11
$\begingroup$

Given the positive integer numbers $n$,prove that $$\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$$

where $\sigma(n)$ is defined as $$\sigma{(N)}=\sum_{d|N}d$$

This problem I think is very interesting, and have nice methods?

Thank you everyone.

and I find some of this enter image description here

enter image description here

or can see:http://www.advancesindifferenceequations.com/content/pdf/1687-1847-2013-84.pdf

$\endgroup$
  • $\begingroup$ Please answer the queries to make your previous question understandable. Or do you want to close them? $\endgroup$ – Did Jul 17 '13 at 11:14
  • $\begingroup$ where is problem,I have edit.Thank you $\endgroup$ – math110 Jul 17 '13 at 11:20
  • 1
    $\begingroup$ @math110: I have rewritten a part of your post and clarified the definition of $\sigma(n)$. Does it look right now? P.S: Your questions are hard but interesting. $\endgroup$ – Isomorphism Jul 17 '13 at 11:59
  • $\begingroup$ What's the source (for the 843rd time)? What makes you think it's true? What makes you think there are nice methods? What do you know about the problem that you aren't telling us? $\endgroup$ – Gerry Myerson Jul 17 '13 at 12:37
  • $\begingroup$ @Isomorphism,Thank you,I think this is very interesting. $\endgroup$ – math110 Jul 17 '13 at 14:20
28
$\begingroup$

This answer is a long one, sorry.

For complex $q$ with $|q|<1$, introduce Ramanujan's functions $P(q),Q(q)$: $$P(q) = 1 - 24\sum_{n=1}^\infty \sigma(n)\,q^n$$ $$Q(q) = 1 + 240\sum_{n=1}^\infty \sigma_3(n)\,q^n$$

The left-hand side of your first formula can be understood as the coefficient of $q^{2^n}$ in the square of the power series $$f(q) = \sum_{m=1}^\infty \sigma(2m-1)\,q^{2m-1} = \frac{1}{48}\left(P(-q)-P(q)\right)$$ So we are interested in $f(q)^2$ expressed as a power series in $q$.

Proposition: We have $$f(q)^2 = \frac{1}{240}\left(Q(q^2)-Q(q^4)\right)$$

Once this is proven, we can conclude that $$f(q)^2 = \sum_{m=1}^\infty \left(\sigma_3(2m-1)\,q^{4m-2} + \left(\sigma_3(2m) - \sigma_3(m)\right)\,q^{4m}\right)$$ Thus, the coefficient of $q^{2^n}$ turns out to be $\sigma_3(1) = 1$ for $n=1$ and $$\sigma_3(2^{n-1}) - \sigma_3(2^{n-2}) = 2^{3(n-1)} = 8^{n-1}$$ for $n>1$, so it is indeed $8^{n-1}$ for all positive integers $n$.

It remains to prove the proposition.

To that end, let $\mathbb{H}$ the complex upper half-plane, and introduce a new independent variable $\tau\in\mathbb{H}$ that determines $$q = \mathrm{e}^{2\pi\mathrm{i}\tau}$$ so $|q|<1$. Furthermore, for positive integer $k$ let $$q_k = \mathrm{e}^{2\pi\mathrm{i}\tau/k}$$ so $q_k^k = q$.

By means of these dependencies, we can consider $q$-series as functions of $\tau$. This leads us to the normalized Eisenstein series $$\begin{aligned} \mathrm{E}_2(\tau) &= P(q) \\ \mathrm{E}_4(\tau) &= Q(q) \end{aligned}$$ I will also use the Jacobi thetanull functions (as functions of $\tau$): $$\begin{aligned} \Theta_{00}(\tau) &= \sum_{m\in\mathbb{Z}}q_2^{m^2} \\ \Theta_{01}(\tau) &= \sum_{m\in\mathbb{Z}}(-q_2)^{m^2} \\ \Theta_{10}(\tau) &= \sum_{m\in\mathbb{Z}} q_8^{(2m+1)^2} = 2\,q_8\sum_{m=0}^\infty q^{m (m+1)/2} \end{aligned}$$

