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Title: partial fraction decomposition of

$$ \frac{x^n}{(x-1)(x-3)(x-5)\cdots(x-(2n-1))} $$

Body

I am working on decomposing the following rational function into partial fractions

$$ \frac{x^n}{(x-1)(x-3)(x-5)\cdots(x-(2n-1))}.$$

The denominator is a product of $ n $ odd linear factors $ (x-1), (x-3), \dots, (x-(2n-1)) $, and the numerator is $ x^n $. I want to express this as a sum of partial fractions in the form

$$ \frac{x^n}{(x-1)(x-3)(x-5)\cdots(x-(2n-1))} = \sum_{k \in \{1, 3, 5, \ldots, 2n-1\}} \frac{A_k}{x-k},$$

where $ A_k $ are constants to be determined. Using the standard method, I know that

$$ A_k = \frac{k^n}{\prod\limits_{\substack{j \neq k \\ j \in \{1, 3, 5, \ldots, 2n-1\}}} (k-j)}.$$

Given the structure of the denominator with odd factors, I am curious:

  1. Can this formula for $A_k$ be simplified further by leveraging the fact that all terms in the denominator are odd?

  2. Are there patterns or a recurrence relation that can be derived for $A_k$?

  3. Is there an alternative method for tackling this decomposition efficiently, especially for large $n$?

Any help in deriving a more explicit or simplified relationship for $A_k $ would be greatly appreciated! The following part of the problem is as follows.

Deduce the relation: $$\sum_{k=0}^n (-1)^k (2k + 1)^n \binom{n}{k} = (-1)^n 2^n n!, \quad \forall n \in \mathbb{N}$$

Rodrigo de Azevedo
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David Manga
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    Welcome to MSE. In this forum you're expected to show some effort in order to get help. Edit your question to include your work on the problem, otherwise your question will likely be downvoted and eventually closed. – jjagmath Jan 04 '25 at 19:22
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    Have you tried some small values of $n$? – jjagmath Jan 04 '25 at 19:23
  • The same way you always do for partial fractions. Are you saying you don't know how to do partial fractions at all? There are probably lots of earlier questions on partial fractions where you'll see how it's done. For example, see https://math.stackexchange.com/questions/20963/integration-by-partial-fractions-how-and-why-does-it-work – Gerry Myerson Jan 04 '25 at 19:23
  • What have you tried? What is your experience with partial fractions? – Maxime Jaccon Jan 04 '25 at 19:27
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    The order of $x^n$ has to be less than the order of the denominator to get the partial fraction form you want. – Paul Jan 04 '25 at 19:36

3 Answers3

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This is a hint in the right direction. Here are the partial fractions for some small values of n. Let $$P_n = \frac{x^n}{(x-1)(x-3)\cdots(x-(2n-1))}$$

$ P_1 = \frac{1}{x-1}+1 \\ P_2 = -\frac{1}{2 (x-1)}+\frac{9}{2 (x-3)}+1 \\ P_3 = \frac{1}{8 (x-1)}-\frac{27}{4 (x-3)}+\frac{125}{8 (x-5)}+1 \\ P_4 = -\frac{1}{48 (x-1)} +\frac{81}{16 (x-3)}-\frac{625}{16 (x-5)}+\frac{2401}{48 (x-7)}+1 \\ P_5 = \frac{1}{384 (x-1)} -\frac{81}{32 (x-3)} +\frac{3125}{64 (x-5)}-\frac{16807}{96 (x-7)}+\frac{19683}{128 (x-9)}+1 $

What patterns do you notice? I certainly notice alternating terms, perfect cubes in the numerator, powers of odd numbers, and some symmetry in the denominators of the terms, and a $1$ as the final term. Try to write a closed form that captures all of this.

The next step, of course, is proving it :)

Hope this helps.

$$P_n = \frac{(-2)^{-n} x^n}{\left(\frac{1}{2}-\frac{x}{2}\right)_n}$$ where $(a)_n$ is the Pochhammer symbol.

Maxime Jaccon
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  • Thanks. Let me rephrase my question. Perhaps if I provide the other part of the problem, my question will be clearer. – David Manga Jan 04 '25 at 19:40
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Hint:

$$\frac{x^5}{(x-1) (x-3) (x-5) (x-7) (x-9)}=1+\frac{1}{{2^4 4!}}\left( (-1)^0\binom{4}{0}\frac{1^5}{x-1} + (-1)^1\binom{4}{1}\frac{ 3^5}{x-3} + (-1)^2\binom{4}{2} \frac{5^5}{x-5} + (-1)^3\binom{4}{3}\frac{7^5}{x-7} + (-1)^4\binom{4}{4} \frac{9^5}{x-9} \right)$$

jjagmath
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  • I see, thank you. I will now proceed using recurrence. – David Manga Jan 04 '25 at 20:01
  • @DavidManga Good, just check my calculations, I'm not completely sure about some details. – jjagmath Jan 04 '25 at 20:05
  • "Yeah, it's definitely that. Just one last question: if we sum all the coefficients, shouldn't we get one? Then, the last question is straightforward, since when we reduce to the same denominator, the coefficient of $x^n$ turns out to be one." – David Manga Jan 04 '25 at 20:18
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It suffices to apply the Lagrange interpolating polynomial to the function $f(x) =x^n$ at the nodes $\{1,3,...,(2n-1) \}$ and these $A_j$ you are seeking correspond to "these $y_j$ multiplying by the constant terms of the basis polynomials $l_j(x)$".

NN2
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