Linear approximation and differentials ( Section 2.9)

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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 10 o ≈ 0.173648 ? They use various methods of approximation, one of which is Taylor polynomial approximation

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Linear approximation and differentials(Section 2.9)

Alex Karassev

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)

y

y = f(x)

f(a)

x

a

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

y

f(a) + f′(a) (x - a)

y = f(x)

f(x)

f(a)

x

a

x

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

y

y = L(x)

L(x)

y = f(x)

f(x)

f(a)

x

a

x

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

y

y = L(x) = 2 + ¼ (x - 4)

2.25

√5

y = √x

2

a = 4

5

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
• 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

y

y = L(x)

L(x)

y = f(x)

dy

f(x)

∆y

f(a)

x

x

a

dx = x – a

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

y

dy

y = L(x)

y = f(x)

dy

dy = f′(a) dx

dx

dx

x

x

a

dx = x – a

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
• Find the differential if a = 4 and x = 5
• Find the differential if a = 9 and x = 8
Application of differentials:estimation of errors
• ProblemThe 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%