Expansion of Functions & Error Approximation

Taylor Series
Consider the expansion of a function 𝑓(𝑥) in terms the power series about any given point ‘𝑎’as:
𝑓(𝑥) = ∑∞ 𝑛
𝑛=0 𝑐𝑛 ℎ , where ℎ = 𝑥 − 𝑎.
If 𝑓(𝑥) is infinitely differentiable about the point ‘𝑎’, then 𝑓(𝑥) can be represented as a special type of series known as
𝑓 (𝑛) (𝑎)
Taylor series. It can be shown by repeated differentiation that 𝑐𝑛 = , where 𝑓 (𝑛) (𝑥) is the 𝑛𝑡ℎ derivative.
𝑛!
ℎ2 ℎ3 ℎ𝑛
Thus 𝑓 (𝑥 ) = 𝑓(𝑎 + ℎ) ≈ 𝑓(𝑎) + ℎ𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ + 𝑓 (𝑛) (𝑎) + ⋯ …①
2! 3! 𝑛!
(𝑥−𝑎)2 (𝑥−𝑎)3 (𝑥−𝑎)𝑛 (𝑛)
or 𝑓 (𝑥 ) ≈ 𝑓(𝑎) + (𝑥 − 𝑎)𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ + 𝑓 (𝑎) + ⋯ , if ℎ = (𝑥 − 𝑎) …②
2! 3! 𝑛!
Typically, Taylor series is used to evaluate a function, if the functional value and all it’s derivatives can be computed at
the given point ‘𝑎’.
Example 1 Expand 𝑓(𝑥) = 2𝑥 3 + 7𝑥 2 + 𝑥 − 6 in powers of (𝑥 − 2)
Solution: Let ℎ = 𝑥 − 2, Then using Taylor’s series expansion as given by ②
(𝑥−2)2 (𝑥−2)3 (𝑥−2)𝑛
𝑓 (𝑥 ) ≈ 𝑓 (2) + (𝑥 − 2)𝑓′(2) + 𝑓′′(2) + 𝑓′′′(2) + ⋯ + 𝑓 (𝑛) (2) + ⋯ ③
2! 3! 𝑛!
Now 𝑓(𝑥) = 2𝑥 3 + 7𝑥 2 + 𝑥 − 6 ⇒ 𝑓(2) = 40
𝑓′(𝑥 ) = 6𝑥 2 + 14𝑥 + 1 ⇒ 𝑓′(2) = 53
𝑓′′(𝑥 ) = 12𝑥 + 14 ⇒ 𝑓′′(2) = 38
𝑓′′′(𝑥 ) = 12 ⇒ 𝑓′′′(2) = 12
𝑓 𝑖𝑣 (𝑥 ) = 0
Using these values in ③ , we get
(𝑥−2)2 (𝑥−2)3
𝑓(𝑥) = 40 + 53(𝑥 − 2) + 38 + 12
2! 3!
 ⇒ 𝑓(𝑥) = 40 + 53(𝑥 − 2) + 19(𝑥 − 2)2 + 2(𝑥 − 2)3
𝜋
Example2 Expand 𝑡𝑎𝑛 𝑥 in powers of (𝑥 − ) upto first four terms.
4
𝜋
Solution: Let ℎ = 𝑥 − , then using Taylor’s series expansion as given in ②
4
𝜋 2 𝜋 3 𝜋 𝑛
𝜋 𝜋 𝜋 (𝑥−4) 𝜋 (𝑥−4) 𝜋 (𝑥−4) 𝜋
𝑓(𝑥 ) ≈ 𝑓 ( ) + (𝑥 − ) 𝑓′ ( ) + 𝑓′′ ( ) + 𝑓′′′ ( ) + ⋯ + 𝑓 (𝑛) ( ) + ⋯ ④
4 4 4 2! 4 3! 4 𝑛! 4
𝜋
Now 𝑓(𝑥) = 𝑡𝑎𝑛 𝑥 ⇒𝑓( ) = 1
4
𝜋
𝑓′(𝑥 ) = 𝑠𝑒𝑐 2 𝑥 ⇒ 𝑓′ ( ) = 2
4
𝜋
𝑓′′(𝑥 ) = 2𝑠𝑒𝑐 2 𝑥 𝑡𝑎𝑛 𝑥 ⇒ 𝑓′′ ( ) = 4
4
𝜋
𝑓′′′(𝑥 ) = 2𝑠𝑒𝑐 4 𝑥 + 4 𝑡𝑎𝑛2 𝑥𝑠𝑒𝑐 2 𝑥 ⇒ 𝑓′′′ ( ) = 16
4

Using these values in ④, we get

𝜋 𝜋
2 8 𝜋
3
𝑓(𝑥) = 1 + 2 (𝑥 − ) + 2 (𝑥 − ) + (𝑥 − )
4 4 3 4

Example 3 Estimate the value of √10 correct to four places of decimal.

