7/28/2019 Solved Problems on Taylor Series
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Q 1 : Using Taylors series, find the values off(x) is shown
below :
(i) f(x) = x1 3x3 + 2x2x+ 4 in the powers of (x 1) and hence
find f
(1.1).
(ii) tan
4
x in the powers ofxand hence find the value of tan (4501) if
1 = 0.01745 radiants.
(i) Taylors series is given by
f(x) = f(a) +f(a)!1
)( ax +f(a)
!2
)( 2ax +f(a)
!3
)( 3ax +
Here a = 1,
Differentiate successivelyw.r.t. x,
Let f(x) =x4 3x3 2x2x+ 4 f(1) = 1
f (x) = 4x3 9x2 4x 1, f (1) = 10
f (x) = 12x2 18x 4 f(1) = 10
f(x) = 24x 18, f (1) = 6
f (x) = 24, f (1) = 24
Putting the values of the derivatives, in the Taylors series, we
get
f(x) = 1 10 (x 1) 2
10
(x 1)2
!3
6
(x 1)3
+ !4
24
(x 1)4
or f(x) = 1 10 (x 1) 5 (x 1)2 (x 1)3 + (x 1)4
Now x = 1.1 = 1 + 0.1
x 1 = 0.1
f(1.1) = 1 10 0.1 5 0.001 + 0.0001
= 2.04090
7/28/2019 Solved Problems on Taylor Series
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(ii) Taylors series is given by
f(x+a) =f(a) +x!1
)(' af+x2
!2
)('' af+x3
!3
)(''' af+
Let f(x+ a) = tan (x+ /4)
Here a =4
Differentiate successivelyw.r.t. x,
f(x) = tanx f
4= 1
f (x) = sec2x, f
4
= 2
f (x)= 2 sec2xtanx f
4
= 4
f (x) = 2 sec4x+ 4 sec2xtan2x f
4= 16
Putting the values of the derivatives, in the Taylors series, we
get
f
4
x
= tan
4
x = 1 + 2x+ 2x2 +3
8x3+
Now we want tan (4501)
Let x+ a = 450 + 1 =4
+ 0.01745
f
01745.0
4
= 1 + 2 (0.01745) + 2 (0.01745)2 +3
8(0.01745)3+
= 1.03552
