math for developers

7/23/2019 math for developers

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0.1 Derivative. Geometrical interpretation of

derivative.

Derivative. Let f(x) =y be a function and aand a+h be two points.

a) h is called the argument increment.

b) f(a+h) f(a) is called function increment.

Definition 0.1.1. Derivative of a function at a point x0 is the limit

limh0

f(a+h) f(a)

h .

The derivative can be written in the following ways:

f(a) = df

dx(a).

Geometrical interpretation of derivative

Figura 1: A picture of a gull.

Line equation

Definition 0.1.2. Let A(xA, yA) and B(xB, yB) two points. Then the lineequation which passed through this points is:

x xA

xB xA=

y yA

yB yA.

The above formula can be written in the following way:

y= yB yA

xB xA x

yB yA

xB xA xA+ yA.

The slop of the line is:

m= yB yA

xB xA.

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0.2 Taylor Polynomial

Let f :I R be a function derivable for ntimes in the point a I. TaylorPolynomial attached to the function f in the point ais:

(Tnf)(x) = f(a) +x a

1! f(a) +. . .+

(x a)n

n! f(n)(a). (1)

1. The Taylor formula for the function f in the neighbourhood of thepoint a is:

f(x) = (Tnf)(x) + (Rnf)(x),

where

2. (Rnf)(x) is the rest in Taylor formula.

3. MacLaurin FormulaIf in the Taylor Formula it will be choosen a= 0then:

f(x) =f(0) +xf(0) +. . .+xn

n!f(n)(0) + (Rnf)(x). (2)

Example 0.2.1. 1. ex 1 + x+ x2

2! +. . .+ x

n

n!;

2. sin x x x3

3! + x

5

5! +. . .+ (1)n x

2n+1(2n+1)!

;

3. cos x 1 x2

2! + x

4

4! +. . .+ (1)n x

2n

(2n)!.

Figura 2: Taylor approximation

Example0.2.2 (Approximation for cos function).

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0.3 Newtons Method

The goal of this section is to find, using an approximation, the solution forthe the following equation:

f(x) = 0.

The function must have the following properties:

a) function must be continuous

b) function must be derivable

c) f(a) f(b)< 0.

Property c) and a) gave the a solution for the equation f(x) = 0. To obtainthe rule for Newtons method is it necessary to write the Taylor series for thefunction f(x) around the point a.

f(x) = f(a) +x a

1! f(a) +

(x a)2

2! f(a) +

(x a)3

3! f(a) +. . . (3)

Retaining only the first and the second terms of above series, it is obtainedthe equation of the tangent in a point and an approximation for the initialfunction. So:

f(x) f(a) + (x a)f

(a). (4)

The steps to apply the Newtons method are:

Step 1. It is chosen a point x1 which is the close to the solution according withan estimate.

Step 2. It is create the tangent line at the point (x1, f(x1) using the equationof the tangent:

y = f(x1) + f(x1)(x x1).

Step 3. The tangent line will intersect the Ox-axis when y = 0. So,

f(x1) + f(x1)(x2 x1) = 0 (5)

f(x1) = f(x1)(x2 x1) (6)

f(x1)

f(x1) = x2 x1 (7)

x2 = x1 f(x1)

f(x1). (8)

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Step 4. It is verified iff(x2) = 0. If the answer is YESthen the algorithm is

closed; if not, the process will repeat with another approximation.

...

The Step 3. can be written in a general way:

xn+1 = xn f(xn)

f(xn). (9)

Example 0.3.1. Use Newtons method to find the fourth approximation, x4,to the root of the following equation

x3 x 1 = 0

starting with x1= 1.

No. f(x) =x3 x 1 f(x) = 3x2 1

n xn f(xn) f(xn) xn+1=xn f(xn)f(xn)

1 1 1 2 1.52 1.5 0.875 5.75 1.3478260873 1.347826087 0.100682173 4.449905482 1.325200399

4 1.325200399 0.002058362 4.268468292 1.3247181745 1.324718174 0.000000924 4.264634722 1.324717957

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