Background on those functions is given in textbooks such as:

  1. (Theta functions) E. T. Whittaker and G. N. Watson, A Course of Modern Analysis; 2nd edition 1915; Merchant Books, reprint 2008, ISBN 1-60386-121-1; 1st edition 1902.
  2. (Eisenstein series) Tom M. Apostol, Modular Functions and Dirichlet Series in Number Theory; 2nd edition 1990; Springer, ISBN 0-387-97127-0; 1st edition 1976.

Let us prove, as a first lemma, the frequently used triplet of identities $$\begin{aligned} 3\,\Theta_{00}^4(\tau) &= 4\,\mathrm{E}_2(2\tau)-\mathrm{E}_2\left(\frac{\tau}{2}\right) \\ 3\,\Theta_{01}^4(\tau) &= 4\,\mathrm{E}_2(2\tau)-\mathrm{E}_2\left(\frac{\tau+1}{2}\right) \\ 3\,\Theta_{10}^4(\tau) &= \mathrm{E}_2\left(\frac{\tau+1}{2}\right) - \mathrm{E}_2\left(\frac{\tau}{2}\right) \end{aligned}$$ the last of which can be translated to $$f(q_2) = \left(\frac{\Theta_{10}(\tau)}{2}\right)^4$$ which gives yet another representation for our function $f$.

The second lemma shall comprise the three identities $$\begin{aligned} \Theta_{00}^8(\tau) &= \frac{1}{15}\left(16\,\mathrm{E}_4(\tau) - \mathrm{E}_4\left(\frac{\tau+1}{2}\right)\right) \\ \Theta_{01}^8(\tau) &= \frac{1}{15}\left(16\,\mathrm{E}_4(\tau) - \mathrm{E}_4\left(\frac{\tau}{2}\right)\right) \\ \Theta_{10}^8(\tau) &= \frac{16}{15}\left(\mathrm{E}_4(\tau)-\mathrm{E}_4(2\tau)\right) \end{aligned}$$ the last of which can ultimately be translated to $$f(q_2)^2 = \frac{1}{240}\left(Q(q)-Q(q^2)\right)$$ which is the proposition we need.

Why should we prove six identities although we only need two of them? Well, it turns out that the three Jacobi thetanulls are best dealt with altogether, otherwise things become less elementary. Furthermore, this is the third occasion within a fortnight that such identities find applications on this site (look here and there), so I think it is useful to have a package of them in a place we can link to. And last but not least, it's fun.

The plan is to find certain (modular) symmetries of expressions containing the above functions of $\tau$; then to use the fact that, up to some constant factor, only one function can have such symmetries. To prepare the ground, let us recall the most basic facts about the above Eisenstein series and Jacobi thetanull functions that we will need.

First, all those functions are holomorphic for $\tau\in\mathbb{H}$ and can be represented as a Maclaurin series in $q_k$ for some $k$, as demonstrated by the definitions given. (This implies that they approach a finite limit as $\Im\tau\to\infty$.)

Second, the thetanull functions $\Theta_{00}(\tau), \Theta_{01}(\tau), \Theta_{10}(\tau)$ are nonzero for every $\tau\in\mathbb{H}$ which implies that their reciprocals are holomorphic over $\mathbb{H}$ too.

