NUMERICAL METHODS
AP Calculus AB
Summer Review 2013
ANALYTICAL METHODS
For as long as you have taken math, you have learned
analytical methods to solving problems
 These methods involve a well-defined procedure and
provide an exact answer


AP Calculus AB, SJHS 2013-2014

even if it’s in terms of a complex or irrational number
Unfortunately, very few “real world” problems are
analytical

Most real world problems involve data sets, and you must
infer information from these sets
2
NUMERICAL METHODS



Still, you must use caution not to take numerical solutions as
absolute answers


AP Calculus AB, SJHS 2013-2014
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Numerical methods are ways of solving math problems
based on data sets, and they do not necessarily require an
analytical method or solution
Numerical methods do have an analytical theory behind
them, usually involving statistics
These are useful since there are no constraints
they are often approximations
The answers are found by trial and error methods, which a
computer algebra system (or “CAS”) can do very quickly
So the computer actually checks which answers work
 The one that works best is “printed” as the answer

3
SYSTEMS OF NONLINEAR EQUATIONS

We remember working with systems of linear
equations
AP Calculus AB, SJHS 2013-2014
These are functions of variables to the first power
 No other powers or special functions were allowed

You may remember that one method of solving was to
draw a graph and find the intersection of the lines
 By extension, we can find the intersection(s) of
nonlinear functions to find the solution(s) of systems
of nonlinear equations
 This is numerical since the solution is usually at a
value we can’t analytically calculate

4
x
ye
EX:
y  x  5

AP Calculus AB, SJHS 2013-2014
First, we should notice that since they both equal y,
we can set the functions equal to each other (Method of
Substitution)
x
e  x  5

The problem is that to solve it, we would need to take
a logarithm to undo the exponential, but we just
trade one problem for another

Which is worse?
e x   x  5


 x  ln   x  5 
Instead, let’s just look at our original problem, and
instead view them as two different functions:
y1  e  x
y2   x  5
5
EXAMPLE CONTINUED
Y1  e ^   X 

Your calculator uses this form!

Type this into your calculator and take a look at the
graph
You can also type this into GraphCalc; this program literally
took me 1min to install and it was ready for use


http://www.graphcalc.com/, click “Download” at the top left menu in
blue, click the first link under Regular Users (GraphCalc.exe), click the
green button that says Download Now! (don’t bother with the ad
behind it
If you’re a Mac, you can try Graphing Calculator 3D for Mac


AP Calculus AB, SJHS 2013-2014

Y2   X  5
http://download.cnet.com/Graphing-Calculator-3D/30002053_4-10896524.html#rateit
Now we’re ready for some numerical methods
6
EXAMPLE CONTINUED

You should see this graph

As you can see, the intersections occur
somewhere around x=-2 and x=5
How to solve it on your calculator






After you typed in the functions, go to CALC (the 2nd of TRACE)
Click Intersect
It will take you back to the graph; move the cursor near the
intersection point (doesn’t have to be exact) & click
It now needs you to confirm this on the other graph, so move to that
same intersection point & click
It will now ask for a guess; depending on which intersection, you
can give it a guess; otherwise just press enter
There’s your first answer; do this for all other intersection points
AP Calculus AB, SJHS 2013-2014
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
How to solve it in Graph Calc
Go to Tools > Equation Solver and click
 Type in e^(-x)=-x+5
 Give it your guess; let’s say, -2
 It gives the answer that is closest to your guess; if you said 7, it
would’ve given you the other intersection point

7
ZEROS
A zero is where a function crosses the x-axis
 For linear functions, just set y = 0
 We learned a general analytical solution, the
Quadratic Formula, that solves the zeros of quadratic
functions
 Occasionally you can easily solve zeros of
exponentials, logs, and trig equations
 What if we have other equations that need zeros?

