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**Unformatted text preview: **1 Lecture 23, w12 1.1 Taylor and Maclaurin Series • Do the examples in Section 11.10 in the book
P
n
• Let's consider a power series f (x) = ∞
n=0 cn (x − a) Note that f (a) = c0 :
f (x) = c0 + c1 (x − a) + c2 (x − a)2 + · · · so f (a) = c0 + c1 0 + c2 0 + · · · = c0
We know that for |x − a| < R where R is P
the radius of convergence of f (x), we
n−1
0
can dierentiate the series to get f (x) = ∞
n=1 ncn (x − a)
So
f 0 (x) = 1c1 + 2c2 (x − a) + 3c3 (x − a)2 + · · · and so f 0 (a) = 1 · c1 =⇒ c1 = f 0 (a) Next, again for |x − a| < R,
∞ X
d 0
f (x) =
f (x) =
n (n − 1) cn (x − a)n−2
dx
n=2
00 so that
f 00 (a) = 2 (2 − 1) c2 =⇒ c2 = f 00 (a)
2! These computations motivate the following theorem:
Theorem 1. If f (x) = P∞ n=0 cn (x − a)n for |x − a| < R for some R > 0, then cn = f (n) (a)
.
n! • Recall that f (0) = f , f (1) = f 0 , f (2) = f 00 , etc.
• In the other direction, if we are given a function
P∞ f (x), wen can still compute cn =
(n)
f (a) /n! and get the power series T (x) = n=0 cn (x − a) This power series is called the Taylor series of f (x) at a
If a = 0 then we also call it the Maclaurin series of f (x)
• Note that this says that if a function has a power series, then it must be the Taylor series So there is only one expression for a power series, the Taylor series
Example 1. Find the Taylor series of f (x) = ex at a = 2
Solution 1. 1 • Note that f 0 (x) = ex , f 00 (x) = ex , and in general f (n) (x) = ex
• So the Taylor coecients are
cn = e2
f (n) (2)
=
n!
n! • Hence the Taylor series at 2 is
T (x) = ∞
X
e2
n=0 n! (x − 2)n • QUESTION: Given a function f (x), is it true that f (x) = T (x), its Taylor series? In general, no
• The answer is a little complicated but we will do some examples to understand
P
n
• Let f (x) be a function and let T (x) = ∞
n=0 cn (x − a) be its Taylor series We can write
f (x) = n
X ci (x − a) + Rn (x) where Rn (x) = f (x) −
i i=0 n
X ci (x − a)i i=0 and Rn (x) is called the nth remainder
Since
∞
T (x) = X n cn (x − a) = lim n→∞ n=0 n
X ci (x − a)i i=0 therefore we should have
lim f (x) = lim n→∞ n→∞ n
X !
ci (x − a)i + Rn (x) i=0 i.e.
f (x) = T (x) + lim Rn (x)
n→∞ So it should be that f (x) = T (x) only when limn→∞ Rn (x) = 0
We state this more precisely in the following theorem
Theorem 2.
R > 0, then f (x) be a function. If limn→∞ Rn (x) = 0 for x ∈ (a − R, a + R)
f (x) = T (x), its Taylor series at a, for x ∈ (a − R, a + R).
Let • In practice to show that limn→∞ Rn (x) = 0 we use the following fact 2 for some Theorem 3.
Rn (x) =
where c is some number between a and f (n+1) (c)
(x − a)n+1
(n + 1)! x. • Additional tools for showing limn→∞ Rn (x) = 0: Show instead limn→∞ |Rn (x)| = 0
Show instead limn→∞ |Rn (x)| ≤ 0
n
Use that limn→∞ xn! = 0 for all x (e.g. use L'Hopital's rule n times)
Example 2. Find the Maclaurin series for cos x and show it equals cos x for all x.
Solution 2.
• Let f (x) = cos x
• First we nd the Maclaurien series, so a = 0
• Using the formula cn = f (n) (0) /n! we nd c0 = cos 0 = 1
d
cos x = − sin x, c1 = − sin 0 = 0
dx
d2
dx2 cos x = − cos x, c2 = − cos 0/2 = −1/2 d3
dx3 cos x = sin x, c3 = sin 0/3! = 0 d4 cos x = cos x, c4 = cos 0/4! = 1/4! dx4 • At the 4th derivative we returned to cos x again, so we see a pattern that leads us to the Maclaurien series ∞ X (−1)n
1 2 1 4
T (x) = 1 − x + x − · · · =
x2n
2
4!
(2n)!
n=0 It'll take some practice for recognizing the formula for the cn
• To show that it always equals cos x, we show that limn→∞ |Rn (x)| = 0 Absolute value is nice because we can ignore (−1)n = ±1
• By our theorem have, for c between 0 and x, (n+1) n+1 (n+1) f
f
(c) |x|
(c)
|Rn (x)| = (x − 0)n+1 =
(n + 1)!
(n + 1)!
• The work above shows that f (n+1) (c) is either ± cos c or ± sin c 3 In either case f (n+1) (c) ≤ 1 • So 1 · |x|n+1
=0
lim |Rn (x)| ≤ lim
n→∞
n→∞ (n + 1)! using two of our earlier stated tools
• This is true for all x so by our theorem f (x) = T (x) for all x Example 3. Find the Maclaurin series for e3x and show it equals e3x for all x
Solution 3.
• Let f (x) = e3x
• First we compute the Maclaurien series c0 = e3·0 = 1
d 3x
dx
e = 3e3x , c1 = 3e3·0 = 3
d
3x
dx
= 32 e3x , c2 = 32 e3·0 /2 = 32 /2
2e
So we see that the Maclaurin series is
2 ∞ X 3n
32 2
xn
T (x) = 1 + 3x + x + · · · =
2
n!
n=0
• Next we show convergence: (n+1) f 3n+1 e3c |x|n+1
(c)
n+1 lim |Rn (x)| = lim (x − 0) = lim
n→∞
n→∞ (n + 1)!
n→∞
(n + 1)!
|3x|n+1
= e3c · 0 = 0
n→∞ (n + 1)! = e3c lim • So the remainders limit to 0 for all x and hence by our theorem f (x) = T (x) for all x Example 4. Find the Maclaurin series of x2 ln (1 + x3 )
Solution 4.
• Its a fact that ln (1 + x) has Maclaurin series
ln (1 + x) = ∞
X
1 (−1)n−1 xn
n Maclaurin series like this, and the ones listed on p. 768, are important to know
Instead of memorizing them be able to compute them from their Taylor coecients
like in the earlier examples 4 • So
2 x ln 1 + x 3 =x 2 ∞
X n n−1 (−1) 1 = x5 − x8 x11
+
− ···
2
3 5 ∞ X
x3n+2
(x3 )
=
(−1)n−1
n
n
1 ...

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- Fall '19
- Maclaurin Series, Power Series, Taylor Series, cos x, Note