Solving expression with multiple summation notations

Question

In the following, I need to solve the expression in terms of $p_o$. I will appreciate any help in this regard.

$$Y = \lim_{J\to100}\sum^{J}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j}{e^\sigma j!(j-m)!}\lim_{Q\to100}\sum^{Q}_{q=0}\frac{(-1)^q}{q!}\Big(\frac{A}{p_o}\Big)^q(\theta)^{2q-(j-m)+1}$$

$Y_i$ is the probability of finding an object in my equation, $\sigma$ is any positive integer. $A$ is a constant which is a combination of a long-term not shown here.

I need to find the solution in terms of $p_o$. Can someone help me?

$$Y = \lim_{J\to100\\Q\to 100}\sum^{J}_{j=0}\sum^{j}_{m=0}\sum^{Q}_{q=0} \frac{\sigma^j}{e^\sigma j!(j-m)!}\frac{(-1)^q}{q!}A^q (\theta)^{2q-(j-m)+1}\frac{1}{p_o^q}$$

Answer 1

As mentioned in the comments above, the equation can be re-written as :

$$Y = \sum^{\infty}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j}{e^\sigma j!(j-m)!}\sum^{\infty}_{q=0}\frac{(-1)^q}{q!}\Big(\frac{A}{p_o}\Big)^q(\theta)^{2q-(j-m)+1}$$

$$Y = \theta \sum^{\infty}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j \theta^{-(j-m)}}{e^\sigma j!(j-m)!}\sum^{\infty}_{q=0}\frac{1}{q!}\Big(\frac{-A\theta^2}{p_o}\Big)^q$$

$$Y = \theta \sum^{\infty}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j \theta^{-(j-m)}}{e^\sigma j!(j-m)!}e^{\frac{-A\theta^2}{p_o}}$$

$$e^{\frac{-A\theta^2}{p_o}} = \frac{Y}{\theta \sum^{\infty}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j \theta^{-(j-m)}}{e^\sigma j!(j-m)!}} $$

Taking log both sides,

$$\frac{-A\theta^2}{p_o} = \log\frac{Y}{\theta \sum^{\infty}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j \theta^{-(j-m)}}{e^\sigma j!(j-m)!}} $$

$$\frac{-A\theta^2}{\log\frac{Y}{\theta \sum^{\infty}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j \theta^{-(j-m)}}{e^\sigma j!(j-m)!}} } = p_o$$

$$p_o = \frac{-A\theta^2}{\log Y - \log {\theta \sum^{\infty}_{j=0}\sum^{j}_{m=0}\frac{\sigma^j \theta^{-(j-m)}}{e^\sigma j!(j-m)!}} }$$

Answer 2

This is just a start. As Ross Millikan observed, you don't need the limits. We have $$ Y = \sum^{100}_{j=0}\sum^{j}_{m=0}\sum^{100}_{q=0} \frac{\sigma^j}{e^\sigma j!(j-m)!}\frac{(-1)^q}{q!}A^q (\theta)^{2q-(j-m)+1}\frac{1}{p_o^q}=\\ e^{-\sigma}\sum^{100}_{j=0}\sum^{j}_{m=0}{\frac{\sigma^j\theta^{m-j+1}}{j!(j-m)!}}\sum^{100}_{q=0}{\frac{(-1)^q}{q!}\left(\frac{A\theta^2}{p_0}\right)^q} $$ The inner sum $\approx e^{-A\theta^2/p_0}$ with an error less than $\frac{1}{101!}\left(\frac{A\theta^2}{p_0}\right)^{101}.$

At least, this reduces the problem from a triple sum to a double sum.