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I have been talking to a few people about when it is ok to use the Maclaurin series instead of using the Taylor series.

I.e.

Let $f(x)=e^{x^2-2x+1}$.

Write down the degree 3 Taylor polynomial for $f(x)$ centered at 1.

Obviously you can use that $p(x)=f(1)+f'(1)(x-1)+f''(1)(x-1)^2/2+f'''(1)(x-1)^3/3!$, but you can also do $e^{(x-1)^2}=\sum(x-1)^{2n}/n!\approx 1+(x-1)^2$.

Why are you able to use the Maclaurin series instead? Given that the series is supposed to be centered at 1? The Maclaurin series is centered at 0. I know that the parabola $y=(x-1)^2$ is centered at 1. So instinctively I think that this is the reason why. But I would like a more detailed understanding of why this is the case.

EDIT:

To go into a little more detail with what the students were asking. With this example, say that instead of it factoring into $e^{(x-1)^2}$ the function was $e^{(x-2)^2}$ and still had it centered at 1, would I be able to use that same idea of using the Maclaurin to answer the question.

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    $\begingroup$ My best answer is: Questions like "when is it ok" or "when are we able" aren't specific enough to answer. (You can do anything you want!) What goal are you trying to accomplish? One can answer questions like "which of these techniques better accomplishes this specific goal". $\endgroup$ – Greg Martin Oct 22 '20 at 17:24
  • $\begingroup$ @GregMartin I mean yes, but specifically I think the people asking this question were wondering WHY it was ok, and I didn't have an answer better than the one you gave. $\endgroup$ – NewbieMather Oct 22 '20 at 17:27
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Let $f(x)=\sum_{k=0}^\infty a_k x^k$ be a function with the given Taylor expansion centered at $0$. Then $f(x-x_0)=\sum_{k=0}^\infty a_k(x-x_0)^k$ is a function with the given Taylor expansion centered at $x_0$.

Now apply this to $f(x)=\exp(x^2)$ and $x_0=1$.

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  • $\begingroup$ Perfect. Thank you. $\endgroup$ – NewbieMather Oct 22 '20 at 17:45
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Because the sum of the power series $\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}$ is $e^x$, for every real number $x$. In particular,$$e^x=1+x+o(x^2)$$near $0$, and therefore$$e^{(x-1)^2}=1+(x-1)^2+o\bigl((x-1)^4\bigr)$$near $0$, from which it follows that the Taylor polynomial of order $3$ of $e^{(x-1)^2}$ centered at $1$ is $1+(x-1)^2$.

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  • $\begingroup$ This is perfect and answers my question. But how would I explain this to Calc 2 students though? $\endgroup$ – NewbieMather Oct 22 '20 at 17:29
  • $\begingroup$ I am sorry, but I am not aware of the syllabus of Calc 2. $\endgroup$ – José Carlos Santos Oct 22 '20 at 17:30
  • $\begingroup$ They have limited understanding of any theory essentially. Basically up to this point they know derivatives, integrals, and now have a basic understanding of Taylor series. $\endgroup$ – NewbieMather Oct 22 '20 at 17:32
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    $\begingroup$ @NewbieMather You are basically changing variables: expanding $e^{(x-1)^2}$ about $x=1$ is the same as expanding $e^{u^2}$ about $u=0$ where $u=x-1$. $\endgroup$ – Ian Oct 22 '20 at 17:41
  • $\begingroup$ That is what I told them too. But then they asked what if the function was $e^{(x-2)^2}$ and still centered at 1. Would we be able to use the Maclaurin series in this case too. And if so why. And I wasn't sure. My instinct was to say no, that at this point you would need to use the Taylor series. But if that is the case I was wondering for the theory of how it worked. $\endgroup$ – NewbieMather Oct 22 '20 at 17:44

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