Derivation of Taylor Series

1. Taylor Series
By
MUJUNGU HERBERT
National Teachers’
College Kabale

2. Taylor Series
Competences
1. Derives and state the Taylor polynomial
2. Obtain Taylor’s series for different functions about specified
values.
3. Approximate some values of the functions

3. Taylor’s Series
Introduction
Let us assume a general polynomial for any function;
f(x) =
∞
n=0
cn(x − a)n
or
f(x) = co + c1(x − a) + c2(x − a)2
+ c3(x − a)3
+ c4(x − a)4
+ ... (1)

4. Taking the general formula,
f(x) = co + c1(x − a) + c2(x − a)2
+ c3(x − a)3
+ c4(x − a)4
+ ... (2)
deriving recursively,
f (x) = c1 + 2c2(x − a) + 3c3(x − a)2
+ 4c4(x − a)3
+ ... (3)
f (x) = 2c2 + 3 · 2 · c3(x − a) + 4 · 3 · c4(x − a)2
+ ... (4)
f (x) = 3 · 2 · c3 + 4 · 3 · 2 · c4(x − a) + ... (5)
f(4)
(x) = +4 · 3 · 2 · c4 + ... (6)
Substituting x = a in Eqns 9, 3, 4, 5 and 6, we get
co = f(a), c1 = f (a), c2 =
f (a)
2
, c3 =
f (a)
3 · 2
and c4 =
f(4)(a)
4 · 3 · 2

5. Taylors series
Substituting the coefficients back in Eqn 1, we get;
f(x) = f(a) + f (a)(x − a) +
f (a)
2
(x − a)2
+
f (a)
3 · 2
(x − a)3
+
f(4)
(a)
4 · 3 · 2
(x − a)4
+ ... (7)
or if you let x = a + h
f(x) = f(a + h) = f(a) + f (a)h +
f (a)
2
h2
+
f (a)
3 · 2
h3
+
f(4)(a)
4 · 3 · 2
h4
+ ... (8)

6. Example 1
Use Taylor’s theorem to expand sin(π
6 + h) in ascending powers of
h.

7. Example 1
Use Taylor’s theorem to expand sin(π
6 + h) in ascending powers of
h.
Solution

8. Example 1
Use Taylor’s theorem to expand sin(π
6 + h) in ascending powers of
h.
Solution
Let f(x) = sin(x) = sin(
π
6
+ h) ⇒ f (π
6 ) = sin(π
6 ) = 1
2
f (x) = cos(x) ⇒ f (π
6 ) = cos(π
6 ) =
√
3
2
f (x) = − sin(x) ⇒ f (π
6 ) = − sin(π
6 ) = −1
2
f (x) = − cos(x) ⇒ f (π
6 ) = − cos(π
6 ) = −
√
3
2
f (4)
(x) = sin(x) ⇒ f (4)(π
6 ) = sin(π
6 ) = 1
2
sin(
π
6
+ h) =
1
2
+
√
3
2
h −
1
4
h2
−
√
3
12
h3
+
1
48
h4
+ ...

9. Example 2
Use Taylor’s theorem to express tan(π
4 + h) as a series in ascending
powers of h as far as h3.

10. Example 2
Use Taylor’s theorem to express tan(π
4 + h) as a series in ascending
powers of h as far as h3.
Solution

11. Example 2
Use Taylor’s theorem to express tan(π
4 + h) as a series in ascending
powers of h as far as h3.
Solution
Let f(x) = tan(x) = tan(
π
4
+ h) ⇒ f (π
4 ) = tan(π
4 ) = 1
f (x) = 1 + tan2
(x) ⇒ f (π
4 ) = 1 + tan2(π
4 ) = 2
f (x) = 2 tan x + 2 tan3
(x) ⇒ f (π
4 ) = 4
f (x) = 2 sec2
(x) + 6 sec2
(x)tan2
(x) ⇒ f (π
4 ) = 16
tan(
π
4
+ h) = 1 + 2h + 2h2
+
8
3
h3

12. Example 3
Use Taylor’s theorem to find the first four terms in the expansion
of cos(x) in ascending powers of ()x − α), where α = arctan(4
3).

