summation of infinite series

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Sudhir Sahoo
Sudhir Sahoo 2021-2-15
评论： Walter Roberson 2021-2-17
How I can impliment this type of infinite series in MATLAB given a function as where ,, and . Here l,γ,R are constant.
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Walter Roberson
Walter Roberson 2021-2-16
Post your code.
In terms of the code I posted earlier, the way forward would be to comment out that line that produces the error message, as I showed in the code how to get outputs without the error; I left the call to double() in to demonstrate that double() itself could not be used for the situation.
Sudhir Sahoo
Sudhir Sahoo 2021-2-17
format long g
syms k;snr = 2;l = 2;
Limit = Inf;
expression2 = symsum([D(k,R)*(G(snr,l) + H(k + 1, snr,l)*(snr/2)^1.5*(l^((2*k + 1)*(m + 2))/sqrt(pi)) )],k,1,Limit);
D = 150;
ev = vpa(expression2, D);
ed = double(ev);
where is function
function [out] = D(k,R)
m = -1/log2(cos(pi/6)); % Here m is $\gamma$
exp_c_k = factorial(2*k -1)/(factorial(k)*factorial(k-1)*2^(2*k - 1));
exp_r = (1 + R)^[((2*m + 4)*k + m + 4)/2] - 1;
exp_denom = pi * R * (2*k + 1)* ((2*m + 4)*k + m + 4);
out = 8*exp_c_k * exp_r/exp_denom;
end
and as
function [out] = H(k,z,l)
%gam, y,z,l are user inputs
y = k;
m = -1/log2(cos(pi/6)); % in paper its used as small gamma.(\gamma)
out1 = (2/z)^y;
out2 = gamma(y) - igamma(y,(l^(-2*(m + 2))/2)*z);
out = out1 * out2;
end
as
function out = G(x,l)
m = -1/log2(cos(pi/6));
out = qfunc(sqrt(l^(-2*(m + 2))*x))
end
This code is simplified version of the expression of the
where ,, and . Here l,γ,R are constant.
when I code this I got the error like this "The following error occurred converting from sym to double:
Unable to convert expression containing remaining symbolic function calls into double array. Argument must be expression that evaluates to number."
Howerver I have tried in this manner for convergence is it correct one ?
Limit = Inf;
for k = 1:Inf
expression2 = [D(k,R)*(G(snr,l) + H(k + 1, snr,l)*(snr/2)^1.5*(l^((2*k + 1)*(m + 2))/sqrt(pi)))];
if abs(expression2) < 10^(-7)
break;
end
end

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回答（2 个）

Walter Roberson
Walter Roberson 2021-2-17
You did not give a value or range for R. My suspicion is that if R is not -1 or less, that the sum is infinite when the upper bound is infinite.
The code you were using required evaluation at over 1500 digits of precision to stableize the output when using 200 terms. However once I looked more closely at the code, I realized that some of it was being evaluated in pure double precision, so the answers it was getting were effectively numeric nonsense.
Unfortunately, the revised version is very slow.
I had to completely invent a value of R for testing purpose. As I had no guidance as to a proper value, I used 7 in the code.
With 200 terms, and using 5000 decimal places, the result is over 10^3000 . As the number of terms goes up, the result goes up quickly.
If you use fewer DIGITS, and the answer you get out for ev is in scientific notation, starting with a digit followed by a period, then you need to increase the DIGITS: whatever result you got out has been compromised. If the result you get out is negative, then you need to increase the DIGITS. If you do get out an actual number without scientific notation, then you might still need to increase the DIGITS
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Sudhir Sahoo
Sudhir Sahoo 2021-2-17
编辑：Sudhir Sahoo 2021-2-17
Sir R value is 1. Actually this is a part of a code of probablility of error calculation in a communication research.
Walter Roberson
Walter Roberson 2021-2-17
Note: going beyond 500 terms for the upper limit will make the calculation unbearably slow. I do not mean a gradual deterioration, I mean a sharp deterioration. vpa(factorial(sym(1001)),2600) is very very slow, but 1000 instead is fast.

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Sudhir Sahoo
Sudhir Sahoo 2021-2-16
format long g
syms k;snr = 2;l = 2;
Limit = Inf;
expression2 = symsum([D(k,R)*(G(snr,l) + H(k + 1, snr,l)*(snr/2)^1.5*(l^((2*k + 1)*(m + 2))/sqrt(pi)) )],k,1,Limit);
D = 150;
ev = vpa(expression2, D);
ed = double(ev);
where is function
function [out] = D(k,R)
m = -1/log2(cos(pi/6)); % Here m is $\gamma$
exp_c_k = factorial(2*k -1)/(factorial(k)*factorial(k-1)*2^(2*k - 1));
exp_r = (1 + R)^[((2*m + 4)*k + m + 4)/2] - 1;
exp_denom = pi * R * (2*k + 1)* ((2*m + 4)*k + m + 4);
out = 8*exp_c_k * exp_r/exp_denom;
end
and as
function [out] = H(k,z,l)
%gam, y,z,l are user inputs
y = k;
m = -1/log2(cos(pi/6)); % in paper its used as small gamma.(\gamma)
out1 = (2/z)^y;
out2 = gamma(y) - igamma(y,(l^(-2*(m + 2))/2)*z);
out = out1 * out2;
end
This code is simplified version of the expression of the
where ,, and . Here l,γ,R are constant.
when I code this I got the error like this "The following error occurred converting from sym to double:
Unable to convert expression containing remaining symbolic function calls into double array. Argument must be expression that evaluates to number."

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