Question

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

Answer 1

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

Answer 2

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

Answer 3

Thanks.

Answer 4

i would setup a matrix to row reduce

Answer 5

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

Answer 6

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

Answer 7

1/10321920

if i dint mistype it up

Answer 8

or 1

lets go with 1

Answer 9

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

Answer 10

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 :)

Answer 11

u actually set up a system of equation, 9 equations with 9 variables...still stuck, how u got f(0) ??

Answer 12

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.

Answer 13

is this close to what you are doing?

Answer 14

thats right

Answer 15

and yes close to what amistre did :)

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

right

Answer 20

is this from a particular class or book?

Answer 21

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

Answer 22

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

Answer 23

do explain

Answer 24

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

Answer 25

setting \(x=0\) we have\[g(0)=0\times f(0)-1=-1\]on the other hand\[-1=g(0)=a (-1)(-2)(-2^2)...(-2^8)\]\[a=2^{-(1+2+...+8)}=2^{-36}\]

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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