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Linearization and Differentials
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Overview
1. Linearization
2. Examples of linearization
3. Example with Mathematica
4. Differential of function
5. Properties of differentials
6. Examples of calculation of differentials of functions
7. Differentials with Mathematica
8. Applications
References
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1. Linearization
Normally, when numerical values of some function
f (x) at given points have to be calculated we meet the
following situations:
 The formula of the function is complicated, such as
=
f ( x) sin( π + x )
 The results of the computations are practically
2
= 0.28571... ≈ 0.286
example:
7
always rounded off, for
 The most of the real numbers are replaced by some
rational number with a given accuracy, for instance
2 ≈ 1.4142 or π ≈ 3.141593 .
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This means that it the most cases we cannot abtain
the exact requested value. Moreover, an approximate
value sufficiently closed to the exact one is fully
acceptable.
In this lesson we will show how to avoid the above
difficulties by approximating the functions by simpler
ones that give the accuracy we want and are easier to
work with.
We will discuss the approximation to every
differentiable function in the neighbourhood of a point,
based on tangent line to the graphics of the function at
the same point. This is called linearization.
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The idea of a linearization of
a part of a function by using the
tangent at some point is seen in
Fig. 1. The tangent t(x) (green
line) is drawn to f (x) for x=a.
In some small interval
(neighbourhood) of a to either
side, denoted by (a-ε, a+ε), we
observe that the values of f and
the tangent line t are very
closed.
y
f (x)
t (x)
(a, f (a))
0
a-ε a a+ε
x
Fig. 1 Tangent t (x) to f
(x) at (a, f (a)) is very
closed to the function for
x near a.
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Equation of the tangent
The tangent to a differentiable real function ƒ(x), at a
point x=a passes through the point (a, ƒ(a)), so its
point-slope equation is:
y = ƒ(a) + ƒ′(a) (x – a).
Thus, this tangent line is the graph of the linear
function
t(x) =ƒ(a) + ƒ′(a) (x – a).
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Definition 1
If ƒ(x) is a differentiable real function at x=a, then
the approximating function
t (x) =ƒ(a) + ƒ′(a) (x – a)
(1)
is called the linearization of f at a.
The approximation
f (x) ≈ t (x)
of f by t is the standard linear approximation of f at
a. The point x=a is the center of the approximation.
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The accuracy of the approximation can be measured
by different formulas.
Definition 2
For simplicity we will use the absolute value of the
difference
=
d
f ( x) − t ( x)
(2)
Notes. The utility of a linearization is its ability to
replace a complicated formula by a simpler one over
some interval of values. A linear approximation
normally loses accuracy away from its center [1].
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2. Examples of linearization
Example 1. Find the linearization of f ( x) = x at
x=1. Estimate the accuracy d.
2
Solution We differentiate:
f ′( x) = 2 x .
For x=1 we have f (1) = 1 and f ′(1) = 2 . The
linearization is
t ( x) =1 + 2( x − 1) =2 x − 1 .
To find the accuracy of approximation we calculate the
values of f and t and compare them regarding (2):
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Approximation True value
t ( x)
x
1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
1.4
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Accuracy
f ( x) =
d f ( x) − t ( x)
1
0
1.1025
0.0025
1.21
0.01
1.3225
0.0225
1.44
0.04
1.5625
0.0625
1.69
0.09
1.8225
0.1225
1.96
0.16
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The same type of accuracy error will give the
approximation to the left side hand of x=1.
From the previous table we observe that the
accuracy is about 0.01 ( two decimal digits) for x in
the interval (0.9, 1.1). But for x=1.4 and more, the
accuracy is greater than 0.1, so not acceptable.
Example 2. Find the linearization of
x
f ( x) =
(3)
x +1
at x=0. Find the interval for an accuracy of
approximation of d=0.001.
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Solution We calculate the first derivative:
2+ x
f ′( x) =
2( x + 1) x + 1 .
