How can I plot the graph for this function?
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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.

5 comentarios
Dyuman Joshi
el 7 de Jun. de 2022
Are you sure it's gamma and not square root?
Gufran Malik
el 8 de Jun. de 2022
Dyuman Joshi
el 8 de Jun. de 2022
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)
Gufran Malik
el 10 de Jun. de 2022
Dyuman Joshi
el 11 de Jun. de 2022
It worked on my pc without any errors, but it took an eternity to run :')
(i3-5th gen, 8 gb ddr3 ram)

Respuesta aceptada
Más respuestas (1)
Gufran Malik
el 17 de Ag. de 2022
0 votos
4 comentarios
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
inner(j) = int(h(s), s, j/m, (j+1)/m)
outer = symsum( gamma(m*x + j) / (gamma(m*x) * 2^j * factorial(j)) * inner(j), j, 0, inf)
f(x) = outer / 2^(m*x)
fs(x) = simplify(f(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
Gufran Malik
el 21 de Ag. de 2022
Gufran Malik
el 21 de Ag. de 2022
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