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Taylor and Maclaurin Series solved questions
Taylor and Maclaurin Series solved questions
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CHAPTER 39Taylor and Maclaurin Series
39.1 Find the Maclaurin series of e*.
Let /(*) = **. Then f("\x) = e' for all «>0. Hence, /'"'(O) = 1 for all n & 0 . Therefore, the
Maclaurin series
39.2 Find the Maclaurin series for sin x.
Let /(;c) = sin X. Then, /(0) = sinO = 0, /'(O) = cosO= 1, /"(O) = -sin 0 = 0, /"'(O) = -cosO= -1,
and, thereafter, the sequence of values 0,1,0, — 1 keeps repeating. Thus, we obtain
39.3 Find the Taylor series for sin x about ir/4.
Let f(x) = sinx. Then /(ir/4) = sin (ir/4) = V2/2, f ' ( i r / 4 ) = cos(irf4) = V2/2, f(ir/4) = -sin (77/4) =
and, thereafter, this cycle of four values keeps repeating.
Thus, the Taylor series for sin* about
39.4 Calculate the Taylor series for IIx about 1.
Let Then, and, in general,
So /(">(1) = (-!)"«!. Thus, the Taylor series is
39.5 Find the Maclaurin series for In (1 - x).
Let/(AT) = In (1 - x). Then,/(0) = 0,/'(0) = -1,/"(0) = -1,/"'(0) = -1 -2,/ < 4 ) = -1 • 2 • 3, and, in general
/ ( ">(0) = -(„ _ i)i Thus, for n > l,/ ( / 0 (0)/n! = -1/n, and the Maclaurin series is
39.6 Find the Taylor series for In x around 2.
Let f(x) = lnx. Then, and, in general,
So /(2) = In 2, and, for n > 1, Thus, the
Taylor series is (x - 2)" = In 2 + i(x - 2) - 4(x - 2)2 + MX - 2)3 - • • •.
39.7 Compute the first three nonzero terms of the Maclaurin series for ecos".
Let f ( x ) = ecos*. Then, f'(x)=-ecos'sinx, /"(*) = e'05* (sin2 *-cos*), /'"(x) = ecos' (sin jc)(3 cos x +
1-sin 2 *), /(4)(;<:) = e<:osj:[(-sin2*)(3 + 2cosA-) + (3cosA: + l-sin 2 Ar)(cosA:-sin 2 jc)]. Thus, /(O) = e,
/'(0) = 0, f"(0) = -e, f"(0) = 0, /(4)(0) = 4e. Hence, the Maclaurin series is e(l - |*2 + |x4 + • • • ) .
340
, TAYLOR AND MACLAURIN SERIES 341
39.8 Write the first nonzero terms of the Maclaurin series for sec x.
I Let /(x) = secx. Then, /'(*) = sec x tan x, f"(x) = (sec *)(! + 2 tan 2 x), /'"(*) = (sec x tan x)(5 + 6 tan 2 x),
fw(x) = 12 sec3 x tan 2 x + (5 + 6 tan2 x)(sec3 x + tan 2 x sec x). Thus, /(O) = 1, /'(O) = 0, /"(O) = 1,
/"'(0) = 0, / ( 4 ) =5. The Maclaurin series is 1+ $x2 + &x* + • • •.
39.9 Find the first three nonzero terms of the Maclaurin series for tan x.
Let /(x) = tan;t. Then /'(*) = sec2*, /"(*) = 2 tan x sec2 x, f'"(x) = 2(sec4 x + 2 tan 2 x + 2 tan 4 x),
/ (*) = 8 (tan x sec2 x)(2 + 3 tan2 x),
(4)
/(5)(*) = 48 tan 2 x sec4 x + 8(2 + 3 tan2 *)(sec4 x + 2 tan 2 x + 2 tan 4 x).
So /(0) = 0, /'(0) = 1, f'(0) = 0, /"'(0) = 2, /< 4 >(0) = 0, / <5) (0) = 16. Thus, the Maclaurin series is
*+ 1*3 + &X3 + ••-.
39.10 Write the first three nonzero terms of the Maclaurin series for sin -1x.
Let /Ot^sirT1*. /'(*) = (I-*2)'1'2, /"(x) = *(1 - Jc2)'3'2, f'"(x) = (1 - x2ys'2(2x2 + 1).
/(4)(*) = 3*(l-*2)-7/2(2*2 + 3), /(5)(*) = 3(1 - * 2 )- 9/2 (3 + 24*2 + 8*4). Thus, /(0) = 0, /'(0) = 1,
/"(O) = 0, /'"(O) = 1, / <4> (0) = 0, /(5>(0) = 9. Hence, the Maclaurin series is x + $x3 + -jex5 + • • -.
x
39.11 If f(x) = 2 an(x - a)" for \x - a\ < r, prove that In other words, if f(x) has a power
ii -o
series expansion about a, that power series must be the Taylor series for/(*) about a.
f(a) = aa. It can be shown that the power series converges uniformly on | * - a | < p < r , allowing
differentiation term by term:
n(n - 1) • • • [/I - (A: - 1)K(* - «)""*- M we let x = a, fm(a) = k(k-l) 1 • ak = k\ • ak. Hence,
39.12 Find the Maclaurin series for
By Problem 38.34, we know that for |*| <!• Hence, by
Problem 39.11, this must be the Maclaurin series for
39.13 Find the Maclaurin series for tan ' x.
By Problem 38.36, we know that for Hence, by
Problem 39.11, this must be the Maclaurin series for tan *. A direct calculation of the coefficients is tedious.
39.14 Find the Maclaurin series for cosh *.
By Problem 38.43, for all *. Hence, by Problem 39.11, this is the Maclaurin series for
cosh *.
39.15 Obtain the Maclaurin series for cos2 *.
Now, by Problem 38.59, and, therefore, cos 2* =
Since the latter series has constant term 1, and
By Problem 39.11, this is the Maclaurin series for cos2 *.
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