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This is something of an extension of this question which is currently unanswered. I've done some rearranging of my initial equation which was originally:

$$ \frac{M_t}{M_\infty} = 1 - \frac{32}{\pi} * \sum_{n=1}^{25}\frac{exp(-q^2_nDt/R^2)}{q_n^2} * \sum_{p=0}^{100} \frac{exp(-(2p+1)^2)\pi^2Dt/H^2)}{(2p+1)^2} $$

and simplified it to perhaps make it easier to answer. My equation is now:

$$ A = \sum_{n=1}^{25}\sum_{p=0}^{100}\frac{(E_nG_p)^D}{F_nI_p} $$

I am trying to solve for D; all other values are known. I'm just a bit stumped on how to isolate D.

Update: Here are some examples of data used in the original equation.

$\frac{M_t}{M_\infty}$ = 0.663228201

$t$ = 1123000

$H$ = 0.003235088

$R$ = 0.001268375

$D$ = 1.13816081338369290421042023809E-13

$q_1$ = 2.404825557695772768621631879326454643124244909145967135706

$q_2$ = 5.520078110286310649596604112813027425221865478782909853757

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    Are any of the values for any of the variables negative? – 2017-02-26
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    @AlgorithmsX No, they aren't. I've updated the question with some data examples. – 2017-02-26
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    Is $E_nG_p$ guaranteed to be less than one or greater than one? If it's less than one, this function decreases continuously as the value of $D$ increases and increases continuously as $D$ increases otherwise. – 2017-02-26
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    In either of the two cases I gave, a binary search would work really well. If not, you might get away with using Newton's Method. – 2017-02-26
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    I know for certain that you can get a numerical answer, but I don’t think it's possible to get an exact answer. Also, it really doesn’t matter whether they're all greater than one or less than one, as having some of them be less than one and some be greater than one will only really get you another solution. – 2017-02-26
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    @AlgorithmsX Yes, I do not expect to get an exact answer. But thank you, I will try using the binary search or Newton's method and see if they work out. Will update with my results. – 2017-02-26
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    I'll post an answer at some point. – 2017-02-26
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    @AlgorithmsX Thank you, that would be much appreciated as well. – 2017-02-26

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