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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!

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If you enclose your $\LaTeX$ in dollar signs it gets rendered. Single dollar signs is inline, double gets display mode. –  Ross Millikan Jun 10 '11 at 15:07

1 Answer 1

up vote 2 down vote accepted

Unless there is unstated dependence on $i$, you can distribute out $H$ and $N_u$. If so you have $$N_c = \frac{N_u}{H}\sum_{i=1}^{max} \frac{1}{i^s}$$ This is a series that can be summed in terms of generalized harmonic numbers, but I am not sure that helps a lot. You can also approximate the sum by an integral $$\sum_{i=1}^{max} \frac{1}{i^s}\approx \int_1^{max} x^{-s} \; dx=\frac{1}{(1-s)}x^{1-s}\bigg| _1^{max}=\frac{1}{(1-s)}(max^{1-s}-1)$$and be pretty close. If you want the exact value you could then search.

Added: for the new version, your $H$ is just the same as the sum above, but then you need to sum again. I would presume the sums are over two different variables, so I will change the $H$ sum to go over $j$ and let $m$=max: $$H=\sum_{i=j}^{m} \frac{1}{j^s}\approx \int_1^{m} x^{-s} \; dx=\frac{x^{1-s}}{1-s}\bigg| _1^{m}=\frac{1}{1-s}\left(m^{1-s}-1\right)$$

Now your new $$N_c=\frac{N_u}{H}\sum_{i=1}^m\frac{i}{i^s}=\frac{N_u}{H}\sum_{i=1}^m\frac{1}{i^{s-1}}$$ and we can use the same approximation, as the sum is the same except for being $s-1$: $$N_c\approx N_u\frac{1-s}{2-s}\frac{m^{2-s}-1}{m^{1-s}-1}$$

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@Ross Millikan, thank you for the reply. Even if i approximate the sum by an integral, i should know the "max" value anyway, is n't it? –  Maurizio Jun 12 '11 at 12:37
    
If you integrate from $1$ the integral is less than the sum. For example, if max=2, the integral includes all the area from 1 to 2, but the sum is the higher value at 1 (times the width=1). But if you add 1 (the value at $i=1$), you have an upper bound as now the integral gets the average over each interval and the sum gets the least. But it could take a lot of terms when max is large to make up the 1, so you could have a range of uncertainty. –  Ross Millikan Jun 12 '11 at 13:15
    
I add a new version of the formula..the first was a bit wrong! –  Maurizio Jun 14 '11 at 10:59
    
@Ross Millikan, I add a revision of your formula about "H"! I have to verify also the second part regarding "Nc"! –  Maurizio Jun 14 '11 at 14:52
    
@Maurizio: I had gotten the $x$ in the denominator instead of the numerator and carried it through. I have corrected it. In your check of $H$ it now comes out 44.72, much closer –  Ross Millikan Jun 14 '11 at 15:39

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