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I wish to approximate the value of $e$ and show this is accurate to at least the sixth digit. I am wondering if in doing so I need to use the Taylor Series for $e^x$ about $1$ or does it suffice to use $\Sigma_{n=0}^{\infty}\frac{x^n}{n!}$ with $x=1$.

From here I believe I then would need to solve $|R_n(x)|<|\frac{f^{n+1}(c)}{(n+1)!}|=|\frac{3}{(n+1)!}|<.0000005$ which would work for $n=10$. Then just calculate $P_{10}(x)$. I have already shown that $e^x<3$ which is how I bounded the remainder function.

Does this work or should I redo the process with the Taylor Series centered about the value which we wish to approximate?

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  • $\begingroup$ You just showed that the remainder, i.e. the difference between the computed value and the true value is within tolerance. What more do you want ? $\endgroup$
    – user65203
    Dec 12 '19 at 16:26
  • $\begingroup$ I was just wondering if I should have used the Taylor Series centered around $x=1$ to estimate it at that value and a reasoning for why $\endgroup$
    – joseph
    Dec 12 '19 at 16:27
  • $\begingroup$ What is the rationale of recomputing the value when you have proven that you have the right one ? $\endgroup$
    – user65203
    Dec 12 '19 at 16:31
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This works. It was possible, because you know the values of $e^x$ and its derivatives, at 0 (they all equal 1). If you centre the series at 1 instead, then you need to know the value of $e^x$ at $x=1$, which is the very thing that you hope to approximate !

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    $\begingroup$ ah perfect thank you so much! So in general, can we approximate $e^x$ with the Taylor Series at $x=0$ for any value of $x$ within the interval of convergence $\endgroup$
    – joseph
    Dec 12 '19 at 16:28
  • $\begingroup$ Yes. And I have good news for you: The radius of convergence is infinite, in this case ! $\endgroup$
    – Simon
    Dec 12 '19 at 16:32
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    $\begingroup$ Thank you so much for clearing that up! Super convenient that the terms go to zero quickly too $\endgroup$
    – joseph
    Dec 12 '19 at 16:34
  • $\begingroup$ You're most welcome ! And, I know, right ?! $\endgroup$
    – Simon
    Dec 12 '19 at 16:36

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