How can i store values into an array while making summation

Hi, im trying to sum this series

I want to make S(in the code) an array to keep all of values from n=N/2 to infinity. Than i will sum them up and obtain the sum of my series. My code also should do the calculation for different N values.At the and i will make a plot of Sum of my series vs N. It is shown in the image also.Thanks!

 clc;

 clear;

 clear all;

 format long

 N=10:2:100;

 for i=1:length(N)

 for n=(N(i)/2)+1:1:10e5

 S(i)=(1/n)^2;

 Stot=sum(S);

 end 

 R(i)=2*pi*Stot;

 end

 loglog(N,R)

[Guillaume]

keep all of values from [...] to infinity.

Unless you have infinite memory, that's going to be difficult!

The sample code you show seems to assume that infinity is 1e6 which is a very small number in the ream of floating point double (whose maximum value is about 1e308) and certainly not infinity.

I do not understand your summation notation and how it relates to the code you've written.

[Zuy]

It should converge actually 1000 is enough for infinity in my case but i couldnt obtain the true graph

[madhan ravi]

explain step by step what you are trying to achieve ? if its summation you don't even have to use loop

[Zuy]

I am trying to sum this series (1/n^2) going from k=(N/2)+1 to k=inf. But at the same time i want to change N because i want to make a plot that shows N vs summation. This plot will show how my summation change according to N.(starting value of summation). Thanks for your help!

[madhan ravi]

   N=10:2:100;
   n=(N./2)+1:1:10e5;
   S=cumsum((1./n).^2);
   plot(N,S(1:numel(N)))

[Zuy]

Thanks!

[madhan ravi]

As always! Anytime :)

[Guillaume]

Beware!

N=10:2:100;
n=(N./2)+1:1:10e5;

Is exactly the same as

N = 10;
n = N/2+1:10e5;

There is no point in passing a vector to the : (colon) operator as it only uses the first element.

Notice that the step 2 of N has been lost. Fortunately, the plot still works because the original step of 2 of N would result in a step of 1 for n. But with any other step for the original N vector, the above would not work.

[madhan ravi]

Ah thank you Guillame how to overcome this issue ? Any insights would be helpful

[Zuy]

Yes unfortunatelly it works in a wrong way :(

[Guillaume]

While you probably could work out a vectorised version that calculate the cumsum of n with a half step and selecting the correct sum depending on the parity of the elements in the N vector, I think you would be better off with an explicit loop over N:

N = 10:5:100
Nsum = zeros(size(N));
for Nidx = 1:numel(N)
   n = N(Nidx)/2+1 : 1e6;
   Nsum(Nidx) = sum(1./n.^2);
end
plot(N, Nsum)

I still have no idea how that relates in any way to the summation image in the question.

[madhan ravi]

wow your intelligent , I thought we could get rid of that loop by any chance , neither I could interpret the image without the question

[Zuy]

Thank you so much!. fk in the summation image is 1/k. Bottom indice of the summation is N/2+1 and top indices is going inf. Also N is changes.

[Zuy]

What if i use symsum for the infinite summation? Like that

 clear;

 clc;

 N = 0:2:70;

 Nsum = zeros(size(N));

 for Nidx = 1:numel(N)

    syms n

    Nsum(Nidx) = symsum(1./4.^n,n,(N(Nidx)/2)+1,inf);

 end

 plot(N,log10( Nsum))

[Guillaume]

Possibly that would work. As I don't have the symbolic toolbox, I can't help with that.

[Zuy]

Thank you so much!!

Answer 1

[madhan ravi]

    Stot(i)=sum(S(i));  %use iterator as an index to avoid overwriting

also see preallocation for making code efficient

[Zuy]

Thanks!

[madhan ravi]

Anytime :)