## mukushla Group Title Suppose $$f(x)$$ is a degree $$8$$ polynomial such that $$f(2^i)=\frac{1}{2^i}$$ for all integers $$0≤i≤8$$. Evaluate $$f(0)$$. one year ago one year ago

1. Mendicant_Bias Group Title

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

2. mukushla Group Title

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

3. Mendicant_Bias Group Title

Thanks.

4. amistre64 Group Title

i would setup a matrix to row reduce

5. amistre64 Group Title

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 Group Title

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

7. amistre64 Group Title

1/10321920 if i dint mistype it up

8. amistre64 Group Title

or 1 lets go with 1

9. amistre64 Group Title

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 Group Title

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 Group Title

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

12. cruffo Group Title

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 Group Title

is this close to what you are doing?

14. mukushla Group Title

thats right

15. mukushla Group Title

and yes close to what amistre did :)

16. cruffo Group Title

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

17. mukushla Group Title

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

18. cruffo Group Title

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

19. mukushla Group Title

right

20. cruffo Group Title

is this from a particular class or book?

21. amistre64 Group Title

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

22. mukushla Group Title

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

23. amistre64 Group Title

do explain

24. mukushla Group Title

$$g(x)$$ 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 Group Title

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 Group Title

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

27. mukushla Group Title

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

28. amistre64 Group Title

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

29. mukushla Group Title

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