Applications of
Differentiation
Learning Outcomes:
• Definitions
• Use definitions to sketch functions.
• Formulate Fermat’s theorem.
• Formulate increasing/Decreasing Test.
• Prove Theorem 15 and the second derivative test.
• Formulate and prove theorems
• Maxima and Minima problems
• Taylor series
Definitions
Even and Odd functions
Even: 𝑓 −𝑥 = 𝑓(𝑥)
It is symmetric about the y-axis
e.g cos −𝑥 = cos(𝑥)
Odd: 𝑓 −𝑥 = −𝑓(𝑥)
Symmetric about the origin
e.g −𝑥 3 = −𝑥 3
Periodic functions
𝑓 𝑥+𝑝 =𝑓 𝑥
Repeats its values .
e.g 𝑓 𝑡 + 2 = 𝑓 𝑡
Definitions
Increasing and decreasing functions
Increasing: If 𝑥 < 𝑦 𝑥, 𝑦 ∈ 𝐼 ⇒ 𝑓 𝑥 ≤ 𝑓(𝑦)
e.g 𝑓 𝑥 = 3𝑥
Decreasing: If 𝑥 < 𝑦 𝑥, 𝑦 ∈ 𝐼 ⇒ 𝑓 𝑥 ≥ 𝑓(𝑦)
e.g 𝑓 𝑥 = 3−𝑥
Absolute Maximum and Absolute minimum (Illustrate with a sketch)
Absolute Max: Highest point of the graph in the interval
Absolute Min: Lowest point of the graph in the interval
Also known as the global extrema
Critical point: The point 𝑐 in an interval such that 𝑓 ′ 𝑐 = 0. Also known as stationary point.
Local Maximum and Local minimum
Local Max: Highest point given by the critical point
Local Min: Lowest point given by the critical point
Also known as the relative maximum and relative minimum
Theorem: Extreme Value Theorem
If 𝑓 is continuous on 𝑎, 𝑏 , then 𝑓 attains an absolute max at 𝑓(𝑐) and absolute min at 𝑓(𝑑) at
some numbers 𝑐 and 𝑑 in [𝑎, 𝑏].
Theorem: Fermat’s Theorem
If 𝑓 has a local minimum or local maximum, then 𝑓 ′ 𝑐 = 0 exists and 𝑓 ′ 𝑐 = 0.
First Derivative Test
Question: Find the absolute extrema and the local extrema for
1. 𝑓 𝑥 = 𝑥 2 − 4𝑥 + 9 on 1, 4 .
2. ℎ 𝑥 = 4𝑥 3 − 39𝑥 2 + 90𝑥 + 2 on 1, 6 .
3. 𝑝 𝑥 = 𝑥 3 on (−∞, ∞)[ans: No global extrema on the given interval]
𝑓 𝑥 = 𝑥 2 − 4𝑥 + 9 on 1, 4
𝑓′ 𝑥 = 0
2𝑥 − 4 = 0 ⇒ 𝑥 = 2 is the critical point.
By the First derivative test, the local minimum is at 𝑥 = 2
x
f(x)
1
6
2
5
4
9
The point (2, 5) is the local minimum and the absolute minimum
The point (4, 9) is the absolute maximum.
+
2
ℎ 𝑥 = 4𝑥 3 − 39𝑥 2 + 90𝑥 + 2 on 1, 6
𝑓′ 𝑥 = 0
12𝑥 2 − 78𝑥 + 90 = 0 ⇒ 𝑥 − 5 2𝑥 − 3 = 0
3
𝑥 = 5 and 𝑥 = 2 are the critical points.
+
3
3
By the First derivative test, the local minimum is at 𝑥 = 5 and local maximum is at 𝑥 = 2. 2
x
f(x)
1
57
3
2
62
5
-23
6
2
The point (5, −23) is the local minimum and the absolute minimum
The point (1.5, 62) is the absolute maximum and local maximum.
-
+
5
Increasing and Decreasing Test
Suppose that 𝑓 is differentiable on the interval 𝐼.
(a) If 𝑓′(𝑥) > 0 for all 𝑥 ∈ 𝐼, then 𝑓 is increasing on 𝐼.
