show me the steps of determine blue

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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show me the steps of determine blue

8.6.6
Processes Involving A and D
This section presents results illustrating certain subtleties in the applications
of the operators A and D. Let f(x) be a function of the continuous variable
x, with
df (x)
Df(x)
(8.249a)
dx
D'f(x) = F(x)+ A, A = constant
(8.249b)
Af(x) = f(x + 1) – f(x).
(8.249c)
It can be shown that
(AD-1)m
= A" D-m
(8.250)
(See Jordan 1979, pp. 200–201.) Note that the result given by equation (8.250)
is nontrivial, in the sense that A and D-1 do not commute, i.e.,
-1
AD-1
(8.251)
400
Difference Equations
This can be seen by calculating (AD-1)f(x) and (D-'A)f(x):
(AD-1)f(x) = A[D-lf(x)] = A[F(x) + A] = F(x+1) – F(x), (8.252)
and
(D-'A)f(x) = D¯f(x+ 1) – f(x)] = F(x +1) – F(x) + A1,
(8.253)
where A1 is an arbitrary constant.
The following gives the results of letting AD-1 operate on several special
f (x)'s:
(i) f(x) = 1
(: (AD-1)1= A(D¬l1) = A(x + A) = (x + 1) – x = 1.
(8.254)
(ii) ƒ(x)= x
C.(AD ')e = A(D 'x) =- A+A)
2
C-(4)
(x + 1)² – (6)
= x +
(8.255)
(iii) f(x) = e®
:. (AD-1)e" = A(D-'e")= A(e* + A)
= e"+1 - eº = (e – 1)eª.
(8.256)
(iv) f(x)=
.: (AD-1)
- + A
= -A
1
1
1
(8.257)
+1
x(x + 1)
x + 1)2
Carrying out the same calculations for the operator DA-1 on these f(x)
gives:
(i) ƒ(x) = 1
(: (DA-1)1 = 1+ p(x).
(8.258)
(ii)
(IDA ")= ==-+ ple).
+ p(x).
2
(8.259)
(ii)
1
(DA-1)
1
+ p(x),
(8.260)
(x + 1)2
x2
Transcribed Image Text:8.6.6 Processes Involving A and D This section presents results illustrating certain subtleties in the applications of the operators A and D. Let f(x) be a function of the continuous variable x, with df (x) Df(x) (8.249a) dx D'f(x) = F(x)+ A, A = constant (8.249b) Af(x) = f(x + 1) – f(x). (8.249c) It can be shown that (AD-1)m = A" D-m (8.250) (See Jordan 1979, pp. 200–201.) Note that the result given by equation (8.250) is nontrivial, in the sense that A and D-1 do not commute, i.e., -1 AD-1 (8.251) 400 Difference Equations This can be seen by calculating (AD-1)f(x) and (D-'A)f(x): (AD-1)f(x) = A[D-lf(x)] = A[F(x) + A] = F(x+1) – F(x), (8.252) and (D-'A)f(x) = D¯f(x+ 1) – f(x)] = F(x +1) – F(x) + A1, (8.253) where A1 is an arbitrary constant. The following gives the results of letting AD-1 operate on several special f (x)'s: (i) f(x) = 1 (: (AD-1)1= A(D¬l1) = A(x + A) = (x + 1) – x = 1. (8.254) (ii) ƒ(x)= x C.(AD ')e = A(D 'x) =- A+A) 2 C-(4) (x + 1)² – (6) = x + (8.255) (iii) f(x) = e® :. (AD-1)e" = A(D-'e")= A(e* + A) = e"+1 - eº = (e – 1)eª. (8.256) (iv) f(x)= .: (AD-1) - + A = -A 1 1 1 (8.257) +1 x(x + 1) x + 1)2 Carrying out the same calculations for the operator DA-1 on these f(x) gives: (i) ƒ(x) = 1 (: (DA-1)1 = 1+ p(x). (8.258) (ii) (IDA ")= ==-+ ple). + p(x). 2 (8.259) (ii) 1 (DA-1) 1 + p(x), (8.260) (x + 1)2 x2
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