Precision when dividing two functions going against zero

4 次查看（过去 30 天）
T
T 2014-7-28
编辑： Matt J 2014-7-28
Hi, I have two functions A=sinc(c1*x)-cos(c2*x) whichs is Zero for x=0. Now I want to calculate A/(c3*x). The expected result is exactly 1 at x=0. But numerically I get 1.05 Is there a way to increase precision in matlab? Format Long didn't work. I want to plot the behaviour of the function for a variety of x values, from large to very small
Thanks!
  2 个评论
Michael Haderlein
Michael Haderlein 2014-7-28
Hm, how did you get 1.05? I'd do something like this (since I don't know your c1,2,3, I set all to 1):
x=logspace(-10,0);
semilogx(x,(sinc(x)-cos(x) )./x)
This way, it converges towards 0 (for x-> 0).
David Young
David Young 2014-7-28
MATLAB computes in double precision by default. It is most unlikely that your problem is anything to do with the precision of the computation. If you show the code you used to get the value of 1.05 it may be possible to explain what is happening.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（2 个）

Matt J
Matt J 2014-7-28
编辑：Matt J 2014-7-28
Use a Taylor expansion of A(x) around x=0 to evaluate A(x)/(c3*x) at x in the near neighborhood of zero.
Patrik Ek
Patrik Ek 2014-7-28
编辑：Patrik Ek 2014-7-28
I think that you should try to do the calculations manually. I did the calculations manually and got that the value should converge towards 0, using a 3rd order taylor expansion. When manually evaluating:
f(x) = ( sinc(c1*x)-cos(c2*x) ) / (c3*x)
the expression I got was:
f(x) = x / ( c3*factorial(3) ) * (3*c2^2-c1^2)
This will converge towards 0 which the expression did for me when x was small enough. Notice that
sinc(c1*x) = sin(c1*x) / (c1*x)
If you plot sinc(x), cos(x), 1-x, 1+x in the same plot (using hold on) around 0 ( (-0.2, 0.2) is enough I think), it becomes quite clear that it should converge towards something small. It can be seen that cos(x) ans sinc(x) is similar at that point, but not the lines.

类别

在 Help Center 和 File Exchange 中查找有关 Lengths and Angles 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by