I am trying to create an array h of size 1001 which can store value of 2 integrals

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Tarun Agarwal 2021-7-25

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评论： Tarun Agarwal 2021-7-25

gl=10;
gr=10;
N=1000;
L=10;
l=10;
h = zeros(1,1000);
syms x;
for i=1:1001
    f =gl*(7/44)*(exp((-1*L*i)/(N*l*cos(x))));
    A=vpa(int(f,-1.57,1.57));
    A
    f =gl*(7/44)*(exp((-1*(L-N)*i)/(N*l*cos(x))));
    B=vpa(int(f,1.57,4.71));
    B
    h(i)=A+B;
end

1- On running instead of value of B I am getting exact expression is printing intead of value of B

2- I want to store all these values in the array h from index 1 to 1001 but its showing error message "Unable to convert expression into double array.".

Plase help and thanks a lot in advance.

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darova 2021-7-25

Try double instead of vpa

Jan 2021-7-25

在 MATLAB Online 中打开

@darova: This does not work also. See:

gl=10;

gr=10;

N=1000;

L=10;

l=10;

syms x;

i = 1;

f = gl * (7 / 44) * (exp((-1 * (L-N) * i) / (N * l * cos(x))));

B = vpa(int(f, x, [1.57, 4.71]))

B =

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Answer 1

Jan 2021-7-25

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https://ww2.mathworks.cn/matlabcentral/answers/885484-i-am-trying-to-create-an-array-h-of-size-1001-which-can-store-value-of-2-integrals#answer_753494

在 MATLAB Online 中打开

gl=10;

gr=10;

N=1000;

L=10;

l=10;

syms x;

i = 1;

f = gl * (7 / 44) * (exp((-1 * (L-N) * i) / (N * l * cos(x))));

B = vpaintegral(f, x, [1.58, 4.71]) % Smaller interval, x > pi/2

B =

3.91702

B = vpaintegral(f, x, [1.57, 4.71]) % Original interval

B =

For cos(x)==0, or x=pi/2 the exponent gets infinite and the integration fails due to the pole.

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Yongjian Feng 2021-7-25

Ah... cos(x) is in the denominator! You are right.

Tarun Agarwal 2021-7-25

@Jan Thank you very much It worked

@darova @Yongjian Feng Thanks a lot .

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Answer 2

Yongjian Feng 2021-7-25

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编辑：Yongjian Feng 2021-7-25

It doens't allow you to go outside (-1.57, 1.57). This is essentially (-PI/2, PI/2). Since your f(x) is a function of cos(x), so your f(x) is also a periodic function. Try to keep the integration range within (-PI/2, PI/2), and it should work.