# Approximating $\sin 100$

If Taylor polynomial for $$\sin(x)$$ is $$\sum_{n=0}^{+\infty} \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}$$. What do I have to do to find what degree of Taylor polynomial I have to use so the error is not greater than $$10^{-4}$$ in approximation of $$\sin(100)$$?

• It's true only around 0. – Archis Welankar Jan 26 '20 at 15:46
• Well, since the series alternates, and decreases for sufficiently large terms, you could just use Sterling's approximation. But surely it is more efficient to use periodicity...$32\pi\approx 100.5309649$ helps. – lulu Jan 26 '20 at 15:48
• Why using this series in practice is a really bad idea: the largest term in the alternating series is about $\approx 100^{100}/100! \approx 10^{42}$. Thus if you are to use the alternating series to compute it you would need to add about $300$ numbers that would be up to $47$ digits long (you also need $5$ decimal digits to get the desired precision). This would not even work on a computer unless you used arbitrary precision numbers. – Winther Jan 26 '20 at 16:03
• Your remainder term is not correct. It needs a factor $x^n$ in it, which is huge and why Winther gets that you need so many terms – Ross Millikan Jan 26 '20 at 17:06
• Without using the value of $\pi$, you can use angle reduction using the double-angle trigonometric identities. Set $x=100/2^6=1.5625$, then compute $\sin x$ and $\cos x$ with an error less than $10^{-8}$ using the Taylor expansions, and then reconstruct the wanted quantities using $\sin(2^{k+1}x)=2\sin(2^kx)\cos(2^kx)$ and $\cos(2^{k+1}x)=\cos^2(2^kx)-\sin^2(2^kx)$ or similar. – Lutz Lehmann Jan 26 '20 at 19:29

It is a(n eventually) converging alternating series, so the alternating series theorem applies. Find the first term after it starts decreasing that is less than $$10^{-4}$$ in magnitude and you are done.
Probably you are expected to look up the error term for the Taylor series. Note that all the derivatives of $$\sin x$$ are less than $$1$$ in magnitude, so you can ignore that.
As the comments point out, you will get there with many fewer terms if you are allowed to center the Taylor series at $$32\pi$$ instead of $$0$$.
• The series you quote is centered at $0$. The closer you can center the series to the point of interest the better because all those $x^i$ terms in your series become $(x-c)^i$ where $c$ is the point you center at. As you can easily calculate trig functions of $32\pi$ that is a good place. – Ross Millikan Jan 26 '20 at 16:04