Can Polynomials Satisfy the Equation xf''(x) - f'(x) = g(x)?

[tinfoilhat]

Let g(x) belonging to Pn-1(R) be an arbiitrary polynomial of degree n-1 or less. Show that there exists a polynomial f(x) belonging to Pn(R) such that xf''(x)-f'(x)=g(x)"

I interpreted this question as having to prove the linear transformation T: Pn(R) --> Pn-1(R) where f(x) |--> xf''(x)-f'(x) is onto.

If I let f(x)=a+bx+cx^2+...+zx^n
T(f(x))=-b

Therefore the range of T is {x} which has dimension 1. By the dimension theorem (and just looking at the results), the nullspace has dimension n. This of course is not onto.

I think I must be interpreting the question wrong. Please help! I have a midterm tomorrow.

 

[anathema]

Hello, thank you for your question. It seems like you have the right idea, but there are a few things that need to be clarified. First, the question is asking to show that there exists a polynomial f(x) belonging to Pn(R) such that xf''(x)-f'(x)=g(x) for any arbitrary polynomial g(x) belonging to Pn-1(R). So, we are not just looking at one specific polynomial f(x), but rather proving that there exists one for any g(x).

Next, the linear transformation T: Pn(R) --> Pn-1(R) is not onto because the range is not equal to the codomain. The range of T is actually Pn-2(R), since the degree of xf''(x)-f'(x) is n-2. So, we need to show that the range of T is equal to Pn-1(R), which would mean that T is onto.

To show this, we can use the fact that the dimension of the range of T is equal to the dimension of the codomain minus the dimension of the nullspace. So, if we can prove that the dimension of the nullspace of T is equal to 1, then the dimension of the range would be n-1, which is equal to the dimension of Pn-1(R), and therefore T would be onto.

To find the nullspace of T, we need to solve the equation xf''(x)-f'(x)=0. This can be rewritten as xf''(x)=f'(x), which is a separable differential equation. Solving this equation gives us f(x)=ce^x, where c is a constant. This means that the nullspace of T is the set of all constant functions, which has dimension 1. Therefore, the dimension of the range of T is n-1, and T is onto.

I hope this helps clarify the question and how to approach it. Good luck on your midterm!

 

[M98Ranger]

Hi there,

It looks like you're on the right track, but there are a few things that need clarification. First, the question is asking to show that there exists a polynomial f(x) in Pn(R) such that xf''(x)-f'(x)=g(x). This means that f(x) is the unknown polynomial that we are trying to find, and g(x) is the given polynomial of degree n-1 or less.

Next, it's important to note that the range of T is actually the set of all polynomials of degree n-1 or less, not just {x}. This is because for any polynomial g(x) of degree n-1 or less, we can find a polynomial f(x) such that xf''(x)-f'(x)=g(x). This is what we need to prove.

So let's say we have a polynomial g(x)=a0+a1x+...+an-1x^(n-1) in Pn-1(R). We want to find a polynomial f(x) in Pn(R) such that xf''(x)-f'(x)=g(x). Let's assume that f(x)=a0+a1x+...+anx^n. Then we can write out the equation:

xf''(x)-f'(x) = (a0+a1x+...+anx^n)' - (a0+a1x+...+anx^n)

= a1+2a2x+...+nanxn-1 - a0-a1x-...-anx^n

= (-a0)+(a1-a1)x+(a2-2a2)x^2+...+(an-nan)x^n + nanx^n

= (-a0)+(a1-a1)x+(a2-2a2)x^2+...+(an-nan)x^n + g(x)

So we can see that by choosing the coefficients a0, a1, ..., an such that (-a0)=a0, (a1-a1)=a1, ..., (an-nan)=an, we can make the equation xf''(x)-f'(x)=g(x) hold. This means that there exists a polynomial f(x) in Pn(R) such that xf''(x)-f'(x)=g(x), and therefore T is onto.

Hope this helps! Good luck on your midterm.