chapter 11 maclaurin series
TRANSCRIPT
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CHAPTER 11 MACLAURIN
SERIES
Focus on Exam 11
1 Let f ( x) = e x sin x f(0) = e0
= 1 f ( x) = ( x cos x + sin x) e x sin x = ( x cos x + sin x) f( x) f (0) = (0 cos 0 + sin 0)(1)
= 0 f ( x) = ( x cos x + sin x) f ( x) + f ( x)(- x sin x + cos x + cos x)
= ( x cos x + sin x) f ( x) + f ( x)(- x sin x + 2 cos x) f ( x) = 0 + 1(2)
= 2Hence, f ( x) = 1 + f (0) x + f (0) x
2
2! +
e x sin x = 1 + 2 x22
+
= 1 + x 2 + [Shown]2 y = ln (cos x)
d yd x
= sin xcos x
= - tan x
d 2 yd x2
= - sec 2 x = - (1 + tan 2 x)
d 3 y
d x3 = - 2 tan x(sec 2 x)
= - 2(- tan x)(- sec 2 x)
d yd x
d 2 yd x2
= - 2 d yd x
d 2 yd x2
[Shown]
d 4 yd x4
= - 2 d yd x
d 3 yd x3
+ d 2 y
d x2 1- 2d
2 yd x22
= - 2 d yd x d 3
yd x3 - 21d 2
yd x222
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ACE AHEAD Mathematics (T) Second Term2
When x = 0, y = ln (cos 0) = ln 1 = 0
When x = 0,
d yd x = - tan 0
= 0When x = 0,
d 2 yd x2
= - (1 + tan 2 0) = - 1
When x = 0,
d 3 yd x3
= - 2 d yd x
d 2 yd x2
= - 2(0)( - 1) = 0
When x = 0,
d 4 yd x4
= - 2 d yd x
d 3 yd x3
- 21d 2 yd x222
= - 2(0)(0) - 2(- 1)2
= - 2
Hence, ln (cos x) = f (0) + f (0) x + f (0) x2
2! + f (0) x
3
3! + f IV x
4
4! +
= 0 + 0 x - x2
2 + 0 x3 - 21 x
4
242 + = - x
2
2 - x
4
12 +
3 y2 = 1 + sin x
2 y d yd x
= cos x
2 yd 2 yd x2
+ d yd x
2 d yd x
= - sin x
2 y d 2
yd x2
+ 21d yd x22
= 1 - y 2
2 yd 2 yd x 2
+ 21d yd x 22
+ y 2 - 1 = 0 [Shown]
2 yd 3 yd x3
+ d 2 y
d x2 2 d y
d x + 41d yd x21
d 2 yd x22 + 2 y d yd x = 0
yd 3 yd x3
+ d yd x
d 2 yd x2
+ 21d yd x21d 2 yd x22 + y d yd x = 0
yd 3 yd x 3 + 31
d yd x 21
d 2 yd x 22 + y
d yd x = 0
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Fully Worked Solution 3
When x = 0, y = 1 + sin 0 = 1
When x = 0,
2(1)d yd x = cos 0
d yd x
= 12
When x = 0,
2(1)d 2 yd x2
+ 211222
+ 12 - 1 = 0
d 2 yd x2
= - 14
When x =
0,(1)
d 3 yd x3
+ 311221- 142+ (1) 12 = 0
d 3 yd x3
= - 18
Hence, 1 + sin x = f (0) + f (0) x + f (0) x2
2! + f (0) x
3
3! +
= 1 + 12
x - 141 x
2
2!2 - 18 1 x3
3!2 + = 1 + x 2 -
x 2
8 - x
3
48 +
4 y = cos x y2 = cos x
2 y d yd x
= - sin x
2 yd 2 yd x2
+ d yd x
2 d yd x
= - cos x
yd 2 yd x2
+ 1d yd x2
2
= - 12
cos x
y d 2 y
d x2 + 1d yd x2
2
= - 12
y2
yd 2 yd x 2
+ 1d yd x 22
+ 12
y 2 = 0 [Shown]
yd 3 yd x3
+ d 2 y
d x2
d yd x
+ 2 d yd x
