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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 - PowerPoint PPT Presentation
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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)
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
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
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%