/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Let \(V\) be the vector space of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 6

Let \(V\) be the vector space of functions which have derivatives of all orders, and let \(D: V \rightarrow V\) be the derivative. What is the kernel of \(D ?\)

Short Answer

Expert verified
The kernel of the derivative operator \(D\) in the vector space of functions with derivatives of all orders is the set of constant functions: \(\operatorname{Ker}(D) = \{ f \in V \mid f(x) = C \text{ for some constant } C\}\).

Step by step solution

01

Finding the kernel

The kernel of an operator is defined as the space of elements which are mapped to the zero element under the operator. In this case, the operator is the derivative operator, \(D\). Recall that a function is in the kernel of \(D\) if and only if its derivative is equal to zero: $$ D(f) = f' = 0, \text{ for } f \in V. $$ Now, we need to find all functions that satisfy this condition.
02

Integrating the zero function

In the above equation, the function \(f'\) is equal to zero. To find what functions have this property, we can integrate both sides of the equation with respect to x: $$ \int f'(x) dx = \int 0 dx. $$ By the fundamental theorem of calculus, we have: $$ f(x) = C, $$ where C is the constant of integration.
03

Conclusion

We have found that any function with the property that its derivative is equal to zero must be of the form \(f(x) = C\), where \(C\) is a constant. Hence, the kernel of the derivative operator, in this case, is the set of constant functions: $$ \operatorname{Ker}(D) = \{ f \in V \mid f(x) = C \text{ for some constant } C\}. $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Vector Space
In mathematics, a vector space is a fundamental concept used to discuss spaces of functions. A vector space consists of a set of elements called vectors, along with operations of vector addition and scalar multiplication that satisfy certain axioms. These axioms ensure that vectors can be added together and multiplied by scalars in a way that mirrors familiar arithmetic.
  • Elements: The elements of a vector space can be functions, numbers, or other objects that can be systematically combined.
  • Operations: Vector addition and scalar multiplication are the operations that define how elements interact within the space.
  • Closure: For any two elements in a vector space, their sum and scalar multiple must also belong to the same space.
When discussing the vector space of functions, these functions are the vectors of the space. The space of functions with derivatives of all orders means that every function can be differentiated indefinitely. This implies a rich structure since every derivative operation should also produce functions within the same space.
Derivatives
In calculus, the derivative represents the rate of change of a function with respect to a variable. It is a crucial concept for understanding how functions behave as their inputs change. Derivatives provide a connection between algebra and geometry because they describe the slope of a function's graph.
  • Notation: The derivative of a function \(f\) is often denoted as \(f'\) or \(\frac{df}{dx}\).
  • Physical Interpretation: In physics, derivatives often represent velocities or rates of change over time.
  • Mathematical Property: A function is differentiable at a point if its derivative exists at that point.
For our exercise, the derivative operator \(D\) differentiates functions within the vector space, mapping them to other functions. The kernel of the derivative operator includes all functions that differentiate to zero, indicating no change and, hence, a constant nature.
Constant Functions
Constant functions hold a significant position in calculus as the simplest form of functions. A constant function is one that remains the same, no matter what value the independent variable takes.
  • Definition: A function \(f(x)\) is constant if \(f(x) = C\) for a constant \(C\).
  • Graph: The graph of a constant function is a horizontal line, indicating a zero slope everywhere.
  • Derivative: The derivative of a constant function is always zero, as there is no change in the function's value with respect to the independent variable.
In the given exercise, the kernel of the derivative operator is composed of these constant functions. This set of functions is important because it lays the foundational understanding of functions that do not change, which aligns with the conclusion that if a function's derivative is zero, the function itself must be a constant.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(F: V \rightarrow W\) be a linear map and assume that the image of \(F\) is all of \(W\), Assume that \(V\) and \(W\) have the same dimension \(n\). Show that the kernel of \(F\) is \(\\{O\\}\).

Let \(L: \mathbf{R}^{n} \rightarrow \mathbf{R}\) be a linear map and \(c\) a number. Show that the set \(\mathcal{S}\) consisting of all points \(A\) in \(\mathbf{R}^{n}\) such that \(L(A)>c\) is convex.

Let \(F\) be a linear map from \(\mathbf{R}^{2}\) into itself such that $$ F\left(B_{1}\right)=(1,1) \quad \text { and } \quad F\left(E_{2}\right)=(-1,2) $$ Let \(S\) be the square whose corners are at \((0,0),(1,0),(1,1)\), and \((0,1) .\) Show that the image of this square under \(F\) is a parallelogram.

(a) Let \(U, W\) be vector spaces. We let \(U \times W\) be the set of all pairs \((u, w)\) with \(u \in U\) and \(w \in W .\) If \(\left(u_{1}, w_{1}\right),\left(u_{2}, w_{2}\right)\) are such pairs, define their sum $$ \left(u_{1}, w_{1}\right)+\left(u_{2}, w_{2}\right)=\left(u_{1}+u_{2}, w_{1}+w_{2}\right) $$ If \(c\) is a number, define \(c(u, w)=(c u, c w) .\) Show that \(U \times W\) is a vector space with these definitions. What is the sero element? (b) If \(U\) has dimension \(n\) and \(W\) has dimension \(m\), whst is the dimension of \(U \times W ?\) Exhibit s basis of \(U \times W\) in terms of a basis for \(U\) and s basis for \(W\). (c) If \(U\) is a subspace of a vector space \(V\), show that the subset of \(V \times V\) consisting of all elements \((u, u)\) with \(u \in U\) is a subspace.

Let \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) be the mapping defined by \(F(x, y)=(x y, y)\). Describe the image under \(F\) of the straight line \(x=2\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.