numerical methods applications of loops: the power of matlab mathematics + coding 1
TRANSCRIPT
Numerical Methods
Applications of Loops:
The power of MATLAB
Mathematics + Coding
1
Numerical MethodsNumerical methods are techniques used to find quantitative solutions
to mathematical problems.
Many numerical methods are iterative solutions – this means that the technique is applied over and over, gradually getting closer (i.e. converging) on the final answer.
Iterative solutions are complete when the answer from the current iteration is not significantly different from the answer of the previous iteration.
“Significantly different” is determined by the program or user – for example, it might mean that there is less than 1% change or it might mean less than 0.00001% change.
2
Root Finding
If we can find where f(x) = 0, then we can find where f(x) equals any value.
To find where f(x) = -2, find the roots of f(x)+2
Essentially, we can shift any function up or down and solve for the roots, so all value finding problems can essentially be adjusted to be root finding problems.
We can turn any equation into a root finding problem.
3
Bracketing Methods
There are 2 basic “bracketing methods” (only 1 of which we’ll discuss).
They both rely on the same basic concept
If you know that f(x1) > 0 and f(x2) < 0, then a root must occur between x1 and x2.
4
More ExamplesES206 Fluid Dynamics
5
fRD
e
f N
5 1.2
7.3lo g0.2
1
Solve for f
The Colebrook Equation
ES206 – Fluid Dynamics
6
fRD
e
f N
5 1.2
7.3lo g0.2
1
“WOW – that’s hard to solve!”
How about this instead:
“By hand, find the square root of 127.3”
Both problems use the same technique to solve: the Bisection Method.
Bisection Method
For the Colebrook Equation, we know that this equation must hold:
Re-arrange and you have a roots problem. Guess for f and if the equation is not satisfied, adjust f and try again.
This is the same as solving for 127.3 = X2 . We made it X2 -127.3 = 0 and found the roots.
7
fRD
e
f N
5 1.2
7.3lo g0.2
1
Bisection Method
The Bisection Method is a very common technique for solving mathematical problems numerically.
Most people have done this even if they didn’t know the name.
The technique involves taking a guess for the desired variable, then applying that value to see how close the result is to a known value.
Based upon the result of the previous step, a better “guess” is made and the test repeated. 8
Bisection Method
Typical human method: Find the square root of 127.3 First guess: 11 (since we know that 112 is 121) 11*11 = 121 too low (looking for 127.3) Adjust: 12 is the new guess 12*12 = 144 too high Adjust: 11.5 (“bisect” the difference) 11.5*11.5 = 132.25 too high Adjust: 11.25 11.25 * 11.25 = 126.5625 too low etc.
9
Bisection MethodLet’s use the programmer’s method – an algorithm:
Define x1 so that f(x1) is greater than 0.
Define x2 so that f(x2) is less than 0.
Evaluate f((x1+x2)/2)
If it is positive, set x1 to (x1+x2)/2
If is it negative, set x2 to (x1+x2)/2
Repeat until you’re happy with x1-x2 10
Pros/Cons with Bisection
Cons Requires f(x1) < 0 and f(x2) > 0.
CAN be hard to find Can require A LOT of iterations.
Pros GUARANTEED to work
11
Newton’s Method
X is the root we’re trying to find – the value where the function becomes 0
Xn is the initial guess (hopefully, a reasonably-close guess!)
Xn+1 is the next approximation using the tangent line
Credit to: http://en.wikipedia.org/wiki/Newton’s_method12
The Big Picture (1/4)
2nd: Go hit the actual curve
At this (xn , yn), calculate the slope of the curve using the derivatives:
m = f’(xn)
14
The Big Picture (1/4)3rd: Using the slope m, the coordinates (Xn , Yn), calculate Xn+1 :
15
Remember: m was obtained from the derivative – only Xn+1 is unknown. What is Yn+1?
The Big Picture (2/4)
Guess again!
TOO FAR
16
The Big Picture (2/4)
Guess again!
TOO FAR
18
The Big Picture (3/4) – Eventually…
Stop the loop, when there is not much difference between the guess and the new value!
CLOSE ENOUGH!
Old estimate
New estimate
19
Newton’s Method
Prepare a MATLAB program that uses Newton’s Method to find the roots of the equation:
y(x) = x3 + 2x2 – 300/x
20
Newton’s Method
f(x) = x3 + 2x2 – 300/x
What does it seem the answer is?
21
Newton’s MethodCloser look…
Actual Root!
