infinite loop and the term within loop again contains infinity
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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?
댓글 수: 5
Walter Roberson
2018년 5월 28일
In that case, as I posted, the sum over m is sign(alpha) times infinity
답변 (2개)
Rik
2018년 5월 27일
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;
댓글 수: 1
Ameer Hamza
2018년 5월 27일
You can also use integral(). The function name integral() appears to be more intuitive as compared to quadgk() as himself mentioned by Mathwork cofounder.
Walter Roberson
2018년 5월 27일
For the case of limit as beta approaches infinity, then under the assumption that m is a non-negative integer, the individual terms are
Pi^2*exp(4*m^2+4*m+1)*alpha
The sum of this over m = 0 to infinity is sign(alpha) * infinity
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