Algenometry

Realize that the maximum answer will be when the a's go towards 1. The second running product will go to zero very rapidly and can be ignored. The first running product will go towards 2^2015 in the limit as a's go toward 1. So then Amax ~ 2^2015, then ln(Amax) = 2015*ln(2) = 1396.69

u can use trigo but i will use logic...expand both.

it is a poly like (x-a)(x-b).....where x is 1

all of the terms with odd powers of 1 will cancel.

now assume all a's are equal(u will soon know Y) thus u have only a^(even numbers)...i hope u inderstand this..

now u can write this as

(1+a)^2015 + (1-a)^2015

since a^2 is increasing contiously as we go from 0 to one..

to have minimum a=0 ..u get the answer 2015 ln2

Substitute $x_i=a_{i}^{2}$ . The bounds stay the same. Differentiate wrt $x_1$ . You get $\prod_{i=2}^{2015} (1+x_{i}) - \prod_{i=2}^{2015} (1-x_{i})$

$x_1\geq0$ , so the product on the right is greater than the product on the left, and the derivative is $\geq0$ . Thus the maximum is obtained when $x_1=1$ . The same can be said for all $x_i$ . Thus the maximum is $2^{2015}$