0
$\begingroup$

I have to find a polynomial with the following characteristics for a problem.

Find a polynomial $p(x)$ such that $$p(-1)=p'(-1)=p''(-1)=p(1)=p'(1)=p''(1)=0$$

I know and understand the process of how to find u(x), But since I don't know which

polynomial behave like this I'm stuck in the problem. I will be so thankfully if

somebody can help me to find this polynomial. The rest of the problem is so easy,

once I identify this polynomial.

$\endgroup$
3
  • $\begingroup$ What is $u(x)$? $\endgroup$
    – Alp Uzman
    Feb 16, 2015 at 0:10
  • $\begingroup$ You can use the facts that for a polynomial $p(x)$ if $x=a$ is a root then it of the form $(x-a)q(x)$ for another polynomial $q(x)$ (remainder theorem). Think, what happens further when we differentiate $(x-a)^2f(x)$ with respect to $x$ and then substitute $x=a$. $\endgroup$
    – P Vanchinathan
    Feb 16, 2015 at 0:26
  • $\begingroup$ @Uzman the problem is Find a Function u(x0 that is Twicely continuously differentiable on (-infinity, infinity) such that u(x)>0 in (-1,1) and u(x)=0 for all the x that are not in (-1,1). and the hint is find the polynomial p(x) such that p(-1)=p'(-1)=p''(-1)=p(1)=p'(1)=p''(1)=0 $\endgroup$
    – Evangelina
    Feb 16, 2015 at 0:43

1 Answer 1

Reset to default
1
$\begingroup$

If you have that $p(a)=0\Rightarrow p(X)=(X-a)q(X)$, a fact which comes from what it is knows as Bezout's little theorem (http://en.wikipedia.org/wiki/Polynomial_remainder_theorem), which can be proved using Euclid's algorithm.

If you have that $p(a)=p'(a)=0\Rightarrow p(X)=(X-a)^2r(X)$. Why? $p(X)(X-a)q(X)\Rightarrow p'(X)=q(X)+(X-a)q'(X)\Rightarrow q(a)=0\Rightarrow q(X)=(X-a)r(X)\Rightarrow p(X)=(X-a)^2r(X)$ and you can go on with the process. Therefore $(X-1)^3(X+1)^3$ check your problem.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .