infinite loop and the term within loop again contains infinity

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math seeker 2018-5-27

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评论： Walter Roberson 2018-5-28

I want to solve a loop given below

prev_val=0; 
  for m=0:inf
    curr_val=(beta/(2*m+1)*pi)*sin((2*m+1)*pi/beta)*exp((2*m+1)^2)*alpha;
    prev_val=prev_val+curr_val;
  endfor
gt=const*prev_val;

I am facing two problems with this. In this, the upper limit of the loop, as per the equation to be solved, is infinity. In addition, the value of beta is again infinity. How can I solve this problem?

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math seeker 2018-5-28

I mean beta is a constant and its value is infinity.

Walter Roberson 2018-5-28

In that case, as I posted, the sum over m is sign(alpha) times infinity

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Answer 1

Rik 2018-5-27

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I found an answer to your question on stackoverflow: if you don't have the symbolic toolbox, quadgk can still solve your integral.

f = @(x) x.*exp(-x);
a = quadgk(f, 0, inf) 
a =
    1.0000

How you can deal with beta being infinity is something I don't have any ideas for. Maybe you can try with just a big number, maybe enlarging it in a while loop to simulate the limit function.

f=@(m) (beta/(2*m+1)*pi)*sin((2*m+1)*pi/beta)*exp((2*m+1)^2)*alpha;

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Ameer Hamza 2018-5-27

You can also use integral(). The function name integral() appears to be more intuitive as compared to quadgk() as himself mentioned by Mathwork cofounder.

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