Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx

Fill in the blank with necessary, sufficient or necessary and sufficient.Continuity is a condition for differentiability.. Solution 3.2 Continuity is a necessary condition for differenti

0 <  < 2 there is no value of δ > 0 such that | sin(1/x1) − sin(1/x2)| <  for all x1, x2 ∈ (0, 1) and |x1− x2| < δ.Thus sin(1/x) is not uniformly continuous in the open interval (0, 1).

Solution 3.6

First consider the function √

x Note that the function √

x −√ξ| <  is |x − ξ| < 2 The function √

x is uniformlycontinuous on the interval (0, 1)

Consider any positive δ and  Note that

1

x − 1

x + δ > for

x < 12

r

δ2+4δ

 − δ

!

Thus there is no value of δ such that

1

x− 1ξ

< for all |x − ξ| < δ The function 1x is not uniformly continuous on the interval (0, 1)

Since f (x) is continuous, e(x, δ) is a continuous function of x on the same closed interval Since continuous functions

on closed intervals are bounded, there is a continuous, increasing function (δ) satisfying,

e(x, δ) ≤ (δ),for all x in the closed interval Since (δ) is continuous and increasing, it has an inverse δ() Now note that

|f (x) − f (ξ)| <  for all x and ξ in the closed interval satisfying |x − ξ| < δ() Thus the function is uniformlycontinuous in the closed interval

3 If an > 0 for all n > 200, and limn→∞an = L, then L is not necessarily positive Consider an = 1/n, whichsatisfies the two constraints.

c Consider a right triangle with hypotenuse of length 1 in the first quadrant of the plane Label the vertices A, B,

C, in clockwise order, starting with the vertex at the origin The angle of A is θ The length of a circular arc ofradius cos θ that connects C to the hypotenuse is θ cos θ The length of the side BC is sin θ The length of acircular arc of radius 1 that connects B to the x axis is θ (See Figure 3.20.)

Considering the length of these three curves gives us the inequality:

= lim

θ→0

 cos θ − 1θ

cos θ + 1cos θ + 1



= lim

θ→0

cos2θ − 1θ(cos θ + 1)



= lim

θ→0

 − sin2θθ(cos θ + 1)



= lim

θ→0

 − sin θθ

lim

θ→0

sin θ(cos θ + 1)

Now we’re ready to find the derivative of sin x.

d Let u = g(x) Consider a nonzero increment ∆x, which induces the increments ∆u and ∆f By definition,

∆f = f (u + ∆u) − f (u), ∆u = g(x + ∆x) − g(x),and ∆f, ∆u → 0 as ∆x → 0 If ∆u 6= 0 then we have

We divide this equation by ∆x and take the limit as ∆x → 0.

 lim

∆x→0

∆f

∆x

+ lim

∆x→0

lim

du

dx+ (0)

 dudx



= dfdu

dudxThus we see that

dx[f (cos(g(x)))] = −f

0

(cos(g(x))) sin(g(x))g0(x)

ddx

1

ddx

1

f (ln x)



= − f

0(ln x)x[f (ln x)]2

d First we write the expression in terms exponentials and logarithms,

dx(x ln x) ln x + exp(x ln x)

1x

dxx

x x

= xxx+x x−1+ ln x + ln2x

e For x > 0, the expression is x sin x; for x < 0, the expression is (−x) sin(−x) = x sin x Thus we see that

Let y(x) = tan x Then y0(x) = 1/ cos2x.

y0(1/2) can have the two values:

y0 12



= ±√1

3.Solution 3.14

Differentiating the equation

x2− xy(x) + [y(x)]2 = 3yields

2x − y(x) − xy0(x) + 2y(x)y0(x) = 0

Solving this equation for y0(x)

y0(x) = y(x) − 2x

2y(x) − x.Now we differentiate y0(x) to get y00(x)

y00(x) = 3x

y(x)−2x 2y(x)−x − y(x)(2y(x) − x)2 ,

f00(x) = 12(x − 2) + 12(x − 6) = 24(x − 4)Since f00(2) = −48 < 0, x = 2 is a local maximum Since f00(6) = 48 > 0, x = 6 is a local minimum.

