Summation of Infinity Terms

S = 0;
tol = eps;
term = inf;
n = 0;
while abs(term) > tol
    term = exp(x*log10(n));
    S = S + term;
    n = n + 1;
end 

It try by this way but I don't know what is wrong

لا

[dpb]

Put the Q? in the space for it, not the title.

We don't solve homework without the poster showing what effort they've put into it first...

[Rewida Hassan]

This is my question and I made edit

[dpb]

Again, put the question in the space supplied for it (unless the submitter also can't edit the title once entered?)

Then use the code button to format the code legibly.

Then, what is the specific Matlab question?

[dpb]

No, what is YOUR specific Matlab question regarding the homework problem? Where did you have trouble?

But, thank you for proper formatting...helps greatly for folks to be able to follow. :)

[Rewida Hassan]

I'm trying to solve this question by this code but it doesn't work , what is wrong in my code in summation to infinity

[dpb]

For starters, x isn't defined...

You probably want to set the tolerance to something > eps for initial testing.

Other than that, looks ok to me; set breakpoint in the debugger and step through to see what happens as you iterate.

[Rewida Hassan]

It's first time to me in Matlab can you write the right code for me please because it's not work

[dpb]

I missed one thing in my first look...look at the summation limits carefully in the definition and what you initialized n as.

Actually, missed the biggie by not paying attention...dunno where you got the expression you're evaluatiing, but that isn't the expression in the assignment at all...

Again, you do need to define x (>1 for series to converge)

[Rewida Hassan]

S = 0;
tol = eps;
term = inf;
n = 1;
If (x<1)
'not valid starting points enter another points'
end
while abs(term) > tol
    term = (1/power (n , x));
    S = S + term;
    n = n + 1;
end

Answer 1

[David Hill]

You need to create a function. The easiest is to use the existing function. It doesn't sound like your instructions prevent this.

function z = myzeta(x)
z=zeta(x);
end

Alternatively, it is easy to determine the stopping point but floating point inaccuracies will affect the result.

function z = myzeta(x)
n=ceil((1/eps)^(1/x));
z=sum((1:n).^(-x));
end

Answer 2

[dpb]

S = 0;
tol = eps;
term = inf;
n = 1;
If (x<1)
'not valid starting points enter another points'
end
while abs(term) > tol
    term = (1/power (n , x));
    S = S + term;
    n = n + 1;
end

Is not bad start for learner...I used similar with few changes and got answers that match those looked up...

S = 0;
tol = 1E-7;       % run quicker; approx single precision accuracy  eps;
term = bigvalue;  % use finite values instead of inf
n = 1;
if (x<=1), error('not valid x--enter another points >1'), end
while abs(term) > tol
  term = (1/power (n , x));
  S = S + term;
  n = n + 1;
end