Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx

A function is continuous on the closed interval [a, b] if the function is continuous for each point x ∈ a, b andlimx→a+yx = ya and limx→b−yx = yb.Discontinuous Functions.. If both the le

defined, limx→ξy(x) exists and limx→ξy(x) = y(ξ) A function is continuous if it is continuous at each point in itsdomain A function is continuous on the closed interval [a, b] if the function is continuous for each point x ∈ (a, b) andlimx→a+y(x) = y(a) and limx→b−y(x) = y(b).

Discontinuous Functions If a function is not continuous at a point it is called discontinuous at that point Iflimx→ξy(x) exists but is not equal to y(ξ), then the function has a removable discontinuity It is thus named because

we could define a continuous function

z(x) =

(y(x) for x 6= ξ,limx→ξy(x) for x = ξ,

to remove the discontinuity If both the left and right limit of a function at a point exist, but are not equal, then thefunction has a jump discontinuity at that point If either the left or right limit of a function does not exist, then thefunction is said to have an infinite discontinuity at that point

Example 3.2.1 sin xx has a removable discontinuity at x = 0 The Heaviside function,

1 for x > 0,

has a jump discontinuity at x = 0 1

x has an infinite discontinuity at x = 0 See Figure 3.3.Properties of Continuous Functions

Arithmetic If u(x) and v(x) are continuous at x = ξ then u(x) ± v(x) and u(x)v(x) are continuous at x = ξ u(x)v(x)

is continuous at x = ξ if v(ξ) 6= 0

Function Composition If u(x) is continuous at x = ξ and v(x) is continuous at x = µ = u(ξ) then u(v(x)) iscontinuous at x = ξ The composition of continuous functions is a continuous function

Figure 3.3: A Removable discontinuity, a Jump Discontinuity and an Infinite Discontinuity

Boundedness A function which is continuous on a closed interval is bounded in that closed interval

Nonzero in a Neighborhood If y(ξ) 6= 0 then there exists a neighborhood (ξ − , ξ + ),  > 0 of the point ξ suchthat y(x) 6= 0 for x ∈ (ξ − , ξ + )

Intermediate Value Theorem Let u(x) be continuous on [a, b] If u(a) ≤ µ ≤ u(b) then there exists ξ ∈ [a, b] suchthat u(ξ) = µ This is known as the intermediate value theorem A corollary of this is that if u(a) and u(b) are

of opposite sign then u(x) has at least one zero on the interval (a, b)

Maxima and Minima If u(x) is continuous on [a, b] then u(x) has a maximum and a minimum on [a, b] That is, there

is at least one point ξ ∈ [a, b] such that u(ξ) ≥ u(x) for all x ∈ [a, b] and there is at least one point ψ ∈ [a, b]such that u(ψ) ≤ u(x) for all x ∈ [a, b]

Piecewise Continuous Functions A function is piecewise continuous on an interval if the function is bounded onthe interval and the interval can be divided into a finite number of intervals on each of which the function is continuous.For example, the greatest integer function, bxc, is piecewise continuous (bxc is defined to the the greatest integer lessthan or equal to x.) See Figure 3.4 for graphs of two piecewise continuous functions

Uniform Continuity Consider a function f (x) that is continuous on an interval This means that for any point ξ

in the interval and any positive  there exists a δ > 0 such that |f (x) − f (ξ)| <  for all 0 < |x − ξ| < δ In general,this value of δ depends on both ξ and  If δ can be chosen so it is a function of  alone and independent of ξ then

Figure 3.4: Piecewise Continuous Functions

the function is said to be uniformly continuous on the interval A sufficient condition for uniform continuity is that thefunction is continuous on a closed interval

If the slope of the secant line has a limit as ∆x approaches zero then we call this slope the derivative or instantaneousrate of change of the function at the point x We denote the derivative by dydx, which is a nice notation as the derivative

is the limit of ∆y∆x as ∆x → 0

∆x may approach zero from below or above It is common to denote the derivative dydx by dxdy, y0(x), y0 or Dy

A function is said to be differentiable at a point if the derivative exists there Note that differentiability impliescontinuity, but not vice versa

