# Thread: Analysis involves a polynomial with mult degree

1. ## Analysis involves a polynomial with mult degree

Q: Let $\displaystyle p: \Re \mapsto \Re$be a polynomial with degree no greater than 5.
Suppose that at some point $\displaystyle x_{0} \epsilon \Re$ $\displaystyle p(x_{0}) = p\prime (x_{0}) = . . . = p^{(5)}(x_{0}) = 0.$
Prove that $\displaystyle p(x) = 0$ $\displaystyle \forall x \epsilon \Re$

By a theorem that I learned in class today, I have -

Proof: Since $\displaystyle \Re$ is an open interval, and $\displaystyle \exists x_{0} \epsilon\Re$ with $\displaystyle p^{(n)}(x_{0})=0$, therefore $\displaystyle \exists z$ between $\displaystyle x_{0}$ and $\displaystyle x$ such that $\displaystyle p(x) = \frac{p^{(5)}(z)(x-x_{0})}{5!}$

Now, I don't know how to prove $\displaystyle p^{(5)}(z)(x-x_{0})$ equals to zero.

Am I on the right path?

Thank you.

KK

Q: Let $\displaystyle p: \Re \mapsto \Re$be a polynomial with degree no greater than 5.
Suppose that at some point $\displaystyle x_{0} \epsilon \Re$ $\displaystyle p(x_{0}) = p\prime (x_{0}) = . . . = p^{(5)}(x_{0}) = 0.$
Prove that $\displaystyle p(x) = 0$ $\displaystyle \forall x \epsilon \Re$

By a theorem that I learned in class today, I have -

Proof: Since $\displaystyle \Re$ is an open interval, and $\displaystyle \exists x_{0} \epsilon\Re$ with $\displaystyle p^{(n)}(x_{0})=0$, therefore $\displaystyle \exists z$ between $\displaystyle x_{0}$ and $\displaystyle x$ such that $\displaystyle p(x) = \frac{p^{(5)}(z)(x-x_{0})}{5!}$

Now, I don't know how to prove $\displaystyle p^{(5)}(z)(x-x_{0})$ equals to zero.

Am I on the right path?

Thank you.

KK
First it is considered old school to use $\displaystyle \Re$ today people use $\displaystyle \mathbb{R}$.

A polynomial of degree at most five is expressible as,
$\displaystyle f(x)=a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$
You are told that,
$\displaystyle f(0)=0$
Thus,
$\displaystyle a_0=0$.
Then,
$\displaystyle f'(x)=5a_5x^4+4a_4x^3+3a_3x^2+2a_2x^1+a_1$
You are told that,
$\displaystyle f'(0)=0$
Thus,
$\displaystyle a_1=0$.
Continue in this manner to get,
$\displaystyle a_0=a_1=...=a_5=0$

3. I don't quite understand this solution.

Yes, now I know all the a equals to zero, but how does that related to prove all $\displaystyle f(x)\epsilon\mathbb{R}$ equals to zero?

And in the assumtion, we only know that $\displaystyle f(x_{0})$ equals to zero for a GIVEN POINT $\displaystyle x_{0}\epsilon\mathbb{R}$, not for all x.

The professor today did suggest that we should use the theorem that I use above, what do you think? Would that be the easiest?

Thanks.

KK

Q: Let $\displaystyle p: \Re \mapsto \Re$be a polynomial with degree no greater than 5.
Suppose that at some point $\displaystyle x_{0} \epsilon \Re$ $\displaystyle p(x_{0}) = p\prime (x_{0}) = . . . = p^{(5)}(x_{0}) = 0.$
Prove that $\displaystyle p(x) = 0$ $\displaystyle \forall x \epsilon \Re$

By a theorem that I learned in class today, I have -

Proof: Since $\displaystyle \Re$ is an open interval, and $\displaystyle \exists x_{0} \epsilon\Re$ with $\displaystyle p^{(n)}(x_{0})=0$, therefore $\displaystyle \exists z$ between $\displaystyle x_{0}$ and $\displaystyle x$ such that $\displaystyle p(x) = \frac{p^{(5)}(z)(x-x_{0})}{5!}$

Now, I don't know how to prove $\displaystyle p^{(5)}(z)(x-x_{0})$ equals to zero.

Am I on the right path?

Thank you.

KK
I would have approached it as follows:As $\displaystyle p$ is a polynomial of degree no greater than 5, there exist constants $\displaystyle a_0, a_1, ..., a_5\ \in\mathbb{R}$ such that:

$\displaystyle p(x)=a_0+a_1\,x+ ..a_5\,x^5$

and that for any $\displaystyle x_0$ we can find constants $\displaystyle b_0, b_1, ..., b_5\ \in \mathbb{R}$ such that:

$\displaystyle p(x)=b_0+b_1\,(x-x_0)+ ..b_5\,(x-x_0)^5$

so:

$\displaystyle p^{(n)}(x_0)=\frac{b_n}{n!},\ \ \ n=1,..,5$

hence if $\displaystyle p^{(n)}(x_0)=0, \ \ n=1,..,5$, then $\displaystyle b_n=0, \ \ n=1,..,5$, and hence $\displaystyle p(x)=0\ \forall \ x\in \mathbb{R}$

Which is essentialy what ImPerfectHacker was saying.

RonL