I don't know if matlab is calculating this properly

Is this calculating correctly? It returns my answer as 1. It is meant to represent this series 1 + 1⁄r + 1⁄r^2 + 1⁄r^3 + … + 1⁄r^n.

function sum = mysum(r,n) 
sum = 1; 
for i = 1:n;     
sum = sum + 1/(r^n);
end

Please help.

Answer 1

[John D'Errico]

MATLAB is calculating what it calculated properly. The issue is, you told it to calculate the wrong thing. Computers are sooooo picky. :)

You don't tell us what values of r that you used. Or what n was when you ran this code. But if you used a large value for n, then yes, it SHOULD return 1. Or it might return inf. Really? What did you write, and why is that?

In fact, you wrote:

1 + 1⁄r^n + 1⁄r^n + 1⁄r^n + … + 1⁄r^n

Your exponent was fixed, at n.

The exponent in what you wrote was CONSTANT, at n. So if n was large, and abs(r ) was relatively large so that 1/abs(r ) is small, then you would see 1 as a result, because the powers will underflow. Or if r is itself small, then you would just get overflows.

The fix is easy, of course. Change n to i in the expression. That is, the exponent of r needs to be the index variable, not the number n itself.

function sum = mysum(r,n) 
sum = 1; 
for i = 1:n;     
  sum = sum + 1/(r^i);
end

And using the name sum for a variable is a really bad idea, as it will cause bugs in your code sometime, when you actually want to use the FUNCTION named sum.

Always avoid using existing function names as variable names.

Answer 2

[madhan ravi]

    Result = mysum(2,10) % just call it like this
    doc function % to know how to use functions 

    function SUM = mysum(r,n)
    SUM = 1; % don’t name variable sum because it will shadow the MATLAB’s inbuilt function
    for ii = 1:n;
        SUM = SUM + 1/(r^ii);
                       % ^^-- have a look here 
    end
    end

[madhan ravi]

I don’t know why you deleted your previous question.

[Patrick Smith]

I am asking if it is calculating properly. I'm asking if the matlab program is working correctly and returning the correct answer for mysum(3,12) of the series equation

[madhan ravi]

No you had a minor mistake, see the edited answer. You can avoid the loop completely by doing:

SUM = 1 + sum(1./r.^(1:n))

[John D'Errico]

Madhan beat me to the punch. :)

One day I'll learn to type faster.

[Patrick Smith]

But how bout the loop itself? I understand that's more effiecient but does my method work correctly? I'm looking for origionality here.

[madhan ravi]

@John D' Errico: :) I wasn't able to spot it at the beginning though. If you would have been my algebra teacher, I would have developed my maths knowledge even further and wouldn't be suffering in few places (just saying :)).

@Patrick Smith: Yes , you just had to change n to i.

[John D'Errico]

Your problem in the loop was the exponent (n) of r, which needs to be the INDEX variable, not the fixed number n.

[John D'Errico]

Please accept the answer from Madhan, as he was first.

[Patrick Smith]

I'm confused which n do I change to an i?

function sum = mysum(r,n)

sum = 1;

for i = 1:i;

sum = sum + 1/(r^n);

end

Do you mean this? or can you send back what you guys mean lol

[madhan ravi]

function SUM = mysum(r,n)
    SUM = 1; 
    for ii = 1:n;
        SUM = SUM + 1/(r^ii);
                       % ^^-- have a look here 
    end
    end

[Patrick Smith]

Perfectt! Thank you! I couldn't figure out why it kept giving me 1 which made not sense.