NaN for zero divide by zero.

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Stian Molvik
Stian Molvik 2011 年 4 月 6 日
Will illustrate with an example, I.e:
a = (0:0.1:1); b = (1-a)*2./(1-a);
Matlab will output the end value in b as NaN, when in reality b should converge to 2 for a = 1.
How can I avoid this?

採用された回答

Matt Fig
Matt Fig 2011 年 4 月 6 日
There are two things to remember when looking at this problem.
1. A function having a limit as the independent variable approaches a point does not imply that the function value at that point is defined. You plugged in for the function value at that point. The fact that you get NAN says nothing about the limit of the function as the independent variable approaches that point.
2. NAN is the correct value when evaluating the function at that point. This is a standard double precision definition.
  4 件のコメント
Matt Fig
Matt Fig 2011 年 4 月 6 日
Yes, if you know the limit, you can do
b(isnan(b)) = 2;
Stian Molvik
Stian Molvik 2011 年 4 月 6 日
Ah. Ok. Hoped there would be an easier way, but still.
Thank you very much everyone for your help! :)

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その他の回答 (3 件)

Walter Roberson
Walter Roberson 2011 年 4 月 6 日
In order to avoid this, you will need to use the symbolic toolbox.
MATLAB is giving the correct numeric answer according to IEEE 754 floating point definitions. MATLAB works with actual values computed on real machines, not with limits or ideal values. See for example this portion of the FAQ

Sean de Wolski
Sean de Wolski 2011 年 4 月 6 日
0/0 is the definition of a nan
doc nan

Stian Molvik
Stian Molvik 2011 年 4 月 6 日
Thanks for the reply Walter.
I haven't had too much experience with Matlab earlier. How would I go about using the symbolic toolbox?
  1 件のコメント
Walter Roberson
Walter Roberson 2011 年 4 月 6 日
Try this:
sym a
b = (1-a)*2/(1-a);
and see what b contains afterwards.

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