BUILDING UP VIRTUAL MATHEMATICS LABORATORY Partnership project LLP-2009-LEO- М P-09, MP 09-05414 Linearization and Differentials 1/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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 2/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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 = π + ( ) sin( ) f x x The results of the computations are practically 2 7 = ≈ 0.28571... 0.286 always rounded off, for example: The most of the real numbers are replaced by some rational number with a given accuracy, for instance ≈ π ≈ or . 2 1.4142 3.141593 3/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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 . 4/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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 a - ε a a + ε x 0 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 . 5/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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 ). 6/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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. 7/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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]. 8/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
2. Examples of linearization = 2 ( ) f x x Example 1. Find the linearization of at x= 1. Estimate the accuracy d . Solution We differentiate: ′ = ( ) 2 f x x . f ′ = and = (1) 1 (1) 2 f For x= 1 we have . 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): 9/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
Approximation True value Accuracy = − ( ) ( ) t x f x x ( ) ( ) d f x t x 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 10/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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) + 1 x at x= 0. Find the interval for an accuracy of approximation of d= 0.001. 11/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
Solution We calculate the first derivative: + 2 x ′ = ( ) f x + + . (4) 2( 1) 1 x x f ′ = = . The (0) 0 (0) 1 f For x= 0 we have and linearization is = ( ) t x x . This way we obtained x = ≈ ( ) f x x + 1 x near the point x= 0. As in the previous example we give the table of accuracy only for x > 0. All 12/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
calculations are done with round-off error of 0.00001. Approximation True value Accuracy = − ( ) ( ) t x f x x ( ) ( ) d f x t x 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 0.03922 0.04 0.00078 0.05 0.04880 0.05 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 13/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
x ∈ (0, 0.05) We see that for 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. 14/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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 (2) 0.38490 f and . 3 3 3 The linearization now is 2 = + ≈ + ( ) (1 ) 0.3849(1 ) t x x x . (5) 3 3 This way we obtained x = ≈ + ( ) 0.3849(1 ) f x x + 1 x near the point x= 2 with a round-off error of 0.00001. 15/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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 = − ( ) ( ) t x f x x ( ) ( ) d f x t x 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 16/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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. 17/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
In Fig. 4 we observe that the linearization (5) is good only in a narrow local region, near the point x= 2. 4 t(x ) 3 f (x ) 2 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. 18/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
3. Example with Mathematica Example 4. Find the linearization of x = ( ) f x near x= 1. + 1 x 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: 19/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
We define also the derivative as a function: 20/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
Now for a= 1 we define the corresponding linearization of the function according to (1): ( ) f x This can be used for approximation of . Finally, lets to draw the graphics by: 21/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
We will obtain: 0.8 0.6 0.4 0.2 0.5 1.0 1.5 2.0 22/37 ------------------ Snezhana Gocheva-Ilieva, Plovdiv University, 2011 ---------------------
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