Turning a product into a sum

Question

Is it possible to change $$\prod_{i=1}^n(1+2a_ib_i),$$ where all elements are contained in an unital associative algebra generated by $a_i,b_i$, $i=1,...,n$, such that $a_ib_i=-b_ia_i$, into a sum which contains the summand $$\sum_{i=1}^n 2a_ib_i?$$

There is no special background for this question it is just about what I am asking.

Thank you very much.

Answer 1

Sure it is.

Have a look at the case $n = 2$:

$(1 + 2a_1b_1)(1 + 2a_2 b_2) = 1 + 2a_1b_1 + 2a_2 b_2 + 4a_1b_1a_2b_2; \tag 1$

have a look at the case $n = 3$:

$(1 + 2a_1b_1)(1 + 2a_2 b_2)(1 + 2a_3 b_3) = (1 + 2a_1b_1 + 2a_2 b_2 + 4a_1b_1a_2b_2)(1 + 2a_3 b_3)$ $= 1 + 2a_1b_1 + 2a_2 b_2 + 4a_1b_1a_2b_2$ $+ 2a_3b_3 + 4a_1b_1a_3b_3 + 4a_2b_2a_3b_3 + 8a_1b_1a_2b_2a_3b_3$ $= 1 + 2a_1b_1 + 2a_2 b_2 + 2a_3b_3 + 4a_1b_1a_2b_2$ $+ 4a_1b_1a_3b_3 + 4a_2b_2a_3b_3 + 8a_1b_1a_2b_2a_3b_3; \tag 2$

we can proceed by induction; suppose we simplify notation by settig

$c_i = a_ib_i; \tag 3$

then we already have

$(1 + 2c_1)(1 + 2c_2) = 1 + 2c_1 + 2c_2 + 4c_1c_2, \tag 4$

and

$(1 + 2c_1)(1 + 2c_2)(1 + 2c_3) = 1 + 2c_1 + 2c_2 + 2c_3$ $+ 4c_1c_2 + 4c_1c_3 + 4c_2c_3 + 8c_1c_2c_3, \tag 5$

so we have established the $n = 1$ and $n = 2$ cases of

$\displaystyle \prod_1^n (1 + 2c_i) = 1 + \sum_1^n 2c_i + H_n, \tag 6$

where $H_n$ is just shorthand for the higher order terms in $\prod_1^n(1 + 2c_i)$, i.e., the ones of degree at least two in the $c_i = a_i b_i$; now if

$\displaystyle \prod_1^k (1 + 2c_i) = 1 + \sum_1^k 2c_i + H_k, \tag 7$

then

$\displaystyle \prod_1^{k + 1} (1 + 2c_i) = \left ( \prod_1^k (1 + 2c_i) \right )(1 + 2c_{k + 1}) = (1 + \displaystyle \sum_1^k 2c_i + H_k)(1 + 2c_{k + 1})$ $= 1 + \displaystyle \sum_1^k 2c_i + H_k + 2c_{k + 1} + \left ( \sum_1^k 2c_i \right )2c_{k + 1} + 2H_k c_{k + 1}$ $= 1 + \displaystyle \sum_1^{k + 1} 2c_i + H_k + 2\sum_1^k c_i c_{k + 1} + 2H_kc_{k + 1} = 1 + \sum_1^{k + 1} 2c_i + H_{k + 1}, \tag 8$

which completes the induction and shows that (6) holds for all $n$.

Note that the property $a_ib_i = -b_ia_i$ was not needed here.

Answer 2

When dealing with problems like this it is useful to consider small examples. This product depends on $n$, so let's look at $n=2,3,\ldots$:

For $n=2$, we have $$ (1+2a_1b_1)(1+2a_2b_2)=1+(2a_1b_1+2a_2b_2)+4a_1b_1a_2b_2 $$ by "foiling". For $n=3$, \begin{align} (1+2a_1b_1)(1+2a_2b_2)(1+2a_3b_3)=&(1+(2a_1b_1+2a_2b_2)+4a_1b_1a_2b_2)(1+2a_3b_3)\\ =&[1+(2a_1b_1+2a_2b_2)+4a_1b_1a_2b_2]\\&+[2a_3b_3+(2a_1b_1+2a_2b_2)(2a_3b_3)+(4a_1b_1a_2b_2)(2a_3b_3)]\\ =&1+(2a_1b_1+2a_2b_2+2a_3b_3)\\&+4(a_1b_1a_2b_2+a_1b_1a_3b_3+a_2b_2a_3b_3)+8a_1b_1a_2b_2a_3b_3. \end{align} Perhaps you can finish from here.