# How to find parameters from an equation

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!

-

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

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 $s-1$: $$N_c\approx N_u\frac{1-s}{2-s}\frac{m^{2-s}-1}{m^{1-s}-1}$$

-

@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
show 4 more comments
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9282333254814148, "perplexity_flag": "middle"}