## anonymous 3 years ago Suppose $$f(x)$$ is a degree $$8$$ polynomial such that $$f(2^i)=\frac{1}{2^i}$$ for all integers $$0≤i≤8$$. Evaluate $$f(0)$$.

1. anonymous

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

2. anonymous

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

3. anonymous

Thanks.

4. amistre64

i would setup a matrix to row reduce

5. amistre64

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

6. amistre64

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

7. amistre64

1/10321920 if i dint mistype it up

8. amistre64

or 1 lets go with 1

9. amistre64

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

10. amistre64

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 :)

11. anonymous

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

12. anonymous

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.

13. anonymous

is this close to what you are doing?

14. anonymous

thats right

15. anonymous

and yes close to what amistre did :)

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. anonymous

right

20. anonymous

is this from a particular class or book?

21. amistre64

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

22. anonymous

finally, key is defining $g(x)=xf(x)-1$

23. amistre64

do explain

24. anonymous

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

25. anonymous

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}$

26. anonymous

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

27. anonymous

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

28. amistre64

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

29. anonymous

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