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a.) Is V closes under scalar multiplication? Explain

a.) Is V closes under scalar multiplication? Explain

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Solve the following problems and show your complete solutions. Write it on a paper and do not type your answer.
1.) Let V be the set of all polynomials of exactly degree 2 with the definitive of addition and scalar multiplication as in example 6. EXAMPLE 6 Another source of examples are sets of polynomials; therefore, we recall some well-known facts about such functions. A polynomial (in 7) is a function that is expressible as p(t) =ant"+an-it +...+at+ao. where a, a.....a, are real numbers and n is a nonnegative integer. If a,, 0. then p(t) is said to have degree n. Thus the degree of a polynomial is the highest power of a term having a nonzero coefficient; p(t) = 2r + 1 has degree 1, and the constant polynomial p(t) = 3 has degree 0. The zero polynomial, denoted by 0. has no degree. We now let P, be the set of all polynomials of degree ≤n together with the zero polynomial. If p(t) and q (1) are in P, we can write p(t) =ant" +an-1" + +a₁ + ao and We define p(1)q(r) as p(1) @q(t) = (an+ba)t" + (a-1 +ba-11-1+...+(a₁ + bi)t + (ao+bo). If e is a scalar, we also define e p(t) as q(t) = bat" + bn-1t+...+bit + bo. cop(t) = (can)t" + (can-1) ++(cap)t + (cap). We now show that P, is a vector space. Let p (1) and q (1), as before, be elements of P; that is, they are polynomials of degreen or the zero polynomial. Then the previous definitions of the operations and show that p(1)q(t) and c O p(1), for any scalar c, are polynomials of degreen or the zero polynomial. That is, p(t)q(t) and c O p(t) are in P, so that (a) and (b) in Definition 4.4 hold. To verify property (1), we observe that q (1) p(1) = (b + a₂)" + (b₁-1+an-1)"1++ (b₁ + a₁)t + (ao+bo). and since a, + b = b; +a; holds for the real numbers, we conclude that p(t) Ⓡ q(t) = q (1) p(1). Similarly, we verify property (2). The zero polynomial is the element 0 needed in property (3). If p(r) is as given previously, then its negative, -p(1), is -ant"-a-1-1-at-ao. We shall now verify property (6) and will leave the verification of the remaining properties to the reader. Thus (c+d) Ⓒ p(t) = (c +d)ant" + (c+d)an-it-++(c+d)ayt +(c+d)ao = cant" + dant" + can-1"-1 +dan-1-1+...+cart + dayt+cao + dao = c(ant" an-1"-+ + a₁ + a) +d(ant"+an-it- + + a₁t+ao) =c0p(1) dop(t). a.) Is V closes under scalar multiplication? Explain
Transcribed Image Text:1.) Let V be the set of all polynomials of exactly degree 2 with the definitive of addition and scalar multiplication as in example 6. EXAMPLE 6 Another source of examples are sets of polynomials; therefore, we recall some well-known facts about such functions. A polynomial (in 7) is a function that is expressible as p(t) =ant"+an-it +...+at+ao. where a, a.....a, are real numbers and n is a nonnegative integer. If a,, 0. then p(t) is said to have degree n. Thus the degree of a polynomial is the highest power of a term having a nonzero coefficient; p(t) = 2r + 1 has degree 1, and the constant polynomial p(t) = 3 has degree 0. The zero polynomial, denoted by 0. has no degree. We now let P, be the set of all polynomials of degree ≤n together with the zero polynomial. If p(t) and q (1) are in P, we can write p(t) =ant" +an-1" + +a₁ + ao and We define p(1)q(r) as p(1) @q(t) = (an+ba)t" + (a-1 +ba-11-1+...+(a₁ + bi)t + (ao+bo). If e is a scalar, we also define e p(t) as q(t) = bat" + bn-1t+...+bit + bo. cop(t) = (can)t" + (can-1) ++(cap)t + (cap). We now show that P, is a vector space. Let p (1) and q (1), as before, be elements of P; that is, they are polynomials of degreen or the zero polynomial. Then the previous definitions of the operations and show that p(1)q(t) and c O p(1), for any scalar c, are polynomials of degreen or the zero polynomial. That is, p(t)q(t) and c O p(t) are in P, so that (a) and (b) in Definition 4.4 hold. To verify property (1), we observe that q (1) p(1) = (b + a₂)" + (b₁-1+an-1)"1++ (b₁ + a₁)t + (ao+bo). and since a, + b = b; +a; holds for the real numbers, we conclude that p(t) Ⓡ q(t) = q (1) p(1). Similarly, we verify property (2). The zero polynomial is the element 0 needed in property (3). If p(r) is as given previously, then its negative, -p(1), is -ant"-a-1-1-at-ao. We shall now verify property (6) and will leave the verification of the remaining properties to the reader. Thus (c+d) Ⓒ p(t) = (c +d)ant" + (c+d)an-it-++(c+d)ayt +(c+d)ao = cant" + dant" + can-1"-1 +dan-1-1+...+cart + dayt+cao + dao = c(ant" an-1"-+ + a₁ + a) +d(ant"+an-it- + + a₁t+ao) =c0p(1) dop(t). a.) Is V closes under scalar multiplication? Explain
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