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Math Help - Finite zeros on interval

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    Finite zeros on interval

    I came up with a theorem.

    Define a super function to be a function defined for all real numbers that is made out of sum/difference or product (not quotient) or composition of: sines,cosines,exponentials,polynomials.

    For example, \sin (e^x + 2x) is a super function. Also \sin (\cos (\sin e^{x^2+x}))) is a super function. But \sin (1/x) is not because 1/x is not one of the basic functions on the list and also because division is not allowed in this defintion.

    Theorem: Let I be a finite interval and let f(x) be a super function. Then unless f(x) is identically zero the equation f(x) = 0 has only finitely many solutions on the interval I.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I came up with a theorem.

    Define a super function to be a function defined for all real numbers that is made out of sum/difference or product (not quotient) or composition of: sines,cosines,exponentials,polynomials.

    For example, \sin (e^x + 2x) is a super function. Also \sin (\cos (\sin e^{x^2+x}))) is a super function. But \sin (1/x) is not because 1/x is not one of the basic functions on the list and also because division is not allowed in this defintion.

    Theorem: Let I be a finite interval and let f(x) be a super function. Then unless f(x) is identically zero the equation f(x) = 0 has only finitely many solutions on the interval I.
    What about
    \prod_{n = 1}^{\infty} \left ( x - \frac{1}{n} \right )

    I believe this fits your category for a superfunction, but has an infinite number of zeros in (0, 1).

    -Dan
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    Quote Originally Posted by topsquark View Post
    What about
    \prod_{n = 1}^{\infty} \left ( x - \frac{1}{n} \right )

    I believe this fits your category for a superfunction, but has an infinite number of zeros in (0, 1).

    -Dan
    Good idea but here is a better example.
    Consider,
    f(x)=\prod_{n=1}^{\infty}  x\left(x^2 - n^2 \pi^2\right)
    Using Weierstrass factorization we can show that,
    f(x) = x\sin \frac{1}{x}.
    Now f(x) is not a super function.

    Maybe I should have said "finite combinations of sum/difference/composition", but it seemed clear enough.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Using Weierstrass factorization we can show that,
    f(x) = x\sin \frac{1}{x}.
    now i'd love to see that. but i'm quite sure i wouln't understand it, so never mind.
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    Quote Originally Posted by Jhevon View Post
    now i'd love to see that. but i'm quite sure i wouln't understand it, so never mind.
    It is hard. If you remember Leonard Euler expanded the sine function into an infinite product. Karl Weierstrass later justifed that and even came up with a theorem. It is an advanced theorem in complex analysis. Trojan.Bagle32.Symantec

    It has to do with the idea that a polynomial can be expanded in terms of its zeros. Now \sin x is an infinite polynomial intuitievely speaking. And its zeros, here x=\pi n are its factors.
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