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Handbook of mathematics for engineers and scienteists part 42 ppsx

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Then x0is called a point of local minimum resp., local maximum of the function f x.. Points of local minimum or maximum are called points of extremum.. A function f x can have an extremu

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6.2.3-3 Methods for interpreting other indeterminate expressions.

1 Expressions of the form 0⋅∞ and ∞–∞ can be reduced to indeterminate expressions

0

0 or ∞ ∞ by means of algebraic transformations, for instance:

u (x)v (x) = u (x)

1/v (x) transformation rule 0 ⋅∞ =⇒ 00,

u (x) – v(x) =

u (x)

1

v (x)



u (x)v(x) transformation rule ∞ – ∞ =⇒ 00

2 Indeterminate expressions of the form 1, 0, 00 can be reduced to expressions of the form 0

0 or ∞ ∞ by taking logarithm and using the formulas ln u v = v ln u =

ln u

1/v

Example 2 Let us calculate the limit lim

x→0(cos x)1/x2.

We have the indeterminate expression 1 We find that

ln lim

x→0(cos x)1/x2= lim

x→0ln(cos x)1/x2= lim

x→0

ln cos x

x2 = lim

x→0

(ln cos x)  (x2) = lim

x→0

(– tan x)

2.

Therefore, lim

x→0(cos x)1/x2= e–1/2= 1

e.

6.2.4 Higher-Order Derivatives and Differentials Taylor’s Formula

6.2.4-1 Derivatives and differentials of higher orders

The second-order derivative or the second derivative of a function y = f (x) is the derivative

of the derivative f  (x) The second derivative is denoted by y  and also by y  xx, d

2y

dx2, f  (x).

The derivative of the second derivative of a function y = f (x) is called the third-order derivative, y  = (y ) The nth-order derivative of the function y = f (x) is defined as the derivative of its (n –1)th derivative:

y(n) = (y(n–1))

The nth-order derivative is also denoted by y(x n), d n y

dx n , f(n) (x).

The second-order differential is the differential of the first-order differential, d2y =

d (dy) If x is the independent variable, then d2y = y (dx)2 In a similar way, one defines differentials of higher orders

6.2.4-2 Table of higher-order derivatives of some elementary functions

(x a)(n) = a(a –1) (a – n +1)x a–n, (a x)(n) = (ln a) n a x,

(ln x)(n)= (–1)n–1 (n –1)! 1

x n, (loga x)(n)= (–1)n–1 (n –1)!

ln a

1

x n, (sin x)(n)= sin



x+ πn 2



, (cos x)(n)= cos



x+ πn 2



,

(sinh x)(n)=

cosh x if n is odd,

sinh x if n is even, (cosh x)

(n)=

cosh x if n is even, sinh x if n is odd.

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6.2.4-3 Rules for calculating higher-order derivatives.

1 Derivative of a sum (difference) of functions:

[u(x) v (x)](n) = u(n) (x) v(n) (x).

2 Derivatives of a function multiplied by a constant:

[au(x)](n) = au(n) (x) (a = const).

3 Derivatives of a product:

[u(x)v(x)]  = u  (x)v(x) +2u  (x)v  (x) + u(x)v  (x),

[u(x)v(x)]  = u  (x)v(x) +3u  (x)v  (x) +3u  (x)v  (x) + u(x)v  (x),

[u(x)v(x)](n)=

n



k=0

C k

n u(k) (x)v(n–k) (x) (Leibnitz formula),

where C n k are binomial coefficients, u(0)(x) = u(x), v(0)(x) = v(x).

4 Derivatives of a composite function:



f (u(x))

= f uu  (u  x)2+ f u  u 

xx,



f (u(x))

= f uuu  (u  x)3+3f 

uu u  x u  xx + f u  u  xxx.

