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mukushla

  • one year 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 \(f(0)\).

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  1. Mendicant_Bias
    • one year ago
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    The text is a little small, what is two to the power of in f(2^[this])?

  2. mukushla
    • one year ago
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    \[\large f(2^i)=\frac{1}{2^i}\]its \(i\)

  3. Mendicant_Bias
    • one year ago
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    Thanks.

  4. amistre64
    • one year ago
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    i would setup a matrix to row reduce

  5. amistre64
    • one year ago
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    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
    • one year ago
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    but f(0) would have amounted to the top right value

  7. amistre64
    • one year ago
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    1/10321920 if i dint mistype it up

  8. amistre64
    • one year ago
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    or 1 lets go with 1

  9. amistre64
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    u actually set up a system of equation, 9 equations with 9 variables...still stuck, how u got f(0) ??

  12. cruffo
    • one year ago
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    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
    • one year ago
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    is this close to what you are doing?

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  14. mukushla
    • one year ago
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    thats right

  15. mukushla
    • one year ago
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    and yes close to what amistre did :)

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

  17. mukushla
    • one year ago
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    there must be a neater way...but i cant see a clue that will lead us to a nice solution

  18. cruffo
    • one year ago
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    they are only asking for the constant, f(0) = a_0

  19. mukushla
    • one year ago
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    right

  20. cruffo
    • one year ago
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    is this from a particular class or book?

  21. amistre64
    • one year ago
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    lol, i see i used "i" instead of "2^i" for my point set

  22. mukushla
    • one year ago
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    finally, key is defining \[g(x)=xf(x)-1\]

  23. amistre64
    • one year ago
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    do explain

  24. mukushla
    • one year ago
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    \(g(x)\) 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
    • one year ago
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    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
    • one year ago
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    \(f(x)\) becomes\[f(x)=\frac{2^{-36} \ (x-1)(x-2)(x-2^2)...(x-2^8)+1}{x}\]

  27. mukushla
    • one year ago
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    then num of last expression has 0 as a root so the coefficient of \(x\) in num will be our answer

  28. amistre64
    • one year ago
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    ill have to review that later when ive got the time to better digest it, but good job nonetheless

  29. mukushla
    • one year ago
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    ok, for checking, final answer is\[2-\frac{1}{2^8}=\frac{511}{256}\]

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