For x=0 we have f (0) = 0 and f ′(0) = 1 . The
linearization is
(4)
t ( x) = x .
This way we obtained
x
=
f ( x)
≈x
x +1
near the point x=0. As in the previous example we
give the table of accuracy only for x >0. All
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calculations are done with round-off error of 0.00001.
Approximation True value
x
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
t ( x)
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Accuracy
f ( x) =
d f ( x) − t ( x)
0.00000
0.00000
0.00995
0.00005
0.01980
0.00020
0.02956
0.00044
0.03922
0.00078
0.04880
0.00120
0.05828
0.00172
0.06767
0.00233
0.07698
0.00302
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We see that for x ∈ (0, 0.05) we obtain accuracy
less than 0.001. The graphs of the function and its
linearization are shown in Fig. 2.
t(x)
0.4
f(x)
0.2
0.4
0.2
0.2
0.2
0.4
0.4
0.6
Fig. 2 Linear approximation (in green color)
of the function (3) near the point x=0.
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Example 3. Find the linearization of the same
function (3) at x=2.
Solution From (3)-(4) at x=2 we find
2
2
′(2)
f (2)
=
≈ 1.15470 and f =
≈ 0.38490 .
3
3 3
The linearization now is
2
t (=
x)
(1 + x) ≈ 0.3849(1 + x)
.
(5)
3 3
This way we obtained
x
f ( x) =
≈ 0.3849(1 + x)
x +1
near the point x=2 with a round-off error of 0.00001.
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The table of accuracy is presented for the interval
[1.7, 2.3]. All calculations are done with round-off
error of 0.00001.
Approximation True value
x
t ( x)
1.7
1.8
1.9
2
2.1
2.2
2.3
1.03923
1.07772
1.11621
1.15470
1.19319
1.23168
1.27017
Accuracy
f ( x) =
d f ( x) − t ( x)
1.03459
0.00464
1.07571
0.00201
1.11572
0.00049
1.15470
0.00000
1.19272
0.00047
1.22984
0.00184
1.26611
0.00406
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The obtained accuracy is less than 0.005.
The graphs of the function (3) and its linearization
(5) in vicinity of x=2 are shown in Fig. 3.
t(x)
1.4
f(x)
1.2
1.0
1.5
2.0
2.5
3.0
Fig. 3 Linear approximation (5) of the
function (3) near the point x=2.
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In Fig. 4 we observe that the linearization (5) is good
only in a narrow local region, near the point x=2.
4
3
2
1
1
2
t(x)
f(x)
2
4
6
8
10
Fig. 4 The accuracy of the linear approximation
(5) away from point x=2 is not acceptable.
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3. Example with Mathematica
Example 4. Find the linearization of
x
f ( x) =
1 + x near x= 1.
Solution We start by a simple Mathematica code
with the definition of the function and the calculation
of its first derivative. The respective Mathematica
input and output are:
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We define also the derivative as a function:
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Now for a=1 we define the corresponding
linearization of the function according to (1):
This can be used for approximation of f ( x) .
Finally, lets to draw the graphics by:
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We will obtain:
0.8
0.6
0.4
0.2
0.5
1.0
1.5
2.0
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4. Differential of a function
In calculus, the differential represents the principal
part of the change in a function y = ƒ(x) with respect
to changes in the independent variable. We note that in
fact, the principal part in the change of a function is
expressed by using the linearization of the function at a
given point. Differentials are often constrained to be
very small quantities.
The notation of differential was introduced by
Gottfried Leibniz (1646-1716).
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Definition 3
If y=f (x) is a differentiable real function in some
interval and dx is a small change of the independent
variable. The differential dy is
dy = f ′( x) dx
(6)
The variable dy is always a dependent variable,
depending on both x and dx. If dx is given a specific
value and x is a particular number in the domain of the
function ƒ, then the numerical value of dy is
determined [1].
dy
Formally, by Leibnitz notation: f '( x) = dx .