5 Derivatives of a parametrically defined function x = x(t), y = y(t):

y = x  t y tt  – y t  x  tt

(x  t 3 , y

= (x  t2y ttt  –3x 

t x  tt y tt  +3y 

t (x  tt)2– x  t y t  x  ttt

(n)= (y(n–1)) t

x  t

6 Derivatives of an implicit function defined by the equation F (x, y) =0:

y = 1

F3

y –F

2

y F xx+2F x F y F xy – F x2F yy

,

y  = 1

F5

y

–F y4F xxx+3F x F y3F xxy–3F2

x F y3F xyy + F x3F y F yyy+3F3

y F xx F xy

–3F x F y2F xx F yy–6F x F y2F xy2 –3F3

x F yy2 +9F2

x F y F xy F yy

,

where the subscripts denote the corresponding partial derivatives

7 Derivatives of the inverse function x = x(y):

x 

yy = – y



xx (y  x)3, x



yyy = –y



xxx (y  x)4 +3(y xx  )2

(y x )5 , x

(n)

y = 1

y 

x [x

(n–1)

y ] x.

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6.2.4-4 Taylor’s formula.

Suppose that in a neighborhood of a point x = a, the function y = f (x) has derivatives

up to the order (n +1) inclusively Then for all x in that neighborhood, the following

representation holds:

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

1! (x – a) +

f  (a)

2! (x – a)

2+· · · + f(n) (a)

n! (x – a)

n + R

n (x), (6.2.4.1)

where R n (x) is the remainder term in the Taylor formula.

The remainder term can be represented in different forms (6.2.4.1):

R n (x) = o[(x – a) n] (Peano),

R n (x) = f

(n+1) a + k(x – a)

(n +1)! (x – a)

n+1 (Lagrange),

R n (x) = f

(n+1) a + k(x – a)

n! (1– k) n (x – a) n+1 (Cauchy),

R n (x) = f

(n+1) a + k(x – a)

n !p (1– k) n+1p (x – a) n+1 (Schl¨omilch and Roche),

R n (x) = 1

n!

 x

a f

(n+1)(t)(x – t) n dt (integral form),

where0 < k <1and p > 0; k depends on x, n, and the structure of the remainder term.

The remainders in the form of Lagrange and Cauchy can be obtained as special cases of the

Schl¨omilch formula with p = n +1and p =1, respectively

For a =0, the Taylor formula (6.2.4.1) turns into

f (x) = f (0) + f (0)

1! x+

f (0)

2! x

2+· · · + f(n)(0)

n! x

n + R

n (x) and is called the Maclaurin formula.

The Maclaurin formula for some functions:

e x=1+ x

1! +

x2

2! +

x3

3! +· · · + x n

n! + R n (x),

sin x = x – x

3

3! +

x5

5! –

x7

7! +· · · + (–1)n x

2n+1 (2n+1)! + R2n+1(x),

cos x =1– x

2

2! +

x4

4! –

x6

6! +· · · + (–1)n x

2n (2n)! + R2n(x).

6.2.5 Extremal Points Points of Inflection

6.2.5-1 Maximum and minimum Points of extremum

Let f (x) be a differentiable function on the interval (a, b) and f  (x) >0(resp., f  (x) <0)

on (a, b) Then f (x) is an increasing (resp., decreasing) function on that interval*.

Suppose that there is a neighborhood of a point x0 such that for all xx0 in that

neighborhood we have f (x) > f (x0) (resp., f (x) < f (x0)) Then x0is called a point of local

minimum (resp., local maximum) of the function f (x).

Points of local minimum or maximum are called points of extremum.

* At some points of the interval, the derivative may vanish.

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6.2.5-2 Necessary and sufficient conditions for the existence of extremum.

NECESSARY CONDITION OF EXTREMUM A function f (x) can have an extremum only at

points in which its derivative either vanishes or does not exist (or is infinite)

FIRST SUFFICIENT CONDITION OF EXTREMUM Suppose that f (x) is continuous in some neighborhood (x0 –δ, x0+δ) of a point x0and differentiable at all points of the neighborhood

except, possibly, x0 If f  (x) >0for x (x0– δ, x0)and f  (x) <0for x (x0, x0+ δ), then x0is a point of local maximum of this function If f  (x) <0for x(x0– δ, x0)and

f  (x) >0for x(x0, x0+ δ), then x0is a point of local minimum of this function

If f  (x) is of the same sign for all xx0, x(x0– δ, x0+ δ), then x0cannot be a point

of extremum

SECOND SUFFICIENT CONDITION OF EXTREMUM Let f (x) be a twice differentiable function in a neighborhood of x0 Then the following implications hold :

