Find accuracy of 0.xx.. using Symsum

Hello!

I was given the following series (1 to inf):

((47^(2*n) *(-1)^n)/ Factorial(2*n),1,inf)

Which eventually gives you cos 47 in radians.

I'm trying to find a way to calculate the index number that gives me the cos 47 accuracy of 0.001 (or less), using symsum. (and not with a 'while' loop that checks every iteraion if cos47 minus current series gives me 0.001 or less).

Thank you very much!

[Walter Roberson]

My symbolic toolbox says that sum would be cos(47)-1 and that you would have to start from 0 instead of 1 to get cos(47)

[Walter Roberson]

Careful, the values get larger until you reach term 47

[Study a]

My bad!yes you are right, it should start from 0. I still can't find a way to find the accuracy (using the while solution I think it was between n=90 to n=100

[Study a]

For those who find this question later on, I found a solution:

syms n;
a= symsum((47^(2*n)*(-1)^n)/factorial(2*n),0,65);
a=vpa(a,4);
b= symsum((47^(2*n)*(-1)^n)/factorial(2*n),0,66);
b=vpa(a,4);
disp(a-b);

this gives you an accuracy of 0.001.

Answer 1

[Study a]

For those who find this question later on, I found a solution:

syms n;
a= symsum((47^(2*n)*(-1)^n)/factorial(2*n),0,65);
a=vpa(a,4);
b= symsum((47^(2*n)*(-1)^n)/factorial(2*n),0,66);
b=vpa(a,4);
disp(a-b);

this gives you an accuracy of 0.001.