22
Ex 1: algorithm, skeleton
% Provide an initial guess for the x-intercept
% Repeat until close enough: compare most recent x-intercept to previous x-intercept
% Find the slope of the tangent line at the point (requires derivative)
% Using the slope formula, find the x intercept of the tangent line
23
Newton’s Method% Define initial difference as very large (force loop to start)
% Repeat as long as change in x is large enough
% Make the “new x” act as the current x
% Find slope of tangent line using f’(x)
% Compute y-value
% Use tangent slope to find the next value: % two points on line are the current point (x, y) % and the x-intercept (x_new, 0)
24
% Define initial difference as very large % (force loop to start)x_n = 1000;x_np1 = 5; % initial guess
% Repeat as long as change in x is large enoughwhile (abs(x_np1 – x_n) > 0.0000001) % x_np1 now becomes the new guess x_n = x_np1;
% Find slope of tangent line f'(x) at x_n m = 3*x_n^2 + 4*x_n + 300/(x_n^2);
% Compute y-valuey = x_n^3 + 2*x_n^2 - 300/x_n;
% Use tangent slope to find the next value: % two points on line: the current point (x_n, y_n) % and the x-intercept (x_np1, 0) x_np1 = ((0 - y) / m) + x_n; end
TOO FAR XnXn+1
? XnXn+1
25
Newton’s Method
Add fprintf’s and see the results:
x y------------------- ------------------- 5.000000000000000 115.000000000000000 3.925233644859813 14.864224007996924 3.742613907949879 0.279824373263295 3.739045154311932 0.000095471294827 3.739043935883257 0.000000000011084
Root of f(x) = x3 + 2x2 – 300/x is approximately 3.7390439 26
Real-World
Numerical Methods are used in CFD, where convergence can take weeks/months to complete (get “close enough”)...
while loops are generally used, since the number of iteration depends on how close the result should be, i.e. the precision.
Practice! Code this, and usea scientific calculator to see how close the results are.
27
Cons
Cons Divide by 0 Error
What happens if x is a maximum or minimum – the slope of the tangent line is 0.
Divergence What if our tangent line
leads us away from our root.
Cycling f(x)=x3-x-3 f’(x)=3x2-1 x0=0
28
xi xi+1
0.0000 -3.0000
-3.0000 -1.9615
-1.9615 -1.1472
-1.1472 -0.0066
-0.0066 -3.0004
-3.0004 -1.9618
-1.9618 -1.1474
-1.1474 -0.0073
-0.0073 -3.0005
-3.0005 -1.9619
Secant Method:
Find the Derivative
Numerical Methods
29
Derivatives, intro.
Another numerical method involves finding the value of the derivative at a point on a curve.
Basically, this method uses secant lines as approximations to the tangent line.
This method is used when the actual symbolic derivative is hard to find, the formula is simply not needed, or the function is prone to change as the main program does its job.
30
Derivatives, big picture
What is the slope at this point? 31
Derivatives (1/5)
Choose a and b to be equidistant from that point, calculate slope 1
Slope 1
X
32
range
range/2 range/2
Derivatives (2/5)
Zoom in on x
Slope 1
xloX
hiX
33
Reduce the range for the second iteration.
Derivatives (3/5)
Find new secant line.
Slope 1
xloX
hiX
34
Derivatives (3/5)
Calculate new slope
Slope 1
xloX
hiX
loY
hiY
Slope 2
35
Derivatives (3/5)
Slope 1
xloX
hiX
Slope 2
Is slope 2 “significantly different” from slope 1?Assume they are different… let’s iterate!...
36
Derivatives (4/5)
Zoom in again..
xloX
hiX
Slope3
Slope 2
Slope 1
37
Derivatives (4/5)
Zoom in again..
xloX
hiX
Slope3
Is Slope2 very different from Slope3?
Iterate while yes!
Slope 2
Slope 1
38
(5/5) Eventually…
Stop the loop when the slopes are no longer “significantly different”
Remember: “significantly different” means maybe a 1% difference? maybe a 0.1% difference? maybe a 0.000001% difference?
