Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A polynomial \(p(x)\) is called even if \(p(-x)=p(x)\) and odd if \(p(-x)=-p(x)\). Let \(E_{n}\) and \(O_{n}\) denote the sets of even and odd polynomials in \(\mathbf{P}_{n}\). a. Show that \(E_{n}\) is a subspace of \(\mathbf{P}_{n}\) and find \(\operatorname{dim} E_{n}\). b. Show that \(O_{n}\) is a subspace of \(\mathbf{P}_{n}\) and find \(\operatorname{dim} O_{n}\).

Short Answer

Expert verified
\(E_n\) and \(O_n\) are subspaces with dimensions \(\lfloor \frac{n}{2}\rfloor + 1\) and \(\lfloor \frac{n+1}{2}\rfloor\) respectively.

Step by step solution

01

Define Polynomial Space

The set \( \mathbf{P}_{n} \) denotes the space of all polynomials of degree \( n \) or less. This space has dimension \( n+1 \) because a polynomial of degree \( n \) can be expressed as \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \). A basis for \( \mathbf{P}_{n} \) is given by \( \{ 1, x, x^2, \ldots, x^n \} \).
02

Define Even and Odd Polynomials

A polynomial \( p(x) \) is even if \( p(-x) = p(x) \) and odd if \( p(-x) = -p(x) \). This means even polynomials are symmetric with respect to the y-axis, while odd polynomials are symmetric with respect to the origin.
03

Prove \(E_{n}\) is a Subspace

To show that \(E_{n}\) is a subspace of \(\mathbf{P}_{n}\), we need to check: (a) the zero polynomial is in \(E_{n}\), (b) the sum of any two elements in \(E_{n}\) is also in \(E_{n}\), and (c) the scalar multiple of any element in \(E_{n}\) is also in \(E_{n}\). Since \(0(-x) = 0(x)\), the zero polynomial is even. If \(p(x), q(x) \, \in E_{n}\), then \((p + q)(-x) = p(-x) + q(-x) = p(x) + q(x)\) so \(p(x) + q(x)\) is in \(E_{n}\), and for a scalar \(c\), \(cp(-x) = cp(x)\). Thus, \(E_{n}\) is a subspace.
04

Find \(\text{dim}(E_{n})\)

An even polynomial of degree \(n\) contains only terms with even exponents (like \(x^0, x^2, x^4, \ldots, x^{k}\) where \(k\) is the largest even number \(\leq n\)). Depending on whether \(n\) is odd or even, the dimension is \(\lfloor \frac{n}{2} \rfloor + 1\).
05

Prove \(O_{n}\) is a Subspace

Similarly, to show \(O_{n}\) is a subspace of \(\mathbf{P}_{n}\): (a) The zero polynomial can be considered trivially odd. (b) If \(p(x), q(x) \, \in O_{n}\), then \((p + q)(-x) = p(-x) + q(-x) = -p(x) - q(x)\) which is \(-(p(x) + q(x))\), so \(p(x) + q(x)\) is in \(O_{n}\), and (c) For a scalar \(c\), \(cp(-x) = -cp(x)\). Thus, \(O_{n}\) is also a subspace.
06

Find \(\text{dim}(O_{n})\)

An odd polynomial of degree \(n\) contains only terms with odd exponents (like \(x, x^3, x^5, \ldots, x^{k}\) where \(k\) is the largest odd number \(\leq n\)). Depending on whether \(n\) is odd or even, the dimension is \(\lfloor \frac{n+1}{2} \rfloor\).
07

Verify Total Dimension

Check that the dimensions sum to \(n+1\): \(\text{dim}(E_{n}) + \text{dim}(O_{n}) = \lfloor \frac{n}{2} \rfloor + 1 + \lfloor \frac{n+1}{2} \rfloor = n+1\), which confirms the dimensions of both \(E_{n}\) and \(O_{n}\) are correct.

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 Vaia!

Key Concepts

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

Even Polynomials
Even polynomials have a symmetry about the y-axis, meaning they look the same on both sides of this axis. A polynomial \( p(x) \) is defined as even if it satisfies the condition \( p(-x) = p(x) \). This characteristic is similar to how even numbers behave, where flipping the sign does not change the result. Examples of even polynomials include constant terms or expressions like \( x^2 \), \( x^4 \), and similar power functions.

This behavior originates from the power of \( x \). If the power of \( x \) is an even number, the polynomial becomes symmetrical around the y-axis. This happens because raising a negative number to an even power results in a positive number, just as raising a positive number does. Thus, even polynomials leverage this property, contributing to many symmetrical designs in geometry and algebra.
Odd Polynomials
Odd polynomials showcase the contrast to even ones, with symmetry about the origin. This type of symmetry means that rotating the graph 180 degrees around the origin doesn't change its appearance. A polynomial \( p(x) \) is odd if \( p(-x) = -p(x) \). This anti-symmetry pattern shares a connection with how odd numbers behave, where reversing the sign gives the negative of the original number.

