How do I calculate an infinite series using a function file with for-end loops?

Given f(x)= [(x^2)/37]+[(x^3)/3!]+[(x^4)/4!]+[(x^5)/5!]+...

Using a function file (function fn=my_fun(x,n)) with for-end loops for several test files.

I'm just having issues defining the function file using the for-end loop. What I have so far is:

function [xval,nterms]=my_series(x,n)

[Youssef Khmou]

hi is the code below fine? i mean your input is : vector X and order N , so as for every point in X we sum N terms defined with general formula .

[Walter Roberson]

Could you confirm that the first term is divided by 37, and not factorial(2) as would match the 3! 4! 5! pattern?

[Youssef Khmou]

that is matter of Initial conditions

Answer 1

[Youssef Khmou]

hi gordon,

you function returns a scalar or vector values ? is this exp(x) Taylor series ?

you can try this light version :

 function [xval]=your_series(x,n)

 % X is vector input
 % n number of iteration
 xval=zeros(size(x));

 order=0:n;

 for ii=1:length(x)
     for t=1:length(order)

         xval(ii)=xval(ii)+(x(ii)^t)/factorial(t);

     end
 end

This function approximates well the EXP FUNCTION, replace the term ( x(ii)^t/t! ) with your general Formula .

[Rick]

Function returns scalar quantity.

This is very helpful in the general procedure, but in defining the xval(ii)=... term, I'm not sure how to replace it with my formula, which has the first term divided by 37, not factorial(2). How do I generalize this formula and input it in the for-end loop for calculation?

[Youssef Khmou]

ok, and in the input is vector ?

[Youssef Khmou]

if it is a scalar then :

 % given scalar x 
 S=0;
 S=(x^2)/37;
 for ii=3:N
 S=S+((x^ii)/factorial(ii));
 end

Answer 2

[Azzi Abdelmalek]

x=2
n=5
f=x^2/37+sum(arrayfun(@(n) x^n/factorial(n),3:n))