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Unread 16-12-2012, 20:19
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Re: numerical computation contest

Quote:
Originally Posted by Richard Wallace View Post
It would be interesting to see a graph comparing the 2nd, 4th, 6th, and 8th order expansions of 1 - sinc(x) plotted vs. x, for 0 < x < 0.05. The range of most interest will be near 1/5280 ....
... or 1/5281

sin(θ)/θ = 5280/5281 = 1 - 1/5281 => 1 - sin(θ)/θ = 1/5281

The Taylor expansions (even the 2nd degree one) are so accurate over the range 0<x<0.05 that you can't see the difference on the graph at that scale. Which maybe was your point

Also attached is the 64 decimal digit precision computation of θ using the zero of the function y = 5280/5281 - sin(θ)/θ and its Taylor expansions.

Finally, notice that the Taylor8 expansion actually gives a more accurate answer than the original function itself, when using only 16 digit precision. Presumably, this is because Taylor8 is numerically more stable.


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Last edited by Ether : 16-12-2012 at 21:00.