Typical examples of odd polynomials involve terms like \( x \), \( x^3 \), \( x^5 \), and so on. Odd-numbered powers of \( x \) create an effect where the negative input is turned into a negative output, mirroring the input-output relationship. This behavior results in graphs that pass through the origin and helps in understanding phenomena that change direction or have opposing views. Recognizing odd polynomials can be crucial to respecting different symmetry properties in algebraic tasks.
Subspace
In the realm of polynomials, a subspace is a smaller, selected set of polynomials within a larger one, adhering to certain conditions. A subset of a polynomial space becomes a subspace if it includes the zero polynomial, if the sum of any two polynomials in it is also part of it, and if a scalar multiple of any polynomial within is included in it.

Both the sets of even and odd polynomials (denoted as \( E_n \) and \( O_n \)) fit this description as subspaces of \( \mathbf{P}_n \), the space of all polynomials of degree \( n \) or less. Confirming \( E_n \) as a subspace involves ensuring every even polynomial behaves like \( e(x) \) under the defined constraints. Likewise, \( O_n \) being a subspace requires the same, but with properties unique to odd polynomials.
  • Includes zero polynomial.
  • Sum of two elements remains in the set.
  • Scalar multiple stays within the set.
Polynomial Dimension
The concept of dimension provides insight into how complex or structured a space can be, indicating the number of vectors needed to form a basis. For polynomial spaces, this corresponds to the highest degree of a polynomial where each term can be represented independently.

In the space \( \mathbf{P}_n \), the dimension is \( n+1 \) since it can accommodate polynomials having terms from \( x^0 \) to \( x^n \). With even polynomials like those in \( E_n \), only terms with even degrees are considered, making their dimension \( \lfloor \frac{n}{2} \rfloor + 1 \). On the other hand, odd polynomial space \( O_n \) relies on terms with odd degrees, resulting in dimensions of \( \lfloor \frac{n+1}{2} \rfloor \).

This dichotomy in dimensions aligns with the nature of combinations that even and odd polynomials create, ultimately ensuring that their sums equal the total dimension of the general polynomial space.

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

Consider the differential equation \(f^{\prime}+a f^{\prime}+b f=g,\) where \(g\) is some fixed function. Assume that \(f_{0}\) is one solution of this equation. a. Show that the general solution is \(c f_{1}+d f_{2}+f_{0}\), where \(c\) and \(d\) are constants and \(\left\\{f_{1}, f_{2}\right\\}\) is any basis for the solutions to \(f^{\prime \prime}+a f^{\prime}+b f=0\). b. Find a solution to \(f^{\prime \prime}+f^{\prime}-6 f=2 x^{3}-x^{2}-2 x\).

Which of the following functions lie in \(\operatorname{span}\left\\{\cos ^{2} x, \sin ^{2} x\right\\} ?(\) Work in \(\mathbf{F}[0, \pi] .)\) a. \(\cos 2 x\) b. 1 c. \(x^{2}\) d. \(1+x^{2}\)

Use Taylor's theorem to derive the binomial theorem: $$(1+x)^{n}=\left(\begin{array}{l}n \\\0\end{array}\right)+\left(\begin{array}{l}n \\\1\end{array}\right) x+\left(\begin{array}{l}n \\\2\end{array}\right) x^{2}+\cdots+\left(\begin{array}{l}n \\\n\end{array}\right) x^{n}$$\ Here the binomial coefficients \(\left(\begin{array}{c}n \\\ r\end{array}\right)\) are defined by $$\left(\begin{array}{l}n \\\r\end{array}\right)=\frac{n !}{r !(n-r) !}$$ where \(n !=n(n-1) \cdots 2 \cdot 1\) if \(n \geq 1\) and \(0 !=1\).

Show that each of the following sets of vectors is independent. a. \(\left\\{1+x, 1-x, x+x^{2}\right\\}\) in \(\mathbf{P}_{2}\) b. \(\left\\{x^{2}, x+1,1-x-x^{2}\right\\}\) in \(\mathbf{P}_{2}\) \(\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{rr}0 & 0 \\ 1 & -1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right]\right\\}\) \(\left\\{\left[\begin{array}{ll}1 & 1 \\ 1 & 0 \\ \text { in } & \mathbf{M}_{22}\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\right\\}\)

\(\mathbf{}\) If \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right\\}\) is independent, show that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{1}+\mathbf{v}_{2}, \ldots, \mathbf{v}_{1}+\mathbf{v}_{2}+\cdots+\mathbf{v}_{k}\right\\}\) is also independent.

See all solutions

Recommended explanations on Economics Textbooks

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.

Sign-up for free