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Numerical Methods
Applications of Loops:
The power of MATLAB
Mathematics + Coding
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Numerical Methods
Numerical 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.
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Ex 1. Newton’s Method
Newton’s Method is used to find the roots of an equation.
“When the curve crosses the x-axis.”
“When f(x)=0.”
Prepare a MATLAB program that uses Newton’s Method to find
the roots of the equation:
y(x) =
3
x
+
2
2x
– 300/x
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Newton’s Method
f(x) = x3 + 2x2 – 300/x
What does it seem
the answer is?
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Newton’s Method
Closer look…
Actual Root!
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Newton’s Method
X is the root we’re trying to find – the value where the function becomes 0
Xn is the initial guess
Xn+1 is the next approximation using the tangent line
Credit to: http://en.wikipedia.org/wiki/Newton’s_method
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The Big Picture (1/4)
1st: guess!
xn
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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)
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The Big Picture (1/4)
3rd: Using the slope m, the
coordinates (Xn , Yn), calculate Xn+1 :
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The Big Picture (2/4)
TOO FAR
Guess again!
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The Big Picture (2/4)
Iterate!
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The Big Picture (2/4)
TOO
FAR
Guess again!
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The Big Picture (3/4) –
X interecept
Eventually…
CLOSE
ENOUGH!
guess
Stop the loop, when there is not much difference between the guess and
the new value!
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Ex 1: algorithm, skeleton
% Provide an initial guess at the answer
% Find the slope of the tangent line at the
point (requires derivative)
% Using the line’s equation, find the x
intercept
% Repeat above until close enough, using the
x intercept as the new “guess”
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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)
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% Define initial difference as very large
% (force loop to start)
Xn+1
TOO FAR
x_n = 1000;
x_np1 = 5; % initial guess
Xn
% Repeat as long as change in x is large enough
while (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);
Xn+1
?
Xn
% Compute y-value
y = 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
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Newton’s Method
Add fprintf’s and see the results:
x
------------------5.000000000000000
3.925233644859813
3.742613907949879
3.739045154311932
3.739043935883257
y
------------------115.000000000000000
14.864224007996924
0.279824373263295
0.000095471294827
0.000000000011084
Root of f(x) = x3 + 2x2 – 300/x is approximately 3.7390439
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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 use
a scientific calculator to see
how close the results are.
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Numerical Methods
Secant Method:
Find the Derivative
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Derivatives, intro.
Another numerical method involves finding the derivative at
a point on a curve.
Basically, this method uses the secant line as an
approximation – bringing it closer and closer to the
tangent line
This method is used when the actual symbolic derivative is
hard to find! f’(x)= 
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Derivatives, big picture
What is the slope at this
point?
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Derivatives (1/5)
Choose a and b to be equidistant from that point, calculate slope 1
Slope 1
X
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Derivatives (1/5)
Reduce the range for the second iteration.
Slope 1
range
X
range/2
range/2
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Derivatives (2/5)
Slope 1
loX
x
hiX
Zoom in on x
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Derivatives (3/5)
Slope 1
loX
x
hiX
Find new secant line.
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Derivatives (3/5)
Slope 1
Slope 2
hiY
loY
loX
x
hiX
Calculate new slope
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Derivatives (3/5)
Slope 1
Slope 2
loX
x
hiX
Is slope 2 “significantly
different” from slope 1?
Assume they are different…
let’s iterate!...
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Derivatives (4/5)
Slope 1
Slope 2
Slope3
loX
x
hiX
Zoom in again..
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Derivatives (4/5)
Slope 1
Slope 2
Slope3
loX
x
hiX
Zoom in again..
Is Slope2 very
different from
Slope3?
Iterate while yes! 29
(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?
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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)
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% Define initial difference as very large
% (force loop to start)
m = 1;
oldm = 2;
Code
% Specify starting range and desired point
range = 10;
x=5;
ctr=0;
% Repeat as long as slope has changed a lot
(or have just started)
while (ctr<3 || (abs((m-oldm)/oldm) > 0.00000001))
% Cut range in half
range = range / 2;
% Compute the new low x value
lox = x – range ;
% Compute the new high x value
hix = x + range ;
% Compute the new low y value
loy = sin(lox) + 3/sqrt(lox);
% Compute the new high y value
hiy = sin(hix) + 3/sqrt(hix);
% Save the old slope
oldm = m;
% Compute the slope of the secant line
m = (hiy – loy) / (hix – lox);
% increment counter
ctr = ctr + 1;
end % end of loop
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Derivatives - the end
Add some fprintf’s and…
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
14:
15:
16:
17:
LOW x
0.000000000000000,
2.500000000000000,
3.750000000000000,
4.375000000000000,
4.687500000000000,
4.843750000000000,
4.921875000000000,
4.960937500000000,
4.980468750000000,
4.990234375000000,
4.995117187500000,
4.997558593750000,
4.998779296875000,
4.999389648437500,
4.999694824218750,
4.999847412109375,
4.999923706054688,
HIGH x
10.000000000000000
7.500000000000000
6.250000000000000
5.625000000000000
5.312500000000000
5.156250000000000
5.078125000000000
5.039062500000000
5.019531250000000
5.009765625000000
5.004882812500000
5.002441406250000
5.001220703125000
5.000610351562500
5.000305175781250
5.000152587890625
5.000076293945313
Slope
-Inf
-0.092478729683983
0.075675505484728
0.130061340134248
0.144575140383541
0.148263340290110
0.149189163013407
0.149420855093148
0.149478792897432
0.149493278272689
0.149496899674250
0.149497805028204
0.149498031366966
0.149498087951815
0.149498102098005
0.149498105634484
0.149498106518877
Difference in slope
-Inf
Inf
0.168154235168711
0.054385834649520
0.014513800249293
0.003688199906569
0.000925822723297
0.000231692079740
0.000057937804284
0.000014485375257
0.000003621401561
0.000000905353954
0.000000226338761
0.000000056584850
0.000000014146190
0.000000003536479
0.000000000884393
Slope of f(x) = sin(x) + 3/sqrt(x) at x=5 is approximately 0.1494981
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Numerical Methods
Bisection Method
A general purpose
mathematical solution finder
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More Examples
ES206 Fluid Dynamics
Solve for f
e

1
2.51

D
 2.0 log

 3.7 R N f
f

The Colebrook Equation




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ES206 – Fluid Dynamics
e
2.51

D
 2.0 log

 3.7 R N f
f

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.
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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.
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Bisection Method










For example: 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.
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Bisection Method

For the Colebrook Equation, we know that this equation
must hold:
e


1
2
.
51

 2.0 log D 
 3.7 R N f 
f



Guess for f and compare the left side value to the right
side value. If they’re not the same, adjust f and try
again.

This is the same as solving for 127.3 = X2 :
we guessed at X and then compared the two sides. 39
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.
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Wrapping Up
We learned about three numerical methods

1. Newton’s Method for finding the roots of an equation


2. Secant line approximation of a tangent point


Requires the symbolic derivative
Finds the numerical value of the derivative at a point
3. Bisection Method


Finds numerical solutions to complex mathematical problems
Very general – essentially “Guess, Test, Adjust”
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