# How to shift a power series to be centered at a?

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. 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!

• 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 $x=\frac {5\pi}4$.
– lulu
Apr 4 at 15:24
• You can shift the sine function by using $\sin(a+h) = \sin(a)\cos(h)+\cos(a)\sin(h)$ Apr 4 at 15:30
• 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? Apr 4 at 15:30
• @Gribouillis How does that help? Where would I go from there? Apr 4 at 15:32
• 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 $x=\pi$. Apr 4 at 15:41

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.