Solution: Here 𝑓(𝑥 ) = √10 = √9 + 1 , taking 𝑎 = 9 and ℎ = 1 , ⸪𝑥 = 𝑎 + ℎ
Using Taylor’s series expansion as given by ① , we have
ℎ2 ℎ3 ℎ𝑛
𝑓(𝑥 ) ≈ 𝑓(𝑎) + ℎ𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ + 𝑓 (𝑛) (𝑎) + ⋯
2! 3! 𝑛!
1 1 1
12 13 14
⇒ (10)2 ≈ (9 + 1)2 = 92 + 1. 𝑓′(9) + 𝑓′′(9) + 𝑓′′′(9) + 𝑓 𝑖𝑣 (9) + ⋯ ⑤
2! 3! 4!
1
Now 𝑓(𝑥) = 𝑥 2 ⇒ 𝑓(9) = 3
−1
1 1
𝑓′(𝑥 ) = 𝑥2 ⇒ 𝑓′(9) = = 0.1667
2 6
 −3
1
𝑓′′(𝑥 ) = − 𝑥 2 ⇒ 𝑓′′(9) = −0.0093
4
−5
3
𝑓′′′(𝑥 ) = 𝑥 2 ⇒ 𝑓′′′(2) = 0.0015
8
⋮
Using these values in ⑤ , we get
1
0.0093 0.0015
(10) = 3 + 0.1667 −
2 + +⋯
2 6
1
⇒ (10)2 = 3 + 0.1667 − .0047 + .0003 + ⋯
= 3.1623 approx
Maclaurin Series
It is the special case of Taylor series about the point zero. Putting 𝑎 = 0 in ② ,
𝑥2 𝑥3 𝑥𝑛
Thus, Maclaurin series of 𝑓(𝑥) is given by: 𝑓(𝑥 ) = 𝑓(0) + 𝑥𝑓′(0) + 𝑓′′(0) + 𝑓′′′(0) + ⋯ + 𝑓 (𝑛) (0) + ⋯⑥
2! 3! 𝑛!
Maclaurin series of some standard functions:
𝑥3 𝑥5 𝑥7
1. sin 𝑥 = 𝑥 − + − +⋯
3! 5! 7!
𝑥2 𝑥4 𝑥6
2. cos 𝑥 = 1 − + − +⋯
2! 4! 6!
𝑥2 𝑥3 𝑥4
3. e𝑥 = 1 + 𝑥 + + + +⋯
2! 3! 4!
𝑥2 𝑥3 𝑥4
4. log(1 + 𝑥) = 𝑥 − + − + ⋯ , |𝑥 | < 1
2 3 4
𝑥2 𝑥3 𝑥4
5. log(1 − 𝑥) = −𝑥 − − − − ⋯, |𝑥 | < 1
2 3 4
𝑚(𝑚−1) 2 𝑚(𝑚−1)(𝑚−2)
6. (1 + 𝑥)m = 1 + 𝑚𝑥 + 𝑥 + 𝑥3 + ⋯
2! 3!
Example 4 Expand 𝑒 𝑥 𝑐𝑜𝑠 𝑥 by Maclaurin series.
