lec30_tuesday.pdf - 1 Lecture 23 w12 1.1 Taylor and Maclaurin Series \u2022 Do the examples in Section 11.10 in the book P n \u2022 Let's consider a power

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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 dierentiate 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 coecients 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 coecients 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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