Example 3.3.1 Consider the derivative of y(x) = x2 at the point x = 1

Example 3.3.2 We can compute the derivative of y(x) = x2 at an arbitrary point x

0.5 1 1.5 2 0.5

1 1.5 2 2.5 3 3.5 4

0.5 1 1.5 2 0.5

1 1.5 2 2.5 3 3.5 4

Figure 3.6: Secant lines and the tangent to x2 at x = 1

Properties Let u(x) and v(x) be differentiable Let a and b be constants Some fundamental properties ofderivatives are:

uv



= v

du

dx − udv dx

dv

dx = u

0

(v(x))v0(x) Chain RuleThese can be proved by using the definition of differentiation

Example 3.3.3 Prove the quotient rule for derivatives.

ddx

uv



= lim

∆x→0

u(x+∆x) v(x+∆x) − u(x)v(x)

v2

Trigonometric Functions Some derivatives of trigonometric functions are:

d

dxtan x =

1cos2x

dxsinh x = cosh x

d

dxarcsinh x =

1(x2+ 1)1/2

d

dxtanh x =

1cosh2x

Example 3.3.5 The inverse function of y(x) = ex is x(y) = ln y We can obtain the derivative of the logarithm fromthe derivative of the exponential The derivative of the exponential is

dy

dx = e

x

.Thus the derivative of the logarithm is

of x and y even when y(x) is defined by an implicit equation

Example 3.4.1 Consider the implicit equation



−1 ± √ 5x

5x2− 4

.One can obtain the same result without first solving for y If we differentiate the implicit equation, we obtain

2x − y − xdy

dx− 2ydy

dx = 0.

We can solve this equation for dydx.

= 5(y − xy

0)(x + 2y)2

Substitute in the expression for y0

= −10(x

2− xy − y2)(x + 2y)2

Use the original implicit equation

(x + 2y)2

3.5 Maxima and Minima

A differentiable function is increasing where f0(x) > 0, decreasing where f0(x) < 0 and stationary where f0(x) = 0

A function f (x) has a relative maxima at a point x = ξ if there exists a neighborhood around ξ such that f (x) ≤ f (ξ)for x ∈ (x − δ, x + δ), δ > 0 The relative minima is defined analogously Note that this definition does not requirethat the function be differentiable, or even continuous We refer to relative maxima and minima collectively are relativeextrema

Relative Extrema and Stationary Points If f (x) is differentiable and f (ξ) is a relative extrema then x = ξ

is a stationary point, f0(ξ) = 0 We can prove this using left and right limits Assume that f (ξ) is a relative maxima.Then there is a neighborhood (x − δ, x + δ), δ > 0 for which f (x) ≤ f (ξ) Since f (x) is differentiable the derivative

Thus we have 0 ≤ f0(ξ) ≤ 0 which implies that f0(ξ) = 0

It is not true that all stationary points are relative extrema That is, f0(ξ) = 0 does not imply that x = ξ is anextrema Consider the function f (x) = x3 x = 0 is a stationary point since f0(x) = x2, f0(0) = 0 However, x = 0 isneither a relative maxima nor a relative minima

It is also not true that all relative extrema are stationary points Consider the function f (x) = |x| The point x = 0

is a relative minima, but the derivative at that point is undefined

First Derivative Test Let f (x) be differentiable and f0(ξ) = 0

• If f0(x) changes sign from positive to negative as we pass through x = ξ then the point is a relative maxima

• If f0(x) changes sign from negative to positive as we pass through x = ξ then the point is a relative minima

• If f0(x) is not identically zero in a neighborhood of x = ξ and it does not change sign as we pass through thepoint then x = ξ is not a relative extrema.

