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Math Help - 2 problems with continous functions

  1. #1
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    2 problems with continous functions

    1) Show that 2^x = 3x for some x element (0,1)

    2) show that any polynomial of odd degree has at least one real root
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    1) Show that 2^x = 3x for some x element (0,1)
    Here's 1)

    Consider the function f(x) = 2^x – 3x

    f(0) = 1
    f(1) = -1

    Therefore, by the intermediate value theorem, there exist some c between 0 and 1, such that:
    f(c) = 0
    that is, for x = c, 0<c<1
    2^x – 3x = 0
    => 2^x = 3x
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    2) show that any polynomial of odd degree has at least one real root
    This is the same idea as Jhevon's solution to the first problem.

    Given y(x) = ax^n + ... + b, where a is not 0 and n is odd, note that
    lim[y(x), x --> -(infinity)] --> (-/+)(infinity) <-- Choice of sign depends on the sign of the coefficient a.
    lim[y(x), x--> (infinity)] --> (+/-)(infinity)

    Since the polynomial function either goes from -(infinity) to (infinity) or from (infinity) to -(infinity) we know it has to cross the x-axis at least once.

    -Dan
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    2) show that any polynomial of odd degree has at least one real root
    Consider the odd degree polynomial, P(x) = ax^n + bx^(n-1) + cx^(n-2) + ..., where n is odd.

    Assume a>0

    Now P(k) = ak^n + bk^(n-1) + ck^(n-2) + ...
    => P(k) = ak^n(1 + [bk^(n-1)]/ak^n + [ck^(n-2)]/ak^n + ...)
    => P(k) = ak^n(1 + b/ak + c/ak^2 + ...)
    => lim{k--> -inf}P(k) = lim{k--> -inf}ak^n(1 + b/ak + c/ak^2 + ...) = lim{k--> -inf}aK^n

    since n is odd,
    => lim{k--> -inf}ak^n = -inf
    So there is some k for which P(k) < 0

    Now P(l) = al^n + bl^(n-1) + cl^(n-2) + ...
    => P(l) = al^n(1 + [bl^(n-1)]/al^n + [cl^(n-2)]/al^n + ...)
    => P(l) = al^n(1 + b/al + c/al^2 + ...)
    => lim{l--> inf}P(l) = lim{l--> inf}al^n(1 + b/al + c/al^2 + ...) = lim{l--> inf}al^n

    since n is odd,
    => lim{l--> inf}al^n = inf
    So there is some l for which P(l) > 0

    Now, the opposte happens for a < 0, that is, P(k)>0 and P(l)<0, either way, 0 is between P(k) and P(l). Thus by the intermediate value theorem, there is some c between k and l such that P(c) = 0. So for x = c, P(x) has a real root

    EDIT: Ah man, Dan beat me to it. I went a lot slower for the second one than he did though, I guess you can use his way
    Last edited by Jhevon; April 14th 2007 at 05:48 PM.
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