=0}=-6$, $\;f'''(x)=6=f'''(0)$, $\;f^{(4)}(x)=0$. 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.

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