Chapter 9: Problem 13
If \(U\) and \(W\) denote the subspaces of even and odd polynomials in \(\mathbf{P}_{n}\), respectively, show that \(\mathbf{P}_{n}=U \oplus W\). (See Exercise 6.3.36.)
Short Answer
Expert verified
Polynomials in \(\mathbf{P}_n\) can be uniquely written as a sum of even and odd polynomials; hence, \(\mathbf{P}_n = U \oplus W\).
Step by step solution
01
Understand Subspaces U and W
Subspaces of polynomials are given: \(U\) consists of even polynomials and \(W\) of odd polynomials. A polynomial \(p(x)\) is even if \(p(x) = p(-x)\) and odd if \(p(x) = -p(-x)\). We want to prove that any polynomial in \(\mathbf{P}_{n}\) can be written uniquely as a sum of an even and an odd polynomial.
02
Consider any polynomial in \(\mathbf{P}_{n}\)
Let \(p(x)\) be any polynomial in \(\mathbf{P}_{n}\). We can always express \(p(x)\) as a sum of its even and odd parts: \(p(x) = e(x) + o(x)\) where \(e(x) = \frac{p(x) + p(-x)}{2}\) and \(o(x) = \frac{p(x) - p(-x)}{2}\).
03
Verify even and odd components
Check that \(e(x)\) is an even polynomial: \(e(x) = \frac{p(x) + p(-x)}{2}\). It follows that \(e(-x) = \frac{p(-x) + p(x)}{2} = \frac{p(x) + p(-x)}{2} = e(x)\), confirming \(e(x)\) is even. Similarly, check that \(o(x)\) is odd: \(o(x) = \frac{p(x) - p(-x)}{2}\). Then \(o(-x) = \frac{p(-x) - p(x)}{2} = -\frac{p(x) - p(-x)}{2} = -o(x)\), confirming \(o(x)\) is odd.
04
Uniqueness of Decomposition
Suppose \(p(x) = e_1(x) + o_1(x) = e_2(x) + o_2(x)\) with \(e_1, e_2\) even and \(o_1, o_2\) odd. Then, \(e_1(x) - e_2(x) = o_2(x) - o_1(x)\). An even function equals an odd function only if both are zero. Therefore, \(e_1(x) = e_2(x)\) and \(o_1(x) = o_2(x)\), proving uniqueness.
05
Conclusion
We have expressed any polynomial \(p(x)\) in \(\mathbf{P}_{n}\) as a sum \(p(x) = e(x) + o(x)\), where \(e(x)\) is even and \(o(x)\) is odd, with this representation being unique. Thus, \(\mathbf{P}_{n} = U \oplus W\).
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
Polynomials can be much easier to understand when we classify them into types based on their behavior. One type is the *even polynomial*. An even polynomial is a function where the same value results when you substitute a number and its negative into the polynomial. Mathematically, we describe it as:
This symmetry implies that it "mirrors" itself perfectly across the vertical axis.
To get the even part of a polynomial \( p(x) \), you calculate \( e(x) = \frac{p(x) + p(-x)}{2} \).
- For an even polynomial, if we have a function \( p(x) \), \( p(-x) = p(x) \).
This symmetry implies that it "mirrors" itself perfectly across the vertical axis.
To get the even part of a polynomial \( p(x) \), you calculate \( e(x) = \frac{p(x) + p(-x)}{2} \).
- Why this works: By adding \( p(x) \) and \( p(-x) \) and then dividing by 2, we cancel out any terms that change signs, leaving us with an even function.
Odd Polynomials
Unlike their even counterparts, *odd polynomials* behave differently when swapping a number for its negative. Their defining feature is:
On the graph, they appear to flip over the x-axis and y-axis simultaneously to land back on the same position.
To extract the odd part of a polynomial \( p(x) \), calculate \( o(x) = \frac{p(x) - p(-x)}{2} \).
- For an odd polynomial \( p(x) \), \( p(-x) = -p(x) \).
On the graph, they appear to flip over the x-axis and y-axis simultaneously to land back on the same position.
To extract the odd part of a polynomial \( p(x) \), calculate \( o(x) = \frac{p(x) - p(-x)}{2} \).
- Calculating in this way nullifies elements that remain the same, emphasizing those that flip in sign.
Subspaces
In mathematics, *subspaces* are a critical concept when discussing structures such as polynomials. A subspace is essentially a subset of a larger space that is itself a space, satisfying certain conditions.
Understanding subspaces implies recognizing how these distinct parts (even and odd) contribute to the entirety of the polynomial structure.
- A subspace must contain the zero vector (or zero polynomial in polynomial terms).
- It must be closed under addition, meaning adding any two elements in the subspace results in another element in this subspace.
- The subspace must be closed under scalar multiplication; multiplying any element in the subspace by a scalar results in another element in the subspace.
Understanding subspaces implies recognizing how these distinct parts (even and odd) contribute to the entirety of the polynomial structure.
Polynomial Uniqueness
*Polynomial uniqueness* refers to the interesting property that any polynomial can be uniquely divided into its even and odd components. In other words, each polynomial has a distinct "fingerprint" made up of an even part and an odd part.
This concept means:
This principle assures that the decomposition of any polynomial into even and odd parts is definitive and conclusive, showing Polynomial substructure's beauty and precision.
This concept means:
- If you express a polynomial \( p(x) \) as \( p(x) = e(x) + o(x) \), where \( e(x) \) is an even polynomial and \( o(x) \) is an odd polynomial, this representation is one-of-a-kind.
- If another form \( e_1(x) \) and \( o_1(x) \) claims to break down the same polynomial but does not match \( e(x) \) and \( o(x) \), it implies \( e_1(x) = e(x) \) and \( o_1(x) = o(x) \).
This principle assures that the decomposition of any polynomial into even and odd parts is definitive and conclusive, showing Polynomial substructure's beauty and precision.