# [SOLVED] Proving that a polynomial is 0

• Mar 18th 2009, 10:34 AM
Moo
[SOLVED] Proving that a polynomial is 0
Hi (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 $\displaystyle P(x)=0$.
• Mar 18th 2009, 12:15 PM
running-gag
Salut (Hi)

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 $\displaystyle P^2(t) \geq 0$
• Mar 19th 2009, 11:33 AM
Moo
Very neat solution (Nod)

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 (Rofl)
• Mar 19th 2009, 02:00 PM
PaulRS
Quote:

Originally Posted by Moo
Very neat solution (Nod)

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 (Rofl)

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 (Rofl)) 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 $\displaystyle p(t)\equiv 0$)
• Mar 20th 2009, 05:42 PM
ThePerfectHacker
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.
• Mar 22nd 2009, 04:25 AM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
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.