Overview 1. Linearization 2. Examples of linearization 3. Example - - PowerPoint PPT Presentation

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BUILDING UP VIRTUAL MATHEMATICS LABORATORY

Partnership project LLP-2009-LEO-МP-09, MP 09-05414

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

( ) sin( ) f x x = + π

• The results of the computations are practically

always rounded off, for example:

2 0.28571... 0.286 7 = ≈

• The most of the real numbers are replaced by some

rational number with a given accuracy, for instance

2 1.4142 ≈

• r

3.141593 π ≈

.

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This means that it the most cases In this lesson we will show how to avoid the above difficulties by approximating the functions by simpler

• nes that give the accuracy we want and are easier to

work with. we cannot abtain the exact requested value. Moreover, an approximate value sufficiently closed to the exact one is fully acceptable. 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

• bserve that the values of f and

the tangent line t are very closed.

t (x) y x (a, f (a)) f (x) a-ε a a+ε

• 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)

• f 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

2

( ) f x x =

at x=1. Estimate the accuracy d. Solution We differentiate:

( ) 2 f x x ′ =

. For x=1 we have

(1) 1 f = and (1) 2 f ′ =

. The linearization is

( ) 1 2( 1) 2 1 t x x x = + − = − .

To find the accuracy of approximation we calculate the values of f and t and compare them regarding (2):

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x Approximation

( ) t x

True value

( ) f x

Accuracy

( ) ( ) d f x t x = −

1 1 1 1.05 1.1 1.1025 0.0025 1.1 1.2 1.21 0.01 1.15 1.3 1.3225 0.0225 1.2 1.4 1.44 0.04 1.25 1.5 1.5625 0.0625 1.3 1.6 1.69 0.09 1.35 1.7 1.8225 0.1225 1.4 1.8 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

( ) 1 x f x x = +

(3) 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 ( ) 2( 1) 1 x f x x x + ′ = + +

. (4) For x=0 we have

(0) f =

and

(0) 1 f ′ = . The

linearization is

( ) t x x =

. This way we obtained

( ) 1 x f x x x = ≈ +

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.

x Approximation

( ) t x

True value

( ) f x

Accuracy

( ) ( ) d f x t x = −

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.00000 0.00995 0.01980 0.02956 0.03922 0.04880 0.05828 0.06767 0.07698 0.00000 0.00005 0.00020 0.00044 0.00078 0.00120 0.00172 0.00233 0.00302

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We see that for

(0, 0.05) x∈

we obtain accuracy less than 0.001. The graphs of the function and its linearization are shown in Fig. 2.

f (x ) t(x ) 0.4 0.2 0.2 0.4 0.6 0.4 0.2 0.2 0.4

• Fig. 2 Linear approximation (in green color)
• f 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) 1.15470 3 f = ≈

and

2 (2) 0.38490 3 3 f ′ = ≈

The linearization now is .

2 ( ) (1 ) 0.3849(1 ) 3 3 t x x x = + ≈ +

. (5) This way we obtained

( ) 0.3849(1 ) 1 x f x x x = ≈ + +

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.

x Approximation

( ) t x

True value

( ) f x

Accuracy

( ) ( ) d f 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 1.03459 1.07571 1.11572 1.15470 1.19272 1.22984 1.26611 0.00464 0.00201 0.00049 0.00000 0.00047 0.00184 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.

f (x ) t(x ) 1.5 2.0 2.5 3.0 1.0 1.2 1.4

• 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

• nly in a narrow local region, near the point x=2.

f (x ) t(x ) 2 4 6 8 10 2 1 1 2 3 4

• 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

( ) 1 x f x x = +

near x= 1. Solution We start by a simple Mathematica code with the definition of the function and the calculation

• f 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:

0.5 1.0 1.5 2.0 0.2 0.4 0.6 0.8

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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]. Formally, by Leibnitz notation:

'( ) dy f x dx =

.

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dx x = ∆

( ) t x a dx +

( , ( )) a f a

a x ( ) t f a dx dy ′ ∆ = =

• 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.

( ) y f x = y ( ) ( ) y f a d x f a ∆ = + −

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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 ′ =

(7)

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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

• f two functions (terms)

4) d(u.v)= v.du +u .dv - differential of a product of two functions (terms) 5)

2

. . u v du u dv d v v −   =    

6)

• differential of a quotient, if dx≠0

( )

( ( )) . d u v x du dv =

• chain rule

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• 6. Examples of calculation of

differentials of functions

Example 5. Let

3

5 2 4 y x x = − +

(a) Find the differential dy. . (b) Find the value of dy when x = 1 and dx =0.1. Solution (a)

( ) ( )

3 2

' 5 2 4 15 2 dy y dx x x dx x dx ′ = = − + = −

(b) Substituting x = 1 and dx =0.1 in (a) we have

( )

2

15.1 2 0.1 1.3 dy = − =

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Example 6. Finding differentials of functions (a)

cos y x x =

(b)

2

1 x y x + =

(c)

sin(3 ) y x =

(d)

ln( ) y x =

Solution (a)

( ) ( )

' cos cos sin dy y dx x x dx x x x dx ′ = = = −

(b)

2 2 4 4

(1 )2 2 ' x x x x x dy y dx dx dx x x − + − − = = =

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(c)

( )

sin(3 ) 3cos(3 ) dy x dx x dx ′ = =

(d)

( )

1 1 1 ln( ) 2 2 ln( ) dy x dx dx x x x ′ = = 1 4 ln( ) dx x x =

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• 7. Differentials with Mathematica

Example 7. Find differentials of the following functions using Mathematica software system: a)

2

x

b)

2 sin

x x

c)

1 1 x x + −

d)

3

ln 2 1 x x + +

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

2

( ) 1.7 2.4 f t t = +

, where t is the number of years. 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 ′ =

( )

2

1.7 2.4 3.4 3.4*3*0.5 5.1 df t dt t dt ′ = + = = =

This way

( )

2

(3.5) (3) 1.7*3 2.4 5.1 17.7 5.1 22.8 f f df ≈ + = + + = + =

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