Example 3.5.1 Consider y = x2 and the point x = 0 The function is differentiable The derivative, y0 = 2x, vanishes

at x = 0 Since y0(x) is negative for x < 0 and positive for x > 0, the point x = 0 is a relative minima See Figure3.7.Example 3.5.2 Consider y = cos x and the point x = 0 The function is differentiable The derivative, y0 = − sin x

is positive for −π < x < 0 and negative for 0 < x < π Since the sign of y0 goes from positive to negative, x = 0 is arelative maxima See Figure 3.7

Example 3.5.3 Consider y = x3 and the point x = 0 The function is differentiable The derivative, y0 = 3x2 ispositive for x < 0 and positive for 0 < x Since y0 is not identically zero and the sign of y0 does not change, x = 0 isnot a relative extrema See Figure 3.7

Figure 3.7: Graphs of x2, cos x and x3

Concavity If the portion of a curve in some neighborhood of a point lies above the tangent line through that point,the curve is said to be concave upward If it lies below the tangent it is concave downward If a function is twicedifferentiable then f00(x) > 0 where it is concave upward and f00(x) < 0 where it is concave downward Note that

f00(x) > 0 is a sufficient, but not a necessary condition for a curve to be concave upward at a point A curve may beconcave upward at a point where the second derivative vanishes A point where the curve changes concavity is called

a point of inflection At such a point the second derivative vanishes, f00(x) = 0 For twice continuously differentiable

functions, f00(x) = 0 is a necessary but not a sufficient condition for an inflection point The second derivative mayvanish at places which are not inflection points See Figure 3.8.

Figure 3.8: Concave Upward, Concave Downward and an Inflection Point

Second Derivative Test Let f (x) be twice differentiable and let x = ξ be a stationary point, f0(ξ) = 0

• If f00(ξ) < 0 then the point is a relative maxima

• If f00(ξ) > 0 then the point is a relative minima

• If f00(ξ) = 0 then the test fails

Example 3.5.4 Consider the function f (x) = cos x and the point x = 0 The derivatives of the function are

f0(x) = − sin x, f00(x) = − cos x The point x = 0 is a stationary point, f0(0) = − sin(0) = 0 Since the secondderivative is negative there, f00(0) = − cos(0) = −1, the point is a relative maxima

Example 3.5.5 Consider the function f (x) = x4 and the point x = 0 The derivatives of the function are f0(x) = 4x3,

f00(x) = 12x2 The point x = 0 is a stationary point Since the second derivative also vanishes at that point thesecond derivative test fails One must use the first derivative test to determine that x = 0 is a relative minima

3.6 Mean Value Theorems

Rolle’s Theorem If f (x) is continuous in [a, b], differentiable in (a, b) and f (a) = f (b) = 0 then there exists apoint ξ ∈ (a, b) such that f0(ξ) = 0 See Figure 3.9

Figure 3.9: Rolle’s Theorem

To prove this we consider two cases First we have the trivial case that f (x) ≡ 0 If f (x) is not identically zerothen continuity implies that it must have a nonzero relative maxima or minima in (a, b) Let x = ξ be one of theserelative extrema Since f (x) is differentiable, x = ξ must be a stationary point, f0(ξ) = 0

Theorem of the Mean If f (x) is continuous in [a, b] and differentiable in (a, b) then there exists a point x = ξsuch that

f0(ξ) = f (b) − f (a)

b − a .That is, there is a point where the instantaneous velocity is equal to the average velocity on the interval

We prove this theorem by applying Rolle’s theorem Consider the new function

g(x) = f (x) − f (a) − f (b) − f (a)

b − a (x − a)Note that g(a) = g(b) = 0, so it satisfies the conditions of Rolle’s theorem There is a point x = ξ such that g0(ξ) = 0

We differentiate the expression for g(x) and substitute in x = ξ to obtain the result

g0(x) = f0(x) −f (b) − f (a)

b − a

Figure 3.10: Theorem of the Mean.

We have assumed that g(a) 6= g(b) so that the denominator does not vanish and that f0(x) and g0(x) are notsimultaneously zero which would produce an indeterminate form Note that this theorem reduces to the regulartheorem of the mean when g(x) = x The proof of the theorem is similar to that for the theorem of the mean

Taylor’s Theorem of the Mean If f (x) is n + 1 times continuously differentiable in (a, b) then there exists apoint x = ξ ∈ (a, b) such that

f (b) = f (a) + (b − a)f0(a) +(b − a)

f (b) = f (a) + (b − a)f0(ξ),

which is just a rearrangement of the terms in the theorem of the mean,

f0(ξ) = f (b) − f (a)

b − a .

