# Precision when dividing two functions going against zero

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T il 28 Lug 2014
Modificato: Matt J il 28 Lug 2014
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!
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Michael Haderlein il 28 Lug 2014
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 il 28 Lug 2014
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.

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### Risposte (2)

Matt J il 28 Lug 2014
Modificato: Matt J il 28 Lug 2014
Use a Taylor expansion of A(x) around x=0 to evaluate A(x)/(c3*x) at x in the near neighborhood of zero.
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Patrik Ek il 28 Lug 2014
Modificato: Patrik Ek il 28 Lug 2014
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.
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Matt J il 28 Lug 2014
Modificato: Matt J il 28 Lug 2014
L'hopital's rule also confirms that the value should go to zero at x=0.

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