0
$\begingroup$

I know the following theorems.

Theorem 1 $:$ For $|x|<1$ we have $$\prod\limits_{k=1}^{\infty} \frac {1} {1-x^k} = 1 + \sum\limits_{k=1}^{\infty} p(k)x^k.$$

Theorem 2 $:$ For $|x|<1$ we have $$\prod\limits_{m=1}^{\infty} (1-x^m) = 1 + \sum\limits_{m=1}^{\infty} (-1)^m\left \{x^{\omega (m)} + x^{\omega(-m)} \right \}.$$

where $\omega (n) = \frac {3n^2-n} {2},\ n \in \Bbb Z$ are called pentagonal numbers.

By the above two theorems we get $$\left (1+\sum\limits_{m=1}^{\infty} (-1)^m\left \{x^{\omega (m)} + x^{\omega(-m)} \right \} \right ) \left ( 1+\sum\limits_{k=1}^{\infty} p(k)x^k \right ) = \left ( \prod\limits_{m=1}^{\infty} (1-x^m) \right ) \left ( \prod\limits_{k=1}^{\infty} \frac {1} {1-x^k} \right ) = 1.$$ This shows that

$$\sum\limits_{k=1}^{\infty} p(k)x^k + \sum\limits_{m=1}^{\infty} (-1)^m\left \{x^{\omega (m)} + x^{\omega(-m)} \right \} + \left ( \sum\limits_{m=1}^{\infty} (-1)^m\left \{x^{\omega (m)} + x^{\omega(-m)} \right \} \right ) \left ( \sum\limits_{k=1}^{\infty} p(k)x^k \right ) = 0.$$

Now if $n \geq 1$ be such that $n$ is not a pentagonal number then there will be no term involving $x^n$ in the infinite sum $\sum\limits_{m=1}^{\infty}(-1)^m \left \{x^{\omega (m)} + x^{\omega(-m)} \right \}.$ Hence by equating the coefficient of $x^n$ to $0$ we get $$p(n) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) + \cdots = 0.$$ But if $n$ is a pentagonal number then there is a term involving $x^n$ in the infinite sum $\sum\limits_{m=1}^{\infty} (-1)^m\left \{x^{\omega (m)} + x^{\omega(-m)} \right \}$ with coefficient $+1$ or $-1.$ Hence in that case we get $$p(n) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) + \cdots = 1\ \text {or}\ -1.$$ Am I right? But in my book the recurrence relations are given same for both the pentagonal numbers and non pentagonal numbers? Where am I doing mistake? Can anybody please point it out?

Thank you very much for your valuable time.

$\endgroup$
  • 1
    $\begingroup$ Your $\sum_k p\left(k\right) x^k$ sum is missing its $k = 0$ addend, which would contribute a $p\left(0\right) = 1$ precisely when $n$ is a pentagonal number (thus making the two cases uniform). $\endgroup$ – darij grinberg May 7 at 16:43
  • $\begingroup$ @darji grinberg I don't get your point. Can you please be more explicit? $\endgroup$ – math maniac. May 7 at 16:45
  • $\begingroup$ You've forgotten a factor of $(-1)^m$ in Theorem 2. $\endgroup$ – Lord Shark the Unknown May 7 at 16:46
  • $\begingroup$ The sum $p(n) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) + \cdots$ will contain a $p(0) = 1$ addend exactly when $n$ is a pentagonal number. But this addend does not arise when you compare coefficients in your equality. (I suggest using \label and \tag, by the way; then I could refer to your equalities more easily.) $\endgroup$ – darij grinberg May 7 at 16:46
  • $\begingroup$ Yeah now I have understood what you are trying to say. Good point @darji grinberg. Thank you very much for your kind help. $\endgroup$ – math maniac. May 7 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.