THE DIFFERENCE CALCULUS 25 Setting k = 0 in the original function and its differences allows us to conclude that Am fo ат т 3 0,1, ....., п. (1.189) m! To illustrate the use of this theorem, consider the function fk = k4. (1.190) Now Afk = 4k3 + 6k? + 4k + 1, A² fk = 12k2 + 24k + 14, (1.191) A³ fk = 24k + 36, Aª fk = 24, %3D and fo = 0, Afo= 1, A² fo =14, A³ fo = 36, A* fo = 24. (1.192) %3D Therefore, from equation (1.189), ao = 0, a1 = 1, а2 3D 7, аз — 6, a4 = 1, || and fk = k4 has the factorial polynomial representation fk = k4 = k(1) + 7k(2) + 6k(3) + k(4). (1.194
THE DIFFERENCE CALCULUS 25 Setting k = 0 in the original function and its differences allows us to conclude that Am fo ат т 3 0,1, ....., п. (1.189) m! To illustrate the use of this theorem, consider the function fk = k4. (1.190) Now Afk = 4k3 + 6k? + 4k + 1, A² fk = 12k2 + 24k + 14, (1.191) A³ fk = 24k + 36, Aª fk = 24, %3D and fo = 0, Afo= 1, A² fo =14, A³ fo = 36, A* fo = 24. (1.192) %3D Therefore, from equation (1.189), ao = 0, a1 = 1, а2 3D 7, аз — 6, a4 = 1, || and fk = k4 has the factorial polynomial representation fk = k4 = k(1) + 7k(2) + 6k(3) + k(4). (1.194
Calculus: Early Transcendentals
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Chapter1: Functions And Models
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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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Explain the deretmaine
Transcribed Image Text:l stc ksa
2:26 AM
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Q5 = (1+4 · 5)(3)
(21)(3) = (21)(20)(19) = 7980,
(1.185)
we would obtain the indicated wrong value, 7980, rather than the correct
value, 4641. So, be careful!
1.7.2 Example B
49
Prove the following (Newton's theorem): If fk is a polynomial of the nth
degree, then it can be written in the form
fk = fo +
1!
Afo .(1) +
A" fo 1(n).
2!
+...+
n!
(1.186)
Assume that fk has the representation
fk
= ao + a1k() + a2k(²) + ·..
+ ank(n).
(1.187)
where ao, a1, ..., an are constants. Differencing fk n times gives
= a1 + 2a2k(1)
+ 3azk(2)
+ па,k(п-1),
+...
A? fk = 2.1. a2 + 3 · 2 · azk(1)
+..+n(n – 1)ank(n-2).
(1.188)
A" fk = ann(n – 1) ... (1).
-
THE DIFFERENCE CALCULUS
25
Setting k
= 0 in the original function and its differences allows us to conclude
that
Am fo
ат
0, 1,
(1.189)
m =
...., n.
т!
To illustrate the use of this theorem, consider the function
fk = k4.
(1.190)
Now
Afk = 4k° + 6k² + 4k + 1,
A² fk = 12k? + 24k + 14,
(1.191)
A³ fk = 24k + 36,
A* fk = 24,
and
fo = 0, Afo = 1,
A² fo = 14, A³ fo = 36, A fo = 24.
(1.192)
Therefore, from equation (1.189),
ао — 0,
a1 = 1,
a2 = 7,
аз — 6,
a4
1,
and fk
k4 has the factorial polynomial representation
fk = k4 = k(1) + 7k(2) + 6k(3) + k(4).
(1.194)
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