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let $f(x)=\frac{e^x-1-x-\frac{x^2}{2}}{x}$, because $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, $f$ can be expressed as $$f(x) = \frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x-\frac{x^2}{2}}{x}=\frac{\sum_{n=3}^\infty \frac{x^n}{n!}}{x}=\sum_{n=0}^\infty \frac{x^{n+2}}{(n+3)!}$$ the power series converge in $(-\infty, \infty)$ because $\lim_{n\to\infty} \sqrt[n]{\frac{1}{(n+3)!}}=0$ and let $f_n(x) = \frac{x^{n+2}}{(n+3)!} \Longrightarrow f'_n(x) = \frac{x^{n+1}}{(n+1)!(n+3)}$, $\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}= \sum_{n=0}^\infty f'_n(x)$ also converge in $(-\infty, \infty)$ (for the same reason), hence $$f'(x) = \sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}$$ by repeating this process once more I get $$f''(x) = \sum_{n=0}^\infty \frac{x^n}{n!(n+3)}$$ and if $x=3$ I get $$\sum_{n=0}^\infty \frac{3^n}{n!(n+3)} = f''(3)$$ which is what was looking for. my problem is that $f$ isn't defined for $x=0$ yet the series does converge for it as $\sum_{n=0}^\infty \frac{0^n}{n!(n+3)}=0$, so was the function $f$ I used wrong? or could it be that I can't differentiate $f$ the way I did?

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3 Answers 3

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Hint

$f$ is not formally defined at $0$. However you can extend it by continuity at $0$.

In particular

$$\lim\limits_{x \to 0} \frac{e^x - 1}{x} = (e^x)^\prime(0) = 1$$

Hence you can extend $f$ by continuity at $0$ with $f(0)=0$.

There is no contradiction and what you did seems OK regarding the computations.

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You are correct that the version of $f$ given by the recipe $\frac{\mathrm{e}^x-1-x-x^2/2}{x}$ is undefined at $x = 0$. However, the limit of this recipe as $x \rightarrow 0$ is $0$, so there is a continuous function, $\hat{f}$, with domain $(-\infty, \infty)$, which agrees with $f$ on $(-\infty, \infty) \smallsetminus \{0\}$ and agrees with $f$'s limit as $x \rightarrow 0$. You have already written a recipe for $\hat{f}$, when you wrote $\sum_{n=0}^\infty x^{n+2}/(n+3)!$.

Since you manipulated the series, $\hat{f}$, with infinite radius of convergence, you need not worry about the defects of the original recipe. Furthermore, $f$ and $\hat{f}$ agree at $x = 3$, the only point at which you intend to evaluate. You would be properly concerned if you were trying to evaluate at $x = 0$, but you are not.

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Try using the following instead so you avoid that mess: $$x^2e^x=\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$$

$$\int x^2e^x \; dx=x^2e^x-2xe^x+2e^x+C=\sum_{n=0}^{\infty} \frac{x^{n+3}}{n!(n+3)}$$ At $x=3$: $$9e^3-6e^3+2e^3-2=5e^3-2=\sum_{n=0}^{\infty} \frac{3^{n+3}}{n!(n+3)}$$

Notice that the series on the right is what you're looking for but multiplied by $3^3$, so divide both sides by $27$.

I understand that this approach is slightly different than yours, but I believe this approach is faster and easier to understand so I thought you might appreciate it. I'm sure you can refer to other answers posted here regarding your confusion with $f(0)$.

$$\boxed{\frac{5e^3-2}{27}}$$

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  • $\begingroup$ No aspect of this Answer responds to the Question's "was the function $f$ I used wrong? or could it be that I can't differentiate $f$ the way I did?" The method proposed here is subject to exactly the same questions. $\endgroup$ May 30, 2020 at 16:07
  • $\begingroup$ The function is defined at $x=0$ in my answer. I was showing an alternate and easier way to calculate the sum. $\endgroup$
    – Ty.
    May 30, 2020 at 16:09
  • $\begingroup$ while this way is, by far, more elegant, I haven't "learnt" integrals with series yet, thus I had to use derivative. $\endgroup$ May 30, 2020 at 16:37

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