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Prove that the differentiation operator \(\frac{d}{d x}\) in the space of polynomials of degree less than \(n\) is nilpotent.

Short Answer

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Question: Prove that the differentiation operator is nilpotent in the space of polynomials of degree less than n. Answer: To prove that the differentiation operator is nilpotent, apply the differentiation operator to a generic polynomial P(x) of degree less than n multiple times, and observe the degree of the resulting polynomial. After applying the differentiation operator n times, the polynomial becomes zero in the space of polynomials of degree less than n. Thus, proving that the differentiation operator is nilpotent.

Step by step solution

01

Representation of a Polynomial

First, let's represent a generic polynomial P(x) of degree less than \(n\) as: $$ P(x)=a_0+a_1 x+a_2 x^2+\cdots+a_{n-1} x^{n-1} $$ Where \(a_0, a_1,\ldots, a_{n-1}\) are constants.
02

Apply the differentiation operator once

Lets apply the differentiation operator to P(x) once: $$ \frac{d P(x)}{d x} = a_1 + 2a_2 x + 3a_3 x^2 + \cdots + (n-1)a_{n-1} x^{n-2} $$ Now the degree of the derivative is one less than that of P(x), i.e., degree less than \((n-1)\).
03

Apply the differentiation operator repeatedly

We continue applying the differentiation operator to the resulting polynomial k times: $$ \frac{d^k P(x)}{d x^k} = \binom{n-k}{1} a_{n-k} x^{n-k-1} +\cdots + (n-k-1) \cdot 2\cdot 1\cdot a_{n-1} $$ The degree of the polynomial continues to decrease as we apply the differentiation operator.
04

Apply the differentiation operator n times

Apply the differentiation operator to P(x) n times: $$ \frac{d^n P(x)}{d x^n} = n(n-1)(n-2)\cdots 1\cdot a_n + \text{ higher degree terms} $$ Since P(x) has a degree less than \(n\), all higher degree terms vanish, and \(a_n\) is also zero. Therefore, we have: $$ \frac{d^n P(x)}{d x^n} = 0 $$ This means that the differentiation operator \(\frac{d}{d x}\) in the space of polynomials of degree less than \(n\), when applied \(n\) times, becomes a zero operator, proving that it is nilpotent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Ordinary Differential Equations (ODEs)
Ordinary differential equations (ODEs) are equations that involve functions and their derivatives. They describe the relationship between a function and its rates of change, which makes them crucial in modeling dynamic systems like population growth, heat transfer, and motion.

In the context of our exercise, understanding ODEs is not just about solving them, but also exploring the properties of differentiation operations within the space of polynomials. Since differentiation is a foundational operation in calculus used to derive an ODE, the concept of a nilpotent differentiation operator gives insight into the behavior of these equations when the differentiation operation is applied consecutively.
The Process of Polynomial Differentiation
Polynomial differentiation is a fundamental concept in calculus, involving the application of the differentiation operator, typically denoted as \( \frac{d}{dx} \), to a polynomial function. The rules for differentiating polynomials are straightforward, as each term of the polynomial, having the form \( a_n x^n \), is differentiated individually.

Rules of Polynomial Differentiation

The power rule, which states that \( \frac{d}{dx} (a_n x^n) = n \times a_n x^{n-1} \), is repeatedly used. When the differentiation operator is applied to any polynomial, the degree of the resulting polynomial reduces by one. This process is central to proving the nilpotency of the differentiation operator in this context - after a finite number of differentiations, specifically \( n \) times, we eventually reach a constant or the zero function.
Foundations of Mathematical Proofs
Mathematical proofs are structured arguments that establish the truth of a mathematical statement. Crafting proofs requires a deep understanding of logic and mathematical principles.

Proving the nilpotent nature of the differentiation operator on polynomials less than degree \( n \) is an exercise in mathematical proof. It involves setting up the premise that applying the operator successively will eventually lead to a polynomial of degree zero, which when differentiated, yields zero. The conclusion from these sequential operations provides a logical and structured demonstration of nilpotency, which is a testament to the importance of mathematical proofs in verifying conceptual understanding and properties of mathematical operations.

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Most popular questions from this chapter

Suppose the Wronskian of two functions is identically zero. Does it follow that these functions are linearly dependent?

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Prove that the derivative has the following properties: $$ \frac{d}{d t}\left(\varphi_{1}+\varphi_{2}\right)=\frac{d \varphi_{1}}{d t}+\frac{d \varphi_{2}}{d t}, \quad \frac{d}{d t}\left(\varphi_{1} \varphi_{2}\right)=\frac{d \varphi_{1}}{d t} \varphi_{2}+\varphi_{1} \frac{d \varphi_{2}}{d t} $$ In particular the derivative of a polynomial with complex coefficients is given by the same formula as in the case of real coefficients. If \(n>1\), it is impossible to multiply two curves with values in \(C^{n}\), How ever, since \(C^{n}\) is a \(C\)-module, it is possible to multiply a curve \(\varphi: I \rightarrow C^{n}\) by a function \(f: I \rightarrow C\) $$ (f \varphi)(t)=f(t) \varphi(t) $$

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