(i) f  (x0) =0 and f  (x0) <0 =⇒ f(x) has a local maximum at the point x0; (ii) f  (x0) =0 and f  (x0) >0 =⇒ f(x) has a local minimum at the point x0 THIRD SUFFICIENT CONDITION OF EXTREMUM Let f (x) be a function that is n times differentiable in a neighborhood of a point x0and f  (x0) = f  (x0) = · · · = f(n–1)(x0) =0,

but f(n) (x0)≠ 0 Then the following implications hold :

(i) n is even and f(n) (x0) <0 =⇒ f(x) has a local maximum at the point x0;

(ii) n is even and f(n) (x0) >0 =⇒ f(x) has a local minimum at the point x0.

If n is odd, then x0cannot be a point of extremum

6.2.5-3 Largest and the smallest values of a function

Let y = f (x) be continuous on the segment [a, b] and differentiable at all points of this

segment except, possibly, finitely many points Then the largest and the smallest values

of f (x) on [a, b] belong to the set consisting of f (a), f (b), and the values f (x i), where

x i (a, b) are the points at which f  (x) is either equal to zero or does not exist (is infinite).

6.2.5-4 Direction of the convexity of the graph of a function

The graph of a differentiable function y = f (x) is said to be convex upward (resp., convex

downward) on the interval (a, b) if for each point of this interval, the graph lies below (resp.,

above) the tangent line at that point

If the function y = f (x) is twice differentiable on the interval (a, b) and f  (x) <0(resp.,

f  (x) > 0), then its graph is convex upward (resp., downward) on that interval (At some points of the interval, the second derivative may vanish.)

Thus, in order to find the intervals on which the graph of a twice differentiable function

f (x) is convex upward (resp., downward), one should solve the inequality f  (x) <0(resp.,

f  (x) >0)

6.2.5-5 Inflection points

An inflection point on the graph of a function y = f (x) is defined as a point (x0, f (x0)) at which the graph passes from one side of its tangent line to another At an inflection point, the graph changes the direction of its convexity

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Suppose that the function y = f (x) has a continuous second derivative f  (x) in some neighborhood of a point x0 If f  (x0) =0and f  (x) changes sign as x passes through the point x0, then (x0, f (x0)) is an inflection point

6.2.6 Qualitative Analysis of Functions and Construction of Graphs

6.2.6-1 General scheme of analysis of a function and construction of its graph

1 Determine the domain in which the function is defined

2 Find the asymptotes of the graph

3 Find extremal points and intervals of monotonicity

4 Determine the directions of convexity of the graph and its inflection points

5 Determine whether the function is odd or even and whether it is periodic

6 Find the points at which the graph crosses the coordinate axes

7 Draw the graph, using the properties 1 to 6

Example Let us examine the function y = ln x

x and construct its graph.

We use the above general scheme.

1 This function is defined for all x such that0< x < + ∞.

2 The straight line x =0 is a vertical asymptote, since lim

x→+0

ln x

x = –∞ We find the oblique asymptotes:

k= lim

x→+∞

y

x = 0 , b= lim

x→+∞ (y – kx) =0

Therefore, the line y =0 is a horizontal asymptote of the graph.

3 The derivative y  = 1– ln x

x2 vanishes for ln x =1 Therefore, the function may have an extremum at

x = e For x ( 0, e), we have y > 0, i.e., the function is increasing on this interval For x(e, +∞), we have

y < 0, and therefore the function is decreasing on this interval At x = e the function attains its maximal value

ymax = 1

e.

One should also examine the points at which the derivative does not exist There is only one such point,

x= 0 , and it corresponds to the vertical asymptote (see Item 1).

4 The second derivative y = 2ln x –3

x3 vanishes for x = e3/2 On the interval ( 0, e3/2), we have y < 0 ,

and therefore the graph is convex upward on this interval For x(e3/2, +∞), we have y > 0 , and therefore

the graph is convex downward on this interval The value x = e3/2corresponds to an inflection point of the

graph, with the ordinate y = 32e–3/2.

5 This function is neither odd nor even, since it is defined only for x >0and the relations f (–x) = f (x)

or f (–x) = –f (x) cannot hold Obviously, this function is nonperiodic.

