Linear approximation and differentials(Section 2.9)

Alex Karassev

Linear approximation

Problem of computation

How do calculators and computers know that √5 ≈ 2.236 or sin 10o ≈ 0.173648 ?

They use various methods of approximation, one of which is Taylor polynomial approximation

A simplest case of Taylor polynomial approximation is linear approximation or linearization

Linear approximation

Equation of tangent line to y=f(x) at a isy = f(a) + f′(a) (x - a)

x

y

y = f(x)f(a)

a

Linear approximation

x

y

y = f(x)f(a)

a

If x is near a, we have:f(x) ≈ f(a) + f′(a) (x - a)

x

f(x)f(a) + f′(a) (x - a)

Linear approximation

x

y

y = f(x)f(a)

a

Function L(x) = f(a) + f′(a) (x - a) is called linear approximation (or linearization)of f(x) at a

x

f(x)

L(x)y = L(x)

Example

Find linearization of f(x) = √x at a Use it to find approximate value of √5

Linearization

)(2

1)( Therefore

2

1)(

))(()()(

)(

axa

axL

xxxf

axafafxL

xxf

Approximation of √5

Find a such that √a is easy to compute a is close to 5

Use linearization at a

Take a = 4 and compute linear approximation

Approximation of √5

236.25 :gives Calculator

25.25 Thus

25.24

1245

4

12)5()5(5

and 44

1 2 4

42

14

:obtain we4 aat Therefore

4 ,5 ),(2

1)()(

Lf

xxxL

axaxa

afxL

Approximation of √5

a = 4

y

y = L(x) = 2 + ¼ (x - 4)

y = √x

5

2√5

2.25

Example

Find approximate value of sin 10o

Example

We measure x in radians So, 10o = 10 (π/180) = π/18 radians Consider f(x) = sin x Find a such that

sin(a) is easy to compute a is near π/18

Take a = 0 and compute linear approximation

Solution

f(x) ≈ f(a) + f′(a) (x - a) = f(0) + f′(0) (x - 0)

f(x) = sin x, f′(x) = (sin x) ′ = cos x

Therefore we obtain:sin x ≈ sin(0) + cos(0) (x - 0) = 0 +1(x – 0) = x

Thus sin x ≈ x (when x is near 0)

For x = π/18 we obtain:sin 10o = sin (π/18) ≈ π/18 ≈ 0.1745

Calculator gives:sin 10o ≈ 0.1736

Differentials

Differentials

Compare f(x) and f(a)

Change in y: ∆y = f(x) – f(a)

f(x) ≈ f(a) + f′(a) (x - a)

Therefore ∆y = f(x) – f(a) ≈ (f(a) + f′(a) (x - a)) – f(a) = f′(a) (x - a)

Let x – a = ∆x = dx

Then ∆y ≈ f′(a) (x - a) = f′(a) dx

Differentials

Definition

dy = f′(a) dx is called the differential of function x at a

Thus, ∆y ≈ dy Note: dx = x – a

Differentials

dy = f′(a) dx

x

y

y = f(x)f(a)

a x

f(x)

y = L(x)

dx = x – a

∆y

L(x)dy

dx

Differential as a linear function

dy = f′(a) dx For fixed a, dy is a linear function of dx

x

y

y = f(x)

a x

y = L(x)

dx = x – a

dy

dx

dy

dy = f′(a) dx

Differential at arbitrary point

We can let a vary Then, differential of function f at any number

x is dy = f′(x) dx For each x, we obtain a linear function with

slope f′(x)

df(x) = f′(x) dx

Differentials and linear approximation

dy = f′(a) dx

∆y = f(x) – f(a)

Therefore f(x) = f(a) + ∆y

∆y ≈ dy

Thus f(x) ≈ f(a) + dy

Example

Let f(x) = √x

1. Find the differential if a = 4 and x = 5

2. Find the differential if a = 9 and x = 8

Solution of 1.

236.02545)4()5()()(

25.4/1142

1y and 145

:obtain we4, a and 5 When x

2

1y Therefore

2

1)(

)( )(

ffafxfy

ddx

axdx

dxa

dx

xxf

dxafdyxxf

Solution of 2.

646.27 gives Calculator

667.2)333.0(9)9(7 Therefore

)( :Note

354.03797)9()7()()(

333.03/1)2(92

1y and 297

:obtain we9, a and 7 When x

2

1y

dyf

dyafΔy f(a) f(x)

ffafxfy

ddx

axdxdxa

d

Application of differentials:estimation of errors

Problem

The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Estimate the maximum possible error in computing the volume of the cube.

Solution

Suppose that the exact length of the edge is x and the "ideal" value is a = 30 cm.

Then the volume of the cube is V(x) = x3

Possible error is the absolute value of the difference between the "ideal" volume and "real" volume:

∆V = V(x) – V(a) ∆V ≈ dV = V'(a) dx dx = ± 0.1 V'(x) = (x3)' = 3x2

∆V ≈ dV = V'(a) dx =3a2 dx = ± 3(30)2 (0.1) = ± 270 cm3

So error = |∆V| ≈ 270 cm3

Relative error = |∆V| / V ≈ 270 / 303 = 0.01 = 1%