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

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

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

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

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

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

    But of course 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 \langle x,y\rangle=\int_0^1 x(t)y(t) ~dt
    y(t)=t^k belongs to the basis of P[X]
    So we have \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 \langle x,y\rangle=\int_0^1 x(t)y(t) ~dt
    y(t)=t^k belongs to the basis of P[X]
    So we have \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 <br />
\left\langle {p_1,p_1} \right\rangle  = 0 \Leftrightarrow p_1 = \bold 0<br />
(the 0 polynomial ) by the definition of inner-product


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

    We define: <br />
\left\langle , \right\rangle :\mathcal{P}_n \left[ t \right] \times \mathcal{P}_n \left[ t \right] \to \mathbb{R}<br />
by: <br />
\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} <br />

    You can check that this is an inner-product.

    Now <br />
B = \left\{ {1,t,...,t^n } \right\}<br />
is a basis of <br />
\mathcal{P}_n \left[ t \right]<br />

    And <br />
p\left( t \right)<br />
is orthogonal to each of them, hence it's orthogonal to all vectors of <br />
\mathcal{P}_n \left[ t \right]<br />
just see that <br />
\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 } <br />
. (in particular to itself and hence p(t)\equiv 0)
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  5. #5
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    This problem has a nice generalization.
    If f:[0,1]\to \mathbb{R} is continous with \int_0^1 f(x)x^t dx = 0 for all t=0,1,2,3,... then 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 f:[0,1]\to \mathbb{R} is continous with \int_0^1 f(x)x^t dx = 0 for all t=0,1,2,3,... then f is identically zero.
    it's a nice and simple application of the Weierstrass approximation theorem.
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