1
$\begingroup$

We are asked to generate the taylor polynomial $P(x)$ for $$ f(x) = \frac{e^{{(x-1)}^2}-1}{(x-1)^{2}} $$ about $x=1$

Using substitution into the known taylor polynomial of $e^{x}$ and further algebraic manipulation, I can find the taylor series expansion: $$ P(x) = \sum_{n=0}^{\infty}\frac{(x-1)^{2n}}{(n+1)!} = 1+ \frac{(x-1)^2}{2!} + \frac{(x-1)^4}{3!}+...$$ which can be confirmed by Wolfram Alpha.

Here, I now want to find $f''(1)$; given that the general taylor series expansion for a function $k(x)$ about $a$ is $$k(a) + k'(a)(x-a) + \frac{k''(a)(x-a)^2}{2!} + \frac{k'''(a)(x-a)^3}{3!} +...$$ Thus should $$\frac{f''(1)}{2!}$$ not equal the coefficient of the second term of $P(x)$? Giving $f''(1) = 1$. However this appears to be incorrect when checked with Wolfram, which gives the second derivative as 'indeterminate'. Can anyone explain why? Help is much appreciated.

$\endgroup$
1
  • $\begingroup$ I would guess that W|A would mark every derivative and even the function itself as indeterminate at $x=1$. $\endgroup$
    – Brad
    Commented Mar 24, 2014 at 1:53

3 Answers 3

1
$\begingroup$

It also says that when asking for $f(1)$.

Technically speaking, your $f$ really is not defined at $x=1$; implicitly you mean to talk about the continuous extension of $f$. Mathematica often does such things automatically, but not always.

You could work around the problem by asking for the limit of $f''(x)$ as $x \to 1$.

$\endgroup$
0
$\begingroup$

When dividing by $(x-1)^2$, you canceled terms on top and the bottom. These terms do not cancel when $x=1$.

$\endgroup$
2
  • $\begingroup$ Ah, does this indicate that the second term of the taylor polynomial (obtained through substitution) is incorrect? As the second derivative at 1 should not exist. $\endgroup$
    – Harrison
    Commented Mar 24, 2014 at 1:53
  • $\begingroup$ Well if you cannot cancel terms then this isn't quite the Taylor series, though it would be equal to the Taylor series outside of $x=1$ if the Taylor series existed, but it does not. $\endgroup$ Commented Mar 24, 2014 at 1:59
0
$\begingroup$

It has the form 0/0. For example, $$f(1)=\frac{e^0-1}{0^2}={0\over 0}$$ You would have to use L'hopital's Rule to evaluate the derivatives exactly, or you can just consult the Taylor series. In other words, it is not true that $$T(x)=f(a)+f'(a)(x-a)+...$$ in all cases. In this case, none of those derivatives even exist. The taylor series gives an estimate for a different function, which would be $$g(x)=\lim_{t\to x}{f(t)}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .