differential calculus: a refresher · differential calculus ii –rafael lópez-monti 1...

Differential Calculus II – Rafael López-MontiGWU Math Camp 2014 1Draft for teaching only, do not cite. Please contact me If you find a typo

Differential Calculus: A refresher

(Part 2)

Math Camp, August 2014

Rafael López-MontiDepartment of Economics – PhD Program

George Washington University

[email protected]

mailto:[email protected]

Differential Calculus II – Rafael López-MontiGWU Math Camp 2014

TOTAL DIFFERENTIAL:

2

Definition: If 𝒛 = 𝒇 𝒙𝟏, 𝒙𝟐, … , 𝒙𝒏 is a function mapping from ℝ𝒏 to ℝand 𝒅𝒙𝟏, 𝒅𝒙𝟐, … , 𝒅𝒙𝒏 are arbitrary numbers, the total differential of 𝒛is given by:

𝒅𝒛 =

𝒊=𝟏

𝒏𝝏𝒇(𝒙𝟏, … , 𝒙𝒏)

𝝏𝒙𝒊. 𝒅𝒙𝒊

Let 𝑔 = 𝐺(𝑥1, 𝑥2, … , 𝑥𝑛), and 𝑥𝑖 = 𝑓𝑖 𝑡1, 𝑡2, … , 𝑡𝑚 for 𝑖 = 1,2, … 𝑛 and

𝑗 = 1,2, … 𝑚 be functions. In order to get𝜕𝑔

𝜕𝑡𝑗we can apply the general

chain rule :

𝜕𝑔

𝜕𝑡𝑗=

𝑖=1

𝑛𝜕𝐺(𝑥1, … , 𝑥𝑛)

𝜕𝑥𝑖.𝜕𝑥𝑖

𝜕𝑡𝑗


Geometric illustration of the definition of the differential for functions of two variables: 𝒛 = 𝒇(𝒙, 𝒚)

Define ∆𝑧 = 𝑓 𝑥1 + 𝑑𝑥1, 𝑥2 + 𝑑𝑥2, … , 𝑥𝑛 + 𝑑𝑥𝑛 − 𝑓 𝑥1, 𝑥2, … , 𝑥𝑛

Thus, ∆𝑧 ≈ 𝑑𝑧 when 𝑑𝑥1 , 𝑑𝑥2 , … , 𝑑𝑥𝑛 are all small enough….

Example: Let 𝑧 = 𝑓 𝑥, 𝑦 = 𝑥2 + 𝑥𝑦 + 𝑦2, find the total differential of z:𝑑𝑧 = 2𝑥 + 𝑦 𝑑𝑥 + 𝑥 + 2𝑦 𝑑𝑦

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TAYLOR APPROXIMATIONS

I. Approximations by polynomials (Mac Laurin’s Formula)

Given the functions 𝑓: 𝑋 → ℝ , and 𝑓 ∈ 𝐶𝑚 (i.e. m times continuouslydifferentiable), suppose that 0 ∈ 𝑋. For some 𝑛 ∈ ℕ, 𝑎𝑛𝑑 𝑛 < 𝑚, we

want to construct an 𝒏𝒕𝒉 degree polynomial, 𝑃𝑛: 𝑋 → ℝ, such that thevalue of 𝑓and its first 𝑛 derivatives evaluated at zero are the same asthe values of 𝑃𝑛 and its first 𝑛 derivative at 0.

We know that:𝑃𝑛 𝑥 = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + ⋯ + 𝑎𝑛𝑥𝑛

Note :𝑃𝑛

′ x = 𝑎1 + 2𝑎2𝑥 + ⋯ + 𝑛𝑎𝑛𝑥𝑛−1

𝑃𝑛′′ x = 2𝑎2 + 3.2. 𝑎3𝑥 + ⋯ + 𝑛(𝑛 − 1)𝑎𝑛𝑥𝑛−2

𝑃𝑛′′′ x = 3.2. 𝑎3 + 4.3.2. 𝑎4𝑥 + ⋯ + 𝑛(𝑛 − 1)(𝑛 − 2)𝑎𝑛𝑥𝑛−3

⋮𝑃𝑛

𝑛 x = 𝑛 𝑛 − 1 𝑛 − 2 . . 2.1. 𝑎𝑛

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Evaluated at x=0:𝑃𝑛 0 = 𝑎0

𝑃𝑛′ 0 = 𝑎1

𝑃𝑛′′ 0 = 2𝑎2

𝑃𝑛′′′ 0 = 3.2. 𝑎3 = 3! 𝑎3

⋮𝑃𝑛

𝑛 0 = 𝑛 𝑛 − 1 𝑛 − 2 . . 2.1. 𝑎𝑛 = 𝑛! 𝑎𝑛

Since we want to find 𝑎𝑗 𝑗=0

𝑛such that first 𝑛 derivatives of 𝑓 are the

same as the first 𝑛 derivatives of 𝑃𝑛, then:𝑃𝑛 0 = 𝑓(0)𝑃𝑛

′ 0 = 𝑓′(0)𝑃𝑛

′′ 0 = 𝑓′′(0)𝑃𝑛

′′′ 0 = 𝑓′′′(0)⋮𝑃𝑛

𝑛 0 = 𝑓𝑛(0)

The sequence 𝑎𝑗 𝑗=0

𝑛that satisfies these equalities is:

𝑎0 = 𝑓(0) and ∀𝑘 ∈ ℕ, 𝑘 ≤ 𝑛 then 𝑎𝑘 =1

𝑘!𝑓𝑘(0)

5


Example:

Consider 𝑓: −0.5,0.5 → ℝ defined 𝑓 𝑥 =1

1+𝑥 2 . Taking the derivatives:

𝑓′ 𝑥 = −2(1 + 𝑥)−3;𝑓′′ 𝑥 = 6(1 + 𝑥)−4. Evaluate them at zero: 𝑓′ 0 = −2; 𝑓′′ 𝑥 = 6. Knowing that 𝑓 0 = 1, then:

𝑷𝟏 𝒙 = 𝟏 − 𝟐𝒙 (first-degree polynomial)

𝑷𝟐 𝒙 = 𝟏 − 𝟐𝒙 +𝟏

𝟐!𝟔𝒙𝟐 (second-degree polynomial)

6

Therefore the polynomial that we are looking for is:

𝑷𝒏 𝒙 = 𝒇 𝟎 +𝟏

𝟏!𝒇′ 𝟎 𝒙 +

𝟏

𝟐!𝒇′′ 𝟎 𝒙𝟐 + ⋯ +

𝟏

𝒏!𝒇𝒏 𝟎 𝒙𝒏

Since 𝑷𝒏 𝒙 and 𝒇 𝒙 are very close to each other and have the samederivatives when 𝒙 is close to 0, we say that 𝑷𝒏 𝒙 is an 𝒏𝒕𝒉 -orderapproximation of 𝒇 𝒙 about 0.

𝒇 𝒙 ≈ 𝒇 𝟎 +𝟏

𝟏!𝒇′ 𝟎 𝒙 +

𝟏

𝟐!𝒇′′ 𝟎 𝒙𝟐 + ⋯ +

𝟏

𝒏!𝒇𝒏 𝟎 𝒙𝒏

Note: This is just an approximation of 𝒇 𝒙 .


Graphically:

Note: This method is limited because it requires that 0 ∈ X, and we can onlyapproximate the function about 0!!

