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The text is a little small, what is two to the power of in f(2^[this])?

\[\large f(2^i)=\frac{1}{2^i}\]its \(i\)

Thanks.

i would setup a matrix to row reduce

but f(0) would have amounted to the top right value

1/10321920
if i dint mistype it up

or 1
lets go with 1

is this close to what you are doing?

thats right

and yes close to what amistre did :)

prob is those values are large, 256^8 = 2^64.... there has to be a way other than brute force.

there must be a neater way...but i cant see a clue that will lead us to a nice solution

they are only asking for the constant, f(0) = a_0

right

is this from a particular class or book?

lol, i see i used "i" instead of "2^i" for my point set

finally, key is defining \[g(x)=xf(x)-1\]

do explain

\(g(x)\) has the roots \(1,2,2^2,...2^8\) so we can write\[g(x)=a \ (x-1)(x-2)(x-2^2)...(x-2^8)\]

\(f(x)\) becomes\[f(x)=\frac{2^{-36} \ (x-1)(x-2)(x-2^2)...(x-2^8)+1}{x}\]

then num of last expression has 0 as a root so the coefficient of \(x\) in num will be our answer

ill have to review that later when ive got the time to better digest it, but good job nonetheless

ok, for checking, final answer is\[2-\frac{1}{2^8}=\frac{511}{256}\]