Polynomial differential operators

In summary: Substituting this value in the original equation, we get:\frac{1}{p^{(m)}(a)}t^me^{at}Therefore, in summary, we have shown that \frac{1}{p(D)}e^{at}=\frac{1}{p^{(m)}(a)}t^me^{at}, where p(D) is a polynomial D operator of degree n>m and a is a m-fold root of p(t)=0.
  • #1
sassie
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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!
 
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  • #2

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
 

1. What is a polynomial differential operator?

A polynomial differential operator is a mathematical operator that involves differentiating a polynomial function. It can be written in the form P(D), where D represents the differential operator and P is a polynomial function.

2. How is a polynomial differential operator different from a regular polynomial?

A polynomial differential operator acts on a function, while a regular polynomial is a function itself. The polynomial differential operator involves differentiation, while a regular polynomial only involves variables and coefficients.

3. What are the applications of polynomial differential operators?

Polynomial differential operators are commonly used in calculus and differential equations to solve problems involving rates of change and continuous functions. They also have applications in physics, engineering, and other scientific fields.

4. How do you solve a polynomial differential equation?

To solve a polynomial differential equation, you can use techniques such as separation of variables, substitution, or integration. The specific method used will depend on the form of the equation and the type of differential operator involved.

5. Can polynomial differential operators be used with non-polynomial functions?

Yes, polynomial differential operators can be used with non-polynomial functions, as long as the function can be written as a combination of polynomial functions. This is because the operator acts on the function as a whole, rather than individual terms within the function.

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