AP Calculus AB, SJHS 2013-2014
This is usually a nonlinear function with a mix of
incompatible function types
 We can solve these using numerical methods

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EX: x  sin x  0
2
This does not have an analytical solution, and
according to the graph there exist two real roots
 We could do a trial and error using the Location
Principal


AP Calculus AB, SJHS 2013-2014
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zeros occur between any positive/negative transition
Luckily, modern computers can quickly do this trial
and error
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EX: x 2  sin x  0 CONTINUED

To solve using the calculator







To solve in Graph Calc
Go to Tools > Equation Solver and click
Type in x^2+sin(x) … if you don’t give another side of the equation,
it will assume you want a zero
 Give it your guess
 It gives the answer that is closest to your guess


AP Calculus AB, SJHS 2013-2014
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type in the function
go to CALC (2nd of TRACE), click Zero
Your calculator doesn’t know which zero to find (there could be
several), so you need indicate where it is; it asks for a “left bound”
then a “right bound”
When it says “Left Bound”, bring the cursor to a point that is left of
the zero (doesn’t have to be too close)
When it says “Right Bound”, bring the cursor to a point that is
right of the zero (doesn’t have to be too close)
It will ask for a guess; you can provide one, or just press Enter
If you get an error message, you might have placed your bounds
around two zeros, so make sure your bounds only surround one
answer
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LINEAR REGRESSIONS
Among the most useful numerical tools is a linear
regression
 Linear data is the most useful, and if you go into
applied science and math you will see that analysis is
often done by “linearizing”
 The “goodness of fit” is represented by the
correlation value, or r2 parameter
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
AP Calculus AB, SJHS 2013-2014
Values near 1 are a good fit
 If r2 = 1, then it is a perfect fit (not likely)

Let’s look at a simple example
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EXAMPLE

1 Make a linear regression of the data set,
with a proper equation.
 2 Is this a good fit?
 3 What is the high school-to-college GPA
factor? What does this mean?
 4 Predict the GPA of a student 1.6 GPA in
high school.

College
1.6
2
1.8
3.3
2.8
2.1
2
2.65
2.25
2.6
3
3.1
3.2
AP Calculus AB, SJHS 2013-2014
A group of students are surveyed during
their freshman year of college. They are
anonymously asked their GPA during
high school and their first year of college.
The results are summarized on the table
(no names).
HS
2
2.25
2.6
2.9
2.65
2.8
3.1
2.9
3.25
3.3
3.6
3.25
4
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EXAMPLE CONTINUED

We can plot this data set in our calculator, or on a
computer we can use Microsoft Excel
Excel: Type in one set in a column, then the other set in
another column
To see the data, click Insert and choose Scatter (just the dots)
 The Excel table was shown last slide


Calculator: go to STAT > Edit… and several lists will
appear
Fill in those lists with data
 To display, go to STAT PLOT (2nd of “Y=”) and turn stat plot on
 Go to ZOOM, then ZoomStat (it’s there, you just need to find it on
the list)

AP Calculus AB, SJHS 2013-2014
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EXAMPLE CONTINUED

To make a linear regression



Don’t forget to click “Display equation” and “Display r2”
at the bottom. If you forget, just click on the line and go to
format trendline
Calculator: STAT > CALC > LinReg(ax+b)
The equation will display
 To pull up the r2, go to VARS > Statistics > EQ > r2

AP Calculus AB, SJHS 2013-2014
Excel: right click on any data point on the graph
(they’ll all light up) and go to “Add Trendline”, and
choose Linear
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EXAMPLE GRAPH THEN LINEAR
REGRESSION
4
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
High School GPA
4
College GPA
3.5
3
2.5
AP Calculus AB, SJHS 2013-2014
College GPA
3.5
2
1.5
y = 0.7187x + 0.3583
R² = 0.4625
1
15
0.5
0
0
0.5
1
1.5
2
2.5
High School GPA
3
3.5
4
EXAMPLE CONTINUED
1 Linear regression with equation & r2
 2 This is an okay fit; r2 in 80 or 90 percent range is
very good, but we got about 46% (a little low)


3 The coefficient is .71; this means your GPA is
expected to be only about 71% of what it was before.


However, the data goes through a lot of points!
You’ll notice this is on average, but several students in
this data set have basically the same, if not higher, GPAs.
It’s a statistical quantity.
4 To answer our last question, we might jump to say:
AP Calculus AB, SJHS 2013-2014

1.6  .71  1.14
However, this point is outside of our data set domain
 We cannot predict anything lower than 2.00 or higher
than 4.00 since that is the extent of our domain

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OTHER REGRESSIONS
Excel has the capability to give regressions for
polynomials (up to x6), exponential, logarithmic, and
power laws
 Calculators like a TI-84 or other graphing calculator
can give polynomial (up to x4), exponential,
logarithmic (natural and common), power law, and
sine regressions

