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Thread: Problem 40

  1. #1
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    Problem 40

    1)This problem is for the younger kids give them a chance, please. Let $\displaystyle A,B$ be randomly chosen from $\displaystyle \{ 1,2,...,2007 \}$ (they can be the same). What is the probability that it is possible to find two real numbers that have their product equal to $\displaystyle A$ and their sum equal to $\displaystyle B$. (By the way, "real", means non-imaginary).

    2)Let $\displaystyle f(x),g(x)$ be continous on $\displaystyle [0,1]$ so that $\displaystyle \int_0^1 x^n f(x) dx = \int_0^1 x^n g(x) dx$ for all $\displaystyle n\geq 0$. Prove that $\displaystyle f(x) = g(x)$. (This is a classic).
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    MHF Contributor kalagota's Avatar
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    f and g are cont on [0,1], then f and g are integrable.

    Let $\displaystyle F_0 = \int_0^1 x^n f(x)dx = \int_0^1 x^n g(x)dx$

    Let $\displaystyle f_0(x) = x^n f(x)$ and $\displaystyle g_0(x) = x^n g(x)$

    then both $\displaystyle f_0, g_0$ are continuous on [0,1], which implies that $\displaystyle F_0$ is differentiable on [0,1], and $\displaystyle F_0'(x) = f_0(x) = g_0(x)$, for all $\displaystyle x \in [0,1]$ (by a corollary to the FTOC)

    $\displaystyle \implies x^n f(x) = x^n g(x)$

    $\displaystyle \implies f(x) = g(x)$. qed
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    I am sorry, I cannot follow.

    Quote Originally Posted by kalagota View Post
    f and g are cont on [0,1], then f and g are integrable.

    Let $\displaystyle F_0 = \int_0^1 x^n f(x)dx = \int_0^1 x^n g(x)dx$
    How is this a well-defined function? Maybe for a particular value of $\displaystyle n$?

    then both $\displaystyle f_0, g_0$ are continuous on [0,1], which implies that $\displaystyle F_0$ is differentiable on [0,1], and $\displaystyle F_0'(x) = f_0(x) = g_0(x)$, for all $\displaystyle x \in [0,1]$ (by a corollary to the FTOC)
    That is not the Fundamental Theorem of Calculus.
    ---

    Remember I am saying that:
    $\displaystyle \int_0^1 f(x)dx = \int_0^1 g(x) dx$, $\displaystyle \int_0^1 xf(x) dx = \int_0^1 xg(x) dx$, $\displaystyle \int_0^1 x^2f(x) dx = \int_0^1 x^2g(x) dx$, ...
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    1)This problem is for the younger kids give them a chance, please. Let $\displaystyle A,B$ be randomly chosen from $\displaystyle \{ 1,2,...,2007 \}$ (they can be the same). What is the probability that it is possible to find two real numbers that have their product equal to $\displaystyle A$ and their sum equal to $\displaystyle B$. (By the way, "real", means non-imaginary).
    1.) $\displaystyle (\frac{1}{2007} + \frac{1}{2007}) \times (\frac{1}{2007} + \frac{1}{2007})$

    = $\displaystyle \frac{4}{4028049}$

    = $\displaystyle \frac{1}{1007012,25}$
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    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    2)Let $\displaystyle f(x),g(x)$ be continous on $\displaystyle [0,1]$ so that $\displaystyle \int_0^1 x^n f(x) dx = \int_0^1 x^n g(x) dx$ for all $\displaystyle n\geq 0$. Prove that $\displaystyle f(x) = g(x)$. (This is a classic).
    Let $\displaystyle P_n(x);\ n \in \mathbb(Z)_+$ be an orthonormal basis of polynomials for $\displaystyle L^2_{[0,1]}$.

    Then by the conditions specified in the problem:

    $\displaystyle \langle P_n, f-g \rangle =0;\ \forall n \in \mathbb{Z}_+$

    which implies that $\displaystyle f(x)-g(x)=0\ a.e. \in [0,1]$ which as f-g is continuous implies $\displaystyle f(x)-g(x)=0\ x\in [0,1]$

    RonL
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    Super Member angel.white's Avatar
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    Quote Originally Posted by janvdl View Post
    1.) $\displaystyle (\frac{1}{2007} + \frac{1}{2007}) \times (\frac{1}{2007} + \frac{1}{2007})$

