Benginning Calculus Lecture notes 8 - linear, quadratic approximation

Beginning Calculus
Applications of Di¤erentiation
- Approximations and Di¤erentials -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,
FSMT - UPSI
(LECTURE SLIDES SERIES)
VillaRINO DoMath, FSMT-UPSI
(DA3) Applications of Di¤erentiation - Approximations and Di¤erentials 1 / 30

Linear Approximation Quadratic Approximation Di¤erentials
Outlines
Linear Approximation
Quadratic Approximation
Use di¤erentials to estimate values.
Compare linear approximations and di¤erentials.

Linear Approximation
De…nition 1
Let y = f (x) be a curve, and (x0, f (x0)) be a point on the curve.The
linear approximation of f near x = x0 (x 0) is
f (x) f (x0) + f 0
(x0) (x x0) (1)
where f (x0) + f 0 (x0) (x x0) is the equation of the tangent line near
x = x0.

Example
The linear approximation of f (x) =
p
x + 3 near x = 1:
f (x) =
p
x + 3, f 0
(x) =
1
2
p
x + 3
f (1) = 2, f 0
(1) =
1
4
f (x) f (x0) + f 0
(x0) (x x0)
= f (1) + f 0
(1) (x 1)
= 2 +
1
4
(x 1)
=
7 + x
4
)
p
x + 3
7 + x
4

Example - continue
-4 -2 0 2 4
1
2
3
x
y
p
x + 3
7 + x
4
only near x = 1.
p
3.98 = 1. 995 0
p
3 + 0.98
7 + 0.98
4
= 1.995
p
4.05 = 2. 012 5
p
3 + 1.05
7 + 1.05
4
= 2.0125
p
8 = 2. 828 4
p
3 + 5
7 + 5
4
= 3

Example
The linear approximation of f (x) = ln x near 1:
f (x) = ln x, f 0
(x) =
1
x
f (1) = 0, f 0
(1) = 1
f (x) f (1) + f 0
(1) (x 1)
= 0 + (1) (x 1)
= x 1
) ln x x 1

Example - continue
ln x x 1
-1 1 2
-1
1
2
x
y

Remark
f 0
(x0) = lim
∆x!0
∆y
∆x
= lim
∆x!0
f (x0 + ∆x) f (x0)
∆x
lim
∆x!0
∆y
∆x
= f 0
(x0)
∆y
∆x
f 0
(x0) (2)
Equation (1) is equivalence to Equation (2).
f (x) f (x0) + f 0
(x0) (x x0) ,
∆y
∆x
f 0
(x0) (3)

Proof of Remark
Proof:
∆y
∆x
f 0
(x0)
∆y f 0
(x0) ∆x
f (x) f (x0) f 0
(x0) (x x0)
f (x) f (x0) + f 0
(x0) (x x0)

Linear Approximations Near 0
f (x) f (0) + f 0
(0) x (4)
sin x :
cos x :
ex :

Geometric Representation of Linear Approximation Near 0
sin x x
-4 -2 2 4
-4
-2
2
4
x
y

cos x 1
-4 -2 2 4
-1.0
-0.5
0.5
1.0
x
y

ex 1 + x
-4 -2 0 2 4
2
4
x
y

More Linear Approximation Near 0
f (x) f (0) + f 0
(0) x
ln (1 + x)
(1 + x)r

Approximate The Values
ln (1.5) = 0.405 47
ln (1.3) = 0.262 36
ln (1.1) = 0.095 31
The approximations get more accurate as x takes the values closer
and closer to 0.

Example - Linear Approximation Near 0
e 3x
p
1 + x
= e 3x
(1 + x) 1/2

Quadratic Approximation
Quadratic approximation is used when linear approximation is not enough.
f (x) f (x0) + f 0
(x0) (x x0) +
f 00 (x0)
2
(x x0)2
(5)

Discussion on Quadratic Approximation near 0
f (x) f (x0) + f 0
(x0) (x x0) +
f 00 (x0)
2
(x x0)2
(6)

Quadratic Approximation Near 0
f (x) f (0) + f 0
(0) x +
f 00 (0)
2
x2
(7)
sin x :
cos x :
ex :

Geometric Representation of Quadratic Approximation
Near 0
sin x x
-4 -2 2 4
-4
-2
2
4
x
y

Near 0
cos x 1
1
2
x2
-4 -2 2 4
-2
-1
1
2
x
y

Near 0
ex 1 + x +
1
2
x2
-4 -2 0 2 4
2
4
x
y

More on Quadratic Approximation Near 0
f (x) f (0) + f 0
(0) x +
f 00 (0)
2
x2
ln (1 + x)
(1 + x)r

Example
Linear approximation of ln (1 + x) near x = 0 :
Quadratic approximation of ln (1 + x) near x = 0 :
Quadratic approximation gives much more accuracy than linear
approximation (near x = 0 ).

Example - Quadratic Approximation Near 0
e 3x
p
1 + x
= e 3x
(1 + x) 1/2

Linear Approximation of e Near 0
ak = 1 +
1
k
k
! e as k ! ∞
Take ln:
ln ak = ln 1 +
1
k
k
= k ln 1 +
1
k
k
1
k
= 1
with x =
1
k
. (Note: as k ! ∞, x ! 0 )
ln ak ! 1 as k ! ∞ near x = 0.
The rate of convergence (how fast ln ak ! 1)
ln ak 1 ! 0

Quadratic Approximation of e Near 0
ln ak = ln 1 +
1
k
k
= k ln 1 +
1
k
k
1
k
1
2k2
= 1
1
2k
ln ak ! 1 as k ! ∞ near x = 0.

Di¤erentials
De…nition 2
Let y = f (x) . The di¤erential of y (or di¤erential of f )is denoted by
dy = f 0
(x) dx
,
dy
dx
= f 0
(x)
Leibniz interpretation of derivative as a ratio of "in…nitesimals".

Use in Linear Approximations
dx replaces ∆x
dy replaces ∆y

Example
Estimate: (64.1)1/3

Benginning Calculus Lecture notes 8 - linear, quadratic approximation

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Benginning Calculus Lecture notes 8 - linear, quadratic approximation