## mukushla 2 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. Mendicant_Bias

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

2. mukushla

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

3. Mendicant_Bias

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. mukushla

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

12. cruffo

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. cruffo

is this close to what you are doing?

14. mukushla

thats right

15. mukushla

and yes close to what amistre did :)

16. cruffo

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

17. mukushla

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

18. cruffo

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

19. mukushla

right

20. cruffo

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. mukushla

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

23. amistre64

do explain

24. mukushla

$$g(x)$$ 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. mukushla

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. mukushla

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

27. mukushla

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. mukushla

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