𝑥2 𝑥3 𝑥𝑛
Solution: We have, 𝑓 (𝑥 ) = 𝑓 (0) + 𝑥𝑓′(0) + 𝑓′′(0) + 𝑓′′′(0) + ⋯ + 𝑓 (𝑛) (0) + ⋯ ⑥
2! 3! 𝑛!
𝑥 0
Here 𝑓(𝑥 ) = 𝑒 𝑐𝑜𝑠 𝑥 ⇒ 𝑓(0) = 𝑒 𝑐𝑜𝑠 0 = 1
𝑓′(𝑥 ) = 𝑒 𝑥 𝑐𝑜𝑠 𝑥 − 𝑒 𝑥 𝑠𝑖𝑛 𝑥 ⇒ 𝑓′(0) = 𝑒 0 𝑐𝑜𝑠 0 − 𝑒 0 𝑠𝑖𝑛 0 = 1
𝑓′′(𝑥 ) = −2𝑒 𝑥 𝑠𝑖𝑛 𝑥 ⇒ 𝑓′′(0) = −2𝑒 0 𝑠𝑖𝑛 0 = 0
𝑓′′′(𝑥 ) = −2𝑒 𝑥 𝑐𝑜𝑠 𝑥 − 2𝑒 𝑥 𝑠𝑖𝑛 𝑥 ⇒ 𝑓′′′(0) = −2𝑒 0 𝑐𝑜𝑠 0 − 2𝑒 0 𝑠𝑖𝑛 0 = −2
𝑓 𝑖𝑣 (𝑥 ) = −4𝑒 𝑥 𝑐𝑜𝑠 𝑥 ⇒ 𝑓 𝑖𝑣 (0) = −4𝑒 0 𝑐𝑜𝑠 0 = −4
Putting these values in ⑥, we get
𝑥 2𝑥 3 4𝑥 4
e cos 𝑥 = 1 + 𝑥 − − −⋯
3! 4!
𝑥3 𝑥4
or e𝑥 cos 𝑥 = 1 + 𝑥 − − +⋯
3 6
𝑥2 𝑥4
Example5 Show that 𝑙𝑜𝑔 𝑠𝑒𝑐 𝑥 = + +⋯
2 12
𝑥2 𝑥3 𝑥𝑛
Solution: We have, 𝑓 (𝑥 ) = 𝑓 (0) + 𝑥𝑓′(0) + 𝑓′′(0) + 𝑓′′′(0) + ⋯ + 𝑓 (𝑛) (0) + ⋯ ⑥
2! 3! 𝑛!
Here 𝑓(𝑥 ) = 𝑙𝑜𝑔 𝑠𝑒𝑐 𝑥 ⇒ 𝑓(0) = 𝑙𝑜𝑔 1 = 0
𝑓′(𝑥 ) = 𝑡𝑎𝑛 𝑥 ⇒ 𝑓′(0) = 𝑡𝑎𝑛 0 = 0
𝑓′′(𝑥 ) = 𝑠𝑒𝑐 2 𝑥 ⇒ 𝑓′′(0) = 𝑠𝑒𝑐 2 0 = 1
𝑓′′′(𝑥 ) = 2𝑠𝑒𝑐 2 𝑥 𝑡𝑎𝑛 𝑥 ⇒ 𝑓′′′(0) = 2𝑠𝑒𝑐 2 0 𝑡𝑎𝑛 0 = 0
𝑓 𝑖𝑣 (𝑥 ) = 2𝑠𝑒𝑐 4 𝑥 + 4 𝑡𝑎𝑛2 𝑥𝑠𝑒𝑐 2 𝑥 ⇒ 𝑓 𝑖𝑣 (0) = 2
𝑥2 𝑥4
Putting these values in ⑥, we get 𝑙𝑜𝑔 𝑠𝑒𝑐 𝑥 = + +⋯
2 12
𝜋 1 𝜃2 𝜃3
Example 6 Show that 𝑠𝑖𝑛 ( + 𝜃) = (1 + 𝜃 − − + ⋯)
4 2
√ 2! 3!
