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Math Help - real analysis question

  1. #1
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    real analysis question

    i need to prove that sin(x)-x=0 has a unique solution in the open interval (0,pi/2). indicate a computational procedure for finding that solution approximately. on the previous parts of the problem i proved that every contraction mapping is continuous and that every contraction on a complete metric space has a unique fixed pt, so we can use these facts. my intuition is that we'll need to set f(x) = sin x and use the contraction mapping idea, but it's hard to prove that sin x is a contraction mapping. help!!!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by squarerootof2 View Post
    i need to prove that sin(x)-x=0 has a unique solution in the open interval (0,pi/2). indicate a computational procedure for finding that solution approximately. on the previous parts of the problem i proved that every contraction mapping is continuous and that every contraction on a complete metric space has a unique fixed pt, so we can use these facts. my intuition is that we'll need to set f(x) = sin x and use the contraction mapping idea, but it's hard to prove that sin x is a contraction mapping. help!!!
    This is not a proof but could you state that sin(x) has a fixed point on (0,2\pi) and say that f(f(f(f(f(f(f(f(f(x)))))))))) yields the fixed point?
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  3. #3
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    hmm do you think that i'm right in saying that proving sin (x) is a contraction mapping would do the job to show this given the previous parts that i've proven? thanks...
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  4. #4
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    Quote Originally Posted by squarerootof2 View Post
    i need to prove that sin(x)-x=0 has a unique solution in the open interval (0,pi/2). indicate a computational procedure for finding that solution approximately. on the previous parts of the problem i proved that every contraction mapping is continuous and that every contraction on a complete metric space has a unique fixed pt, so we can use these facts. my intuition is that we'll need to set f(x) = sin x and use the contraction mapping idea, but it's hard to prove that sin x is a contraction mapping. help!!!
    since you said the open interval the statement is not true becuase

    0 \notin (0,\frac{\pi}{2})

    Here is why

    consider f:[0,\pi/2] \to \mathbb{R} f(x)=\sin(x)-x \mbox{ then } f'(x)=\cos(x)-1

    note that \cos(x) < 1 \mbox{ } \forall x \in (0,\pi/2)

    so the derivative is always negative on the above interval.

    Now by the Mean Value Theorem we have
    let a=0 and b=x

    f'(c)(b-a)=f(b)-f(a) \iff f'(c)(x)=f(x)

    since the derivative is always negative and x is positive we know that f(x) is negative on the interval i.e f(x) < 0

    so finally we get

    f(x)=\sin(x)-x \iff \sin(x)-x < 0 \iff \sin(x) < x

    by the last they they are never equal, so no solution exits.

    I hope this helps.
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  5. #5
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    isnt there a way to apply the intermediate theorem to show that this solution exists? because the way the instructor asked the question, he implied that there is a solution... might be a mistake on the instructor's part...
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    Quote Originally Posted by squarerootof2 View Post
    isnt there a way to apply the intermediate theorem to show that this solution exists? because the way the instructor asked the question, he implied that there is a solution... might be a mistake on the instructor's part...
    Here is another way to think about it

    \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...=x+\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}

    x-\sin(x)=x-\left( x+\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} \right)=-\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}

    The only way that sum will equal zero is if x is zero.

    Here is a graph as well

    real analysis question-g.jpg

    The only solution is x=0
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by squarerootof2 View Post
    isnt there a way to apply the intermediate theorem to show that this solution exists? because the way the instructor asked the question, he implied that there is a solution... might be a mistake on the instructor's part...
    If it was on a different interval yes, but not on the one given above.

    Any open interval containing the origin would have exactly one solution.
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