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$|q|\lt1$

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

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.

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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. –  Barry Cipra May 2 '13 at 21:32
    
@Barry Cipra Thanks a lot for the link. It is very helpful. –  Mathlover May 6 '13 at 12:39
    
@Mathlover, glad to help. But it looks to me that whynot has answered your questions. –  Barry Cipra May 6 '13 at 14:17
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1 Answer

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.

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