Solution: By Taylor’s expansion, we have
ℎ2 ℎ3 ℎ𝑛
𝑓 (𝑥 ) = 𝑓(𝑎 + ℎ) ≈ 𝑓(𝑎) + ℎ𝑓′(𝑎) + 𝑓′′(𝑎) + 𝑓′′′(𝑎) + ⋯ + 𝑓 (𝑛) (𝑎) + ⋯ …①
2! 3! 𝑛!
ℎ2 ℎ3
⸫ 𝑠𝑖𝑛 𝑥 = 𝑠𝑖𝑛(𝑎 + ℎ) ≈ 𝑠𝑖𝑛 𝑎 + ℎ 𝑐𝑜𝑠 𝑎 + (−𝑠𝑖𝑛𝑎) + (−𝑐𝑜𝑠𝑎) + ⋯
2! 3!
𝜋
Taking 𝑎 = and ℎ = 𝜃
4
𝜋 𝜋 𝜋 𝜃2 𝜋 𝜃3 𝜋
𝑠𝑖𝑛 ( + 𝜃) = 𝑠𝑖𝑛 + 𝜃𝑐𝑜𝑠 + (−𝑠𝑖𝑛 4 ) + (−𝑐𝑜𝑠 4 ) + ⋯
4 4 4 2! 3!
𝜋 1 𝜃2 𝜃3
or 𝑠𝑖𝑛 ( + 𝜃) = (1 + 𝜃 − − + ⋯)
4 √2 2! 3!
Approximation of Errors
Consider a function 𝑦 = 𝑓(𝑥), then if 𝛿𝑥 be a small change in x and 𝛿𝑦 be the resulting change in 𝑦,
𝑑𝑦
then 𝛿𝑦 = 𝛿𝑥 approximately.
𝑑𝑥
Example 7 Find the change in the total surface area of a right circular cone when
(i) the radius is constant but there is a small change in the altitude
(ii) the altitude is constant but there is a small change in the radius.
Solution: Let S be the total surface area of the cone , then S = 𝜋𝑟 2 + 𝜋𝑟√𝑟 2 + ℎ2
𝑑𝑆
(i) The radius 𝑟 is constant, and altitude ℎ changes, then 𝛿𝑆 = 𝛿ℎ
𝑑ℎ
𝑑𝑆 𝜋𝑟 𝜋𝑟ℎ
Now =0+ . 2ℎ =
𝑑ℎ 2√𝑟 2 +ℎ 2 √𝑟 2 +ℎ2
𝑑𝑆 𝜋𝑟ℎ
∴ 𝛿𝑆 = 𝛿ℎ = 𝛿ℎ approximately.
𝑑ℎ √𝑟 2 +ℎ2
𝑑𝑆
(ii) The altitude ℎ is constant, and radius 𝑟 changes, then 𝛿𝑆 = 𝛿𝑟
𝑑𝑟
𝑑𝑆 2𝜋𝑟 2 𝜋(2𝑟 2 +ℎ2 )
Now = 2𝜋𝑟 + 𝜋√𝑟 2 + ℎ2 + = 2𝜋𝑟 +
𝑑𝑟 2√𝑟 2 +ℎ2 √𝑟 2 +ℎ2
 𝑑𝑆 𝜋(2𝑟 2 +ℎ2 )
∴ 𝛿𝑆 = 𝛿𝑟 = (2𝜋𝑟 + ) 𝛿𝑟 approximately.