(b) If 𝑓′(𝑥) < 0 for all 𝑥 ∈ 𝐼, then f is decreasing on 𝐼.
(c) If 𝑓′(𝑥) = 0 for all 𝑥 ∈ 𝐼, then f is constant on 𝐼.
Use the increasing decreasing Test on 𝑓 𝑥 = 𝑥 2 − 3𝑥 + 1 and 𝑓 𝑥 = 𝑥 3 − 12𝑥
𝑓 𝑥 = 𝑥 2 − 3𝑥 + 1
Increasing:
𝑓′ 𝑥 > 0
3
2𝑥 − 3 > 0 ⇒ 𝑥 >
2
Decreasing:
𝑓′ 𝑥 < 0
3
2𝑥 − 3 < 0 ⇒ 𝑥 <
2
Theorem 15
Suppose that 𝑓 and 𝑔 are differentiable on the interval 𝐼. If 𝑓 ′ 𝑥 = 𝑔′ 𝑥 ∀𝑥 ∈ 𝐼, then a
constant 𝑐 can be found such that 𝑓 𝑥 − 𝑔 𝑥 = 𝑐 ∀𝑥 ∈ 𝐼.
Proof[In class]
Concavity
Concave up: 𝑓 ′ is increasing i.e 𝑓"(𝑥) > 0
Concave down: 𝑓′ is decreasing i.e 𝑓"(𝑥) < 0
Point of inflection: 𝑓"(𝑥) = 0.
Point of inflection means change in concavity
Second derivative Test
1. 𝑓 ′ 𝑐 = 0 and 𝑓"(𝑐) > 0 ⇒Local minimum
2. 𝑓 ′ 𝑐 = 0 and 𝑓"(𝑐) < 0 ⇒Local maximum
Question: Use the second derivative Test to determine any local extrema in the function:
𝑓 𝑥 = 2𝑥 3 − 12𝑥 2 .
…
ℎ 𝑥 = sin 𝑥 + cos 𝑥 on 0, 2𝜋
Question: Use the second derivative Test to determine any local extrema in the function: 𝑓 𝑥 = 2𝑥 3 − 12𝑥 2 .
𝑓 ′ 𝑥 = 0 ⇒ 6𝑥 2 − 24𝑥 = 0 ⇒ 𝑥 𝑥 − 4 = 0 ⇒ 𝑥 = 0 𝑜𝑟 𝑥 = 4 are critical points
𝑓"(𝑥) = 12𝑥 − 24.
𝑓"(0) = −24 < 0 ⇒ local maximum is at 𝑥 = 0 .
𝑓"(4) = 24 > 0 ⇒ local minimum is at 𝑥 = 4
Therefore 0, 𝑓 0 = (0, 0) is the local maximum and
4, 𝑓 4 = (4, −64) is the local minimum.
ℎ 𝑥 = sin 𝑥 + cos 𝑥 on 0, 2𝜋
𝜋
5𝜋
ℎ′ 𝑥 = 0 ⇒ 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥 = 0 ⇒ 𝑐𝑜𝑠𝑥 = 𝑠𝑖𝑛𝑥 ⇒ tan 𝑥 = 1 ⇒ 𝑥 = 4 𝑎𝑛𝑑 𝑥 = 4 are the critical points.
ℎ"(𝑥) = −sin 𝑥 − cos 𝑥
𝜋
𝜋
ℎ"( 4 ) = − 2 < 0 local max is at 𝑥 = 4
5𝜋
5𝜋
ℎ"( 4 ) = 2 > 0 local min is at 𝑥 = 4
Concavity
Y-axis
B
A
C
D
E
G
F
H
I
J
A: Concave downwards
B: Concave downwards
C: Concave downwards
D: Concave upwards
E: Concave upwards
F: Concave upwards
G: Concave downwards
H: Concave downwards
I: Concave downwards
J: Concave upwards
NB: Point of inflection is between concave upwards and concave downwards or
between concave upwards and concave downwards.