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y
y = f ( x)
∆=
y f (a + d )x− f (a )
=
∆t f ′=
(a )dx dy
(a, f (a ))
dx = ∆x
t ( x)
a
a + dx
x
Fig. 5 Geometrically, the differential dy is exactly the change
∆t in the linearization of f when x=a changes by a small
amount dx=∆x.
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More precisely, by Fig. 5 and (1) for x=a+dx we
calculate:
∆t= t (a + d x) − t (a )
= f (a ) + f ′(a ) ( (a + d )x− a ) − f (a )
= f ′(a )dx
With (6) it is also used the notation:
df = f ′( x) dx
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Gottfried Wilhelm Leibniz
(1646-1716)
Lebnitz is world-famous
mathematician and philosopher,
born in Germany.
It is considered that Lebnitz developed the
principles of calculus independently of Isaac
Newton.
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5. Properties of differentials
They are analogous to the derivative rules.
1) dc=0 - differential of a constant
2) d(cu)= cdu=0 - product with a constant
3) d(u±v)= du ± dv - differential of a sum or difference
of two functions (terms)
4) d(u.v)= v.du +u .dv - differential of a product of two
functions (terms)
 u  v.du − u.dv
d =
2
- differential of a quotient, if dx≠0
5)  v 
v
6) d ( u (v( x)) ) = du. dv - chain rule
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6. Examples of calculation of
differentials of functions
Example 5. Let y = 5 x − 2 x + 4 .
(a) Find the differential dy.
(b) Find the value of dy when x = 1 and dx =0.1.
3
Solution
dy= y ' dx=
(5x
(
)
3
)
′
− 2 x + 4 dx=
(15x − 2) dx
2
(a)
(b) Substituting x = 1 and dx =0.1 in (a) we have
dy = 15.1 − 2 0.1 = 1.3
2
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Example 6. Finding differentials of functions
1+ x
(b) y = x 2
(a) y = x cos x
(d) y = ln( x )
(c) y = sin(3 x)
Solution
dy y=
' dx
(a)=
′
x
cos
x
=
(
) dx
( cos x − x sin x ) dx
−x − 2x
x − (1 + x)2 x
=
=
dy y=
' dx
dx
dx
4
4
(b)
x
x
2
2
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(c) dy
(d)
sin(3 x) )′ dx
(=
(
dy
=
3cos(3 x)dx
)
′
=
ln( x ) dx
1
1
1 1
dx
2 ln( x ) x 2 x
dx
4 x ln( x )
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7. Differentials with Mathematica
Example 7. Find differentials of the following
functions using Mathematica software system:
a) x
2
2
b) x sin x
x +1
c) x − 1
d) ln
x + 2x + 1
3
Solution In Mathematica the internal function for
differentials has the form:
Dt[expression]
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For solving Example 7 we obtain the following
results:
a)
b)
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c)
d)
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8. Applications
Application of differential in economics:
Example 8. It is established that the “1000 Ballons”
Ltd. company accumulated its income f in millions
dollars according to the empirical expression
=
f (t ) 1.7t + 2.4 , where t is the number of years.
2
Find the expected next income at moment t +dt for
t=3 years and dt=0.5 year.
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Solution Using Definition 3 we have the increment
df = f ′(t ) dt
(
)
′
df =1.7t + 2.4 dt =3.4t dt =3.4*3*0.5 =5.1
2
This way
f (3.5) ≈ f (3) +=
df
(1.7 *3
2
)
+ 2.4 +=
5.1
17.7 + 5.1 =
22.8
Or the expected income of the company will be
about $ 22.8 million.
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References:
[1] G. B. Thomas, M. D. Weir., J. Hass, F. R. Giordano,
Thomas’ Calculus including second-order differential
equations, 11 ed., Pearson Addison-Wesley, 2005.
[2] http://www.wolfram.com/products/mathematica/index.html
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