Third, we have the following useful symmetries: Let $\tau' = \frac{-1}{\tau}$, then $$\begin{aligned} \mathrm{E}_2(\tau+1) &= \mathrm{E}_2(\tau) & \mathrm{E}_2(\tau') &= \frac{6\tau}{\pi\,\mathrm{i}} + \tau^2\,\mathrm{E}_2(\tau) \\ \mathrm{E}_4(\tau+1) &= \mathrm{E}_4(\tau) & \mathrm{E}_4(\tau') &= \tau^4\,\mathrm{E}_4(\tau) \\ \Theta_{00}(\tau+1) &= \Theta_{01}(\tau) & \Theta_{00}(\tau') &= \sqrt{-\mathrm{i}\tau}\,\Theta_{00}(\tau) \\ \Theta_{01}(\tau+1) &= \Theta_{00}(\tau) & \Theta_{01}(\tau') &= \sqrt{-\mathrm{i}\tau}\,\Theta_{10}(\tau) \\ \Theta_{10}(\tau+1) &= \sqrt{\mathrm{i}}\,\Theta_{10}(\tau) & \Theta_{10}(\tau') &= \sqrt{-\mathrm{i}\tau}\,\Theta_{01}(\tau) \end{aligned}$$

The fourth useful fact seems to be rarely mentioned in introductory texts, but it is easy to conclude from the stuff covered there. Concretely, the coefficients of the Eisenstein $q$-series, that is, the divisor power sums $\sigma_k(n)$, have a certain multiplicativity property that translates into functional equations for the Eisenstein series. Particularly, we have: $$\begin{aligned} \mathrm{E}_2\left(\frac{\tau}{2}\right) + \mathrm{E}_2\left(\frac{\tau+1}{2}\right) + 4\,\mathrm{E}_2(2\tau) &= 6\,\mathrm{E}_2(\tau) \\ \mathrm{E}_4\left(\frac{\tau}{2}\right) + \mathrm{E}_4\left(\frac{\tau+1}{2}\right) + 16\,\mathrm{E}_4(2\tau) &= 18\,\mathrm{E}_4(\tau) \end{aligned}$$ This allows us to express an Eisenstein series with argument $\frac{\tau+1}{2}$ in terms of the Eisenstein series with arguments $\frac{\tau}{2}$, $\tau$, and $2\tau$. Why would we want to do that? Because the latter can easily be subjected to the transformation $\tau\to\tau'$. Doing so and substituting back, we find the remarkable symmetries $$\begin{aligned} \mathrm{E}_2\left(\frac{\tau'+1}{2}\right) &= \frac{12\tau}{\pi\,\mathrm{i}} + \tau^2\,\mathrm{E}_2\left(\frac{\tau+1}{2}\right) \\ \mathrm{E}_4\left(\frac{\tau'+1}{2}\right) &= \tau^4\,\mathrm{E}_4\left(\frac{\tau+1}{2}\right) \end{aligned}$$

Now that the basics are in place, consider the functions $$\begin{aligned} U_{00}(\tau) &= \frac {4\,\mathrm{E}_2(2\tau)-\mathrm{E}_2\left(\frac{\tau}{2}\right)} {3\,\Theta_{00}^4(\tau)} \\ U_{01}(\tau) &= \frac {4\,\mathrm{E}_2(2\tau)-\mathrm{E}_2\left(\frac{\tau+1}{2}\right)} {3\,\Theta_{01}^4(\tau)} \\ U_{10}(\tau) &= \frac {\mathrm{E}_2\left(\frac{\tau+1}{2}\right) - \mathrm{E}_2\left(\frac{\tau}{2}\right)} {3\,\Theta_{10}^4(\tau)} \end{aligned}$$ Proving the first lemma is equivalent to showing that $U_{00}, U_{01}, U_{10}$ are all equal to the constant $1$.

As the Eisenstein series and the Jacobi thetanulls are holomorphic over $\mathbb{H}$, and the Jacobi thetanulls are furthermore nonzero on $\mathbb{H}$, we find that $U_{00}, U_{01}, U_{10}$ are holomorphic over $\mathbb{H}$ as well.

