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I have a following expression $$ f=\Pi_{i=1}^n \left[ 1 + \frac{1}{t} +\frac{ (m+i)}{t^2} +\frac{ (m+i)^2}{t^3} + \cdots \right] $$ here $m,n,t$ are positive integers.

I want to obtain a series expansion for $f$ as $1/t,1/t^2,1/t^3,\cdots$ up to maybe $1/t^6$. A general expression for any order is not needed.

I used

f = \!\(
\*UnderoverscriptBox[\(\[Product]\), \(i = 1\), \(n\)]\((1 + 
\*FractionBox[\(1\), \(t\)] + 
\*FractionBox[\((m + i)\), \(t^2\)] + 
\*FractionBox[\(\((m + i)\)^2\), \(t^3\)])\)\)

Series[f, {t, Infinity, 3}, Assumptions -> Element[m, Integers]]

Mathematica 7.0 gives me a complicated expression including Arg, Floor, Csc, etc.

Is there any way to obtain simple expression? I can do low order by hand, but for high order, it is increasingly complicated..

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  • $\begingroup$ Welcome to Mathematica.SE! Could you please add the actual code you evaluated before the Series function? % just refers to the prior evaluation, and we don't know exactly what you've done. $\endgroup$
    – Verbeia
    Oct 16, 2014 at 4:11
  • $\begingroup$ Thanks. It is just the expression for f. f = \!( *UnderoverscriptBox[([Product]), (i = 1), (n)]((1 + *FractionBox[(1), (t)] + *FractionBox[((m + i)), (t^2)] + *FractionBox[(((m + i))^2), (t^3)]))) $\endgroup$
    – user26143
    Oct 16, 2014 at 4:12
  • $\begingroup$ I have replaced % as f. In the input, only the expression up to $t^{-3}$ inside multiplication was used. $\endgroup$
    – user26143
    Oct 16, 2014 at 4:14
  • $\begingroup$ I have added f, copied from mathematica. I do not know what is the best way to present it on stackexchange though... $\endgroup$
    – user26143
    Oct 16, 2014 at 4:18

2 Answers 2

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EDIT: Added verification of Product

f can be expressed in closed form:

f[m_, n_, t_] = Product[1 +
   Sum[(m + i)^k/t^(k + 1), {k, 0, Infinity}],
  {i, 1, n}]

(m - t)/(m + n - t)

Verifying the Product (proof by induction)

s[i_] = 1 + Sum[(m + i)^k/t^(k + 1), {k, 0, Infinity}]

1 - 1/(i + m - t)

f[m, 1, t] == s[1] &&
  f[m, n + 1, t] == f[m, n, t]*s[n + 1] //
 Simplify

True

Your desired expansion is

Series[f[m, n, t], {t, Infinity, 4}] // Simplify

enter image description here

The coefficients are

SeriesCoefficient[f[m, n, t], {t, Infinity, k}]

enter image description here

Verifying the expansion

f[m, n, t] == 1 + Sum[n (m + n)^(k - 1)/t^k, {k, 1, Infinity}] // Simplify

True

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  • $\begingroup$ Thanks for your answer. Excuse me, I do not know how the closed form expression was derived. When I sum the infinity series, I got $$1 + \frac{1}{t-m-i}$$. How to show $$\Pi_{i=1}^n 1 + \frac{1}{t-m-i} = \frac{ m-t}{m+n-t} $$? $\endgroup$
    – user26143
    Oct 16, 2014 at 5:30
  • $\begingroup$ Evaluating Product[1 + 1/(t - m - i), {i, 1, n}] in Mma gives (m-t)/(m+n-t) $\endgroup$
    – Bob Hanlon
    Oct 16, 2014 at 5:41
  • $\begingroup$ I mean the details of derivation, not how mathematica gives the answer... $\endgroup$
    – user26143
    Oct 16, 2014 at 5:50
  • $\begingroup$ Got it, just write the formula explicitly, $\Pi_{i=1}^n 1 + \frac{1}{t-m-i} = \frac{ t-m}{t-m-1} \frac{t-m-1}{t-m-2} \frac{t-m-2}{t-m-3} \cdots \frac{t-m-(n-1)}{t-m-n}$ and cancel the same terms on denominator and numeriator $\endgroup$
    – user26143
    Oct 16, 2014 at 6:34
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Assuming your product is one that Mathematica doesn't know the closed form for, you could use my answer from Infinite product with O[x] here:

Unprotect[Series];

Series[Product[a_, b_], pt_] := Exp[MapAt[Sum[#, b]&, Series[Log[a], pt], 3]]
Series[Inactive[Product][a_, b_], pt_] := Series[Unevaluated[Product[a, b]], pt]

Protect[Series];

Then:

f = Inactive[Product][1 + 1/t + (m + i)/t^2 + (m + i)^2/t^3, {i, 1, n}];

Simplify @ Series[f, {t, Infinity, 3}] //TeXForm

$1+\frac{n}{t}+\frac{n (m+n)}{t^2}+\frac{n (m+n)^2}{t^3}+O\left(\left(\frac{1}{t}\right)^4\right)$

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