Find the polynomials P in R[X]

• geoffrey159
In summary, the problem is solved if the sum of the exponents of the coefficients of the polynomials is zero.
geoffrey159

Homework Statement

Find the polynomials P in R[X] such that
## X(X+1) P'' +(X+2) P' -P = 0 ##​

Homework Equations

If P is a solution, I assume ## P = \sum_{n=0}^N a_nX^n##

The Attempt at a Solution

I get the answer ##\{\alpha (X+2),\alpha\in\mathbb{R}\} ##, but I've done many calculation errors before getting it right.
Is there a way to solve that problem without too much calculation ?

Last edited:
geoffrey159 said:

Homework Statement

Find the polynomials P in R[X] such that
## X(X+1) P'' +(X+2) P' -P = 0 ##​

Homework Equations

If P is a solution, I assume ## P = \sum_{n=0}^N a_nX^n##

The Attempt at a Solution

I get the answer ##\{\alpha (X+2),\alpha\in\mathbb{R}\} ##, but I've done many calculation errors before getting it right.
Is there a way to solve that problem without too much calculation ?
When you find a solution to a differential equation, the first thing you should do is to check your solution. Have you don't this?

BTW, what does the notation R[X] mean?

Your answer does seem to work, but what is needed for higher order polynomials?

You might find it useful to use the form:
## P = \sum_{n=0}^N a_nX^n,\\
P' = \sum_{n=0}^{N-1} (n+1)a_{n+1}X^{n},\\
P'' = \sum_{n=0}^{N-2} (n+2)(n+1)a_{n+2}X^{n} ##

Yes I checked that it works, after the calculations, I get something like (minus the errors :-))

##0 = 2a_1 - a_0 + (N^2 -1) a_N X^N + \sum_{n=1}^{N-1} ( (n^2-1)a_n + (n+1)(n+2) a_{n+1}) X^n ##.

So if P has degree N, then N = 1 and ##2a_1 = a_0##.

(R[X] : Polynomials of a single variable with real coefs)

So, to do this quickly depends on your definition of quick.
## X(X+1) P'' +(X+2) P' -P = 0 ##

## P = \sum_{n=0}^N a_nX^n,\\
P' = \sum_{n=0}^{N-1} (n+1)a_{n+1}X^{n},\\
P'' = \sum_{n=0}^{N-2} (n+2)(n+1)a_{n+2}X^{n} ##
Breaking apart the components:
## X^2 P'' +XP''+X P' +2P' - P = 0 ##
Gives:
## \sum_{n=0}^{N-2} (n+2)(n+1)a_{n+2}X^{n+2}\\
+ \sum_{n=0}^{N-2} (n+2)(n+1)a_{n+2}X^{n+1}\\
+ \sum_{n=0}^{N-1} (n+1)a_{n+1}X^{n+1}\\
+\sum_{n=0}^{N-1} 2(n+1)a_{n+1}X^{n}\\
-\sum_{n=0}^N a_nX^n =0##
Then, group all the terms by powers of X:
##X^N (N^2-N+N-1)a_N =0##
for n = 2 to N-1: ##X^n [a_n(n^2-n+n-1) + a_{n+1}(n^2+n+2n+2)] = 0 ##
##X(2a_2+4a_2 +a_1-a_1)=0 ##
##1(2a_1-a_0 ) = 0 ##
Then it should be easy to see that for ##N>1,a_N =0## in order to push the highest order term to zero.
Quickly, you will see the solution collapse down to what you found.

I think that anytime you are playing around with indices on sums, it is easy to make mistakes...but I don't think it takes a tremendous amount of time or skill to gather exponents.

In fact, you could have just used the expansion for the highest order term to tell you that you could not have a solution of order 2 or greater, then the challenging stuff in the middle would be unnecessary.

geoffrey159
RUber said:
In fact, you could have just used the expansion for the highest order term to tell you that you could not have a solution of order 2 or greater, then the challenging stuff in the middle would be unecessary.

Yes, that's a shortcut I could have used. And it is way more elegant. Thank you !

Last edited:

1. What is a polynomial?

A polynomial is an expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication. The variables can only have non-negative integer exponents.

2. What is R[X] in the context of polynomials?

R[X] refers to the set of polynomials with coefficients in the real numbers (R). This means that all the coefficients in the polynomial are real numbers.

3. How do you find the degree of a polynomial?

The degree of a polynomial is the highest exponent of the variable in the expression. For example, in the polynomial 3x^2 + 5x + 1, the degree is 2 as it is the highest exponent of x.

4. What are the different methods for finding the roots of a polynomial?

There are several methods for finding the roots of a polynomial, including the Rational Root Theorem, Descartes' Rule of Signs, and the Quadratic Formula. Additionally, graphing the polynomial can also help in finding the roots.

5. How do you solve a polynomial equation?

To solve a polynomial equation, you need to find the roots of the polynomial. This can be done by factoring, using one of the methods mentioned in question 4, or by using a graphing calculator. Once you have the roots, you can substitute them back into the equation to check if they satisfy the original equation.

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