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Thread: [SOLVED] Proving that a polynomial is 0

  1. #1
    Moo
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    [SOLVED] Proving that a polynomial is 0

    Hi

    Challenging (more or less) question. I put it in this subforum as to avoid giving hints on the way to solve it.
    My friend told me a method, but I'm curious to know if there are others.


    So we have $\displaystyle P$ a polynomial such that its degree $\displaystyle \text{deg}(P)\leq n$, $\displaystyle n \in \mathbb{N}^*$
    If $\displaystyle \int_0^1 P(t)t^k ~dt=0 \quad \forall k \in \mathbb{N} ~:~ 0\leq k\leq n$,
    prove that $\displaystyle P(x)=0$.
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  2. #2
    MHF Contributor
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    Salut

    Let $\displaystyle P(t)=\sum_{k=0}^{n}\:a_k\:t^k$

    $\displaystyle \int_0^1 P(t)t^k ~dt=0 \quad \forall k \in \mathbb{N} ~:~ 0\leq k\leq n$

    $\displaystyle \sum_{k=0}^{n}\:a_k \int_0^1 P(t)t^k ~dt=0$

    $\displaystyle \int_0^1 P(t)\:\sum_{k=0}^{n}\:a_k t^k ~dt=0$

    $\displaystyle \int_0^1 P^2(t) ~dt=0$

    But of course $\displaystyle P^2(t) \geq 0$
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  3. #3
    Moo
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    Very neat solution


    This is how a teacher told my friend :
    Define the scalar product $\displaystyle \langle x,y\rangle=\int_0^1 x(t)y(t) ~dt$
    $\displaystyle y(t)=t^k$ belongs to the basis of $\displaystyle P[X]$
    So we have $\displaystyle \langle x,y\rangle=0$, for any y in the basis.
    Hence x=0.

    I don't know which theorem was used though
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Moo View Post
    Very neat solution


    This is how a teacher told my friend :
    Define the scalar product $\displaystyle \langle x,y\rangle=\int_0^1 x(t)y(t) ~dt$
    $\displaystyle y(t)=t^k$ belongs to the basis of $\displaystyle P[X]$
    So we have $\displaystyle \langle x,y\rangle=0$, for any y in the basis.
    Hence x=0.

    I don't know which theorem was used though
    Because it implies it is orthogonal to itself. (exactly what running-gag showed)

    Remember that $\displaystyle
    \left\langle {p_1,p_1} \right\rangle = 0 \Leftrightarrow p_1 = \bold 0
    $ (the 0 polynomial ) by the definition of inner-product


    Okay, let $\displaystyle
    \mathcal{P}_n \left[ t \right]
    $ be the space of polynomials of degree less or equal than $\displaystyle n$ over R. (our field will be R)

    We define: $\displaystyle
    \left\langle , \right\rangle :\mathcal{P}_n \left[ t \right] \times \mathcal{P}_n \left[ t \right] \to \mathbb{R}
    $ by: $\displaystyle
    \left\langle {a\left( t \right),b\left( t \right)} \right\rangle = \int_0^1 {a\left( t \right) \cdot b\left( t \right)dt}
    $

    You can check that this is an inner-product.

    Now $\displaystyle
    B = \left\{ {1,t,...,t^n } \right\}
    $ is a basis of $\displaystyle
    \mathcal{P}_n \left[ t \right]
    $

    And $\displaystyle
    p\left( t \right)
    $ is orthogonal to each of them, hence it's orthogonal to all vectors of $\displaystyle
    \mathcal{P}_n \left[ t \right]
    $ just see that $\displaystyle
    \left\langle {p\left( t \right),b_0 + \sum\limits_{k = 1}^n {b_k \cdot t^k } } \right\rangle = b_0 \cdot \left\langle {p\left( t \right),1} \right\rangle + b_k \cdot \sum\limits_{k = 1}^n {\left\langle {p\left( t \right),t^k } \right\rangle }
    $. (in particular to itself and hence $\displaystyle p(t)\equiv 0$)
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  5. #5
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    This problem has a nice generalization.
    If $\displaystyle f:[0,1]\to \mathbb{R}$ is continous with $\displaystyle \int_0^1 f(x)x^t dx = 0$ for all $\displaystyle t=0,1,2,3,...$ then $\displaystyle f$ is identically zero.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    This problem has a nice generalization.
    If $\displaystyle f:[0,1]\to \mathbb{R}$ is continous with $\displaystyle \int_0^1 f(x)x^t dx = 0$ for all $\displaystyle t=0,1,2,3,...$ then $\displaystyle f$ is identically zero.
    it's a nice and simple application of the Weierstrass approximation theorem.
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