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I've been instructed to find the values of x for which the function $f(x) = e^{-2x}$ may be approximated by the Maclaurin series $1-2x+2x^2-\frac{4}{3}x^3$ with an error of less than 0.001, but no interval is given.

How should I choose my interval so as to appropriately calculate $\max|f^{(4)}|$?

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You have to work backwards essentially, of course you can find an expression for $f^{(4)}$. Once this is done, you should find the requirement on $x$ such that so that the error term $\frac{f^{(4)}(\xi)x^4}{4!}$ is less than $0.001$ in absolute value. It will be of the form $|x|<\delta$ for some $\delta>0$.

So what is $f^{(4)}(x)$?

$f^{(4)}(x) = 16 e^{-2x}$.

How do we bound this? We will have some negative $x$, so we are forced to say that $\max f^{(4)}(x) = 16e^{2\delta}$ on our symmetric interval $|x|<\delta$.

So whenever $|x|<\delta$, $\frac{|f^{(4)}(\xi)||x^4|}{4!} < \frac{16e^{2\delta}\delta^4}{24}$. For this to be less than $0.001$, that's the same as asking for $e^{2\delta}\delta^4 \leq 0.0015$. We know it will be increasing, so we can just solve $e^{2\delta}\delta^4 = 0.0015$, which gives $\delta>0.17$, so $\delta = 0.17$ will do the trick.

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    $\begingroup$ But in the Lagrange error bound, to maximize error it is $\max |f^{(4)}|$ which is what I think OP is asking about $\endgroup$ – Andrew Li Apr 22 at 1:06
  • $\begingroup$ In this case, $f^{(4)}(0)=16$, which gives a $\delta$ that's a little too narrow $\endgroup$ – HandsomeGorilla Apr 22 at 1:06
  • $\begingroup$ Yeah good point I will put some stuff on this. $\endgroup$ – George Dewhirst Apr 22 at 1:10
  • $\begingroup$ @GeorgeDewhirst I don't think that's quite right. Here's a plot I developed in desmos that shows an interval which is substantially larger. I'm not sure what to make of this. $\endgroup$ – HandsomeGorilla Apr 22 at 12:56
  • $\begingroup$ So your interval is $\delta = 0.2$ that's not that different $\endgroup$ – George Dewhirst Apr 22 at 15:55

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