One can use Taylor’s theorem to approximate functions with polynomials Consider an infinitely differentiable function

f (x) and a point x = a Substituting x for b into Equation 3.1 we obtain,

f (x) = f (a) + (x − a)f0(a) +(x − a)

If the last term in the sum is small then we can approximate our function with an nth order polynomial

f (x) ≈ f (a) + (x − a)f0(a) + (x − a)

So if the derivative term, f(n+1)(ξ), does not grow to quickly, the error for a certain value of x will get smaller withincreasing n and the polynomial will become a better approximation of the function (It is also possible that thederivative factor grows very quickly and the approximation gets worse with increasing n.)

Example 3.6.1 Consider the function f (x) = ex We want a polynomial approximation of this function near the point

x = 0 Since the derivative of ex is ex, the value of all the derivatives at x = 0 is f(n)(0) = e0 = 1 Taylor’s theoremthus states that

0.5 1 1.5 2 2.5

0.5 1 1.5 2 2.5

0.5 1 1.5 2 2.5

0.5 1 1.5 2 2.5

Figure 3.11: Four Finite Taylor Series Approximations of ex

Note that for the range of x we are looking at, the approximations become more accurate as the number of termsincreases

Example 3.6.2 Consider the function f (x) = cos x We want a polynomial approximation of this function near the

point x = 0 The first few derivatives of f are

f(n)(x) =

((−1)n/2cos x for even n,(−1)(n+1)/2sin x for odd n

Since cos(0) = 1 and sin(0) = 0 the n-term approximation of the cosine is,

cos x = 1 − x

2

2! +

x44! − x

6

6! + · · · + (−1)

2(n−1) x2(n−1)(2(n − 1))! +

x2n(2n)!cos ξ.

Here are graphs of the one, two, three and four term approximations

-1 -0.5

0.5 1

-1 -0.5

0.5 1

-1 -0.5

0.5 1

-1 -0.5

0.5 1

Figure 3.12: Taylor Series Approximations of cos xNote that for the range of x we are looking at, the approximations become more accurate as the number of termsincreases Consider the ten term approximation of the cosine about x = 0,

Note that for any value of ξ, | cos ξ| ≤ 1 Therefore the absolute value of the error term satisfies,

|R| =

x2020! cos ξ

Figure 3.13: Plot of x20/20!

Note that the error is very small for x < 6, fairly small but non-negligible for x ≈ 7 and large for x > 8 The tenterm approximation of the cosine, plotted below, behaves just we would predict

The error is very small until it becomes non-negligible at x ≈ 7 and large at x ≈ 8

Example 3.6.3 Consider the function f (x) = ln x We want a polynomial approximation of this function near the

-10 -5 5 10

-2 -1.5 -1 -0.5 0.5

Figure 3.14: Ten Term Taylor Series Approximation of cos xpoint x = 1 The first few derivatives of f are

0.5 1 1.5 2 2.5 3

-6 -5-4-3-2-1 1 2

0.5 1 1.5 2 2.5 3

-6 -5 -4 -3 -2 -1

12

Figure 3.15: The 2, 4, 10 and 50 Term Approximations of ln x

Below are plots of the 2, 4, 10 and 50 term approximations

Note that the approximation gets better on the interval (0, 2) and worse outside this interval as the number of termsincreases The Taylor series converges to ln x only on this interval

Example 3.6.4 Suppose you sample a function at the discrete points n∆x, n ∈ Z In Figure 3.16 we sample thefunction f (x) = sin x on the interval [−4, 4] with ∆x = 1/4 and plot the data points

-1 -0.5

0.5 1

Figure 3.16: Sampling of sin x

We wish to approximate the derivative of the function on the grid points using only the value of the function on

-6 -5 -4 -3 -2 -1< /small>

1< /small>2

Figure 3 .15 : The 2, 4, 10 and 50 Term Approximations of ln x

Below are plots of the 2, 4, 10 ... 21< /span>

0.5 1. 5 2.5 3< /small>

-6 -5-4 -3< /small>-2 -1 2

0.5 1. 5... ξ

Figure 3 . 13 : Plot of x20/20!

Note that the error is very small for x < 6, fairly small but non-negligible for x ≈ and large for x > The tenterm approximation