7


II. Taylor Approximation

We generalize the previous method by using the Taylor Polynomial.Let 𝑥 ∈ 𝑋, the 𝒏𝒕𝒉 degree polynomial about 𝒙 is a function 𝑃𝑛, 𝑥: 𝑋 → ℝ:

𝑃𝑛, 𝑥 𝑥 = 𝑎0 + 𝑎1(𝑥 − 𝑥) + 𝑎2(𝑥 − 𝑥)2+⋯ + 𝑎𝑛(𝑥 − 𝑥)𝑛

Following the same procedure as in the previous section:𝑃𝑛, 𝑥

′ x = 𝑎1 + 2𝑎2(𝑥 − 𝑥) + ⋯ + 𝑛𝑎𝑛(𝑥 − 𝑥)𝑛−1

𝑃𝑛, 𝑥′′ x = 2𝑎2 + 3.2. 𝑎3(𝑥 − 𝑥) + ⋯ + 𝑛(𝑛 − 1)𝑎𝑛(𝑥 − 𝑥)𝑛−2

𝑃𝑛, 𝑥′′′ x = 3.2. 𝑎3 + 4.3.2. 𝑎4(𝑥 − 𝑥) + ⋯ + 𝑛(𝑛 − 1)(𝑛 − 2)𝑎𝑛(𝑥 − 𝑥)𝑛−3

⋮𝑃𝑛, 𝑥

𝑛 x = 𝑛 𝑛 − 1 𝑛 − 2 . . 2.1. 𝑎𝑛

Now evaluate 𝑃𝑛, 𝑥 𝑥 and its nth derivatives at 𝑥 = 𝑥:

𝑃𝑛, 𝑥 𝑥 = 𝑎0

𝑃𝑛, 𝑥′ 𝑥 = 𝑎1

𝑃𝑛, 𝑥′′ 𝑥 = 2𝑎2

𝑃𝑛, 𝑥′′′ 𝑥 = 3.2. 𝑎3 = 3! 𝑎3 𝑃𝑛, 𝑥

𝑛 𝑥 = 𝑛 𝑛 − 1 𝑛 − 2 . . 2.1. 𝑎𝑛 = 𝑛! 𝑎𝑛

8


As before, we need to find a sequence 𝑎𝑗 𝑗=0

𝑛, such that the value of 𝑓

and its first 𝑛 derivatives evaluated at 𝑥 are the same as the values of𝑃𝑛, 𝑥 and its first 𝑛 derivative at same point. Thus, we find a similar

sequence:

𝑎0 = 𝑓( 𝑥) and ∀𝑘 ∈ ℕ, 𝑘 ≤ 𝑛 then 𝑎𝑘 =1

𝑘!𝑓𝑘( 𝑥)

9

As a result, we obtain the 𝒏𝒕𝒉degree Taylor polynomial approximation of𝒇 about 𝒙:

𝒇 𝒙 ≈ 𝒇 𝒙 +𝟏

𝟏!𝒇′ 𝒙 𝒙 − 𝒙 +

𝟏

𝟐!𝒇′′ 𝒙 𝒙 − 𝒙 𝟐 + ⋯ +

𝟏

𝒏!𝒇𝒏 𝒙 𝒙 − 𝒙 𝒏 = 𝑻𝒇,𝒏, 𝒙 𝒙

The error (or remainder), denoted as 𝑹𝒇,𝒏, 𝒙 𝒙 , of the 𝒏𝒕𝒉degree Taylorpolynomial approximation of 𝒇 about 𝒙 is:

𝑹𝒇,𝒏, 𝒙 𝒙 = 𝒇 𝒙 − 𝑻𝒇,𝒏, 𝒙 𝒙


Example:

Find the third degree Taylor Polynomial for 𝒇 𝒙 =𝟏

𝒙𝟐 about 𝒙 = −𝟏:

𝒇(𝒊) 𝒙 Evaluated at 𝑥 = −1

𝒇(𝟎) 𝒙 =𝟏

𝒙𝟐𝒇(𝟎) −𝟏 = 𝟏

𝒇(𝟏) 𝒙 = −𝟐𝟏

𝒙𝟑𝒇(𝟏) −𝟏 = 𝟐 = 𝟐!

𝒇(𝟐) 𝒙 = 𝟐. (𝟑)𝟏

𝒙𝟒𝒇(𝟐) −𝟏 = 𝟐. 𝟑 = 𝟑!

𝒇(𝟑) 𝒙 = −𝟐. (𝟑). (𝟒)𝟏

𝒙𝟓𝒇(𝟑) −𝟏 = 𝟐. 𝟑 . 𝟒 = 𝟒!

𝒇 𝒙 ≈ 𝟏 +𝟏

𝟏!𝟐 (𝒙 + 𝟏) +

𝟏

𝟐!𝟔(𝒙 + 𝟏)𝟐+

𝟏

𝟑!𝟐𝟒(𝒙 + 𝟏)𝟑

We can also find the pattern for the Taylor series in this case:

𝒇 𝒙 =𝟏

𝒙𝟐 =

𝒏=𝟎

∞𝒇 𝒏 (−𝟏)

𝒏!(𝒙 + 𝟏)𝒏 =

𝒏=𝟎

∞𝒏 + 𝟏 !

𝒏!(𝒙 + 𝟏)𝒏=

𝒏=𝟎

∞

(𝒏 + 𝟏)(𝒙 + 𝟏)𝒏

10


Taylor expansion with two independent variables:

Let 𝑓: ℝ2 ⟶ ℝ, and consider a point ( 𝑥, 𝑦) that belongs to the domain.The function 𝑓 𝑥, 𝑦 can be expressed as the 𝒏𝒕𝒉 degree Taylorpolynomial about ( 𝑥, 𝑦) plus the error term 𝑅𝑓,𝑛,( x, y) 𝑥, 𝑦 :

𝒇 𝒙, 𝒚 = 𝒇 𝒙, 𝒚 + 𝒇𝒙 𝒙, 𝒚 𝒙 − 𝒙 + 𝒇𝒚 𝒙, 𝒚 𝒚 − 𝒚

+𝟏

𝟐!𝒇𝒙𝒙 𝒙, 𝒚 𝒙 − 𝒙 𝟐 + 𝒇𝒚𝒚 𝒙, 𝒚 𝒚 − 𝒚 𝟐 + 𝟐𝒇𝒙𝒚 𝒙, 𝒚 𝒙 − 𝒙 𝒚 − 𝒚

+𝟏

𝟑![𝒇𝒙𝒙𝒙 𝒙, 𝒚 𝒙 − 𝒙 𝟑 + 𝒇𝒚𝒚𝒚 𝒙, 𝒚 𝒚 − 𝒚 𝟑 + 𝟑𝒇𝒙𝒙𝒚 𝒙, 𝒚 𝒙 − 𝒙 𝟐 𝒚 − 𝒚 + 𝟑𝒇𝒙𝒚𝒚 𝒙, 𝒚 (𝒙 − 𝒙) 𝒚 − 𝒚 𝟐]

+ ⋯ + 𝑹𝒇,𝒏,( 𝒙, 𝒚) 𝒙, 𝒚

Work in groups with the following example:

Find the fourth degree Taylor Polynomial for the function:𝒇 𝒙 = 𝒙𝟑 − 𝟏𝟎𝒙𝟐 + 𝟔 about 𝒙 = 𝟑

11


INTEGRATION:

Example:

𝒂𝒙𝟐 + 𝒍𝒏 𝒙 𝒅𝒙 = 𝒂 𝒙𝟐𝒅𝒙 + 𝒍𝒏𝒙 𝒅𝒙

= 𝒂𝒙𝟑

𝟑+ 𝒙. 𝒍𝒏 𝒙 − 𝒙 + 𝑪 = 𝑭 𝒙 + 𝑪

Checking:

𝒅(𝑭 𝒙 +𝑪)

𝒅𝒙= 𝒂𝒙𝟐 + 𝒙.

