$\sum_\eta \exp(b*(\eta_1 + ...+\eta_n)) = (1+e^b)^n$.

Question

I'm wondering why this is a correct relation, $\eta$ is the collection of $\{\eta_1,...,\eta_n \}$; ${\eta_1,...,\eta_n }$ are independent and the given relation is : $\sum_\eta \exp(b*(\eta_1 + ...+\eta_n))= \sum_{\eta_1} \sum_{\eta_2} ...\sum_{\eta_n} \exp(b*(\eta_1 + ...+\eta_n)) = (1+e^b)^n$. Every $\eta_i = \{1,0\}$

Thanks in advance

In your sum over $\eta$ are there any dependencies between say $\eta_1$ and $\eta_2$? What is the range over $\eta$ that you are summing? What have you tried? — Kitter Catter, Jun 18 '19 at 18:24
Should I interpret $\sum_{\eta} = \sum_{\eta_1} \sum_{\eta_2}\cdots \sum_{\eta_n}$ ? — Kitter Catter, Jun 18 '19 at 18:26
Have you tried $(\sum_{\eta_i} e^{b \eta_i})$ for each $\eta_i$ which gives $(\sum_{\eta_i} e^{b \eta_i})^n$? — Kitter Catter, Jun 18 '19 at 18:29
I think I'm still missing something here since that sum doesn't work out. Are you sure there aren't any relationships between the different $\eta_i$? (eg you can only have 3 1 and the rest are -1 or half are +1 and half are +1) — Kitter Catter, Jun 18 '19 at 18:35

score 0 · Answer 1 · answered Jun 18 '19 at 20:25

0

At present what you have is unlikely to be true. To test this out you can look at the case of $n=1$ then your sum is $$\sum_{\eta_1} e^{b \eta_1} = e^b+e^{-b} \neq e^{b}+1$$

Possibly your $\eta$ need to be from $\{0,1\}$. Or there is some other relationship/constraint on the $\eta$

answered Jun 18 '19 at 20:25

Kitter Catter

1,938

That's possible since our lecturer, wasn't the most precise person... – Mari3 Jun 19 '19 at 05:01

$\sum_\eta \exp(b*(\eta_1 + ...+\eta_n)) = (1+e^b)^n$.

1 Answers1