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Linearization and Differentials

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------1/37 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 fx( )= sin(π + x ) . The results of the computations are practically 2 =0.28571... ≈ 0.286 always rounded off, for example: 7 . 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 . ------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------3/37

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 y a part of a function by using the f (x) t (x) tangent at some point is seen in

Fig. 1. The tangent t(x) (green (a, f (a)) 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 0 a-ε a a+ε x observe that the values of f and Fig. 1 Tangent t (x) to f the tangent line t are very (x) at (a, f (a)) is very closed. 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).

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------6/37 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.

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------7/37 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 − tx () (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 fx()= x2 at x=1. Estimate the accuracy d.

Solution We differentiate: fx′()= 2 x. For x=1 we have f (1)= 1 and f ′(1)= 2 . The linearization is

tx( )=+ 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 Accuracy x tx() fx() d= f() x − tx () 1 1 1 0 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 x fx()= x +1 (3) at x=0. Find the interval for an accuracy of approximation of d=0.001.

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------11/37 Solution We calculate the first derivative: 2 + x fx′()= 2(xx++ 1) 1 . (4) For x=0 we have f (0)= 0 and f ′(0)= 1. The linearization is

tx()= x.

This way we obtained x fx()= ≈ x x +1 near the point x=0. As in the previous example we give the table of accuracy only for x >0. All

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------12/37 calculations are done with round-off error of 0.00001.

Approximation True value Accuracy x tx() fx() d= f() x − tx () 0 0 0.00000 0.00000 0.01 0.01 0.00995 0.00005 0.02 0.02 0.01980 0.00020 0.03 0.03 0.02956 0.00044 0.04 0.04 0.03922 0.00078 0.05 0.05 0.04880 0.00120 0.06 0.06 0.05828 0.00172 0.07 0.07 0.06767 0.00233 0.08 0.08 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 0.4 0.2 0.4 0.6

Fig. 2 Linear approximation (in green color) of the function (3) near the point x=0.

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------14/37 Example 3. Find the linearization of the same function (3) at x=2.

Solution From (3)-(4) at x=2 we find 2 2 f (2)= ≈ 1.15470 f ′(2)= ≈ 0.38490 3 and 33 . The linearization now is 2 tx( )= (1 +≈ x ) 0.3849(1 +x ) 33 . (5) This way we obtained x fx( )=≈+ 0.3849(1x ) x +1 near the point x=2 with a round-off error of 0.00001. ------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------15/37 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 Accuracy x tx() fx() d= f() x − tx () 1.7 1.03923 1.03459 0.00464 1.8 1.07772 1.07571 0.00201 1.9 1.11621 1.11572 0.00049 2 1.15470 1.15470 0.00000 2.1 1.19319 1.19272 0.00047 2.2 1.23168 1.22984 0.00184 2.3 1.27017 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 t (x) 2 f(x) 1

2 4 6 8 10 1 2

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 fx()= 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:

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------20/37 Now for a=1 we define the corresponding linearization of the function according to (1):

This can be used for approximation of fx().

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

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------22/37 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).

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------23/37 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 fx'( ) = Formally, by Leibnitz notation: dx . ------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------24/37 y

y= fx()

∆=y fa( + dx ) − fa ()

∆=t f′() a dx = dy (afa , ( ))

dx= ∆ x

tx()

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 ta( + dx ) − ta () =fa() + f′ ()( a( a + dxa ) −−) fa () = 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.

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------27/37 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 = 5) v v2 - differential of a quotient, if dx≠0 6) d( u( v ( x ))) = du . dv - chain rule

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------28/37 6. Examples of calculation of differentials of functions

Example 5. Let yx=53 −+ 24 x. (a) Find the differential dy. (b) Find the value of dy when x = 1 and dx =0.1.

Solution ′ dy= y' dx = 5 x32 −+ 2 x 4 dx = 15 x − 2 dx (a) ( ) ( ) (b) Substituting x = 1 and dx =0.1 in (a) we have dy =15.12 −= 2 0.1 1.3 ( ) ------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------29/37 Example 6. Finding differentials of functions 1+ x = = y 2 (a) yxcos x (b) x (c) yx= sin(3 ) (d) yx= ln( )

Solution ′ (a) dy= y' dx = ( x cos x) dx=( cos x − x sin x) dx

x22−+(1 xx )2 − x −2 x dy= y' dx = dx= dx (b) xx44

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------30/37 ′ (c) dy= (sin(3 x )) dx= 3cos(3 x ) dx

′ 1 11 dy= ( ln( x ) ) dx = dx (d) 2 ln(x ) xx2 1 = dx 4xx ln( )

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------31/37 7. Differentials with Mathematica

Example 7. Find differentials of the following functions using Mathematica software system: 2 2 a) x b) xxsin x +1 3 c) x −1 d) lnxx++ 2 1

Solution In Mathematica the internal function for differentials has the form: Dt[expression]

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------32/37 For solving Example 7 we obtain the following results:

a)

b)

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------33/37 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 ft( )= 1.7 t + 2.4 , 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.

------Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ------35/37 Solution Using Definition 3 we have the increment ′ df= f() t dt ′ df=+==1.7 t2 2.4 dt 3.4 t dt 3.4*3*0.5 = 5.1 ( ) This way f(3.5)≈ f (3) += df ( 1.7*32 + 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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