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Maclaurins Series :
f ( x ) = f (0) + x f (o)+!2
2 x f (0 ) +!3
3 x f (0 ) + . +!n
xn f n(o) + .
1) Series for e x
Let f ( x ) = ex, so that f (0) = 1
f ( x ) = ex, f (0) = 1
f ( x ) = ex, f (0) = 1
f ( x ) = ex, f (0) = 1
Substituting these values with Maclaurin s series, i.e
f ( x ) = f (0) +!1
)0(' f x +!2
)0(" f x 2 +!3
)0("' f x 3 +
We get ex = 1!1
x +!2
2 x +!3
3 x +
We can also find the following series using the above series
a : Replacing x by x in above series
e x = 1!1
x +!2
2 x !3
3 x
b : Replacing x by ix in above series
eix
= 1 + !1ix
+ !2
22 xi+ !3
33 xi+ !4
44 xi+ !5
55 xi +
c : Now ax = ex log a , replacing x by x log a in above series,
we get
a x = 1 +!1
log a x +!2
2 x (log a )2 +!3
3 x (log a )3 +
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2) Series for cosh x and sinh x
cosh x =2
x x ee
sinh x =2
x x ee
e x = 1 +!1
x +!2
2 x +!3
3 x +!4
4 x +!5
5 x +
e x = 1 !1
x +!2
2 x !3
3 x +!4
4 x !5
5 x +
Adding and subtracting the series for e x and e x
We get 2 cosh x = e x + e x
= 2
...!4!2
142 x x
or cosh x = 1 +!2
2x
+!4
4x
+
2 sinh x = e x e x = 2
... x x x
!5!3
53
or sinh x = x +!3
3x
+!5
5x
+
3) Series for cos x and sin x
cos x =2
ixix ee , sin x =2
ixix ee
We have seen earlier the series of eix while finding out the
series for ex
eix = 1 +!1
ix +!2
22 xi +!3
33 xi +!4
44 xi +!5
55 xi +
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But by Eulers formula e ix = cos x + i sin x
Therefore equation the real and imaginary parts of
cos x + i sin x =
... x x
!4!21
42
+ i
... x x x
!5!3
53
(Nothing that i 2 = 1, i 3 = 1, i 4 = 1 )
cos x = 1 !2
2x
+!4
4x
sin x = x !3
3x
+!5
5x
4) Series for tan x and tanh x
Since tan x = x x
cossin
tan x =...
!5!4!21
...!7!5!3642
753
x x x
x x x x
[using the sine and cos series]
We perform division as follows
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1!2
2 x +!4
4 x
...!6
6 x x !3
3 x +!5
5 x !7
7 x +
31
x x 3+125 x 5+
31517 x 7+
x !2
3 x +!4
5 x ...!6
7 x
+
31 x 3
301 x 5 +
8401 x 7 +
31 x 3
6
5 x +72
7 x +
+
12
5 x 5
315
4 x 7 +
125 x 5
245 7 x +
+
31517 x 7 +
Thus tan x = x +3
1 x 3 +
152
x 5 +315
17 x 7 +
Similarly, using tanh x = x x
coshsinh
We can show that tanh x = x 3
1 x 3 +
152 x 5
315
17 x 7 +
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5) Series for log (1 + x ) and log (1 x )
Expansion of log (1 + x ).Let f ( x ) = log (1 + x ),
,1
1)('
x x f 2)1(
1)("
x x f , 3)1(
12)('"
x x f ,
. f (0) = 0, f (0) = 1, f (0) = 1.
f (0) = 2!, f iv (0) = 3! f v (0) = 4!
Putting these values in below Maclaurins series,
f ( x ) = f (0) + xf (o)+!2
2 x f (0 ) +!3
3 x f (0 ) + . +!n
xn f n(o) + .
....)!3(!4
!2!3
)1(!2
)1log(432 x x x
x x
....5432
)1log(5432
x x x x x x
Expansion of log (1 x ).
Changing x to x in above expansion, we get,
....5432
)1log(5432
x x x x x x
6) Series for tan 1 x and cot 1 x
Let f ( x ) = tan 1 x
So that f ( x ) = 211 x
= (1 + x 2) 1
= 1 x 2 + x 4 x 6 + (using the binomial theorem)
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Integrating both sides, we get
f ( x ) = x 3
3 x +5
5 x 7
7 x + + c
To find the value of the constant, put x = 0 in f ( x ) = tan 1
x , so that f (0) = tan 1 0 = 0
0 = c
tan 1 x = x 3
3x
+5
5x
7
7x
+
Now since cot 1 x =2
tan 1 x , the series for cot 1 x is given by
cot 1 x =2
( x
3
3 x +5
5 x 7
7 x + .)
7) Series for sin 1 x and cos 1 x
Let f ( x ) = sin 1 x
So that f ( x ) =21
1
x= (1 x 2) 1/2
or f ( x ) = 1 +21 x 2 +
4231 x 4 +
642531 x 6 +
(using binomial theorem)
Integrating both sides, we get
f ( x ) = x +21
3
3 x +4.231
5
5 x +642531
7
7 x + + c
Put x = 0 in f ( x ) = sin 1 x , to get
f (0) = sin 10 = 0
0 = c
sin 1 x = x +2
1
3
3x
+42
31
5
5x
+642531
7
7x
+ 0
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Since we know that
cos 1 x =2
sin 1 x
The series for cos 1 x can be written as
cos 1 x =2
...
7642531
54231
321 753 x x x x
8) Expansion of (1 + x ) m .
Since f ( x ) = (1 + x )m f (0) = (1 + x )m f ( x ) = m(1 + x )m
1 f (0) = m
f ( x ) = m(m 1) (1 + x )m 2, f (0) = m(m 1) f ( x ) = m( m 1)
(m 2) (1 + x ) m 3 f (0) = m(m 1) (m 2)
If x < 1, then by Maclaurins series,
i.e f ( x ) = f (0) + xf (o)+!2
2 x f (0 ) +!3
3 x f (0 ) + . +!n
xn f n(o) + .
Hence, ....!3
)2)(1(!2
)1(1)1( 32 x
mmm x
mmmx x m