X-axis
Sketching a curve
Information required:
1. Domain of the function
2. Intercepts
3. Symmetry (even/odd/periodic)
4. Asymptotes
Vertical asymptote If 𝑥 = 𝑎 or lim± 𝑓(𝑥) =
𝑥→𝑎
±∞
Horizontal asymptote if 𝑦 = 𝐿 given by
lim 𝑓 𝑥 = 𝐿
𝑥→±∞
5. Increasing/Decreasing
6. Local (relative) extrema
7. Concavity
Sketch 𝑓 𝑥
𝑥2
= 2 . Note: 𝑓 ′ 𝑥
𝑥 +3
=
6𝑥
−18(𝑥−1)(𝑥+1)
, 𝑓"(𝑥) =
𝑥 2 +3 2
𝑥 2 +3 2
Domain: 𝑥 ∈ ℝ
Intercepts: x-int: (0, 0), y-int: (0, 0)
Symmetry: 𝑓 −𝑥 =
−𝑥 2
𝑥2
= 𝑥 2+3 = 𝑓(𝑥) (even) symmetric about the y-axis
−𝑥 2 +3
Asymptotes: There is no vertical asymptote.
𝑥2
𝑥 2 /𝑥 2
1
lim 𝑥 2 +3 = lim 𝑥 2 /𝑥2+3/𝑥2 = lim 1+3/𝑥 2 = 1. Therefore 𝑦 = 1is the horizontal asymptote.
𝑥→∞
𝑥→∞
𝑥→∞
Increasing/Decreasing
Increasing: 𝑓 ′ 𝑥 > 0 ⇒
6𝑥
′ 𝑥
>
0
⇒
𝑥
>
0.Decreasing:
𝑓
2
2
𝑥 +3
Local extrema: By the first derivative test, local min is at 𝑥 = 0 .
Concavity
Concave upwards: 𝑓"(𝑥) > 0 ⇒ −1 < 𝑥 < 1
Concave downwards: 𝑓"(𝑥) < 0 ⇒ 𝑥 < −1 𝑜𝑟 𝑥 > 1
Point of inflection: 𝑓"(𝑥) = 0 ⇒ 𝑥 = −1 𝑜𝑟 𝑥 = 1
1
1
1
6𝑥
< 0 ⇒ 𝑥 2+3 2 < 0 ⇒ 𝑥 < 0 .
𝑓 −1 = 𝑓 1 = 4 , −1, 4 𝑎𝑛𝑑 1, 4 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛.
Sketching a curve
Sketch the graph of a function that satisfies the following conditions:
• 𝑓 ′ 2 = 0, 𝑓 2 = −1, 𝑓 0 = 0
• 𝑓 ′ 𝑥 < 0 if 0 < 𝑥 < 2
• 𝑓 ′ 𝑥 > 0 if 𝑥 > 2
• 𝑓"(𝑥) < 0 if 0 ≤ 𝑥 < 1 or if 𝑥 > 4
• 𝑓"(𝑥) > 0 if 1 < 𝑥 < 4
• lim 𝑓 𝑥 = 1
•
𝑥→∞
𝑓 −𝑥 = 𝑓 𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥.
𝑓 ′ 2 = 0 ⇒ 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑎𝑡 𝑥 = 2.
𝑓 2 = −1, 𝑓 0 = 0 ⇒ 2, −1 , 0, 0 𝑎𝑟𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ. 2, −1 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡.
𝑓 ′ 𝑥 < 0 if 0 < 𝑥 < 2 ⇒ 𝑓 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑎𝑡 𝑡ℎ𝑖𝑠 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
𝑓 ′ 𝑥 > 0 if 𝑥 > 2 𝑓 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑎𝑡 𝑡ℎ𝑖𝑠 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
𝑓"(𝑥) < 0 if 0 ≤ 𝑥 < 1 or if 𝑥 > 4 ⇒ 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑𝑠 𝑎𝑡 𝑡ℎ𝑒𝑠𝑒 𝑡𝑤𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
𝑓"(𝑥) > 0 if 1 < 𝑥 < 4 ⇒ 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝𝑤𝑎𝑟𝑑𝑠 𝑎𝑡 𝑡ℎ𝑖𝑠 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
𝑥 = 1 𝑎𝑛𝑑 𝑥 = 4 𝑎𝑟𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
lim 𝑓 𝑥 = 1 ⇒ 𝑦 = 1 𝑖𝑠 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒
𝑥→∞
𝑓 −𝑥 = 𝑓 𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ⇒ 𝑓 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑎𝑛𝑑 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠
Questions
Minima and Maxima problems
Learning Outcomes:
• Solve practical optimization problems
• Explain and compute production
level, cost function, marginal cost,
average cost, Revenue, marginal
revenue, profit and marginal profit.
