In Exercises 45-51, we define the first difference 8f of a function f by Sf (x) = f(x + 1) - f(x). then 8P = (x +1). Then apply Exercise 48 to conclude that п(n + 1) 1+2+3+.+n = 45. Show that if f(x) = x2, then 8f(x) = 2.x + 1. Calculate 8f for f(x)= x and f (x) = x'. 2 46. Show that 8(10*)= 9· 10 and, more generally, that 8(b*)= (b – 1)b*. 50. Calculate 8(x³), 8(x²), and 8(x). Then find a polynomial P of degree 3 such that 8P = (x + 1)2 and P(0) = 0. Conclude that P(n) = 12 + 22 +.+ n². 47. Show that for any two functions f and g, 8(ƒ + g) = 8f + 8g and 8(c f) = c8(f ), where c is any constant. 51. This exercise combined with Exercise 48 shows that for all whole numbers k, there exists a polynomial P satisfying Eq. (1). The solution requires the Binomial Theorem and proof by induction (see Appendix C). (a) Show that 8(x*+') = (k + 1)x* + · . . . where the dots indicate terms involving smaller powers of x. (b) Show by induction that there exists a polynomial of degree k + 1 with leading coefficient 1/(k + 1): 48. Suppose we can find a function P such that 8P(x)= (x+ 1)* and P(0) = 0. Prove that P(1) = 1*, P(2) = 1* + 2*, and, more generally, for every whole number n, P(n) = 1* + 2* + . ..+ n* k+l +.. k +1 49. Show that if P(x) = x(x + 1) Р(х) %3D such that 8P = (x + 1)* and P(0) = 0.
In Exercises 45-51, we define the first difference 8f of a function f by Sf (x) = f(x + 1) - f(x). then 8P = (x +1). Then apply Exercise 48 to conclude that п(n + 1) 1+2+3+.+n = 45. Show that if f(x) = x2, then 8f(x) = 2.x + 1. Calculate 8f for f(x)= x and f (x) = x'. 2 46. Show that 8(10*)= 9· 10 and, more generally, that 8(b*)= (b – 1)b*. 50. Calculate 8(x³), 8(x²), and 8(x). Then find a polynomial P of degree 3 such that 8P = (x + 1)2 and P(0) = 0. Conclude that P(n) = 12 + 22 +.+ n². 47. Show that for any two functions f and g, 8(ƒ + g) = 8f + 8g and 8(c f) = c8(f ), where c is any constant. 51. This exercise combined with Exercise 48 shows that for all whole numbers k, there exists a polynomial P satisfying Eq. (1). The solution requires the Binomial Theorem and proof by induction (see Appendix C). (a) Show that 8(x*+') = (k + 1)x* + · . . . where the dots indicate terms involving smaller powers of x. (b) Show by induction that there exists a polynomial of degree k + 1 with leading coefficient 1/(k + 1): 48. Suppose we can find a function P such that 8P(x)= (x+ 1)* and P(0) = 0. Prove that P(1) = 1*, P(2) = 1* + 2*, and, more generally, for every whole number n, P(n) = 1* + 2* + . ..+ n* k+l +.. k +1 49. Show that if P(x) = x(x + 1) Р(х) %3D such that 8P = (x + 1)* and P(0) = 0.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:In Exercises 45-51, we define the first difference 8f of a function f by
Sf (x) = f(x + 1) - f(x).
then 8P = (x +1). Then apply Exercise 48 to conclude that
п(n + 1)
1+2+3+.+n =
45. Show that if f(x) = x2, then 8f(x) = 2.x + 1. Calculate 8f for
f(x)= x and f (x) = x'.
2
46. Show that 8(10*)= 9· 10 and, more generally, that 8(b*)=
(b – 1)b*.
50. Calculate 8(x³), 8(x²), and 8(x). Then find a polynomial P
of degree 3 such that 8P = (x + 1)2 and P(0) = 0. Conclude that
P(n) = 12 + 22 +.+ n².
47. Show that for any two functions f and g, 8(ƒ + g) = 8f + 8g and
8(c f) = c8(f ), where c is any constant.
51. This exercise combined with Exercise 48 shows that for all whole
numbers k, there exists a polynomial P satisfying Eq. (1). The solution
requires the Binomial Theorem and proof by induction (see Appendix C).
(a) Show that 8(x*+') = (k + 1)x* + · . . . where the dots indicate terms
involving smaller powers of x.
(b) Show by induction that there exists a polynomial of degree k + 1 with
leading coefficient 1/(k + 1):
48. Suppose we can find a function P such that 8P(x)= (x+ 1)* and
P(0) = 0. Prove that P(1) = 1*, P(2) = 1* + 2*, and, more generally, for
every whole number n,
P(n) = 1* + 2* + . ..+ n*
k+l +..
k +1
49. Show that if
P(x) =
x(x + 1)
Р(х) %3D
such that 8P = (x + 1)* and P(0) = 0.
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