How to calculate derivative and then apply limit in matlab

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Ravikiran Mundewadi 2020-4-22

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评论： Ravikiran Mundewadi 2020-4-24

How to calculate diff((x/(exp(x)-1)),x,n) and then apply the limit at x=0 Where n=11,12,13,14,15 I am getting upto 1 to 10 but not from 11 onwards... please anybody help me...

回答（2 个）

Deepak Gupta 2020-4-22

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编辑：Deepak Gupta 2020-4-23

在 MATLAB Online 中打开

Hi Ravikiran,

I have used a for loop.

syms f(x) x; 
f(x) = x/(exp(x)-1);
g = f;
limitg = sym(zeros(15, 1));
for n = 1:15
    g = diff(g);
    limitg(n) = limit(g, x, 0);
end

Code may throw error of "Devide by Zero".

Thanks,

Deepak

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Deepak Gupta 2020-4-23

编辑：Deepak Gupta 2020-4-23

Ameer, Ravikiran's claim may be correct but not sure how's he getting to the result. Using simplify results will be different because simplify reduces the expression by removing less significant terms.

Using simplify, we can calculate results upto n=15 and ans is :

[ -1/2, 1/6, 0, -1/30, 0, 1/42, 0, 0, 0, 0, 0, 0, 0, 0, 0]

But without using simplify upto n = 10, ans is:

[ -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66]

So you can see results vary with and without using simplify.

In my understanding, For longer expressions, i.e. in this case expression for n=11 is

11242*exp(2*x))/(exp(x) - 1)^3 - (615780*exp(3*x))/(exp(x) - 1)^4 + (9003720*exp(4*x))/(exp(x) - 1)^5 - (56133000*exp(5*x))/(exp(x) - 1)^6 + (180789840*exp(6*x))/(exp(x) - 1)^7 - (325987200*exp(7*x))/(exp(x) - 1)^8 + (332640000*exp(8*x))/(exp(x) - 1)^9 - (179625600*exp(9*x))/(exp(x) - 1)^10 + (39916800*exp(10*x))/(exp(x) - 1)^11 - (11*exp(x))/(exp(x) - 1)^2 - (x*exp(x))/(exp(x) - 1)^2 + (2046*x*exp(2*x))/(exp(x) - 1)^3 - (171006*x*exp(3*x))/(exp(x) - 1)^4 + (3498000*x*exp(4*x))/(exp(x) - 1)^5 - (29607600*x*exp(5*x))/(exp(x) - 1)^6 + (129230640*x*exp(6*x))/(exp(x) - 1)^7 - (322494480*x*exp(7*x))/(exp(x) - 1)^8 + (479001600*x*exp(8*x))/(exp(x) - 1)^9 - (419126400*x*exp(9*x))/(exp(x) - 1)^10 + (199584000*x*exp(10*x))/(exp(x) - 1)^11 - (39916800*x*exp(11*x))/(exp(x) - 1)^12

Matlab fails to find convergence point and hence returns the expression itself.

Ameer Hamza 2020-4-23

编辑：Ameer Hamza 2020-4-23

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I get the same answer, with or without simplify(). I am using R2020a, and from recent questions on this forum, I have noticed that they have improved Symbolic toolbox in this release.

syms x
y = x/(exp(x)-1);
for i=1:15
    Dy = simplify(limit(diff(y, i), x, 0));
    if Dy == 0
        fprintf('i = %d,\t Dy = 0\n', i);
    else
        [n, d] = rat(Dy);
        fprintf('i = %d,\t Dy = %d/%d\n', i, n, d);
    end
end

Result:

i = 1,	 Dy = -1/2
i = 2,	 Dy = 1/6
i = 3,	 Dy = 0
i = 4,	 Dy = -1/30
i = 5,	 Dy = 0
i = 6,	 Dy = 1/42
i = 7,	 Dy = 0
i = 8,	 Dy = -1/30
i = 9,	 Dy = 0
i = 10,	 Dy = 5/66
i = 11,	 Dy = 0
i = 12,	 Dy = 0
i = 13,	 Dy = 0
i = 14,	 Dy = 0
i = 15,	 Dy = 0

Ameer Hamza 2020-4-23

It turns out OP's claim is correct: https://www.wolframalpha.com/input/?i=limit+x-%3E0+d%5E12%2Fdx%5E12+x%2F%28e%5Ex-1%29

This is one of those situations where limitations of MATLAB symbolic toolbox become visible.

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Ameer Hamza 2020-4-23

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在 MATLAB Online 中打开

As discussed in the comment to Deepak's answer. This is a limitation of of MATLAB's symbolic engine. MATLAB calculates the limit for n-th order derivatives for n>10 to be zero

syms x
y = x/(exp(x)-1);
for i=1:15
    Dy = simplify(limit(diff(y, i), x, 0));
    if Dy == 0
        fprintf('i = %d,\t Dy = 0\n', i);
    else
        [n, d] = rat(Dy);
        fprintf('i = %d,\t Dy = %d/%d\n', i, n, d);
    end
end

Result

i = 1,	 Dy = -1/2
i = 2,	 Dy = 1/6
i = 3,	 Dy = 0
i = 4,	 Dy = -1/30
i = 5,	 Dy = 0
i = 6,	 Dy = 1/42
i = 7,	 Dy = 0
i = 8,	 Dy = -1/30
i = 9,	 Dy = 0
i = 10,	 Dy = 5/66
i = 11,	 Dy = 0
i = 12,	 Dy = 0
i = 13,	 Dy = 0
i = 14,	 Dy = 0
i = 15,	 Dy = 0

But Wolfram Alpha is able to calculate the limit: https://www.wolframalpha.com/input/?i=limit+x-%3E0+d%5E12%2Fdx%5E12+x%2F%28e%5Ex-1%29. Even this FEX submission by John: https://www.mathworks.com/matlabcentral/fileexchange/20058-adaptive-numerical-limit-and-residue-estimation is not able to converge to a limit. I guess there is nothing much you can do about in MATLAB, other than writing your own closed-form solution if it is possible.