• Compute minimum cost, maximum
revenue and profit.
Process
Draw a suitable sketch to assess the scenario
• Find the formula expressing the quantity that must be minimised or
maximised
• Determine the minimum or maximum value of this function
•
𝐴 = 𝑥𝑦
Solution
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟: 200 = 4𝑥 + 2𝑦 ⇒ 𝑦 = 100 − 2𝑥
𝐴𝑟𝑒𝑎: 𝐴 = 𝐿𝐵 = 2𝑥𝑦 = 2𝑥 100 − 2𝑥 = 200𝑥 − 4𝑥 2
To find the maximum:
𝐴′ 𝑥 = 0
200 − 8𝑥 = 0 ⇒ 𝑥 = 25𝑚
and 𝑦 = 100 − 2 25 = 50m
Therefore 𝐴 = 2 25 50 𝑚2
Find two numbers whose sum is 23 and whose product is a
maximum.
Sum 𝑥 + 𝑦 = 23 ⇒ 𝑦 = 23 − 𝑥
Product: 𝑃 = 𝑥𝑦 = 𝑥 23 − 𝑥 = 23𝑥 − 𝑥 2
To find the maximum product: 𝑃′ 𝑥 = 0
23
23 − 2𝑥 = 0 ⇒ 𝑥 = 2 (local maximum)
23
23
and 𝑦 = 23 − 2 = 2
If 1200 𝑐𝑚2 of material is available to make a box with a square base
and an open top, find the largest possible volume of the box.
𝑉 = 𝑙2 ℎ = 𝑏2 ℎ
𝐴 = 2𝑙ℎ + 2𝑤ℎ + 𝑙𝑤
𝐴 = 2𝑏ℎ + 2𝑏ℎ + 𝑏2 = 4𝑏ℎ + 𝑏2 (Square box)
1200 = 4𝑏ℎ + 𝑏
2
1200−𝑏2
⇒ ℎ = 4𝑏
1200𝑏−𝑏3
= 400
To find the maximum volume: 𝑉 ′ 𝑏
1200−𝑏
𝑉 = 𝑏2 ℎ = 𝑏2 4𝑏
2
=0
1200−3𝑏2
2 =0⇒𝑏 =
=
0
⇒
1200
−
3𝑏
400
1200 20 − 20 3
Therefore 𝑉 =
= 4000𝑐𝑚3
400
400 = 20𝑐𝑚
Cost Revenue and Profit
Goal: Minimum cost and Maximum profit
Definitions:
• 𝑥 − 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠
• 𝐶 𝑥 = 𝑓𝑖𝑥𝑒𝑑 𝑐𝑜𝑠𝑡 + 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 (Cost
function)
• 𝑝 𝑥 − 𝑃𝑟𝑖𝑐𝑒 𝑜𝑟 𝑑𝑒𝑚𝑎𝑛𝑑
• 𝑅 𝑥 = 𝑥𝑝 𝑥 [Revenue]
• 𝑃 𝑥 = 𝑅 𝑥 − 𝐶(𝑥) [Profit]
• Marginal the derivative of a function
𝐶 𝑥
• Average cost:
𝑥
A company that manufactures washing machines, spends R60000.00 each week
on overheads. The direct cost to manufacture one washing machine is R250.00.
Each washing machine is R400.00. How many washing machines must a
company manufacture to break even?
Total cost: 𝐶 𝑥 = 𝐹𝑖𝑥𝑒𝑑 𝑐𝑜𝑠𝑡 + 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 = 60 000 + 250𝑥
Revenue: R x = 𝑥𝑝 𝑥 = 400𝑥
Break-even: 𝑅 𝑥 = 𝐶 𝑥 = 400𝑥 = 60 000 + 250𝑥 ⇒ 150𝑥 = 60 000
𝑥 = 400 washing machines
Let 𝑝 𝑥 = 20 − 𝑥 and 𝐶 𝑥 = 2𝑥 + 56, 𝑤ℎ𝑒𝑟𝑒 0 ≤ 𝑥 ≤ 20 be the demand and
cost functions respectively.
