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I just came from a final exam where in one question I was asked to derive the Taylor Series for $f(x)=e^{2x}$ centered at $x=1$. I came up with the following:

$$e^{2x}=\sum_{n=0}^{\infty}\frac{2^ne^2}{n!}(x-1)^n$$

I was really confused though, because I was certain that the Maclaurin Series for $e^x$ was $$\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

The problem I found with this, is I wondered where the $e$ went in the Maclaurin Series... Isn't a Maclaurin Series just a special case of the Taylor Series with the center at a=0? If that were the case, wouldn't a Taylor Series expansion of $e^x$ about $a=0$ be $$\sum_{n=0}^{\infty}\frac{e}{n!}x^n$$

Why is there no $e$ in the Maclaurin Series?


Addition: I was asked to show my derivation for $e^{2x}$...

\begin{align} f(x)&=e^{2x}\\ f'(x)&=2e^{2x}\\ f''(x)&=2^2e^{2x}\\ f^{(n)}(x)&=2^ne^{2x}\\ f^{(n)}(a)&=2^ne^2&&x=a=1\\ f(x)&=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\\ &=\sum_{n=0}^{\infty}\frac{2^ne^2}{n!}(x-1)^n\\ \end{align}

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  • $\begingroup$ @AWertheim The question on the exam was $e^{2x}$ $\endgroup$ – agent154 Aug 8 '13 at 16:16
  • $\begingroup$ agent154, please forgive my erroneous comments. $\endgroup$ – Alex Wertheim Aug 8 '13 at 16:18
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    $\begingroup$ The Taylor series of $e$ at $a=0$ is actually $$\sum_{n=0}^{\infty}\frac{e^0}{n!}x^n$$ does this answer your question? ;) $\endgroup$ – N. S. Aug 8 '13 at 16:26
  • $\begingroup$ @N.S. Indeed - I guess I can chalk this question up to lack of sleep... Exams will do that. $\endgroup$ – agent154 Aug 8 '13 at 16:28
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Your reasoning would be correct if you remembered that $e^0 = 1$...

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  • $\begingroup$ Ah yes - brain farts are wonderful aren't they? I was putting $x=1$ in the equation in my head without thinking anything wrong of it. $\endgroup$ – agent154 Aug 8 '13 at 16:26
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The coefficient for $(x-a)^n$ in the Taylor series of $f(x)$ is:

$$\frac{f^{(n)}(a)}{n!}$$

Compute a few terms of this for $a=0$ and $a=1$ to see why an $e$ shows up in the latter and not the former.

You are right to say that a Maclaurin series is a special type of Taylor series, namely centered at $0$, but this doesn't say much about how similar our coefficients should look. Evaluating derivatives at differenct values may give us very different looking coefficients. Centering $\sin x$ at $0$ makes the even valued coefficients vanish, while centering at $\pi/2$ makes the odd valued coefficients vanish.

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To calculate the Taylor series of $f(x)=e^{2x}$ at $x=1$ we can apply Taylor's Theorem directly. However, an easier way is to use what you already know about the exponential and use some algebra. $$ e^{2x} = e^{2(x-1+1)} = e^2e^{2(x-1)} = e^2 \sum_{n=0}^{\infty} \frac{1}{n!}(2(x-1))^n = e^2 \sum_{n=0}^{\infty} \frac{2^n}{n!}(x-1)^n$$ done.

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  • $\begingroup$ This is wrong, James. $\endgroup$ – Pedro Tamaroff Aug 8 '13 at 16:25
  • $\begingroup$ @Peter Tamaroff Ah, thanks. Will fix. $\endgroup$ – James S. Cook Aug 8 '13 at 16:27

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