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Consider the polynomial $$ p(x) = -514-462 x+359 x^2+1129 x^3+165 x^4+490 x^5-418 x^6+497 x^7-227 x^8+60 x^9-10 x^{10}, $$ whose root $A\approx 3.14$ is very close to $\pi$: $$|A-\pi|=2.0746\times 10^{-33}.$$

Let's say I am given as input the eleven integer coefficients above, and nothing else. Is there a way to do calculate $\sin A$ that doesn't use multiple-precision floating-point arithmetic, just the regular double-precision arithmetic? (With multiple-precision arithmetic this is trivial, so I would like to know if there are any methods or ideas that could apply here with this restriction in place.)

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  • $\begingroup$ So the question is "Given a polynomial, compute the sine of its roots (approximately) without computing the roots"? $\endgroup$ – Christian Clason Oct 29 '14 at 22:08
  • $\begingroup$ @ChristianClason Yes, but that's only numerically tricky when the root is close to a multiple of $\pi$, otherwise simply $\sin A$ will work well. $\endgroup$ – Kirill Oct 29 '14 at 22:15
  • $\begingroup$ Maybe it would help if you described what you are doing now, and how precisely this fails for you? $\endgroup$ – Christian Clason Oct 29 '14 at 22:30
  • $\begingroup$ @ChristianClason I came across this polynomial while looking for algebraic relations almost satisfied by $\pi$. I realized that it's tricky to compute $|A-\pi|$ and $\sin A$ without access to high-precision arithmetic, and I couldn't think of a way to do it directly. So the problem is already self-contained in that sense; there's already a trivial way to do it (with high-precision arithmetic), and I am curious whether I am missing some interesting idea that I don't know about yet. $\endgroup$ – Kirill Oct 29 '14 at 23:23
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It probably depends on how much accuracy you want.

Call $\pi - A = d$. If you know $d$, then you could use $\sin(\pi - x) = \sin(x)$, and evaluate $\sin(d)$. It's very close to zero, and a first-order Taylor series yields $\sin(d) \approx d$. (Python's math.sin(2e-33) yields 2 \times 10^{-33} to at least 10 significant digits.) Additional terms in the Taylor series will be negligible, but arbitrary precision arithmetic will be more accurate.

To calculate $d$, you can replace $x$ in your polynomial with $\pi - d$ and solve for it using a standard root finder; I used Newton's method and got approximately -2.075e-33. Of course, this method makes use of the knowledge that the root of interest of the original polynomial is close to $\pi$, but it's a standard type of trick in numerical analysis.

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  • $\begingroup$ I'm not sure I understand what you mean: $fl(A)=fl(\pi)$, and using double-precision $\sin$ will not give the right answer. Computing $\pi-A$ in double precision gives exactly $0$. I am not assuming that I have already computed $|A-\pi|$. $\endgroup$ – Kirill Oct 29 '14 at 20:47
  • $\begingroup$ Based on your question, I thought you might have access to an accurate value of $\pi - A$. If you don't, this answer won't work. $\endgroup$ – Geoff Oxberry Oct 29 '14 at 21:01

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