d 2 y
d x2 + 1
2 2 y d y
d x = 0
yd 3 yd x3
+ 3d yd x
d 2 yd x2
+ y d yd x
= 0
yd 4 yd x4 +
d 3 yd x3
d yd x + 3
d yd x
d 3 yd x3 +
d 2 yd x2 3
d 2 yd x2 + y
d 2 yd x2 +
d yd x
d yd x = 0
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ACE AHEAD Mathematics (T) Second Term4
y d 4 y
d x4 + 4 d y
d x d
3 yd x3
+ 31d 2 yd x222
+ y d 2 y
d x2 +1d yd x2
2
= 0
When x = 0,y = cos 0 = 1
When x = 0,
2(1) d yd x
= - sin 0
d yd x
= 0
When x = 0,
(1)d 2 yd x2
+ 02 + 12
(1) 2 = 0
d 2 yd x2
= - 12
When x = 0,
(1)d 3 yd x3
+ 3(0) 1- 122+ (1)(0) = 0
d 3 yd x3
= 0
When x = 0,
(1)d 4 yd x4
+ 4(0)(0) + 31- 1222
+(1)1- 122 + 02 = 0
d 4 yd x4
= - 14
Hence, cos x = f (0) + f (0) x + f (0) x2
2! + f (0) x
3
3! + f IV(0) x
4
4! +
= 1 - 12
1 x22 2 - 14 1 x4
242 + = 1 - x
2
4 - x
4
96 +
5 y = sin - 1 xsin y = x
(cos y)
1d y
d x2 = 1
1 - sin 2 y 1d yd x2 = 11 - x2 1d yd x2 = 1
(1 - x2)1d yd x22
= 1
(1 - x2) 21d yd x2d 2 y
d x2 + 1d yd x2
2
(- 2 x) = 0
(1 - x 2) d2
yd x 2 - x
d yd x = 0 [Shown]
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Fully Worked Solution 5
(1 - x2) d 3 y
d x3 + d
2 yd x2
(- 2 x) - x d 2 y
d x2 - d y
d x = 0
When x = 0, y = sin -1 0 = 0
When x = 0,1 - 021d yd x2 = 1
d yd x
= 1
When x = 0,
(1 - 02) d 2 y
d x2 - (0)(1) = 0
d 2 yd x2 = 0
When x = 0,
(1 - 02) d 3 y
d x3 + (0)[ -2(0)] - (0)(0) - 1 = 0
d 3 yd x3
= 1
Hence, sin - 1 x = f (0) + f (0) x + f (0) x2
2! + f (0) x
3
3! +
= x - x 3
6 +
6 0.4
0.1e2 cos x d x =
0.4
0.1 11 + x + x2
2! + x
3
3!211 - x2
2!2d x =
0.4
0.1 11 - x2
2 + x - x
3
2 + x
2
2 + x
3
6 2d x =
0.4
0.1 11 + x - x3
3 2 d x
= 3 x + x2
2 - x4
1240.4
0.1
= 0.4 + 0.42
2 - 0.4
4
12 - 10.1 + 0.1 22 - 0.1
4
12 2 = 0.373
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ACE AHEAD Mathematics (T) Second Term6
7 0.5
0 sin x1 - x d x
= 0.5
0(sin x)(1 - x)- 1 d x
= 0.5
0 1 x - x3
3 2[1 + - 1C 1(- x) + - 1C 2(- x)2 + - 1C 3(- x)3] d x
= 0.5
0 1 x - x3
3 2(1 + x + x2
+ x3
) d x
= 0.5
0 1 x + x2 + x3 - x3
3 2d x =
0.5
0 1 x + x2 + 2 x3
3 2d x = 3 x22 + x
3
3 + x
4
6 400.5
= 0.52
2 + 0.5
3
3 + 0.5
4
6 - 0
= 0.177
8 e2 x ln 1 - 2 x = e2 x312 ln (1 - 2 x)4 = 1
2 31 + 2 x + (2 x)
2
2! + (2 x)
3
3! + (2 x)
4
4! + 4 3-2 x - (- 2 x)
2
2 + (- 2 x)
3
3 - (- 2 x)
4
4 + 4
= 12
11 + 2 x + 4 x22 + 8 x3
6 + 16 x
4
24 + 21-2 x - 4 x22 - 8 x
3
3 - 16 x
4
4 + 2
= 12
