# Thread: Analysis involves a polynomial with mult degree

1. ## Analysis involves a polynomial with mult degree

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

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

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

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

Am I on the right path?

Thank you.

KK

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

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

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

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

Am I on the right path?

Thank you.

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

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

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 $f(x)\epsilon\mathbb{R}$ equals to zero?

And in the assumtion, we only know that $f(x_{0})$ equals to zero for a GIVEN POINT $x_{0}\epsilon\mathbb{R}$, 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 $p: \Re \mapsto \Re$be a polynomial with degree no greater than 5.
Suppose that at some point $x_{0} \epsilon \Re$ $p(x_{0}) = p\prime (x_{0}) = . . . = p^{(5)}(x_{0}) = 0.$
Prove that $p(x) = 0$ $\forall x \epsilon \Re$

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

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

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

Am I on the right path?

Thank you.

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

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

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

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

so:

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

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

Which is essentialy what ImPerfectHacker was saying.

RonL