13. Example 3
Use Taylor’s theorem to find the first four terms in the expansion
of cos(x) in ascending powers of ()x − α), where α = arctan(4
3).
Solution
If α = arctan(4
3), then cos(α) = 3/5 & sin(α) = 4/5

14. Example 3
Use Taylor’s theorem to find the first four terms in the expansion
of cos(x) in ascending powers of ()x − α), where α = arctan(4
3).
Solution
If α = arctan(4
3), then cos(α) = 3/5 & sin(α) = 4/5
Also Let f(x) = cos(x) ⇒ f (α) = cos(α) = 3/5
f (x) = − sin(x) ⇒ f (α) = −4
5
f (x) = − cos x ⇒ f (α) = −3
5
f (x) = sin x ⇒ f (α) = 4
5
f (4)
(x) = cos x ⇒ f (4)(α) = 3
5
cos(x) =
3
5
−
4
5
(x − α) −
3
10
(x − α)2
+
12
15
(x − α)3

15. Example 4
Using the first 3 terms of the expansion in Example 1, obtain a
value for sin 31o to five significant figures, taking
√
3 as 1.7321
and 1o as 0.01745 radians.

16. Example 4
Using the first 3 terms of the expansion in Example 1, obtain a
value for sin 31o to five significant figures, taking
√
3 as 1.7321
and 1o as 0.01745 radians.
Solution
Recall sin(π
6 + h) = 1
2 +
√
3
2 h − 1
4h2 −
√
3
12 h3 + 1
48h4 + ...
So we let h = 1o
= 0.01745 radians

17. Example 4
Using the first 3 terms of the expansion in Example 1, obtain a
value for sin 31o to five significant figures, taking
√
3 as 1.7321
and 1o as 0.01745 radians.
Solution
Recall sin(π
6 + h) = 1
2 +
√
3
2 h − 1
4h2 −
√
3
12 h3 + 1
48h4 + ...
So we let h = 1o
= 0.01745 radians
sin(31o
) =
1
2
+
1.7321
2
∗0.01745−
1
4
∗0.017452
−
1.7321
12
∗0.017453
Finally, sin(31o
) = 0.51503

18. Trial Questions
1. Apply Taylor’s theorem to expand ln(x) in ascending powers
of (x − e) as far as the term in (x − e)4,
2. Apply Taylor’s theorem to expand cosec(x) in ascending
powers of (x − π
2 ) as far as the term in (x − π
2 )4
3. Find the Taylor series for f(x) = ex about x = 0.
4. (a) Use Taylor’s theorem to expand cos(π
3 + h) in ascending
powers of h upto the h3
term.
(b) Taking
√
3 as 1.7321 and 5.5 as 0.09599 radians, find the
value of cos 54.5o
to three decimal places.
5. State Taylor’s theorem for the exapnsion of f(a + h) in
ascending powers of h. Prove that the first four terms in the
Taylor expansion of arctan(1 + h) are
1
4
π +
1
2
h −
1
4
h2
+
1
12
h3
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19. Summary
1. General Polynomial
f(x) = co + c1(x − a) + c2(x − a)2
+ c3(x − a)3
+ c4(x − a)4
+ ... (
2. Taylor’s Series
f(x) = f(a) + f (a)(x − a) +
f (a)
2
(x − a)2
+
f (a)
3 · 2
(x − a)3
+
f(4)(a)
4 · 3 · 2
(x − a)4
+ ...(10)
3. Steps for obtaining Taylor’s series.
(a) Obtain successive derivaties of the the function.
(b) Then determine the values of the derivatives at a given point.
(c) Finally use these values to write the Taylor’s series of the
desired function.

20. Thank You for Your Attention
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