2
$\begingroup$

$|q|\lt1$

$A(q)=\sum \limits_{k=0}^\infty q^{2^k}$

Easily We can see that $$A(q)=q+A(q^2)\tag 1$$

Let's assume we redefine $A(q)$ as below

$A(q)=-\sum \limits_{k=1}^\infty c_k \ln{(1-q^k)}$

I would like to find $c_k$

Please see my attempt to find it.

$A(q^2)=-\sum \limits_{k=1}^\infty c_k \ln{(1-q^{2k})}=-\sum \limits_{k=1}^\infty c_k \ln{(1-q^{k})}-\sum \limits_{k=1}^\infty c_k \ln{(1+q^{k})}$

$A(q^2)=A(q)-\sum \limits_{k=1}^\infty c_k \ln{(1+q^{k})}$

If we use equation 1 then

$\sum \limits_{k=1}^\infty c_k \ln{(1+q^{k})}=q \tag2$

If we get derivative both sides

$\sum \limits_{k=1}^\infty k.c_k\cfrac{q^k}{1+q^{k}}=q \tag3$

$\sum \limits_{k=1}^\infty k.c_k q^k(1-q^{k}+q^{2k}-q^{3k}+.....)=q $

$\sum \limits_{k=1}^\infty k.c_k q^k-\sum \limits_{k=1}^\infty k.c_k q^{2k}+\sum \limits_{k=1}^\infty k.c_k q^{3k}-...=q $

If we find few terms of $c_k$

$c_1=1$

$2c_2-c_1=0$ ----> $c_2=\frac{1}{2}$

$3c_3+c_1=0$ ----> $c_3=-\frac{1}{3}$

$4c_4-2c_2-c_1=0$ ----> $c_4=\frac{1}{2}$

$5c_5+c_1=0$ ----> $c_5=-\frac{1}{5}$

$6c_6-3c_3+2c_2-c_1=0$ ----> $c_6=-\frac{1}{6}$

$7c_7+c_1=0$ ----> $c_7=-\frac{1}{7}$

$8c_8-4c_4-2c_2-c1=0$ ----> $c_8=\frac{1}{2}$

$9c_9+3c_3+c1=0$ ----> $c_9=0$

$10c_{10}-5c_5+2c_2-c_1=0$ ----> $c_{10}=-\frac{1}{10}$

$11c_{11}+c_1=0$ ----> $c_{11}=-\frac{1}{11}$

$12c_{12}-6c_6+4c_4-3c_3-2c_2-c_1=0$ ----> $c_{12}=-\frac{1}{6}$

I got an interesting result via using $c_k$

$\int_{0}^{1} \frac{A(q)}{q} dq=2 \tag3$

$\int_{0}^{1} \frac{-\sum \limits_{k=1}^\infty c_k \ln{(1-q^{k})}}{q} dq= -\sum \limits_{k=1}^\infty c_k \int_{0}^{1} \frac{ \ln{(1-q^{k})}}{q} dq\tag4$

$$1-q^{k}=e^{-u}$$

$ \sum \limits_{k=1}^\infty \frac{c_k}{k} \int_{0}^{\infty} \frac{ u e^{-u}}{1-e^{-u}} du=2\tag5$

$ \sum \limits_{k=1}^\infty \frac{c_k}{k} \int_{0}^{\infty} u (e^{-u}+e^{-2u}+e^{-3u}+e^{-4u}+...) du=2\tag6$

$ (1+\frac{1}{2^2}+\frac{1}{3^2}+...)\sum \limits_{k=1}^\infty \cfrac{c_k}{k} =2\tag7$

$ \cfrac{\pi^2}{6}\sum \limits_{k=1}^\infty \cfrac{c_k}{k} =2\tag8$

$\sum \limits_{k=1}^\infty \cfrac{c_k}{k} =\cfrac{12}{\pi^2}\tag9$

$1+\frac{1}{4}-\frac{1}{9}+\frac{1}{8}-\frac{1}{25}-\frac{1}{36}-\frac{1}{49}+....=\frac{12}{\pi^2}\tag{10}$

I summarized my results and statements. I am trying to prove 3th,4th,5th statements

  1. if $k>2$ and it is prime number then $c_{k}=-\frac{1}{k}$
  2. if $k$ is $2^{m_0}p^{m_1}$ where $p>2$ and prime number ;$m_1>1$ and $m_0$ is non-negative integer then $c_{k}=0$ (example is $k=9,18,25,27,36,49,50,63,98,99,100$, need prove it)

  3. (for now without proof) It seems $c_{k}=\frac{1}{2}$ if $k=2^m$ where m is positive integer. I am very near to proof the statement. I will edit if I prove it.

  4. (for now without proof) my statement is if $ k=2^{m_0}.p_1^{m_1}.p_2^{m_2}.p_3^{m_3}...p_n^{m_n}$ where $p_1 , p_2 ,..,p_n$ are primes bigger than 2 and $m_0,m_1,m_2,m_3...m_n>0$ then $c_{k}=(-1)^n\frac{1}{2.p_1.p_2.p_3...p_n}$ (example is $k=6,10,12,24$, I need to prove it)
  5. (for now without proof) my statement is if $ k=p_1^{m_1}.p_2^{m_2}.p_3^{m_3}...p_n^{m_n}$ where $p_1 , p_2 ,..,p_n$ are primes bigger than 2 and $m1,m2,m3...m_n>0$ and integer then $c_{k}=(-1)^n\frac{1}{p_1.p_2.p_3...p_n}$ (example is $k=15,21,105$. I need to prove it)

Can my 2nd ,3rd, 4th and 5th statements be true? EDIT: Thanks to @Barry Cipra for the link and table for first 100 terms. I extended statement 2

Please help to prove or disprove them.

If I find the general rule of $c_k$, $e^{A(q)}$ can be expressed as product terms of $(1-q^k)^{c(k)}$.

Thanks for advises and helps.

$\endgroup$
3
  • 1
    $\begingroup$ I don't know if this will be of any help, but writing your $c_k$s in the form $a(k)/k$ and plugging the sequence 1,-1,-1,2,-1,-1,-1,4,0,-1,-1,-2 into the OEIS leads to oeis.org/A067856 which seems to satisfy your statement 3. $\endgroup$ May 2, 2013 at 21:32
  • $\begingroup$ @Barry Cipra Thanks a lot for the link. It is very helpful. $\endgroup$
    – Mathlover
    May 6, 2013 at 12:39
  • $\begingroup$ @Mathlover, glad to help. But it looks to me that whynot has answered your questions. $\endgroup$ May 6, 2013 at 14:17

1 Answer 1

5
$\begingroup$

In your equation $A(q)=−\sum_{k\ge 1} c_k \ln(1−q^k)$, expand the logarithm. You get $$ A(q)= \sum_{m=1} \frac{q^m}{m}\sum_{k\vert m} kc_k. $$ By Mobius inversion, you get $$ c_k = \frac1k \sum_{d\vert k} \mu(k/d) \delta(d) $$ where $\delta(m)=m$ if $m$ is a power of 2, and 0 otherwise. The expression for $c_k$ can be written as $$ c_k =\frac1k\sum_{u=0}^{v_2(k)}\mu(k/2^u) 2^u, $$ where $v_2(k)$ is the $2$-adic valuation of $k$.

It is easy to compute this sum (use the multiplicativity of $\mu$) and if I am not mistaken, $c_k= \frac{\mu(k)}k$ if $k$ is odd and $c_k=\frac{\mu(\ell)}{2\ell}$ if $k=2^{v_2(k)}\ell$ is even. Or something like that.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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