2πr −128

r2 = 0,2πr3− 128 = 0,

r = 4

3

√

π.The second derivative of the surface area is f00(r) = 2π + 256r3 Since f00(√ 34

π) = 6π, r = √ 34

π is a local minimum of

f (r) Since this is the only critical point for r > 0, it must be a global minimum

The cup has a radius of √ 34

π cm and a height of √ 34

π

The first few terms in the Taylor series of sin(x) about x = 0 are

sin x ≈ x − x

3

6 +

x5120has a maximum error of 50401 ≈ 0.000198 Using this polynomial to approximate sin(1),

To see that this has the required accuracy,



= lim

x→0

 sin x6x



= lim

x→0

hcos x6i

= 16

= 0

lim

x→0

csc x − 1

x



= 0

ln

lim

x→+∞



1 + 1x

x

= lim

x→+∞

ln



1 + 1x

x

= e

d It takes four successive applications of L’Hospital’s rule to evaluate the limit.

lim

x→0

csc2x − 1

= lim

x→0

2 − 2 cos2x + 2 sin2x2x2cos2x + 8x cos x sin x + 2 sin2x − 2x2sin2x

= lim

x→0

8 cos x sin x12x cos2x + 12 cos x sin x − 8x2cos x sin x − 12x sin2x

It is easier to use a Taylor series expansion

lim

x→0

csc2x − 1

a ln xx



= exp

lim

x→∞

a/x1

bx

= eab

Fill in the blank with necessary, sufficient or necessary and sufficient.

Continuity is a condition for differentiability

Differentiability is a condition for continuity

Existence of lim∆x→0f (x+∆x)−f (x)∆x is a condition for differentiability

3.12 Quiz Solutions

Solution 3.1

A function y(x) is said to be continuous at x = ξ if limx→ξy(x) = y(ξ)

Solution 3.2

Continuity is a necessary condition for differentiability

Differentiability is a sufficient condition for continuity

Existence of lim∆x→0f (x+∆x)−f (x)∆x is a necessary and sufficient condition for differentiability

Solution 3.6

If f (x) is n + 1 times continuously differentiable in (a b) then there exists a point x = ξ ∈ (a b) such that

f (b) = f (a) + (b − a)f0(a) +(b − a)

Consider limx→0(sin x)sin x This is an indeterminate of the form 00 The limit of the logarithm of the expression

is limx→0sin x ln(sin x) This is an indeterminate of the form 0 · ∞ We can rearrange the expression to obtain anindeterminate of the form ∞∞ and then apply L’Hospital’s rule

lim

x→0

ln(sin x)1/ sin x = limx→0

Chapter 4

Integral Calculus

4.1 The Indefinite Integral

The opposite of a derivative is the anti-derivative or the indefinite integral The indefinite integral of a function f (x)

Z

f (x) dx = f (x)

While a function f (x) has a unique derivative if it is differentiable, it has an infinite number of indefinite integrals, each

of which differ by an additive constant

Zero Slope Implies a Constant Function If the value of a function’s derivative is identically zero, dfdx = 0,then the function is a constant, f (x) = c To prove this, we assume that there exists a non-constant differentiablefunction whose derivative is zero and obtain a contradiction Let f (x) be such a function Since f (x) is non-constant,there exist points a and b such that f (a) 6= f (b) By the Mean Value Theorem of differential calculus, there exists a

point ξ ∈ (a, b) such that

f0(ξ) = f (b) − f (a)

b − a 6= 0,which contradicts that the derivative is everywhere zero

Indefinite Integrals Differ by an Additive Constant Suppose that F (x) and G(x) are indefinite integrals

Thus we see that F (x) − G(x) = c and the two indefinite integrals must differ by a constant For example, we have

R sin x dx = − cos x + c While every function that can be expressed in terms of elementary functions, (the exponent,logarithm, trigonometric functions, etc.), has a derivative that can be written explicitly in terms of elementary functions,the same is not true of integrals For example, R sin(sin x) dx cannot be written explicitly in terms of elementaryfunctions

Properties Since the derivative is linear, so is the indefinite integral That is,

Z(af (x) + bg(x)) dx = a

Z

f (x) dx + b

Zg(x) dx

For each derivative identity there is a corresponding integral identity Consider the power law identity, dxd(f (x))a =a(f (x))a−1f0(x) The corresponding integral identity is

Z(f (x))af0(x) dx = (f (x))

Figure 4.1: Plot of ln |x| and 1/x.

