how do solve it?
just integrate three times and you're done
[math] f\left( x \right) = \prod\limits_{n = 0}^\infty {\left( {1 + {x^{{4^n}}}} \right)} [/math]
>>7716751
Find the convergence value of the function when x = 3/8. insert that into the next function then flip
>>7716780
i dont know how to do that stuff. also the problem set said at the top in big letters you do not need technical knowledge to solve the problems and the first couple you sure didnt
btw the answer is 9/64 guys
That infinite product is just the geometric series in disguise, i.e. can shown to be 1/(1-x^2), and so f^-1(8/[5f(3/8)])=f^-1(8/[5(64/55)])=f^-1(11/8)=sqrt(3/11)
>>7716751
just plug in all numbers between 0 < x < 1 and and see which one gives you [math] \frac{8}{5f(\frac{3}{8})} [/math] duh
I looked up the solution online and it's pretty silly :/
>i will eventually have to learn this
Oh the live of an engineering student
>>7716892
>dumb enough to think engineers have to learn competition math
I mean I shouldn't have expected more but...
>>7716881
wheres the solution??
>>7716913
thats incorrect though. the answer is 9/64
>>7716926
says who?
>>7716952
says the answers you fool
http://www.isinj.com/arml/ARML%201995%20-%202003.pdf
>>7716751
Well I got the answer but my solution needs refinement.
So to start we have:
f(x) = prod k from 0->inf of 1+x^(2^(2k))
So f(x^2) = prod k from 0->inf of 1+x^(2^(2k+1))
Notice f(x) covers even powers and f(x^2) odd powers so together:
f(x) f(x^2) = prod k from 0->inf of 1+x^(2^k)
Now define g(x) = f(x) f(x^2).
Evaluating g(x^2) increments the power of 2 so g(x) = (1+x) g(x^2).
Repeatedly expanding this yields:
g(x) = g(x^2) + x g(x^2)
g(x) = g(x^4) + x g(x^4) + x^2 g(x^4) + x^3 g(x^4)
g(x) = g(x^8) + x g(x^8) + x^2 g(x^8) + x^3 g(x^8) + ... + x^7 g(x^8)
And so on.
If we repeat this process the argument to g(x) approaches 0 since 0<x<1.
Since g(0) = 1 and g(x) is increasing this gives g(x) = 1+x+x^2+... = 1/(1-x)
Now back to the original problem. We have:
f(x) f(x^2) = 1/(1-x)
x^2 = f^-1(1/(1-x) 1/f(x))
Setting x to 3/8:
(3/8)^2 = f^-1(1/(1-3/8) 1/f(3/8))
9/64 = f^-1(8/(5 f(3/8))