𝟏

𝒙+ 𝒍𝒏𝒙 − 𝟏

then 𝒅(𝑭 𝒙 +𝑪)

𝒅𝒙= 𝒂𝒙𝟐 + 𝒍𝒏 𝒙

12

Definition: The indefinite Integrals can be defined as:

𝒇 𝒙 𝒅𝒙 = 𝑭 𝒙 + 𝑪 ⇔ 𝑭′ 𝒙 = 𝒇(𝒙)

I. Indefinite Integrals


Integration by Parts:

𝒇 𝒙 𝒈′ 𝒙 𝒅𝒙 = 𝒇 𝒙 𝒈 𝒙 − 𝒇′ 𝒙 𝒈 𝒙 𝒅𝒙

Proof: Left as an exercise.

Example: Integrate

𝑥. 𝑙𝑛 𝑥 𝑑𝑥

Let 𝒇 𝒙 = 𝒍𝒏 𝒙 𝑎𝑛𝑑 𝒈′ 𝒙 = 𝒙

Then 𝒇′ 𝒙 = 𝟏 𝒙 and 𝒈 𝒙 = 𝒙 𝒅𝒙 = 𝒙𝟐

𝟐 [See the Integration Table provided]

Applying the definition of integration by parts:

𝑥. 𝑙𝑛 𝑥 𝑑𝑥 =𝑥2

2𝑙𝑛𝑥 −

1

𝑥.𝑥2

2𝑑𝑥

=𝑥2

2𝑙𝑛𝑥 −

1

4𝑥2 + 𝐶

13


II. Definite Integrals

14

If 𝒇: [𝒂, 𝒃] → ℝ is differentiable and 𝒇′: [𝒂, 𝒃] → ℝ is integrable, then

𝒂

𝒃

𝒇′ 𝒙 𝒅𝒙 = 𝒇 𝒃 − 𝒇(𝒂)

FIRST FUNDAMENTAL THEOREM OF CALCULUS:

Suppose that 𝒇: [𝒂, 𝒃] → ℝ is integrable. Define 𝑭: 𝒂, 𝒃 → ℝ by ∀𝒙 ∈ [𝒂, 𝒃]:

𝑭 𝒙 = 𝒂

𝒙

𝒇 𝒕 𝒅𝒕

If 𝒇 is continuous at 𝒙 ∈ 𝑿, then:

𝑭′ 𝒙 = 𝒇( 𝒙)

SECOND FUNDAMENTAL THEOREM OF CALCULUS:


Recalling:

𝐀 𝒙 = 𝒂

𝒙

𝒇 𝒕 𝒅𝒕 ⟹ 𝑨′ 𝒙 = 𝒇 𝒙

𝐀 𝒙 = 𝒙

𝒃

𝒇 𝒕 𝒅𝒕 ⟹ 𝑨′ 𝒙 = −𝒇 𝒙

The shaded area is 𝑨 𝒙 = 𝒂

𝒙𝒇 𝒕 𝒅𝒕, and the derivative of the area function

𝑨 𝒙 is 𝑨′ 𝒙 = 𝒇 𝒙 by the SFTC

Graphically:

15


Properties of Definite Integrals:

1. b

a

a

bdxxfdxxf )()(

2. a

aaFaFdxxf 0)()()(

3. c

b

c

a

b

adxxfdxxfdxxf )()()( for cba

4. ( ) ( ) [ ( ) ( )]b b b

a a af x dx g x dx f x g x dx

5. b

a

b

adxxfkdxxkf )()(

16

We can also apply integration by parts with definite integrals:

𝒂

𝒃

𝒇 𝒙 . 𝒈′ 𝒙 = (𝒇 𝒙 . 𝒈 𝒙 )𝒂

𝒃−

𝒂

𝒃

𝒇′ 𝒙 . 𝒈(𝒙)


Example: Evaluate

𝟔

𝟎

(𝟐 + 𝟓𝐱)𝒆𝟏𝟑𝒙𝒅𝒙

Step 1. Pick f(x) and g’(x) to get an easier integral to evaluate in the Integrationby Parts formula.

