text: ## 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$2-\frac{1}{2^8}=\frac{511}{256}$
explanation: The document contains a math question and a detailed discussion leading to an answer. The question is about evaluating \( f(0) \) for a polynomial \( f(x) \) of degree 8 that satisfies \( f(2^i) = \frac{1}{2^i} \) for \( 0 \leq i \leq 8 \). The discussion includes various approaches and ultimately provides the answer \( f(0) = \frac{511}{256} \).
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distilabel_metadata.role: system
distilabel_metadata.content: ## 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$2-\frac{1}{2^8}=\frac{511}{256}$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed discussion leading to an answer. The question is about evaluating \( f(0) \) for a polynomial \( f(x) \) of degree 8 that satisfies \( f(2^i) = \frac{1}{2^i} \) for \( 0 \leq i \leq 8 \). The discussion includes various approaches and ultimately provides the answer \( f(0) = \frac{511}{256} \).
  
  Label: question-and-answer