By considering the given series expansions of their constituents, we find that $U_{00}, U_{01}, U_{10}$ can be represented as Maclaurin series in $q_2$: $$\begin{aligned} U_{00} &= \frac{4\,P(q^2)-P(q_2)}{3\,\Theta_{00}^4} = \frac{3 + \mathrm{O}(q_2)}{3 + \mathrm{O}(q_2)} = 1 + \mathrm{O}(q_2) \\ U_{01} &= \frac{4\,P(q^2)-P(-q_2)}{3\,\Theta_{01}^4} = \frac{3 + \mathrm{O}(q_2)}{3 + \mathrm{O}(q_2)} = 1 + \mathrm{O}(q_2) \\ U_{10} &= \frac{P(-q_2) - P(q_2)}{3\,\Theta_{10}^4} = \frac{48 q_2 + \mathrm{O}(q_2^3)}{48 q_2 + \mathrm{O}(q_2^3)} = 1 + \mathrm{O}(q_2^2) \end{aligned}$$

Using the symmetries presented above, we find after some calculation $$\begin{aligned} U_{00}(\tau+1) &= U_{01}(\tau) & U_{00}(\tau') &= U_{00}(\tau) \\ U_{01}(\tau+1) &= U_{00}(\tau) & U_{01}(\tau') &= U_{10}(\tau) \\ U_{10}(\tau+1) &= U_{10}(\tau) & U_{10}(\tau') &= U_{01}(\tau) \end{aligned}$$

Together this implies that the functions $$\begin{aligned} C_1 &= U_{00} + U_{01} + U_{10} \\ C_2 &= U_{00}\,U_{01} + U_{00}\,U_{10} + U_{01}\,U_{10} \\ C_3 &= U_{00}\,U_{01}\,U_{10} \end{aligned}$$ are holomorphic over $\mathbb{H}$, representable as Maclaurin series in $q_2$, and invariant under the transformations $\tau\to\tau+1$ and $\tau\to\tau'$. The first invariance also implies that the $q_2$-series only involve integer powers of $q = q_2^2$.

This means that $C_1,C_2,C_3$ are so-called entire modular forms of weight zero. However, as shown e.g. in the text by Apostol (reference 2 above), the only entire modular forms of weight zero are constants. From this very fact we can now draw our conclusions.

Hence each $C_i$ must equal the constant term of its $q$-series, so $$C_1 = 3\qquad C_2 = 3\qquad C_3 = 1$$ independent of $\tau$.

Recall that, by Vieta's formulae, the solutions of the equation $$U^3 - C_1\,U^2 + C_2\,U - C_3 = 0$$ are precisely $U\in\{U_{00}, U_{01}, U_{10}\}$. Now, since the equation amounts to $(U-1)^3 = 0$, we conclude that $$U_{00} = U_{01} = U_{10} = 1 = \mathrm{const.}$$ Thus, the first lemma is proven.

In the very same manner, we can easily tackle the second lemma. You know what? I leave that to you, it's refreshing.

Edit: Typo corrections.

$\endgroup$
  • 4
    $\begingroup$ Where will this be published? $\endgroup$ – Gerry Myerson Jul 17 '13 at 23:17
  • 1
    $\begingroup$ Thanks ccorn for the detailed explanation and links to my questions. Although I am bit novice with the modular forms, and I know the properties of thetanulls via Jacobi/Ramanujan approach, I find your approach very simple. $\endgroup$ – Paramanand Singh Jul 18 '13 at 3:36
6
$\begingroup$

It is difficult to add anything to such a nice answer. So take this more as a comment, too long for the the comment field:
There is another elementary proof, using only an arithmetic identity, which generalizes a classical formula of Liouville. For details, see the article "Elementary Evaluation of Certain Convolution Sums Involving Divisor Functions" by Huard et al.On page $21$ there is Theorem $5$, which yields $$ \sum_{k\le N/2} \sigma(2k-1)\sigma(N-2k+1)=\frac{1}{24}\left( 5\sigma_3(N)-21\sigma_3(N/2)+16\sigma_3(N/4)\right)+(1 − 6N)(σ(N) − 3σ(N/2) + 2σ(N/4))). $$ For $N=2^n$ there result follows also (with some manipulations given in theorem $9$). The proof is completely elementary, only using the generalization of Liouville’s Formula, which is Theorem $1$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.