Plot conflicts with mathematical calculation

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Yumeng Yin
Yumeng Yin 2022-4-26
评论： Yumeng Yin 2022-4-26
I have a model function shown as below:
F=((b.*A-(x-a).*B)./((x-a).^2+b^2-A^2-B^2))+e;
From the mathematical calculation, I calculate its first derivative, and get [B(x-a-bA/B)^2-(b^2-B^2)(A^2+B^2)/B]/[(x-a)^2+b^2-A^2-B^2]^2,
so if b^2>B^2 and B>0,then the first derivative has the possibility to be zero, which gives the extreme points. However, if I set the parameters to be b<B and B>0, I can still get maximum and minimum values shown below.
I am not sure what mistakes I make. Another point is that if I set the range to be (0,2000,50000) and then set it to be (1000,1500,12500), it is not a zoomed-in picture as what I would expect. Actually the values change a lot, which is shown below.
Thanks a lot for all the advice. Below is my script. Thanks.
a=2000;
b=700;
A=800;
B=790;
e=10;
x=linspace(0,2000,50000);
F=((b.*A-(x-a).*B)./((x-a).^2+b^2-A^2-B^2))+e;
plot(x,F)

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DGM
DGM 2022-4-26
编辑：DGM 2022-4-26
Ran in:
There are two poles. The function is undefined when the denominator is zero. The only reason it looks like the extrema vary is simply an artifact of the discrete locations at which the function is being evaluated.
If you evaluate the function at the exact location (at least as close as we can get with FP rounding), you'll get a good illustration:
a=2000;
b=700;
A=800;
B=790;
e=10;
% evaluate F at the calculated location
x = a - (A^2 + B^2 - b^2)^(1/2); % 1120.17
% x = a + (A^2 + B^2 - b^2)^(1/2); % 2879.83
F = ((b.*A-(x-a).*B)./((x-a).^2 + b^2 - A^2 - B^2)) + e
F = Inf
x = x+eps(x); % make the smallest possible step to the right
F = ((b.*A-(x-a).*B)./((x-a).^2 + b^2 - A^2 - B^2)) + e
F = -2.6952e+15
What are the maximum and minimum values? They're the same as they are at every sort of singularity. They're +Inf and -Inf. Does it really matter what the extrema are in that vicinity?
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Walter Roberson
Walter Roberson 2022-4-26
If the denominator is 0 then there are four possibilities:
  • the numerator might be negative. In this case the function value is negative infinity
  • the numerator is positive. In this case the function value is positive infinity
  • the numerator is 0. In this case you need to take the limits to figure out what the situation is
  • the numerator is complex valued. In this case the result is a combination of the above
Anywhere that the denominator is defined and non-zero, you can use the derivative of the numerator to determine the critical points (but whether they correspond to maxima or minima can depend on the sign of the denominator)
Yumeng Yin
Yumeng Yin 2022-4-26
Thanks a lot, yes, I agree with you. After carefully adjusting the parameters, I can get the s-shape as what I expect. Thank you both.

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