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Find the value of f(2)

  1. May 10, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If f(x) is a polynomial function satisfying 2+f(x)f(y)=f(x)+f(y)+f(xy), x,y belongs to R and if f(2)=5, then find the value of f(f(2))

    2. Relevant equations

    3. The attempt at a solution
    The question clearly seeks the value of f(5). I put x=0 and y=2. Then
    2+f(0)f(2)=f(0)+f(2)+f(0)
    2+5f(0)=2f(0)+5
    f(0)=1

    Now I put x=0 and y=5
    2+f(0)f(5)=f(0)+f(5)+f(0)
    f(5)=f(5)
    :confused:
     
  2. jcsd
  3. May 11, 2013 #2

    ehild

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    f(x) is a polynomial function. What does it mean?



    ehild
     
  4. May 11, 2013 #3

    haruspex

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    Consider a substitution f(x) = g(x) + c. Can you find a value of c that simplifies the equation?
     
  5. May 11, 2013 #4

    utkarshakash

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    Nothing special I can think of.
     
  6. May 11, 2013 #5

    Office_Shredder

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    I'm going to second this suggestion
     
  7. May 11, 2013 #6

    ehild

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    f(x) is a polynomial function, of form f(x)=a0+a1x+a2x2+a3x3+...

    What relations do you get for the coefficients from the given equation and data?


    ehild
     
    Last edited: May 11, 2013
  8. May 11, 2013 #7

    utkarshakash

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    I still don't know the degree of polynomial
     
  9. May 11, 2013 #8

    utkarshakash

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    Ok following your method I arrive at this

    2+g(x)g(y)+(c-1){g(x)+g(y)}=3c-c2+g(xy)
     
  10. May 12, 2013 #9

    haruspex

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    Right, so what value of c will simplify that greatly?
     
  11. May 12, 2013 #10

    ehild

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    That is you need to figure out. From the condition f(2)=5 you get a relation between f(y) and f(2y), and that can be fulfilled with polynomials of a certain degree.

    ehild
     
    Last edited: May 12, 2013
  12. May 12, 2013 #11

    utkarshakash

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    The only number I can think of is 1
     
  13. May 12, 2013 #12

    haruspex

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    Right, so what equation do you get for g()? When you have that, suppose α is a root of g(x). What other root(s) can you then deduce?
     
    Last edited: May 12, 2013
  14. May 12, 2013 #13

    ehild

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    I show my way as I think it is quite simple and straightforward.

    2+f(x)f(y)=f(x)+f(y)+f(xy)

    From f(2)=5 follows: 2+5f(x)=5+f(x)+f(2x) ----> -3+4f(x)=f(2x)*

    f(x) is a polynomial f(x)=a0+a1x+a2x2+....+akxk+...

    Plug into * and compare the coefficients of powers of x on both sides

    -3+4(a0+a1x+a2x2+....+akxk+...)=a0+2a1x+4a2x2+....+2kakxk+...


    -3+4a0=a0
    4a1=2a1
    4a2=4a2
    .
    .
    .
    4ak=2kak

    What is the degree of the polynomial?

    ehild
     
  15. May 12, 2013 #14

    haruspex

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    Ah, but mine is so elegant :wink:
     
  16. May 12, 2013 #15

    Dick

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    Elegant and cute, but less obvious.
     
  17. May 12, 2013 #16

    ehild

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    I must be dumb but still do not know your solution.:mad:

    ehild
     
  18. May 12, 2013 #17

    Dick

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    I saw it. g(x)g(y)=g(xy) means if x is root of g then xy must be a root of g for ANY y. Severely limits the choice of roots. I think this is little too subtle.
     
  19. May 12, 2013 #18

    ehild

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    I reached here, but what after? f(x)=1+xh(x). But it is obvious as f(x) is polynomial, and f(0)=1 (obtained by the OP already). Find the possible root of h?

    ehild
     
    Last edited: May 13, 2013
  20. May 13, 2013 #19

    Office_Shredder

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    I'm not sure what h is supposed to be here. But once you get to the post you quoted you say "ah, g(x) must be xk" and use f(2) = 5 to figure out what k is
     
  21. May 13, 2013 #20

    ehild

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    I do not see that "ah":mad: Perhaps I stick to my version too much which gives the degree at once.

    ehild
     
    Last edited: May 13, 2013
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