f(x)=symsum(2^(-m*x)*(gamma(m*x+j-1+1)/gamma(m*x-1+1))*h(j/m),j.0.Inf)
where h(j/m)=(j/m)^2+1, m=100.
Please help me out.

댓글 수: 5

Are you sure it's gamma and not square root?
@Dyuman Joshi. Yes sir, it's gamma function.
The code for the sum is as follows. Though, in this form, it takes to much time to compute.
If you analyse it on pen-paper you might get an reduced form which is easier/faster to compute.
syms x j
m=100;
2^(-m*x)*symsum(gamma(m*x+j)*((j/m)^2+1)/(gamma(m*x)*2^j*factorial(j)),j,0,Inf)
Thanks @Dyuman Joshi but this is giving error.
It worked on my pc without any errors, but it took an eternity to run :')
(i3-5th gen, 8 gb ddr3 ram)

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 채택된 답변

Torsten
Torsten 2022년 6월 12일
편집: Torsten 2022년 6월 12일
m = 100;
h = @(x) x.^2+1;
x = 0:0.01:10.0;
tol = 1e-16;
for i = 1:numel(x)
Lm(i) = fun_Lm(m,h,x(i),tol);
end
plot(x,Lm)
function value = fun_Lm(m,h,x,tol)
value = 0.0;
error = Inf;
j = 0;
summand = 1.0;
while error > tol
value = value + summand;
error = abs(summand);
summand = summand * (m*x+j) * 1/2 * 1/(j+1) * h((j+1)/m)/h(j/m);
j = j + 1;
end
value = value/2^(m*x) ;
end

추가 답변 (1개)

Gufran Malik
Gufran Malik 2022년 8월 17일

0 개 추천

How can we plot this function? @Torsten Please help me with this

댓글 수: 4

You woiuld need a fixed value for m in order to plot that, unless you wanted to plot a surface with a range of m values.
It simplifies surprisingly far
syms h(x)
syms j m s x positive
h(x) = x^2 + 1
h(x) = 
inner(j) = int(h(s), s, j/m, (j+1)/m)
inner(j) = 
outer = symsum( gamma(m*x + j) / (gamma(m*x) * 2^j * factorial(j)) * inner(j), j, 0, inf)
outer = 
f(x) = outer / 2^(m*x)
f(x) = 
fs(x) = simplify(f(x))
fs(x) = 
m = 100;
H = @(lb,ub) (1/3*ub^3+ub) - (1/3*lb^3+lb);
x = 0:0.01:10.0;
tol = 1e-16;
Lm = zeros(size(x));
for i = 1:numel(x)
Lm(i) = fun_Lm(m,H,x(i),tol);
end
Lm_Walter_Roberson = (9*m*x+1+3*m^2+3*(m*x).^2)/(3*m^3);
plot(x,[Lm;Lm_Walter_Roberson])
function value = fun_Lm(m,H,x,tol)
value = 0.0;
error = Inf;
j = 0;
summand = H(0/m,1/m);
while error > tol
value = value + summand;
error = abs(summand);
summand = summand * (m*x+j) * 1/2 * 1/(j+1) * H((j+1)/m,(j+2)/m)/H(j/m,(j+1)/m);
j = j + 1;
end
value = value/2^(m*x) ;
end
@Torsten Thank you so much.
@Walter Roberson Thank you so much.

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질문:

2022년 6월 7일

편집:

2022년 8월 21일

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