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DERIVATIVES June AND DIFFERENTIAL EQUATIONS
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WHAT IS A DERIVATIVE?
Derivative of f(x) at x=a is given by f'(a)
y=f(x) f(a) a
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SIGNIFICANCE OF DERIVATIVE
In the limit, derivative at x=a is equal to the slope m of the tangent line at a.
y=f(x) f(a) a
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APPLICATIONS OF DERIVATIVE
The most widely used application of derivative is in finding the extremum (max or min) points of a function. If a function has a local extremum at a number c then either f'(c)=0 or f'(c) doesn't exist
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CRITICAL NUMBERS
A number c in the domain of a function f is a critical number of f if either f'(c)=0 or f'(c) does not exist
f'(c)=0
x=c
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DERIVATIVES IN MATLAB:diff
MATLAB computes two differentials, dy and dx, using diff, to arrive at the derivative If x and y are input array of numbers dy=diff( y) dx=diff(x) Then, yprime=diff(y)./diff(x)
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How diff works
Take the array y=[3 5 7 5 9]; diff(y) simply is the pairwise differences, i.e. diff(y)=[53 75 57 95] which is equal to diff(y)=[2 2 2 4] Naturally, there is one fewer term in diff(y) 7 5/30/99 1999BG Mobasseri than it is in y
Using diff to find derivatives
Let y=[10 25 30 50 10] for x=[1 2 3 4 5] Then One fewer component diff(y)=[15 5 20 40] diff(x)=[1 1 1 1 1] Dividing termbyterm yprime=[15 5 20 40]
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WORKING WITH humps
humps is a builtin MATLAB function, like peaks and is given by
1 1 y= + -6 2 2 [( - .3) + 0.01] [( - .9) + .04 ]
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HOW IT LOOKS
100 90 80 70 60 50 40 30 20 10 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
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Try it!:derivative of humps
First, we must know how humps was created in order to know dx x=0:0.01:1;%dx in this case is 0.01 y=humps(x); Now use dy=diff(y); yprime=dy./dx; dx=diff(x);
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PLOT OF dy/dx
100 80 60 HUMPS
Function peaks
40 20 0 0 1000 500 0 -500 -1000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Deriv. goes to 0
FIRST DERIVATIVE OF HUMPS
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Finding critical points:theory vs. application
To find max. or min.we should look for instances where y'=0. This is where the difference between textbook methods and the real world shows up. To see for yourself look for instances where yprime=0.
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Where are the zeros?
If you did the search right, you will realize that nowhere in yprime you can actually see a zero Since there are no points where y' is exactly zero, we should look for points of transition + + from positive to negative, i.e. sign changes
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Zero crossing and sign transition
In Matlab, all derivatives are discrete. Some are positive, some negative but no zeros crossing Zero
Since derivative goes from + to , we can infer that it went to zero somewhere in between
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FINDING ZERO CROSSINGS
There are two ways to find where a function crosses zero; 1. Numerical 2. Algebraic(polynomial fitting)
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ZC USING ARRAY OPERATION
To find where a sign transition takes place, multiply two consecutive numbers in the derivative array. If there is a sign change, product the is negative [10 15 16 9 4 3 4 9]
X X X X X X X
Zero crossing
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MATLAB's WAY
Let x be an n element array[x1,x2,...,xn]. To generate the zero crossing array, do y=x(1:n1).*x(2:n) Algebraically, this is what is happening y=[x x x x ,..., x x ]
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Try it!
Evaluate humps in the range x=0 to 1 in increments of 0.01 This gives you a 1x101 array. Find yprime. This is now a 1x100 array Find where in this 101 positions sign changes occur
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LOCATING SIGN CHANGES
Just look for instances of negative sign in the zero crossing array
800 600 400 200 0 -200 -400 -600 -800 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1st=0.29 2nd=0.63 3rd=0.88
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CLOSER LOOK
800 600 400 200 0 -200 -400 -600 -800 0.2
Sign change
0.22
0.24
0.26
0.28
0.3
0.32
0.34
0.36
0.38
0.4
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SECOND ZC
50 40 30 20 10 0 -10 -20 -30 -40 -50 0.55 0.6 0.65 0.7 0.75
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THIRD ZC
50 40 30 20 10 0 -10 -20 -30 -40 -50 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95
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Differential Equations
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Simplest Diff.Eq
A first order differential equation is of the form y '=dy/dx=g(x,y) We are looking for a function y such that its derivative equals g(x,y)
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FEW EXAMPLES
Find y that such that y '=3x
y `=y y y `=(2x)cos
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SOLVING ODE's USING MATLAB
MATLAB solves ordinary DE in two ways: 1):numerical and 2):symbolic For numerical solution use [x,y]=ode23(`function',a,b,initial) See next slide for usage
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EXPLAINING ode23
Here are the elements of [x,y]=ode23(`function',a,b,initial) function this is g(x,y) in y '=g(x,y). Must be written as a separate MATLAB function a left point of the interval b right point of the interval initial initial condition, i.e. y(a)
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