39
Derivatives - Algorithm% Define initial difference as very large % (force loop to start)
% Specify starting range % Repeat as long as slope has changed “a lot”
% Cut range in half
% Compute the new low x value
% Compute the new high x value
% Compute the new low y value
% Compute the new high y value
% Compute the slope of the secant line % end of loop
Let’s do this for f(x) = sin(x) + 3/sqrt(x)
40
Code% Define initial difference as very large % (force loop to start)m = 1; oldm = 2;
% Specify starting range and desired pointrange = 10; x=5; ctr=0;
% Repeat as long as slope has changed a lot while (ctr<3 || (abs((m-oldm)/oldm) > 0.00000001))
% Cut range in halfrange = range / 2;
% Compute the new low x valuelox = x – range ;
% Compute the new high x valuehix = x + range ;
% Compute the new low y value loy = sin(lox) + 3/sqrt(lox);
% Compute the new high y valuehiy = sin(hix) + 3/sqrt(hix);
% Save the old slopeoldm = m;
% Compute the slope of the secant linem = (hiy – loy) / (hix – lox);
% increment counterctr = ctr + 1;
end % end of loop
(or have just started)
41
Derivatives - the endAdd some fprintf’s and…
LOW x HIGH x Slope Difference in slope 1: 0.000000000000000, 10.000000000000000 -Inf -Inf 2: 2.500000000000000, 7.500000000000000 -0.092478729683983 Inf 3: 3.750000000000000, 6.250000000000000 0.075675505484728 0.168154235168711 4: 4.375000000000000, 5.625000000000000 0.130061340134248 0.054385834649520 5: 4.687500000000000, 5.312500000000000 0.144575140383541 0.014513800249293 6: 4.843750000000000, 5.156250000000000 0.148263340290110 0.003688199906569 7: 4.921875000000000, 5.078125000000000 0.149189163013407 0.000925822723297 8: 4.960937500000000, 5.039062500000000 0.149420855093148 0.000231692079740 9: 4.980468750000000, 5.019531250000000 0.149478792897432 0.000057937804284 10: 4.990234375000000, 5.009765625000000 0.149493278272689 0.000014485375257 11: 4.995117187500000, 5.004882812500000 0.149496899674250 0.000003621401561 12: 4.997558593750000, 5.002441406250000 0.149497805028204 0.000000905353954 13: 4.998779296875000, 5.001220703125000 0.149498031366966 0.000000226338761 14: 4.999389648437500, 5.000610351562500 0.149498087951815 0.000000056584850 15: 4.999694824218750, 5.000305175781250 0.149498102098005 0.000000014146190 16: 4.999847412109375, 5.000152587890625 0.149498105634484 0.000000003536479 17: 4.999923706054688, 5.000076293945313 0.149498106518877 0.000000000884393
Slope of f(x) = sin(x) + 3/sqrt(x) at x=5 is approximately 0.1494981 42
Integration
• Some functions can’t be integrated symbolically
• Other functions are just REALLY ugly to do it
Trapezoidal Rule
• Approximate the curve with an easy to integrate first order polynomial - This ends up being a trapezoid.
Trapezoidal Rule• Trapezoidal Rule works out as:
Composite Trapezoidal
• Can we just combine multiple trapezoids?• Yes - Instead of 1 large trapezoid, we’ll use multiple smaller trapezoids
and add up their areas (and their errors)
x y Area
0 00.2910
1 0.58200.9171
2 1.25211.3334
3 1.41471.3044
4 1.19411.0210
5 0.8480
Sum: 4.8669
Can We Go Smaller?
• You can decrease the size of your trapezoids, but they are still always going to be approximating a curved line using a linear formula.
• With a width of 1, the value is 4.8669.• Error = 0.0302
• With a width of 0.5, the value is 4.8924• Error = 0.0074
0.0 0.00000.0482
0.5 0.19270.1937
1.0 0.58200.3878
1.5 0.96940.5554
2.0 1.25210.6624
2.5 1.39730.7030
3.0 1.41470.6874
3.5 1.33500.6323
4.0 1.19410.5544
4.5 1.02370.4679
5.0 0.8480
4.8924
Simpson’s 1/3 Rule
• At some point, we can’t cut our width anymore. We need to actually approximate the line better.
• Trapezoidal Method approximated each pair of consecutive points with a linear line (1st order polynomial).
• What if we approximate every three points with a 2nd order polynomial?
• Error = 0.0011
0.0 0.0000
0.22550.5 0.1927
1.0 0.5820
0.95191.5 0.9694
2.0 1.2521
1.37602.5 1.3973
3.0 1.4147
1.32483.5 1.3350
4.0 1.1941
1.02284.5 1.0237
5.0 0.8480
4.9010
What about 3rd & 4th Order?
• Simpson’s 3/8th Rule - 3rd Order
• Requires an EVEN number of points• Boole’s Rule - 4th Order
Method h Estimate Absolute Error
Evaluations of f(x)
Math Operation
s
Trapezoidal 1 4.8669 0.0302 6 10+4
Trapezoidal 0.5 4.8924 0.0074 11 20+9
Simpson’s 1/3 1 4.901 0.0011 11 30+4
Simpson’s 1/3 0.5 4.9 0.0001 21 60+9
Simpson’s 3/8 1 4.9004 0.0005 16 90+4
Simpson’s 3/8 0.5 4.8999 0 31 180+9
Boole’s 1 4.8999 0 21 130+4Math Operations = Number of +,-,*,/ to calculate segment + Number of Additions to add up all segments
Wrapping Up
Very small amount of code to solve a mathematical problem
MATLAB can iterate 5 times, or 1,000 times, or a million!
The more precision needed, the more iterations, the more time!
Try to code it. See it you can find the slope of any equation.
58
Wrapping Up
We learned about four numerical methods 1. Bisection Method
Finds numerical solutions to complex mathematical problems Very general – essentially “Guess, Test, Adjust”
2. Newton’s Method for finding the roots of an equation Requires the symbolic derivative
3. Secant line approximation of a tangent point Finds the numerical value of the derivative at a point
4. Integration using Trapezoidal and Simpson’s rules59
More…
Numerical Methods is often a separate class at some universities.
Other Common Topics are: Solving large systems of equations Polynomial Approximation Curve Fitting Optimization ODE/PDE
Chapra and Canale’s “Numerical Methods for Engineers” is considered “the resource”
Also: http://nm.mathforcollege.com/
60