1. Find the marginal cost
2. Express the revenue in terms of 𝑥 and find the marginal revenue.
3. Determine the break-even points algebraically.
4. Graph cost function and revenue function on the same set of axes.
5. Determine profit and marginal profit
6. Evaluate marginal profit at 𝑥 = 7 and 𝑥 = 11 and interpret the results.
1. 𝐶 ′ 𝑥 = 2.
2. 𝑅 𝑥 = 𝑥𝑝 𝑥 = 𝑥 20 − 𝑥 = 20𝑥 − 𝑥 2 . 𝑅′ 𝑥 = 20 − 2𝑥.
3. 𝑅 𝑥 = 𝐶 𝑥 = 20𝑥 − 𝑥 2 = 2𝑥 + 56 ⇒ 𝑥 2 − 18𝑥 + 56 = 0
𝑥 − 14 𝑥 − 4 = 0 ⇒ 𝑥 = 14 𝑜𝑟 𝑥 = 4
4.
5. 𝑃 𝑥 = 𝑅 𝑥 − 𝐶 𝑥 = 20𝑥 − 𝑥 2 − 2𝑥 + 56 = −𝑥 2 + 18𝑥 − 56. 𝑃′ 𝑥 = −2𝑥 + 18
6. 𝑃′ 7 = −2 7 + 18 = 4 > 0 𝑃𝑟𝑜𝑓𝑖𝑡 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝑃′ 11 = −2 11 + 18 = −4 < 0 (𝑃𝑟𝑜𝑓𝑖𝑡 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔)
Sigma Notation
Σ is a Greek letter meaning “sum”.
Suppose you have a list of real numbers, say, 𝑎𝑚 , 𝑎𝑚+1 , 𝑎𝑚+2 , … , 𝑎𝑛 where 𝑚 ≤
𝑛.
Then 𝑛𝑖=𝑚 𝑎𝑖 = 𝑎𝑚 + 𝑎𝑚+1 + ⋯ + 𝑎𝑛
e.g
6 3
3
3
3
𝑖
=
4
+
5
+
6
4
3
4
5
6
23
Express + + + + ⋯ + in sigma notation
7
8
9
10
27
23 𝑖
21 𝑖+2
𝑖=3 𝑖+4 or 𝑖=1 𝑖+6
Properties
1.
2.
𝑛
𝑛
𝑐𝑎
=
𝑐
𝑖
𝑖=𝑚
𝑖=𝑚 𝑎𝑖
𝑛
𝑛
(𝑎
±
𝑏
)
=
𝑖
𝑖=𝑚 𝑖
𝑖=𝑚 𝑎𝑖 ±
Beware!!!
𝑛
𝑖=𝑚 𝑎𝑖 𝑏𝑖 ≠
𝑎𝑖
𝑛
𝑖=𝑚 𝑏 ≠
𝑖
𝑛
𝑖=𝑚 𝑎𝑖
𝑛
𝑖=𝑚 𝑎𝑖
𝑛
𝑖=𝑛 𝑏𝑖
𝑛
𝑖=𝑚 𝑏𝑖
𝑛
𝑖=𝑚 𝑏𝑖
Formulas
1.
2.
3.
4.
5.
