Block 4 Nonlinear Systems Lesson 14 – The Methods of Differential Calculus The world is not only nonlinear but is changing as well 1 Narrator: Charles Ebeling University of Dayton

Some Most Important Topics Slopes and Derivatives Newton-Raphson Higher Order Derivatives Taylor Series Partial Differentiation These are excellent topics. The exceptional student will want to stay awake for this. 2

The Slope of a line passing through a curve 3

The Slope of a line tangent at a point on the curve 4

What is the Derivative? The derivative is one of the two central concepts of calculus. The derivative gives the slope of the line tangent to a function at a point. In this way, derivatives can be used to determine many geometrical properties of the function such as concavity or convexity. The derivative provides a mathematical formulation of the instantaneous rate of change; it measures the rate at which the function's value changes as the function's argument changes. 5

Why do we care? Analyzing the behavior of functions increasing versus decreasing convex versus concave The basis for nonlinear optimization finding local max and min Algorithmic development finding roots of nonlinear equations Newton-Raphson method Deriving empirical models curve-fitting using least-squares Approximating complex functions Taylor series expansion … and I need you to take the first partial derivatives, set them equal to zero, and find my maximum profit point. 6

Let’s compute our first derivative! recall the binomial theorem? 7

The Notation All of the following are equivalent when y = f(x): 8

Three Rules to live by These are terrific rules but do you have a problem that can use them? 9

Let’s work a problem Find the equation of the line tangent to y = f(x) = 2x 2 + 2x + 3 at the point (1,7) f’(x) = (2)2x 1 + 2x = 4x + 2 f’(1) = 6 Using the point-slope form of a straight line: y – 7 = 6(x – 1) or y = 6x

The Chain Rule I hope you don’t talk about the chain rule. I hate the chain rule! y=f[u(x)] 11

Alternate approach 12

The Engineer-Manager’s very own Table of Derivatives – the top 5 With these 5 rules and the chain rule, you can solve 91.4% of all the engineering management differential calculus problems. 13

Table of Derivatives – the next 5 As an ENM student, I have had only an occasional use of the next 5 rules but seldom numbers 9 and

Let’s work another problem 15

and some more… 16 The successful student will work the drill problems

Our First Application of Derivatives Analyzing Functions Analyze this! 17

Some Well Known Facts Given f(x) is continuous and differentiable in the interval (a,b), then If f’(x) = 0, x  (a,b), then f(x) is constant If f’(x) > 0, x  (a,b), then f(x) is strictly increasing If f’(x) < 0, x  (a,b), then f(x) is strictly decreasing

Convex and Concave Functions If f’(x) is increasing, x  (a,b), then f(x) is convex in the interval (a,b) If f’(x) is decreasing, x  (a,b), then f(x) is concave in the interval (a,b) 19

Example of Some Well Known Facts f(x) = x 3 – 2x + 1 and f’(x) = 3x 2 – 2 20

Our Second Application of the Derivative Finding Your Roots… 21

Newton-Raphson Method Problem: solve for x: f(x) = 0 when f(x) is not linear or quadratic. Assume f(x) is continuous on the interval [a,b] and f(a) and f(b) have opposite sign, then the equation f(x) = 0 has at least one real root between a and b. 22

Newton-Raphson Method -2 x 2 x 1 f(x 1 ) 23

Newton-Raphson Method - 3 Repeat the above in order to obtain a better approximation To find the solution to f(x) = 0 Recursion formula 24

Newton-Raphson Method - 4 Find the root of f(x) = x 4 – 4x + 1 = 0 in the interval (0,1) f(0) = 1 and f(1) = -2 (a sign change, therefore at least one root) f’(x) = 4x start 25

A Profit Example Given the following profit function, determine the break even point : P(x) = 96x -.01x 2 –750 x.4 –100,000 P’(x) = x –(.4)(750)x.4-1 = x -300x -.6 For the break- even point, find P(x) = 0: 26

A Profit Example x = 1382 x =

Those problems are child’s play. Can’t you do something with a little more substance? Yes, how about second order derivatives? I bet you can’t work those? 28

Higher Order Derivatives First derivative y’f’(x) Second derivative y’’f’’(x) Third derivative y’’’f’’’(x) Fourth derivative Y (4) f (4) Since a derivative is a function, it can have its own derivative 29

An Example f(x) = 6x 3 – 12x 2 + 6x – 2 f’(x) = 18x 2 – 24x + 6 f’’(x) = 36x – 24 f’’’(x) = 36 f (4) = f (5) = … = 0 30

Our Third Application of Derivatives Approximating Complex Functions 31

Taylor’s Series Approximations Taylor’s Theorem: Let a be any point on the real line and n be any non- negative integer. Then for any value of x such that f(x) and its first n+1 derivatives exists and are continuous at every point between a and x inclusive 32

The implication of Taylor’s theorem is that you can approximate a wide range of functions at a point with a polynomial. Hmmm…This could useful? 33

Example: f(x) = 100/x 2 f ’ (x) = -200/x 3 f’’(x) = 600/x 4 f’’’(x)= -2400/x 5 f iv (x) = 12000/x 6 approximate f(x) at x = 1 f(x)  100 – 200(x-1) + 300(x-1) 2 – 400(x-1) 3 +(12000/24) (x-1) 4 Taylor Series Example 34

Partial Derivatives We double our pleasure as we go from a single variable function to a 2 variable function. 35

Partial Derivatives To find f x (x,y), treat y as a constant and differentiate f with respect to x in the usual way. To find f y (x,y) treat, x as a constant and differentiate f with respect to y in the usual way. 36

Example Problems 37

More Notation 38

One Last Topic – Higher Order Partial Derivatives Hurry, he is going to find all of the second partial derivatives of a two- variable function. ENM students hurrying to class. 39

Here it is… 1 st partials 2 nd partials 40

Taylor Series revisited – 2 variables f(x,y) = f(x 0,y 0 ) + f x (x 0,y 0 ) (x – x 0 ) + f y (x 0,y 0 ) (y – y 0 ) + ½ f xx (x 0,y 0 ) (x – x 0 ) 2 + f xy (x 0,y 0 ) (x – x 0 ) (y – y 0 ) + ½ f yy (x 0,y 0 ) (y – y 0 ) 2 + higher order terms A second order approximation of f(x,y) in the neighborhood about the point (x 0,y 0 ): 41

Matrix-Vector Representation f(x,y) = f(x 0,y 0 ) + f x (x 0,y 0 ) (x – x 0 ) + f y (x 0,y 0 ) (y – y 0 ) + ½ f xx (x 0,y 0 ) (x – x 0 ) 2 + f xy (x 0,y 0 ) (x – x 0 ) (y – y 0 ) + ½ f yy (x 0,y 0 ) (y – y 0 ) 2 Hessian gradient 42

This cries out for an example… 43

This concludes an adventure in differentiation Next time – optimize with the best! The adventurous student will now work all the problem exercises. 44