6 The graph of this function does not cross the y-axis, since for x =0 the function is undefined Further,

y= 0only if x =1, i.e., the graph crosses the x-axis only at the point (1 , 0 ).

7 Using the above results, we construct the graph (Fig 6.5).

6.2.6-2 Transformations of graphs of functions

Let us describe some methods which in many cases allow us to construct the graph of a function if we have the graph of a simpler function

1 The graph of the function y = f (x) + a is obtained from that of y = f (x) by shifting the latter along the axis Oy by the distance|a| For a >0the shift is upward, and for a <0

downward (see Fig 6.6 a).

2 The graph of the function y = f (x + a) is obtained from that of y = f (x) by shifting the latter along the Ox by the distance|a| For a >0the shift is to the left, and for a <0to

the right (see Fig 6.6 b).

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x y

e

0.2

1/e

3/2

0.2 0.4

y=ln x

x

Figure 6.5 Graph of the function y = lnx

x .

3 The graph of the function y = –f (x) is obtained from that of y = f (x) by symmetric reflection with respect to the axis Ox (see Fig 6.6 c).

4 The graph of the function y = f (–x) is obtained from that of y = f (x) by symmetric reflection with respect to the axis Oy (see Fig 6.6 d).

5 The graph of the function y = kf (x) for k >1is obtained from that of y = f (x) by extending the latter k times from the axis Ox, and for0< k <1by contracting the latter1/k

times to the axis Ox The points at which the graph crosses the axis Ox remain unchanged (see Fig 6.6 e).

6 The graph of the function y = f (kx) for k >1 is obtained from that of y = f (x)

by contracting the latter k times to the axis Oy, and for0 < k <1by extending the latter

1/k times from the axis Oy The points at which the graph crosses the axis Oy remain unchanged (see Fig 6.6 f ).

7 The graph of the function y =|f (x)|is obtained from that of y = f (x) by preserving the parts of the latter for which f (x)≥ 0and symmetric reflection, with respect to the axis

Ox , of the parts for which f (x) <0(see Fig 6.6 g).

8 The graph of the inverse function y = f–1(x) is obtained from that of y = f (x) by symmetric reflection with respect to the straight line y = x (see Fig 6.6 h).

6.2.7 Approximate Solution of Equations

(Root-Finding Algorithms for Continuous Functions)

6.2.7-1 Preliminaries

For a vast majority of algebraic (transcendental) equations of the form

f (x) =0, (6.2.7.1)

where f (x) is a continuous function, there are no exact closed-form expressions for the

roots

When solving the equation approximately, the first step is to bracket the roots, i.e., find

sufficiently small intervals containing exactly one root each Such an interval [a, b], where the numbers a and b satisfy the condition f (a)f (b) <0(which is assumed to hold in what follows), can be found, say, graphically

The second step is to compute successive approximations x n[a, b] (n =1,2, ) to the desired root c = lim

n→∞ x n, usually by one of the following methods.

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O O

O O

O O O

O x

x

x x

x x

x x

y y

y

y ( ) a

( )c

( )e

( )g

( )b

( )d

( )f

y

y y

y=| ( )|f x

y=f x( )

y=f x( )

y=f x( )

y=f x( )

y=f x( )

y=kf x( )

y=f kx( )

y= -f x( )

y=f x( )

y=f x( )

y=f x( )

y=f x( + )a

a > 0

0 <k <1

0 <k <1

a < 0

a < 0

y= -f( x)

y=f ( )x y

x

Figure 6.6 Transformations of graphs of functions.

6.2.7-2 Bisection method

To find the root of equation (6.2.7.1) on the interval [a, b], we bisect the interval If

f

a + b

2



= 0, then c = a + b

2 is the desired root If f

a + b

2



≠ 0, then of the two intervals

*

a, a + b2

+

and

*a + b

2 , b

+

we take the one at whose endpoints the function f (x)

has opposite signs Now we bisect the new, smaller interval, etc As a result, we obtain either an exact root of equation (6.2.7.1) at some step or an infinite sequence of nested

intervals [a1, b1], [a2, b2], such that f (a n )f (b n) <0 The root is given by the formula

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