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Let's say that I want to approximate $\sin{(4)}$ Using a the first three non-zero terms of a power series. If I'm centered at $x=0$, I get a bad approximation, as you can see in the image.

enter image description here

How do I get a better approximation, not by adding more terms to the power series, but by finding a different power series centered at $x=4$? If I just use the Taylor Series, the first term would be $\sin{(4)}$, but that defeats the purpose - I can't use $\sin{(4)}$ if I'm trying to approximate $\sin{(4)}$. Thank you!

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    $\begingroup$ Well, you probably don't mean "centered at $x=4$"...you want it to be centered at a value near $4$, a value for which you can compute the power series. $x=\pi$ isn't a bad start. Or $x=\frac {5\pi}4$. $\endgroup$
    – lulu
    Apr 4 at 15:24
  • $\begingroup$ You can shift the sine function by using $\sin(a+h) = \sin(a)\cos(h)+\cos(a)\sin(h)$ $\endgroup$
    – Gribouillis
    Apr 4 at 15:30
  • $\begingroup$ So it's not possible to center it exactly at $x=4$ unless I already know the value of $\sin{(4)}$? So I pick a value close to $4$ that is a know sin value? $\endgroup$
    – user3925803
    Apr 4 at 15:30
  • $\begingroup$ @Gribouillis How does that help? Where would I go from there? $\endgroup$
    – user3925803
    Apr 4 at 15:32
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    $\begingroup$ The first term of the Taylor series for $f(x)$ centered at $x=a$ is $f(a)$. In deriving the standard Taylor series for $\sin x$, we use that $\sin 0 = 0$ and that $\cos 0 = 1$. The Taylor series is good for approximating a function in the vicinity of the center, In addition to needing to know the value, it also won't look very nice (or be easy to calculate) if centered at $x=4$. Looking at your graph to see how good 5 terms (two are zero so they are normally left out) approximate $\sin$, I would suggest going for a series centered at $x=\pi$. $\endgroup$
    – Henrik supports the community
    Apr 4 at 15:41

1 Answer 1

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Per the suggestions in the comments, I have expanded the series around $\pi$. Including terms up through $x^3$ seems to approximate $\sin 4$ pretty well already.

The even-numbered terms are multiples of $\sin \pi = 0$, so I have omitted them from this display.

graph of first three nonzero terms of Taylor series for sin(x) around x=\pi graph of sin(x), sum of terms up to x^3, sum of terms up to x^5

Interactive hilarity

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