0

$$ \frac1{(T-t)A_0^2\Delta}\sum_{\{q>p\}, \{p\}}\frac{\tau_p^{z_1-1}}{\Gamma(z_1)}\hat u(t_q-t_p-1)\frac{\tau_q^{z_2-1}}{\Gamma(z_2)} $$

Hi, can anyone help me with this summation?

does $\{q>p\}, \{p\}$ mean $(p=1, q=2, 3, 4...) + (p=2, q=3, 4, 5...) + (p=3, q=4, 5, 6...) + \dots$ and so on?

I tried this method, but the calculation doesn't seem to be correct.

I am not good at mathematics but I need to understand this summation to perform a calculation in Chemistry. Thank you!

Brian M. Scott
  • 616,228
Hsin-Yen Lee
  • 1
  • 1
    I think that we’ll need more context: what’s the general term inside the summation? – Brian M. Scott Jul 09 '13 at 04:06
  • Hello, please take a look at this picture,http://www.flickr.com/photos/ta24/9245974252/

    suppose p = 1500, and q = 1500 too.

     – Hsin-Yen Lee Jul 09 '13 at 04:08
  • I had hoped to be able to tell more from the expression, but I’m afraid that one would need to know more of the background: the intent simply isn’t completely clear just from the notation. If I had to guess, I’d guess that $q$ is a fixed value and that the sum is over all values of $p$ less than that fixed value (and greater than some physically defined minimum), but that would be just a guess, and I have absolutely no confidence in it. – Brian M. Scott Jul 09 '13 at 04:17

1 Answers1

0

There's nothing in the sum, so you're obviously missing something.

However, sometimes you might see the following notation:

$$\sum_{j \ge k}^n x_n.$$

This means nothing by itself, but contextually, we take the sum over all values of $j$ for which $j \ge k$. Typically, you'll see a summation sign like this inside some other notation that relies on indexing, for instance:

$$\sum_{k=1}^n \sum_{j \ge k}^n x_n.$$

So here, when $k=1$, you add $x_1, x_2, \ldots, x_n$.

When $k=2$, you add $x_2, x_3, \ldots, x_n$, and so on...

Emily
  • 35,688
  • 6
  • 93
  • 141