11 + 2 x + 2 x2 + 43 x3 + 23 x4 + 21- 2 x - 2 x2 - 83 x3 - 4 x4 + 2 = 1
2 1- 2 x - 2 x2 - 83 x3 - 4 x4 - 4 x2 - 4 x3 - 163 x4 - 4 x3 - 4 x4 - 83 x4 + 2
= 12
1- 2 x - 6 x2 - 323 x3 - 16 x4 - 2 = - x - 3 x 2 - 16
3 x 3 - 8 x 4 -
When e 2 x ln 1 - 2 x = e15 ln 4
5,
1 - 2 x = 45
x = 110
Hence, e15 ln 4
5 = - 1
10 - 3
100- 16
3000 - 8
10 000
= - 1021
7500
[Shown]
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Fully Worked Solution 7
9 f ( x) = cosec 12 + x2f ( x) = - cosec 12 + x2cot 12 + x2f ( x) = - cosec 1
2
+ x23- cosec 2 12
+ x24+ cot 12
+ x23cosec 12
+ x2cot 12
+ x24 = cosec 3 12 + x2 + cot 2 12 + x2 cosec 12 + x2 = cosec 3 12 + x2+3cosec 2 12 + x2- 14 cosec 12 + x2 = cosec 3 12 + x2 + cosec 3 12 + x2 - cosec 12 + x2 = 2 cosec 3 12 + x2 - cosec 12 + x2 f ( x ) = 2[f ( x )]3 - f ( x ) [Shown]
f ( x) = 6[f ( x)]2 f ( x) - f ( x)f IV ( x) = 6[f ( x)]2 f ( x) + f ( x) 12 f ( x) f ( x) - f ( x)
= 6[ f ( x)]2 f ( x) + 12 f ( x)[f ( x)]2 - f ( x)
When x = 0,
f (0) = cosec 12 + 02 = 1 f (0) = - cosec 12 + 02cot 12 + 02 = 0
f (0) = 2(1) 3 - 1 = 1
f (0) = 6(1) 2(0) - 0 = 0
f IV
(0) = 6(1)2
(1) + 12(1)(0)2
- 1 = 5
f ( x) = f (0) + f (0) x + f (0) x2
2! + f (0) x
3
3! + f IV(0) x
4
4! +
cosec 12 + x2 = 1 + x 2
2 + 5 x
4
24+
Hence, cosec 12 + 1122 1 + 12 111222
+ 524
111224
1.00348
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ACE AHEAD Mathematics (T) Second Term8
10 1 - cos x x sin x
=1 - 11 x22! + x
4
4! + 2
x1 x x33! + x5
5! + 2
=
x2
2 - x
4
24 +
x2 - x4
6 + x
6
120 +
=12
- x4
24 +
1 - x2
6 + x
4
120 +
Dividing each term by x2
Hence, lim x0
1 - cos x x sin x
=12
- 0 +
1 - 0 + 0 +
= 12
11 Let f ( x) = ln (1 + sin x)
f ( x) = cos x1 + sin x
f ( x) = (1 + sin x)(- sin x) - cos x (cos x)(1 + sin x)2
= - sin x - sin2 x - cos 2 x
(1 + sin x)2
= - sin x - (sin2 x + cos 2 x)
(1 + sin x)2
= - sin x - 1(1 + sin x)2
= - (sin x + 1)(1 + sin x)2
= - 11 + sin x
= -(1 + sin x)- 1f ( x) = (1 + sin x)- 2 cos x
= - cos x(1 + sin x)2
When x = 0,f (0) = ln (1 + sin 0)
= 0f (0) = cos 01 + sin 0
= 1
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Fully Worked Solution 9
f (0) = - 11 + sin 0
= - 1
f (0) = cos 0(1 + sin 0) 2
= 1f ( x)= f (0) + f (0) x + f (0) x
2
2!+ f (0) x
3
3!+
ln (1 + sin x) = x - x 2
2 + x
3
6 +
lim x0
ln (1+ sin x) - x
x2= lim
x0 1 x x22 + x
3
6 + 2 x
x2
= lim x0
- x2
2 + x
3
6 +
x2
= lim x0 1- 12 + x6 + 2
= - 12