𝑑𝑟 √𝑟 2 +ℎ2
Example 8 If 𝑎 , 𝑏 , 𝑐 are the sides of the triangle ABC and S is the semi- perimeter, show that if there is a small error 𝛿𝑐
in the measurement of side 𝑐 then the error 𝛿∆ in the area ∆ of the triangle is given by
∆ 1 1 1 1
𝛿∆ = ( + + − ) 𝛿𝑐
4 𝑆 𝑆−𝑎 𝑆−𝑏 𝑆−𝑐
Solution: We know that ∆2 = 𝑆(𝑆 − 𝑎)(𝑆 − 𝑏)(𝑆 − 𝑐)
Taking log of both sides, we get 2𝑙𝑜𝑔∆= 𝑙𝑜𝑔 𝑆 + log(𝑆 − 𝑎) + log(𝑆 − 𝑏) + log (𝑆 − 𝑐)
2 𝑑∆ 1 𝑑𝑆 1 𝑑(𝑆−𝑎) 1 𝑑(𝑆−𝑏) 1 𝑑(𝑆−𝑐)
Differentiating both the sides w.r.t. 𝑐 , we get = + + +
∆ 𝑑𝑐 𝑆 𝑑𝑐 𝑆−𝑎 𝑑𝑐 𝑆−𝑏 𝑑𝑐 𝑆−𝑐 𝑑𝑐
(𝑎+𝑏+𝑐) 2 𝑑∆ 11 1 1 1
Also, 𝑆 = , ⸫ = + + −
2 ∆ 𝑑𝑐 𝑆2 2(𝑆−𝑎) 2(𝑆−𝑏) 2(𝑆−𝑐)
𝑑∆ ∆ 1 1 1 1
⇒ = ( + + − )
𝑑𝑐 4 𝑆 (𝑆−𝑎) (𝑆−𝑏) (𝑆−𝑐)
𝑑∆ ∆ 1 1 1 1
⇒ 𝛿∆= 𝛿𝑐 = ( + + − ) 𝛿𝑐
𝑑𝑐 4 𝑆 (𝑆−𝑎) (𝑆−𝑏) (𝑆−𝑐)

Example 9 If 𝑇 = 2𝜋√(𝑙⁄𝑔) , find the error in T corresponding to 2% error in 𝑙 where 𝑔 is constant.

𝑑𝑇
Solution: Error in T is given by 𝛿𝑇 = 𝛿𝑙
𝑑𝑙
𝑑𝑇 2𝜋 1
Now =
𝑑𝑙 √𝑔 2√𝑙
𝜋 1
∴ 𝛿𝑇 = 𝛿𝑙
√𝑔 √𝑙
𝛿𝑇 𝜋 𝛿𝑙 √𝑔 1 𝛿𝑙
⇒ = =
𝑇 √𝑔 √𝑙 2𝜋√𝑙 2 𝑙
𝛿𝑇 1 𝛿𝑇
⇒ ( . 100) = ( . 100)
𝑇 2 𝑇
i.e. percentage error in 𝑇 is half of the percentage error in 𝑙
 ⸫ corresponding to 2% error in 𝑙 , percentage error in T is 1%.

Exercise
1. Expand 𝑡𝑎𝑛−1 𝑥 in powers of (𝑥 − 1).
𝜋 1 1 1
Ans. 𝑡𝑎𝑛−1 𝑥 = + (𝑥 − 1) − (𝑥 − 1)2 + (𝑥 − 1)3 + ⋯
4 2 4 12
11
2. Using Taylor’s theorem find the approximate value of 𝑓 ( ) where 𝑓(𝑥 ) = 𝑥 3 + 32 + 15𝑥 − 10
10
Ans. 11.461
𝑥2 𝑥3 𝑥4
3. Show that log(1 + 𝑠𝑖𝑛𝑥 ) = 𝑥 − + − +⋯
2 6 12
𝜋 2 8 10
4. Show that 𝑡𝑎𝑛 ( + 𝑥) = 1 + 2𝑥 + 2𝑥 + 𝑥3
+ 𝑥 4 + ⋯ and hence find tan 46o
4 33
Ans. 1.0355
5. A soap bubble of radius 2cm shrinks to radius 1.9 cm. Find the decrease in volume and surface area.
Ans. -5.024 cm3 and -5.024cm2
6. If log104 = 0.6021, find the approximate value of log10404.
Ans. 2.61205
7. Let A, B and C be the angles of a triangle opposite to the sides a, b and c respectively. If small errors 𝛿𝑎, 𝛿𝑏 and
𝑎
𝛿𝑐are made in the sides then show that 𝛿𝐴 = (𝛿𝑎 − 𝛿𝑏 𝑐𝑜𝑠𝐶 − 𝛿𝑐 𝑐𝑜𝑠𝐵) where ∆ is the area of the triangle.
2∆