𝑛
𝑖=1 1 = 𝑛
𝑛
𝑖=1 𝑐 = 𝑐𝑛
𝑛(𝑛+1)
𝑛
𝑖=1 𝑖 =
2
𝑛(𝑛+1)(2𝑛+1)
𝑛
2
𝑖=1 𝑖 =
6
2
𝑛
𝑛+1
𝑛
3 =
𝑖
𝑖=1
2
Simplify and evaluate the following:
2
a) 100
3
+
2𝑖
𝑖=1
b) 𝑛𝑖=1 𝑖(𝑖 − 1)(𝑖 + 3)
a)
𝑛
2
=
𝑖=1 3 + 2𝑖
𝑛
2
𝑖=1 9 + 12𝑖 + 4𝑖
b)
𝑛
𝑖=1 𝑖(𝑖 − 1)(𝑖 + 3) =
=
𝑛
2
𝑖=1 𝑖 𝑖 + 2𝑖 − 3
𝑛
𝑖=1 9 +
=
𝑛
𝑖=1 12𝑖 +
𝑛
3
2
𝑖=1 𝑖 + 2𝑖 − 3𝑖
𝑛
2
𝑖=1 4𝑖 = 9
𝑛
𝑛
𝑛
2
𝑖=1 1 + 12 𝑖=1 𝑖 + 4 𝑖=1 𝑖
𝑛(𝑛+1)
𝑛(𝑛+1)(2𝑛+1)
= 9𝑛 + 12 2 + 4
6
2𝑛(𝑛+1)(2𝑛+1)
= 9𝑛 + 6𝑛 𝑛 + 1 +
3
4𝑛3 +24𝑛2 +47𝑛
=
(subt n=100)
3
𝑛
𝑛
𝑛
3
2
𝑖=1 𝑖 + 2 𝑖=1 𝑖 − 3 𝑖=1 𝑖
𝑛(𝑛+1) 2
𝑛(𝑛+1)(2𝑛+1)
𝑛(𝑛+1)
=
+
2
−
3
2
6
2
4
3
2
3𝑛 +14𝑛 −3𝑛 −14𝑛
=
12
=
Factorial Notation
Definition: For each natural number 𝑛, the quantity 𝑛 − 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑎𝑙 denoted by
𝑛!, is the product of all natural numbers from 1 to 𝑛, that is 𝑛! = 𝑛 𝑛 − 1 (𝑛 −
Taylor series
If we want to estimate the value of a function at a certain point 𝑥, we can use
the Taylor series which is a representation of a function as an infinite sum of
terms.
Formula:
nth degree-Taylor polynomial centred around 𝑎
𝑘 𝑎 𝑥−𝑎 𝑘
𝑓
𝑇𝑛 = 𝑛𝑘=0
𝑘!
𝑓′(𝑎)(𝑥−𝑎)
𝑓"(𝑎)(𝑥−𝑎)2
𝑓3 𝑎 𝑥−𝑎 3
𝑓𝑛 𝑎 𝑥−𝑎 𝑛
=𝑓 𝑎 +
+
+
+ ⋯+
1!
2!
3!
𝑛!
e.g Find the Taylor series of 𝑓 𝑥 = cos(3𝑥) centred at 𝑎 = 0
𝑓 𝑥 = cos 3𝑥 ,
𝑓 ′ 𝑥 = −3 sin 3𝑥 ,
𝑓"(𝑥) = −9cos(3𝑥),
𝑓 ′′′ 𝑥 = 27 sin 3𝑥 ,
𝑓 𝑖𝑣 𝑥 = 81 cos 3𝑥 ,
𝑇𝑛 𝑥
𝑓(0) = cos 0 = 1.
𝑓′ 0 = 0.
𝑓′′(0) = −9.
𝑓 ′′′ 0 = 0.
𝑓 𝑖𝑣 0 = 81.
𝑓′(𝑎)(𝑥−𝑎)
𝑓"(𝑎)(𝑥−𝑎)2
𝑓3 𝑎 𝑥−𝑎 3
𝑓𝑛 𝑎 𝑥−𝑎 𝑛
=𝑓 𝑎 +
+
+
+ ⋯+
1!
2!
3!
𝑛!
2
4
9
81
9𝑥
81𝑥
−1 𝑖 3𝑥 2𝑖
= 1 + 0 − 2! 𝑥 2 + 0 + 4! 𝑥 4 + ⋯ = 1 − 2! + 4! + ⋯ = ∞
𝑖=0
2𝑖 !
Definitions
𝑛 − 𝑡ℎ 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑡𝑒𝑟𝑚 𝑜𝑓 𝑓 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎 ∶ 𝑅𝑛 𝑥 = 𝑓 𝑥 − 𝑇𝑛 (𝑥)
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑇𝑒𝑟𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑥: |𝑅𝑛 𝑥 |
Taylor’s Theorem
Suppose that 𝑛 + 1 −times differentiable in each point of an open interval 𝐼 containing 𝑎.
