# Homework Help: Summation evaluation

1. May 23, 2010

### sara_87

1. The problem statement, all variables and given/known data

How do i evaluate the following sum

$$\sum_{i=1}^{50}\frac{1}{(100-i)^{1/2}}$$

2. Relevant equations

3. The attempt at a solution

i havent a clue on how to do this, can someone please give me a hint?
thank you

2. May 23, 2010

### LCKurtz

I think you don't, at least not in terms of a nice closed formula. You can evaluate it with something like Maple or approximate it with an integral. What is the background for where this problem came from?

3. May 23, 2010

### sara_87

thanks,
it's just a part of a numerical problem.
I used matlab to find the sum :)

but, if in the denominator, i had:
100a-i

where a is constant, how would i evaluate that? is there a formula?
(matlab surely wouldnt work here).

4. May 23, 2010

### Lancelot59

I agree. There isn't any formula I know of that will solve that, and I can't see any way of re-arranging it into a suitable form.

5. May 23, 2010

### Tedjn

If a is a constant, write your Matlab function to take it in as a parameter.

6. May 23, 2010

### sara_87

thanks both.

so, ur saying that if i evaluate it in matlab then it should give me an answer in terms of a?

7. May 23, 2010

### Count Iblis

You can also use the Euler-Maclaurin sum formula.

8. May 24, 2010

### vaibhav1803

Euler maclaurin wont fit easily here, its for sums that can be approximated to integrals, if i'm not wrong,
here the domain of the sum is really small, plus due to the square root makes it worse (see graph of 1/sqrt(x), it apparently ceases to fall/falls very slowly after a certian interval)

hence i suggest use of recursive algorithms to find it, the only way to get the accurate sum from a mathematical standpoint.

9. May 24, 2010

### Lancelot59

Or you could just add up all the terms.

10. May 24, 2010

### Count Iblis

Euler-Maclaurin works fine in this case. As long as the start and end points at which the summand is evaluated are not close to singularities, you'll get an asymptotic series that converges to quite close to the answer before it starts to diverge.

11. May 24, 2010

### Count Iblis

Answer obtained by adding up the numbers on my antique HP-28s calculator

5.97865123618

Answer obtained by Euler-Maclaurin summation, also using my HP-28s calculator

5.97865123616

The difference is only due to roundoff errors. I tried to minimize that by starting with the smallest terms and adding the largest terms at the end. For the Euler-MacLaurin summation, I only needed to go as far as the B_4/4! f^(3) term. I did include the next term, but that did not contribute to the result.

12. May 25, 2010

### sara_87

thank you very much,
this helps a lot
:)

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