Polynomial differential operators

[sassie]

Homework Statement

p(D) is a polynomial D operator of degree n>m. Suppose a is a m fold root of p(t)=0, but not a (m+1) fold root.

Verify that [tex]\frac{1}{p(D)}e^{at}=\frac{1}{p^{(m)}(a)}t^me^{at}[/tex]

where [tex]p^{(m)}(t)[/tex] is the [tex]m^{th}[/tex] derivative of p(t).

Homework Equations

For this question, we were told to use the exponential shift formula: [tex]\frac{1}{p(D)}e^{at}=e^{at}\frac{1}{p(D+a)}[/tex].

Also, [tex]p(D)\frac{1}{p(D)}e^{at}=e^{at}[/tex]

Then somehow I am meant to show that p(D)x(some value)=[tex]e^{at}[/tex] but I'm not sure of what to do from here.

Also, because n>m, p(a) does not equal 0.

The Attempt at a Solution

As above.Your help is very much appreciated!

 

[mighty2000]

Thank you for your question. Let me first clarify the notation used in the problem. The operator p(D) represents a polynomial function of the derivative operator D, and the degree of this polynomial is denoted by n. The variable t represents the independent variable of the polynomial function. The notation p^{(m)}(t) represents the m^th derivative of the polynomial p(t).

Now, let's begin the solution. We are given that a is a m-fold root of p(t)=0, which means that p(a)=p'(a)=...=p^{(m-1)}(a)=0, but p^{(m)}(a) is not equal to 0. This information is crucial in our proof.

Using the exponential shift formula, we have:

\frac{1}{p(D)}e^{at}=e^{at}\frac{1}{p(D+a)}

Now, substituting t=0 in the above equation, we get:

\frac{1}{p(D)}=e^{a\times 0}\frac{1}{p(D+a)}

Simplifying this, we get:

\frac{1}{p(D)}=\frac{1}{p(D+a)}

Multiplying both sides by p(D), we get:

1=p(D+a)p(D)

Now, let us apply the operator p(D) on both sides of the given equation:

p(D)\frac{1}{p(D)}e^{at}=p(D)e^{at}\frac{1}{p(D+a)}

Using the property of the operator p(D) on the left side, we get:

e^{at}=p(D+a)e^{at}\frac{1}{p(D+a)}

Simplifying this, we get:

e^{at}=e^{at}

Therefore, we have shown that p(D)e^{at}=e^{at}. Now, let's substitute t=0 in this equation:

p(D)e^{a\times 0}=e^{a\times 0}

Simplifying this, we get:

p(D)=1

Now, let us substitute t=0 in the given equation:

\frac{1}{p(D)}e^{a\times 0}=\frac{1}{p^{(m)}(a)}0^me^{a\times 0}

Simplifying this, we get:

\frac{1}{p(D)}=\frac{1