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Suppose \(f(x)\) is a degree \(8\) polynomial such that \(f(2^i)=\frac{1}{2^i}\) for all integers \(0≤i≤8\). Evaluate \(f(0)\).

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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 \(i\)

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Other answers:

i would setup a matrix to row reduce
rref{{0^8,0^7,0^6,0^5,0^4,0^3,0^2,0,1,1/2^0}, {1^8,1^7,1^6,1^5,1^4,1^3,1^2,1,1,1/2^1}, {2^8,2^7,2^6,2^5,2^4,2^3,2^2,2,1,1/2^2}, {3^8,3^7,3^6,3^5,3^4,3^3,3^2,3,1,1/2^3}, {4^8,4^7,4^6,4^5,4^4,4^3,4^2,4,1,1/2^4}, {5^8,5^7,5^6,5^5,5^4,5^3,5^2,5,1,1/2^5}, {6^8,6^7,6^6,6^5,6^4,6^3,6^2,6,1,1/2^6}, {7^8,7^7,7^6,7^5,7^4,7^3,7^2,7,1,1/2^7}, {8^8,8^7,8^6,8^5,8^4,8^3,8^2,8,1,1/2^8}} the wolf cannot accept that many characters into their input box
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
you are essentially matching a P8(x) to 2^(-x) and at that many point so close together, we would expect it to be close enough to 2^0
and ... now that i read it again .... f(0) is a point given in the interval that is set to 1/2^0 to begin with .... i need my mtDew :)
u actually set up a system of equation, 9 equations with 9 variables...still stuck, how u got f(0) ??
i have the same question since 2^i cannot equal 0 . I was thinkinf you would need the solve the system to find all tge coefficients of the polynomial.
is this close to what you are doing?
1 Attachment
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
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\[g(x)=a \ (x-1)(x-2)(x-2^2)...(x-2^8)\]
setting \(x=0\) we have\[g(0)=0\times f(0)-1=-1\]on the other hand\[-1=g(0)=a (-1)(-2)(-2^2)...(-2^8)\]\[a=2^{-(1+2+...+8)}=2^{-36}\]
\(f(x)\) 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\[2-\frac{1}{2^8}=\frac{511}{256}\]

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