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Calculate the following sum?

Hi every one, I failed to do the following expression and make it give accurate results if any one can help me I will be glade. I attached my expression in a pic "want this" and my trial as "my trial". the correct answer must equal 0.119 when a=1, b=10, m=3, n=6. thanks a lot in advance.

a = 1
b = 10
m = 3
n = 6
a^1 b^n (Sum[
    Sum[Sum[Sum[(-1)^(k + v - n + m + 1)
         If[k == 0, 1, 
         SeriesCoefficient[Series[(-Log[1 - x])^k, {x, 0, 30}], 
          p + k]] If[n - k - 2 == 0, 1, 
         SeriesCoefficient[
          Series[(-Log[1 - x])^(n - k - 2), {x, 0, 30}], 
          q + (n - k - 2)]]
        Binomial[n - m - 1, k] Binomial[b - 1, 
         v] (-PolyGamma[0, -1 + 1/a - k + n + q] + 
           PolyGamma[0, 2/a + n + p + q + v])/(a (1 + k + p + v) + 
           1), {q, 0, 30 - (n - k - 2)}], {p, 0, 30 - k}], {v, 0, 
      b - 1}], {k, 0, n - m - 1}])/((m - 1)! (n - m - 1)!)
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POSTED BY: Ramy Aldallal
12 Replies

I found the solution for the problem. the problem was when the value of k was 0 the coefficient will not equal 1 but the whole expression must be found from the start for a value of k that will start from 1 and an expression when the value of k is 0. yet I failed to solve it using MATHEMATICA but by doing the above I succeed to get the correct result. thank you all for your precious time and opinions.

POSTED BY: Ramy Aldallal

And yes I separated the coefficients and found that they give what I expect from them.

POSTED BY: Ramy Aldallal

Yes, after 20 terms the expression gives almost the same answers. This expression is originally an integration expression. So from the integration we get the value 0.119. But the integration is very hard when a > 1. That is why I tried to get a numerical form for soing so.

POSTED BY: Ramy Aldallal

I suggest you separately define those coefficient functions a[p_, k_]:= ... and b[q_, n_, k_] : = ....

Then evaluate them for sample values of their arguments to see if you're getting what you expect.

If so, then make your entire expression much simpler to check by using a[p, k] and b[q, n, k] in them rather than the much longer, harder-to-read expressions you have now.

P.S. Why do you believe the correct answer should be 0.119 (approximately)?

Are you sure that the terms in the power series involved decrease in magnitude quickly enough so that 30 terms suffice?

POSTED BY: Murray Eisenberg

Yes. I get the same result but its not the correct result it should not exceed 1 for any value of a, b, m and n. Its accurate result must equal 0.119

POSTED BY: Ramy Aldallal

By taking one (or both) of the parameters a and b to be approximate reals, e.g., a=1. rather than a=1, or by taking N of the entire expression, I get with Mathematica (12.1) the value 1.38785.

POSTED BY: Murray Eisenberg

Aha. you can be sure Sir that If someone answered my problem correctly I will mention the correct answer here. And I am sorry I did not know community guideline. thanks a lot

POSTED BY: Ramy Aldallal
Posted 4 years ago

Crossposted here.

POSTED BY: Rohit Namjoshi

Is it a problem sir?

POSTED BY: Ramy Aldallal
Posted 4 years ago

Hi Ramy,

The "problem" is that readers that are unaware of the crosspost could spend time on a question that may have already been answered. Please read the community guidelines.

If a question is cross-posted to other forums, this should be clearly indicated. A link to the other posting is also strongly recommended (and can be inserted after the fact by editing one's question). Such links save readers time, help to avoid duplication of effort and are quite useful for disseminating good responses.

POSTED BY: Rohit Namjoshi

Please enter your typed Mathematica code, not a photographic image, in your post. This will allow us to copy it and paste it into a Mathematica notebook and try it to see what's happening.

It is utterly unreasonable to expect folks here to have to retype what you already, evidently, had typed into a notebook.

If you don't know how to do it, click the ? icon for help. But the essence is that you can copy your code from a Mathematica notebook, paste that into the edit box here; then select all the code and click the button in the toolbar at the top that looks lie this: < > . That will automatically indent the entire selected block and have it appear as typed code.

POSTED BY: Murray Eisenberg

Thank you so much. I edit it with the code. please take a look at it.

POSTED BY: Ramy Aldallal
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