# Introduction to Bayesian statistics - Bolstad M.

ISBN 0-471-27020-2

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Higher Derivatives

The second derivative of a differentiable function f (x) at a point x = c is the derivative of the derivative function f '(x) at the point. The second derivative is given by

f "(c)=lim^ f '(c + h) — f '(c)'

h— 0 h

if it exists. If the second derivative exists for all points x in an interval, then f "(x) is the second derivative function over the interval. Other notation for the second derivative at point c and for the second derivative function are

f //(c) =f (2)(c) = dxf/(x)

and f (2)(x) = dx^f (x).

Similarly the kth derivative is the derivative of the k — 1th derivative function

fW(c) = ,—0 (f^ + hh — f(k-1)<c>)

if it exists.

x=c

Critical Points

For a function f (x) that is differentiable over an open interval (a, b), the derivative function f' (x) is the slope of the curve y = f (x) at each x-value in the interval. This gives a method of finding where the minimum and maximum values of the function occur. The function will achieve its minimum and maximum at points where the derivative equals 0. When x = c is a solution of the equation

f '(x) = 0,

c is called a critical point of the function f(x). The critical points may lead to local maximum or minimum, global maximum or minimum, or they may be points of inflection. A point of inflection is where the function changes from being concave to convex, or vice versa.

Theorem 5 First derivative test: If f (x) is a continuous differentiable function over an interval (a, b) having derivative function f '(x) which is defined on the same interval. Suppose c is a critical point of the function. By definition, f '(c) = 0.

1. The function achieves a unique local maximum at x = c if, for all points x that are sufficiently close to c

when x < c then f '(x) > 0 and

when x > c then f '(x) < 0.

2. Similarly the function achieves a unique local minimum at x = c if, for all points x that are sufficiently close to c

when x < c then f '(x) < 0 and

when x > c then f '(x) > 0.

284 INTRODUCTION TO CALCULUS

3. The function has a point of inflection at critical point x = c if, for all points x that are sufficiently close to c, either when x < c then f-(x) < 0 and

when x > c then f-(x) < 0,

or

when x < c then f-(x) > 0 and

when x > c then f-(x) > 0.

At a point ofinflection, the function either stops increasing, and then resumes increasing, or it stops decreasing, and then resumes decreasing.

For example, the function f (x) = x3, and its derivative f-(x) = 3 x x2 are shown in Figure A.6. We see that the derivative function f-(x) = 3x2 is positive for x < 0, so the function f (x) = x3 is increasing for x < 0. The derivative function is positive for x > 0 so the function is also increasing for x > 0. However at x = 0, the derivative function equals 0, so the original function is not increasing at x = 0. Thus the function f (x) = x3 has a point of inflection at x = 0.

Theorem 6 Second derivative test: If f (x) is a continuous differentiable function over an interval (a, b) having first derivative function f -(x) and second derivative function f (2)(x) both defined on the same interval. Suppose c is a critical point of the function. By definition, f '(c) = 0.

1. The function achieves a maximum at x = c if f(2) (c) < 0

2. The function achieves a minimum at x = c if f(2) (c) > 0

INTEGRATION

The second main use of calculus is finding the area under a curve using integration. It turns out that integration is the inverse of differentiation. Suppose f (x) is a function defined on an interval [a, b]. Let the function F(x) be an antiderivative of f (x). That means the derivative function F-(x) = f (x). Note that the antiderivative of f (x) is not unique. The function F(x) + c will also be an antiderivative of f (x). The antiderivative is also called the indefinite integral.

The Definite Integral: Finding the Area under a Curve

Suppose we have a nonnegative1 continuous function f (x) definedonaclosedinterval [a, b]. f (x) > 0 for all x e [a, b]. Suppose we partition the the interval [a, b] using

the partition x0,xi,..., xn, where x0 = a and xn = b and xi < xi+i. Note that

the partition does not have to have equal length intervals. Let the minimum and maximum value of f (x) in each interval be

li = sup f (x) and mi = inf f (x)

х£[х^_1,х^] xG[xi-i,xi]

1The requirement that f (x) be nonnegative is not strictly necessary. However since we are using the definite integral to find the area under probability density functions that are nonnegative, we will impose the condition.

INTRODUCTION TO CALCULUS 285

Figure A.6 Graph of f (x) = x3 and its derivative. The derivative function is negative where the original function is increasing, and it is positive where the original function is increasing We see the original function has a point of inflection at x = 0.

where sup is the least upper bound, and inf is the greatest lower bound. Then the area under the curve y = f (x) between x = a and x = b lies between the lower sum

n

Lxo,...,xn ^ ^ 1 x (xi xi-1)

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