Note the absolute value signs This is because dxd ln |x| = 1x for x 6= 0 In Figure4.1is a plot of ln |x| and 1x to reinforcethis

Z2x(x2+ 1)2 dx

= 12

Z du

u2

= 12

−1u

2(x2+ 1).Example 4.1.2 Consider

I =

Ztan x dx =

Zsin xcos xdx.

By choosing f (x) = cos x, f0(x) = − sin x, we see that the integral is

I = −

Z − sin xcos x dx = − ln | cos x| + c.

Change of Variable The differential of a function g(x) is dg = g0(x) dx Thus one might suspect that for

d

dx.Now we’re ready to start The derivative of the left side of Equation 4.1 is

ddξZ

f (ξ) dξ = f (ξ)

Next we differentiate the right side,

ddξ

Z

f (g(x))g0(x) dx = dx

dξ

ddx

= 12

Zsin ξ dξ

Integration by Parts The product rule for differentiation gives us an identity called integration by parts We startwith the product rule and then integrate both sides of the equation.

d

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

0

(x)v(x) + u(x)v0(x)Z

(u0(x)v(x) + u(x)v0(x)) dx = u(x)v(x) + cZ

u0(x)v(x) dx +

Zu(x)v0(x)) dx = u(x)v(x)Z

u(x)v0(x)) dx = u(x)v(x) −

Zv(x)u0(x) dxThe theorem is most often written in the form

Z

u dv = uv −

Z

v du

So what is the usefulness of this? Well, it may happen for some integrals and a good choice of u and v that the integral

on the right is easier to evaluate than the integral on the left

Example 4.1.4 Consider R x ex dx If we choose u = x, dv = ex dx then integration by parts yields

Z

x ex dx = x ex−

Z

ex dx = (x − 1) ex.Now notice what happens when we choose u = ex, dv = x dx

2x

2ex dxThe integral gets harder instead of easier

When applying integration by parts, one must choose u and dv wisely As general rules of thumb:

• Pick u so that u0 is simpler than u.

• Pick dv so that v is not more complicated, (hopefully simpler), than dv

Also note that you may have to apply integration by parts several times to evaluate some integrals

4.2 The Definite Integral

The area bounded by the x axis, the vertical lines x = a and x = b and the function f (x) is denoted with a definiteintegral,

Z b a

f (x) dx

The area is signed, that is, if f (x) is negative, then the area is negative We measure the area with a divide-and-conquerstrategy First partition the interval (a, b) with a = x0 < x1 < · · · < xn−1 < xn = b Note that the area under thecurve on the subinterval is approximately the area of a rectangle of base ∆xi = xi+1− xi and height f (ξi), where

ξi ∈ [xi, xi+1] If we add up the areas of the rectangles, we get an approximation of the area under the curve SeeFigure 4.2

Z b a

(cf (x) + dg(x)) dx = c

Z b a

f (x) dx + d

Z b a

g(x) dx

One can also divide the range of integration

Z b a

f (x) dx =

Z c a

f (x) dx +

Z b c

f (x) dx

We assume that each of the above integrals exist If a ≤ b, and we integrate from b to a, then each of the ∆xi will benegative From this observation, it is clear that

Z b a

f (x) dx = −

Z a b

f (x) dx

If we integrate any function from a point a to that same point a, then all the ∆xi are zero and

Z a a

(b − a)m ≤

Z b a

f (x) dx ≤ (b − a)M

Since

x=0

= 1,

a1 = 11!

ddx

 1 + x + x2

(x − 1)2



x=0

=



1 + 2x(x − 1)2 −2(1 + x + x

2)(x − 1)3



x=0

= 3,

b0 = 10!

 1 + x + x2

x2



x=1

= 3,

b1 = 11!

ddx

 1 + x + x2

x2

 ... class="text_page_counter">Trang 14

f00(x) = 12 (x − 2) + 12 (x − 6) = 24( x − 4) Since f00(2) = ? ?48 < 0, x = is a local maximum... data-page="26">

Figure 4 .1: Plot of ln |x| and 1/ x.

Note the absolute value signs This is because dxd ln |x| = 1< /sup>x for x 6= In Figure4.1is a plot of ln |x| and 1< /sup>x...

)|x =1< /sub> = 3,

a1< /small> = 1< /sup >1!

(x − 1) 2 + 1< /sup>

x −

dx

2(x − 1) 2 − 3

(x − 1) +