    = $\displaystyle \frac{4}{4028049}$

    = $\displaystyle \frac{1}{1007012,25}$
    Hmm, that is much simpler than the method I thought of in my head (I'm a bit too embarrassed to say what my method was, but I'll just say it involves a new pack of pencils, a full ream of paper, a gallon of coffee, a carton of cigarettes, and no obligations for several days.)
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by angel.white View Post
    Hmm, that is much simpler than the method I thought of in my head (I'm a bit too embarrassed to say what my method was, but I'll just say it involves a new pack of pencils, a full ream of paper, a gallon of coffee, a carton of cigarettes, and no obligations for several days.)
    Haha, doesn't mean my method is right you know
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Let $\displaystyle P_n(x);\ n \in \mathbb(Z)_+$ be an orthonormal basis of polynomials for $\displaystyle L^2_{[0,1]}$.

    Then by the conditions specified in the problem:

    $\displaystyle \langle P_n, f-g \rangle =0;\ \forall n \in \mathbb{Z}_+$

    which implies that $\displaystyle f(x)-g(x)=0\ a.e. \in [0,1]$ which as f-g is continuous implies $\displaystyle f(x)-g(x)=0\ x\in [0,1]$

    RonL
    i always have trouble proving things that seem to me to be obvious.

    would it be wrong to do this?

    $\displaystyle \int_0^1 x^n f(x)~dx = \int_0^1 x^n g(x)~dx$

    $\displaystyle \Rightarrow \int_0^1 x^n f(x)~dx - \int_0^1 x^n g(x)~dx = 0$

    $\displaystyle \Rightarrow \int_0^1 x^n [f(x) - g(x)]~dx = 0$

    Obviously, the only way this integral can be zero for all x and all n is if $\displaystyle f(x) - g(x) = 0$. the result follows immediately.



    Now i can see where the problem here would be, the "obviously" part.
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post

    That is not the Fundamental Theorem of Calculus.
    ---
    i think i clearly state that the reason is from a corollary to FTOC and not to FTOC itself..

    anyways, i am not really good at this..

    EDIT: now i get what you mean.. thx for that remark..
    Last edited by kalagota; Nov 11th 2007 at 01:32 AM.
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    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    i always have trouble proving things that seem to me to be obvious.

    would it be wrong to do this?

    $\displaystyle \int_0^1 x^n f(x)~dx = \int_0^1 x^n g(x)~dx$

    $\displaystyle \Rightarrow \int_0^1 x^n f(x)~dx - \int_0^1 x^n g(x)~dx = 0$

    $\displaystyle \Rightarrow \int_0^1 x^n [f(x) - g(x)]~dx = 0$

    Obviously, the only way this integral can be zero for all x and all n is if $\displaystyle f(x) - g(x) = 0$. the result follows immediately.



    Now i can see where the problem here would be, the "obviously" part.
    These proofs are thinly disguised use of the Stone-Weierstrass theorem.

    RonL
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    Quote Originally Posted by CaptainBlank View Post
    These proofs are thinly disguised use of the Stone-Weierstrass theorem.
    You got the main part right! Weierstrass theorem is the secret trick here.

    Quote Originally Posted by Jhevon
    Now i can see where the problem here would be, the "obviously" part.
    It is really not that obvious.

    Quote Originally Posted by janvdl View Post
    1.) $\displaystyle (\frac{1}{2007} + \frac{1}{2007}) \times (\frac{1}{2007} + \frac{1}{2007})$

    = $\displaystyle \frac{4}{4028049}$

    = $\displaystyle \frac{1}{1007012,25}$
    How did you get that? My solution is much longer.
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How did you get that? My solution is much longer.
    Haha, i guess what matters is if we got the same answer? And if mine is right?
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    Quote Originally Posted by janvdl View Post
    Haha, i guess what matters is if we got the same answer? And if mine is right?
    I have no idea what the answer is, I only created the problem and thus know how to solve it. I did not fully solve it.
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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
    You got the main part right! Weierstrass theorem is the secret trick here.
    are you saying we should approximate f(x) - g(x) with some other polynomial, say h(x), and show that we must have h(x) = 0 ?
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    Quote Originally Posted by Jhevon View Post
    are you saying we should approximate f(x) - g(x) with some other polynomial, say h(x), and show that we must have h(x) = 0 ?
    Hint: If $\displaystyle f(x)$ is continous on $\displaystyle [0,1]$ then there exists a sequence of polynomial $\displaystyle p_n(x)$ that converge uniformly to $\displaystyle f(x)$. (That is the Stone-Weierstrass theorem).
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