the question could be "stupid" but i don't know if it is feasible or not, please don't kill me :)
EDIT WITH NEW FORMULAS!
I have an equation like this: (unfortunatly in my first Q&A i cannot upload images for "spam" reason, I post latex version of the formula hoping it is understandable. Otherwise which kind of representation can i use?)
$N_c = \sum_{i=1}^{max} \bigg( \frac{1}{i^s H} * N_u * i\bigg)$
where, $H = \sum_{i=1}^{max} \frac{1}{i^s}$
If all the variables are known except one:
- "$max$" is UNKNOWN
it is possible to find the "$max$" parameter? Or it is mathematically impossible?
Thank you very much.
ADDED (based on Ross Millikan answer): Practical example based on you approximations to verify the correcteness of "H" formula. If i set:
- $s = 0.5$
- $max = 500$ (in this case i know also the "max" value, i want only to verify the integral)
we have:
$H=\sum_{i=j}^{m} \frac{1}{j^s}\approx \int_1^{m} x^{-s} \; dx=\frac{1}{(1-s)x^{1-s}}\mid _1^{m}=\frac{1}{1-s}\left(1-\frac{1}{m^{1-s}}\right) = $ $ = \frac{1}{1-0.5}\left(1-\frac{1}{500^{1-0.5}}\right) = 2 \times 0.95 = 1.91$
But if I calculate the original summatory we obtain:
$H=\sum_{i=j}^{m} \frac{1}{j^s} = 43.28$
I think there is something strange in the integral's "evolution". I think we can expand it in this way:
$H=\sum_{i=j}^{m} \frac{1}{j^s}\approx \int_1^{m} x^{-s} \; dx=\frac{x^{1-s}}{(1-s)}\mid _1^{m}= \frac{500^{0.5}}{1-0.5} - \frac{1^{0.5}}{1-0.5} = 44.72 - 2 = 42.72$
that is a better approximation. What do you think? I don't know if it is correct, this is a "new world" for me!