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Find the Taylor series for $f(x)=e^x$ about the point $c=3$. Then simplify the series and show how it could have been obtained directly from the series for $f$ about $c=0$.

Taylor's Theorem:

$$f(x)=\sum_{k=0}^n{1\over k!}f^{(k)}(c)(x-c)^k+E_n(x) \\ \text{where}\quad E_n(x)={1\over (n+1)!}f^{n+1}(\xi)(x-c)^{n+1}$$

I'm not sure how to do this. Any solutions or hints are greatly appreciated. Should I even be using the error term $E_n(x)$ when I do this? I'm confused about that.

I know that $e^3+e^3(x-3)+{1\over 2}e^3(x-3)^2+\cdots$ is the Taylor series for $f(x)=e^x$ at $c=3$ and that $1+x+{1\over 2!}x^2+{1\over 3!}x^3+\cdots$ is the Taylor series at $c=0$. How do I simplify the first sum to obtain the second?

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You've already calculated your answer: The series is given by $\sum \limits_{n = 0}^\infty \frac{e^3}{n!} (x-3)^n$.

You don't ned to transform this in any way to the Taylor-series around $0$. However, you can easily deduce the Taylor series at $3$ from the one at $0$ by using the identity $e^x = e^{x-3} e^3$.

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  • $\begingroup$ What about the error term? $\endgroup$ – 1233dfv Aug 29 '15 at 0:45
  • $\begingroup$ The error term is not part of the series. $\endgroup$ – Dominik Aug 29 '15 at 2:27
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HINT

The question wants you to expand your series for $\mathrm{e}^x$, close to $x=3$ and show that it gives the same as series for $\mathrm{e}^x$, close to $x=0$. In other words:

$$\mathrm{e}^3 + \mathrm{e}^3(x-3)+\frac{1}{2!}\mathrm{e}^3(x-3)^2 + \frac{1}{3!}\mathrm{e}^3(x-3)^3 \cdots \equiv 1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$$

Let's start by finding all of the constant terms.The constant terms on the LHS are given by $$\mathrm{e}^3-3\mathrm{e}^3+\frac{1}{2!}\mathrm{e}^3(-3)^2+\frac{1}{3!}\mathrm{e}^3(-3)^3+\cdots+\frac{1}{k!}\mathrm{e}^3(-3)^k+\cdots$$ \begin{eqnarray*} &=&\mathrm{e}^3\left(1+(-3)+\frac{(-3)^2}{2!} + \frac{(-3)^3}{3!}+\cdots+\frac{(-3)^k}{k!}+\cdots\right) \\ \\ &=& \mathrm{e}^3\left.\left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\cdots+\frac{x^k}{k!}+\cdots\right)\right|_{x=-3} \\ \\ &=&\mathrm{e}^3\left(\mathrm{e}^{-3}\right) \\ \\ &=& 1 \end{eqnarray*} This is good news. If we expand the series at $x=3$, then the constant term is $1$. This is what we wanted. Now we need to show that if we expand the $x=3$ series then the $x$-coeffient will be $1$, the $x^2$-coefficient $\frac{1}{2!}$, etc. You will be able to get a general formula for the $x^k$-coefficient by using the binomial expansion.

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