Posers and Puzzles

Posers and Puzzles

  1. Standard memberroyalchicken
    CHAOS GHOST!!!
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    29 Dec '03 02:10
    Also not mine.

    Given a function f:R->R such that f(xy) =f(x)f(y) and f(x+y) = f(x) + f(y) for all real x and y, give, with proof, and explicit expression for f(x).
  2. Standard memberTheMaster37
    Kupikupopo!
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    29 Dec '03 09:55
    Originally posted by royalchicken
    Also not mine.

    Given a function f:R->R such that f(xy) =f(x)f(y) and f(x+y) = f(x) + f(y) for all real x and y, give, with proof, and explicit expression for f(x).
    f = Id?

    Id(xy) = xy = Id(x)Id(y)
    Id(x+y) = x+y = Id(x)+Id(y)
  3. DonationAcolyte
    Now With Added BA
    Loughborough
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    29 Dec '03 13:57
    Originally posted by royalchicken
    Also not mine.

    Given a function f:R->R such that f(xy) =f(x)f(y) and f(x+y) = f(x) + f(y) for all real x and y, give, with proof, and explicit expression for f(x).
    f = 0 works as well. Do you want us to find the set of all such functions?
  4. DonationAcolyte
    Now With Added BA
    Loughborough
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    29 Dec '03 14:311 edit
    OK, I think I can do this. Note that '>' means 'greater than or equal to' because I can't get the character to display properly. Assume f(x) is nonzero for some x.

    Then f = id:

    f(x)f(1)=f(x) => f(1) = 1 => f(2) =2
    f(2)f(0)=2f(0)=f(0) => f(0) = 0
    f(-1) + f(1) = f(0) = 0 => f(-1) = -1

    => f(n) = n for all integers n

    f(q)f(p/q) = qf(p/q) = f(p) = p

    => f(p/q) = p/q for all rationals p/q

    Let a>0. Then f(a) = f(sqrt(a))^2 > 0

    Let c>b. Then f(c) = f(c-b) + f(b) > f(b) since f(c-b)>0

    => f is increasing

    Now given any real number x, we can find sequences of rationals t(n) above and u(n) below which tend to x. f(t(n)) > f(x) > f(u(n)) => t(n) > f(x) > u(n) for all n

    => f(x) = x. QED
  5. Standard memberroyalchicken
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    29 Dec '03 19:51
    Well done 😀! Although f(x) = 0 also works, as you said.
  6. Standard memberroyalchicken
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    29 Dec '03 20:07
    Now, work out what the title means.
  7. Joined
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    02 Jan '04 21:50
    Originally posted by royalchicken
    Now, work out what the title means.
    no pun intended?
  8. Standard memberroyalchicken
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    02 Jan '04 22:45
    Right on. But why the 'odd one' bit?
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