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Math Help - quadratic equ. problem no.4 quadratic equ. problem no.4

  1. #1
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    quadratic equ. problem no.4 quadratic equ. problem no.4

    quadratic equ. problem no.4
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    Lord of certain Rings
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    No I dont think I understand the second step in your proof.

    You say f(x) - x = 0
    Then you say f(f(x)) - x = 0 ???
    How

    The idea is functions obey "axiom of substitution",(AOS)
    that is if x=y \Rightarrow f(x) = f(y)

    So if f(x) = x for some x,
    Then by AOS, f(f(x)) = f(x) = x for those x
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Isomorphism View Post
    No I dont think I understand the second step in your proof.

    You say f(x) - x = 0
    Then you say f(f(x)) - x = 0 ???
    How
    in fact, he is right? didn't you notice it? he already omitted something there..
    if i'll just follow the argument, it would be like this..

     f(x) - x = 0 \implies f(x) = x

    now, from your AOS,  f(f(x)) = f(x) = x
    therefore,  f(f(x))  - x = 0


    Quote Originally Posted by Isomorphism View Post
    The idea is functions obey "axiom of substitution",(AOS)
    that is if x=y \Rightarrow f(x) = f(y)

    So if f(x) = x for some x,
    Then by AOS, f(f(x)) = f(x) = x for those x
    hehe, you call it ""axiom of substitution",(AOS)".. first time to see/hear it.. i simply say "function is well-defined"..
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    Lord of certain Rings
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    Quote Originally Posted by kalagota View Post
    in fact, he is right? didn't you notice it?





    hehe, you call it ""axiom of substitution",(AOS)".. first time to see/hear it.. i simply say "function is well-defined"..
    Yes, they are the same things. But if I tell people that it is well-defined, they think it is trivial. Somehow "well-defined" is often confused with "well, defined"

    But if I say it obeys axiom of substitution, it looks like functions follow some kind of law. Makes it look more important

    P.S:LOL, I recently learned that term from Terence Tao's Book on Real Analysis. He builds the number system and introduces this as a basic axiom . That funny name stuck me as something important, so i use it repeatedly to demonstrate wisdom
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  5. #5
    MHF Contributor kalagota's Avatar
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    Isomorphism,
    oic.. i have not seen his book yet.. anyhow, thanks for that info..

    afeasfaerw23231233,
    if you say that [2] equals [1], why don't you apply your result to [b]?

    i don't know if this is the proper proof, but i'll write it anyway..

    substitute \alpha to  f(f(x)) - x and show that it is equal to 0..

    here it is..
    substitution:  f(f(\alpha)) - \alpha

    since \alpha is a root of f(x) - x = 0 , we have f(\alpha) - \alpha = 0 \implies f(\alpha) = \alpha .

    thus,  f(f(\alpha)) - \alpha = f(\alpha) - \alpha = 0.

    therefore \alpha is a root of  f(f(x)) - x = 0 .. (similarly,  \beta is a root.)

    for [b], this is a hint..  f(x + y) \neq f(x) + f(y) ..
    for example, f(x) = a^x and f(1 + 2) = a^{1+2} = a^3 = a^1a^2 \neq a^1 + a^2 = f(1) + f(2)

    got it? so, your [b] should be  f(f(x)) - x = f(x^2 - 3x + 2) - x = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 - x

    just expand it and simplify..
    Last edited by kalagota; December 27th 2007 at 03:16 AM.
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    Lord of certain Rings
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    Since we already know roots of f(x)=x are also roots of f(f(x)) = x,
    Solve f(x) = x, i.e. x^2 - 3x + 2 = x. That is x^2 - 4x + 2 = 0 \Rightarrow (x-2)^2 = 2 \Rightarrow x = 2 \pm \sqrt 2.

    f(f(x)) - x = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 - x

    And it is a quartic equation. But we already know two roots of it by (i), so divide the quartic by x^2 - 4x + 2, you will get a quadratic quotient. Solve that quadratic to get the remaining 2 roots

    Quote Originally Posted by kalagota View Post
    for [b], this is a hint.. f(x + y) \neq f(x) + f(y)..
    for example, [/tex]f(x) = a^x [/tex] and f(1 + 2) = a^{1+2} = a^3 = a^1a^2 \neq a^1 + a^2 = f(1) + f(2).
    Hey I did not get this... Is it some slick solution with tremendous insight???
    Thanks
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  7. #7
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Since we already know roots of f(x)=x are also roots of f(f(x)) = x,
    Solve f(x) = x, i.e. x^2 - 3x + 2 = x. That is x^2 - 4x + 2 = 0 \Rightarrow (x-2)^2 = 2 \Rightarrow x = 2 \pm \sqrt 2.

    f(f(x)) - x = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 - x

    And it is a quartic equation. But we already know two roots of it by (i), so divide the quartic by x^2 - 4x + 2, you will get a quadratic quotient. Solve that quadratic to get the remaining 2 roots



    Hey I did not get this... Is it some slick solution with tremendous insight???
    Thanks
    which one you didn't get?
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  8. #8
    Lord of certain Rings
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    Quote Originally Posted by kalagota View Post
    which one you didn't get?
    What is the connection with f(x) + f(y) \neq f(x+y) and the problem??
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  9. #9
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Isomorphism View Post
    What is the connection with f(x) + f(y) \neq f(x+y) and the problem??
    oic.. in fact, nothing.. there is no connection..

    i'm just wondering where the "2" go after composition..
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