We can be sure, that $$(1-a_{1}x)(1-a_{2}x)(1-a_{3}x)\cdots=1-(a_{1}+a_{2}+a_{3}+\cdots)x+(a_{1}a_{2}+a_{1}a_{3}+\cdots)x^2-(a_{1}a_{2}a_{3}+\cdots)x^3+\cdots$$ Other words, $$\prod\limits_{k=1}^{\infty}(1-a_{k}x)=1-A_{1}x+A_{2}x^2-A_{3}x^3+\cdots$$ Let create a function $$B(n)=\sum\limits_{k=1}^{\infty}a_{k}^{n}$$ Then we can say, that $$B(1)=A_{1}, B(2)=B(1)A_{1}-2A_{2}=A_{1}^{2}-2A_{2}$$ $$B(3)=B(2)A_{1}-B(1)A_{2}+3A_{3}=A_{1}^{3}-3A_{1}A_{2}+3A_{3}$$ $$B(4)=B(3)A_{1}-B(2)A_{2}+B(1)A_{3}-4A_{4}=A_{1}^{4}-4A_{1}^{2}A_{2}+4A_{1}A_{3}+2A_{2}^{2}-4A_{4}$$ In general $$B(n)=(-1)^{n-1}nA_{n}+\sum\limits_{k=1}^{n-1}(-1)^{k-1}B(n-k)A_{k}$$ So if we create a function $$C(n)=(a_{1}a_{2})^n+(a_{1}a_{3})^n+(a_{2}a_{3})^n+\cdots$$ or $$D(n)=(a_{1}a_{2}a_{3})^{n}+\cdots$$ which uses infinite sums of $A_{2}$ and $A_{3}$, how can we find it in general (also not only for this two, but for any $A_{m}$)?

If I made some mistakes, sorry for my English.

  • $\begingroup$ I am not quite sure what you are asking here. Didn't you just describe the coefficients yourself? $\endgroup$ – Fimpellizieri Jan 29 '18 at 19:19
  • $\begingroup$ Can you be sure that each of the $A_i$'s converge? $\endgroup$ – Doughnut Pump Jan 29 '18 at 19:20
  • $\begingroup$ @DoughnutPump Is this even relevant? $\endgroup$ – Fimpellizieri Jan 29 '18 at 19:22
  • $\begingroup$ @Fimpellizieri, thank you for answer! I describe only one case, which use $A_{1}$, and interested in others. $\endgroup$ – user514787 Jan 29 '18 at 19:25
  • 1
    $\begingroup$ Is your question about 'How do I write these expressions in a compact form?' $\endgroup$ – Fimpellizieri Jan 29 '18 at 19:34

If $S_{k,n}$ is the sum of the $n$-th powers of products of $k$ terms $a_i$, then you can write

$$S_{k,n}=\sum_{1\leq i_1<i_2<\ldots<i_k} {\left(a_{i_1}\cdot a_{i_2}\cdot \ldots\cdot a_{i_k}\right)}^n$$



and let $A_{k}(m)$ be the coefficient of $[x^k]\Big(f_m(x)\Big)$. So, using your notation we'd have $A_j=A_j(1)$.

Let $B_n(m)=\sum_{k\geq 1}a_k^{mn}$. So,using your notation, we'd have $B(n)=B_1(n)=B_n(1)$.

Observe that the relations you found between $B$ and $A$ are valid in greater generality with


Moreover, note that in your notation $C(n)=A_2(n)$ and $D(n)=A_3(n)$, and so on.

  • $\begingroup$ Sorry, now I think you don't understand me right. I need form like $B(n)$ after words "In general" for $C(n)$, $D(n)$ and any other cases. $\endgroup$ – user514787 Jan 29 '18 at 19:42
  • $\begingroup$ See the edit. I still find your question much too vague, but eh.] $\endgroup$ – Fimpellizieri Jan 29 '18 at 20:07

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.