For each 𝑥 ∈ 𝐼, there exists at least one point 𝑐 between 𝑥 and 𝑎 such that
𝑅𝑛 𝑥
𝑓 𝑛+1 𝑐 𝑥−𝑎 𝑛+1
=
𝑛+1 !
Question
Consider the function 𝑓 𝑥 = ln 𝑥.
1. Determine the third-degree Taylor polynomial 𝑇3 (𝑥) about 𝑎 = 2.
2. Determine the remainder term 𝑅3 (𝑥).
3. Determine the absolute remainder term in 𝑥 = 1: |𝑅3 1 |
𝑓 𝑥 = ln 𝑥,
1
𝑓′ 𝑥 = 𝑥 ,
𝑓 2 = ln 2.
1
𝑓 ′ 2 = 2.
𝑓 ′′ 𝑥 = − 2 ,
𝑓 ′′ 2 = − .
2
1
𝑥
𝑓 ′′′ 𝑥 = 𝑥 3 ,
𝑇𝑛 𝑥 = 𝑓 𝑎
2
1
4
𝑓 ′′′ 2 = 8.
𝑓′(𝑎)(𝑥−𝑎)
𝑓"(𝑎)(𝑥−𝑎)2
𝑓3 𝑎 𝑥−𝑎 3
𝑓𝑛 𝑎 𝑥−𝑎 𝑛
+
+
+
+ ⋯+
1!
2!
3!
𝑛!
2
−1
3
2
𝑥−2
𝑥−2
1
1
1 𝑥−2 2
8
= ln 2 + 2 𝑥 − 2 + 4 2!
𝑓𝑛+1 𝑐 𝑥−𝑎 𝑛+1
=
= R3 x
𝑛+1 !
+
3!
= ln 2 + 2 𝑥 − 2 −
𝑓4 𝑐 𝑥−2 4
2. 𝑅𝑛 𝑥
=
4!
−6
4 𝑐 𝑥−2 4
𝑥−2 4
6
𝑓
− 𝑥−2 4
4
𝑐
4
Note that 𝑓 𝑥 = − 𝑥 4, so
= 24 = 4𝑐 4
4!
− 1−2 4
−1
1
3. 𝑅3 (1) = 4𝑐 4 = 4𝑐 4 = 4𝑐 4
4∙2!
2
+ 8∙3! 𝑥 − 2 3
Maclaurin series
This is the special case of the Taylor series of a function where 𝑎 = 0 and is in the form
𝑘
𝑘
∞ 𝑓 0 𝑥
=𝑓 0
𝑘=0
𝑘!
or
𝑓′(0)
𝑓"(0)𝑥 2
𝑓3 0 𝑥 3
+ 1! 𝑥 + 2! + 3! + ⋯
Finding the maximum error in the estimation is as follows: |𝑓 𝑥 − 𝑀𝑎𝑐𝑙𝑎𝑢𝑟𝑖𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛|
e.g. Find the error in the point 𝑥 = 1 if 𝑓 𝑥 = cos 𝑥 and 𝑔 𝑥
𝐸𝑟𝑟𝑜𝑟 = 𝑓 𝑥 − 𝑔 𝑥
= cos 1
1𝑛
− 𝑛!
= | cos 1
xn
= 𝑛! is the Maclaurin function.
𝑥𝑛
| Note that lim 𝑛! = 0.
𝑥→∞
EXTRAS
•The Taylor series depends on the point around which it is calculated.
•The value of the function is equal to its Taylor series in a certain interval, then
we must have
𝑎𝑛+1
𝑛→∞ 𝑎𝑛
lim
< 1.
𝑎𝑛+1
𝑛→∞ 𝑎𝑛
• The series converges if lim
< 1 and the radius of convergence is 1.
• The value of the function is equal to its Taylor series for all real numbers if either the remainder approaches 0 if
𝑎
𝑛 → ∞(𝑅𝑛 𝑥 = 0) or if lim 𝑎𝑛+1 = 0 and then the radius of convergence is infinity.
𝑛→∞
𝑛