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Math Help - COntinuity

  1. #1
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    COntinuity

    Hey guys,

    I'm trying to prove that f(x)=x^4 is continuous on R, using the e-delta argument, but I can't find a way to break up x^4-a^4 effectively in the forms of x-a. ANybody have any idea here?

    I appreciate any hints and help!

    Thanks a lot!
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  2. #2
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    Re: COntinuity

    x^4 - a^4 = (x-a)(a^3+xa^2+x^2a+x^3)

    That help? If thats insufficient, maybe prove using similar identities by induction that if x^n is continuous, then x^(n+1) is too? Sorry if this doesn't help, I don't know much about the kind of stuff your trying to do
    Last edited by SworD; September 6th 2012 at 08:16 PM.
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  3. #3
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    Re: COntinuity

    Fix a>0.
    Pick any epsilon > 0.
    Choose some positive delta < min { epsilon/(15a^3), a }.
    Note that x^4 - a^4 = ( x - a )( x^3 + ax^2 + (a^2)x + a^3 )
    When |x-a|<delta, have a-delta < x < a+delta, so, since 0< delta < a, get 0 < x < 2a.
    Then | x^3 + ax^2 + (a^2)x + a^3 | = x^3 + ax^2 + (a^2)x + a^3 < (2a)^3 + a(2a)^2 + (a^2)(2a) + a^3 = 15 a^3.
    Thus |x-a|<delta implies |x^4 - a^4| = | x - a | * | x^3 + ax^2 + (a^2)x + a^3 | < | x - a | * ( 15 a^3 ) < (delta)( 15 a^3 ) < (epsilon/(15a^3))( 15 a^3 ) = epsilon
    Thus 0<|x-a|<delta implies |x^4 - a^4| < epsilon.
    Thus lim x->a of x^4 exists and equals a^4.
    Thus the function f defined by f(x)=x^4 is continuous at x=a for all a>0.
    Since f is even, f is also continuous for x < 0 ( lim x->(-a) f(x) = lim t->a f(-t) = lim t->a f(t) = f(a) = f(-a) ).
    Showing f is continuous at 0 is very simple (use delta = the fourth root of epsilon).
    Thus f is continuous on all of R.
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  4. #4
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    Re: COntinuity

    there is no reason to put any restriction on a.

    what we CAN do is restrict how far away from a we let x be. for example, we can require \delta \leq 1.

    then:

    |x - a| < 1

    but |x| - |a| ≤ |x - a| < 1, so:

    |x| < 1 + |a|.

    therefore:

    |x^3 + ax^2 + a^2x + a^3| \leq |x^3| + |ax^2| + |a^2x| + |a^3|

     = |x|^3 + |a||x|^2 + |a|^2|x| + |a|^3 < (1 + |a|)^3 + |a|(1 + |a|)^2 + |a|^2(1 + |a|) + |a|^3

     = (1 + 3|a| + 3|a|^2 + |a|^3) + |a|(1 + 2|a| + |a|^2) + |a|^2 + |a|^3 + |a|^3

     = 1 + 4|a| + 6|a|^3 + 4|a|^3. for convenience let us call this number L.

    then, letting \delta = \min\{\frac{\epsilon}{L},1\}, we have:

    |x^4 - a^4| = |x - a||x^3 + ax^2 + a^2x + a^3| < \left(\frac{\epsilon}{L}\right)(L) = \epsilon.

    a similar (but much more awkward to write down) proof shows that:

    f(x) = x^n is continuous at any real number a, for any natural number n.
    Last edited by Deveno; September 6th 2012 at 10:27 PM.
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