𝒇 𝒙 = 𝟐 + 𝟓𝒙 𝑎𝑛𝑑 𝒈′ 𝒙 = 𝒆𝟏𝟑𝒙

Therefore, 𝒇′ 𝒙 = 𝟓 and 𝒈 𝒙 = 𝒆𝟏

𝟑𝒙 𝒅𝒙 = 𝟑. 𝒆

𝟏

𝟑𝒙

Step 2. Plug into the Integration by Parts formula:

6

0

(2 + 5x)𝑒13𝑥𝑑𝑥 = ( 2 + 5𝑥 . 3. 𝑒

13𝑥)

6

0

− 6

0

5. (3. 𝑒13𝑥)𝑑𝑥

Step 3. Evaluate the new integral and substitute in the formula

6

0

(2 + 5x)𝑒13𝑥𝑑𝑥 = ( 2 + 5𝑥 . 3. 𝑒

13𝑥)

6

0

−( 45. 𝑒13𝑥)

6

0

Step 4. Take care of the limits, what we found is an area.

6

0

(2 + 5x)𝑒13𝑥𝑑𝑥 = ( 15𝑥. 𝑒

13𝑥 − 39𝑒

13𝑥)

6

0

= 0 − 39 − 90𝑒2 − 39𝑒2 = −𝟒𝟏𝟓. 𝟖𝟒

17


Differentiation under the integral sign:

18

The Leibniz’s formula:

𝒅

𝒅𝒙 𝒖(𝒙)

𝒗(𝒙)

𝒇 𝒙, 𝒚 𝒅𝒚 = −𝒇 𝒙, 𝒖 𝒙𝒅𝒖

𝒅𝒙+ 𝒇 𝒙, 𝒗 𝒙

𝒅𝒗

𝒅𝒙+

𝒖(𝒙)

𝒗(𝒙) 𝝏

𝝏𝒙𝒇 𝒙, 𝒚 𝒅𝒚

Note: Here the interval of the integral is a function of x

Suppose 𝒇 𝒙, 𝒚 is a function on the rectangle 𝑹 = 𝒂, 𝒃 𝒙[𝒄, 𝒅], and𝝏𝒇

𝝏𝒚𝒙, 𝒚 is

continuous on 𝑹. Then:𝒅

𝒅𝒚 𝒂

𝒃

𝒇 𝒙, 𝒚 𝒅𝒙 = 𝒂

𝒃 𝝏𝒇

𝝏𝒚𝒙, 𝒚 𝒅𝒙

THEOREM:

Note: a and b are independent of x.


𝛀𝑨

𝒇 𝒙, 𝒚 𝒅𝒙𝒅𝒚 = 𝒂

𝒃

𝒖(𝒙)

𝒗(𝒙)

𝒇 𝒙, 𝒚 𝒅𝒚 𝒅𝒙

The double integral of a function 𝒇 𝒙, 𝒚over the region 𝜴𝑨

Multiple integrals:Definition of the double integral of 𝒇 𝒙, 𝒚 over a rectangle 𝑹 = 𝒂, 𝒃 𝒙[𝒄, 𝒅]:

𝑹

𝒇 𝒙, 𝒚 𝒅𝒙𝒅𝒚 = 𝒂

𝒃

𝒄

𝒅

𝒇 𝒙, 𝒚 𝒅𝒚 𝒅𝒙 = 𝒄

𝒅

𝒂

𝒃

𝒇 𝒙, 𝒚 𝒅𝒙 𝒅𝒚

19

The two iterated integrals are equal for continuous functions.

FUBINI’S THEOREM:

differential calculus: a refresher · differential calculus ii –rafael lópez-monti 1...

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