To graph these, go to Y=, then press VARS > Statistics >
EQ > ReEq

AP Calculus AB, SJHS 2013-2014
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This menu also stores all constants from your regression
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WHAT KIND OF REGRESSION IS THIS?
y
0
2
3
6
7
8
1
10
12
8
7
8
Given any data, you cannot assume it is a linear set.
First graph the points, then try to eyeball what it
looks like. In this case, it is difficult to tell.
AP Calculus AB, SJHS 2013-2014
x
Graph
15
10
5
0
0
5
10
18
BUT IT DOESN’T LOOK LINEAR!
Linear Regression
Graph
14
12
12
y = 0.3581x + 6.1149
R² = 0.0912
10
10
8
8
6
6
4
4
2
2
0
0
2
4
6
8
10
0
0
2
4
6
8
10
AP Calculus AB, SJHS 2013-2014
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The r2 is too low, and it clearly doesn’t go
through many points. We’ll try another one.
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THE QUADRATIC IS BETTER
Graph
Quadratic Regression
14
12
12
10
10
8
8
6
6
4
4
y = -0.4609x2 + 4.1328x + 2.2031
R² = 0.7273
2
2
0
0
0
2
4
6
8
10
0
2
4
6
You can see that the quadratic is a better fit.
 The r2 is much better.
8
10
AP Calculus AB, SJHS 2013-2014
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
But still not very good...
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TRY A CUBIC REGRESSION
Cubic Regression
Graph
14
14
12
12
10
10
8
8
y = 0.1249x3 - 1.9052x2 + 8.1158x + 0.9076
R² = 0.9912
4
4
2
2
0
0
0
2
4
6
8
10
0
2
4
6
This r2 is nearly perfect! Looks like a good fit.
 Problem: the equation is very complicated.
8
10
AP Calculus AB, SJHS 2013-2014
6
6
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In this case, it’s better to have a very good answer
even if it takes a little more copying.
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NO NEED TO GO TOO FAR
Quartic Regression
Graph
14
14
12
12
10
8
8
6
6
4
4
2
2
0
y = 0.0174x4 - 0.1646x3 - 0.4275x2 + 5.9245x + 0.9893
R² = 0.999
0
0
2
4
6
8
10
0
2
4
6
8
We can take it a step further and get a slightly
better r2, but ask yourself, is the more complex
formula that much better?
 Let’s stick with the cubic in this case

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AP Calculus AB, SJHS 2013-2014
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22
SEQUENCES AND SERIES

A sequence is a set of numbers that follow a pattern


Typically, a problem begins with finding the pattern
and expressing it numerically
Sequences use a notation similar to function notation

The variable is placed as a subscript to the sequence name
AP Calculus AB, SJHS 2013-2014
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A series is the summation of some or all numbers of a
sequence
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EXAMPLE OF A SEQUENCE
The average temperatures for each month in a city
in Arizona are:
Some samples from this series are:
January
m=1
T1 = 66
July
m=7
T7 = 84
November
m =11
T11 = 71
AP Calculus AB, SJHS 2013-2014
Tm  66,71,75,76,76,80,84,83,83,77,71,68
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EXAMPLE OF A SERIES
A small business has the following revenues.
Rn  11320,10240,15770,7880,9990,12540,6500,19430,13010,8850,10060,11210
We model the year profits as a series, which means we add up all
revenue numbers (from the first to the twelfth months), taking only
one tenth of each
12
1
Rn
n 1 10
P
We say:
“P equals the sum from n equals 1 to 12 of one-tenth R-n”
AP Calculus AB, SJHS 2013-2014
90% of that money goes towards expenses, so the
business profits one-tenth of the revenues. How
much does the business profit in one year?
P  1132  1024  1577  788  999  1254  650  1943  1301  885  1006  1121
P  13680
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FINDING PATTERNS OF SEQUENCES


Arithmetic:
There is a common difference between each number
Geometric:
There is a common ratio between each number.
Exponential:
term.
The numbers relate to an exponent of the sequence
Alternating:
Numbers alternate positive and negative.
Even:
component.
All numbers are even, or have a strictly even
Odd:
All numbers are odd, or have a strictly odd component.
Recursive:
Each term is dependent on the value of one or more
AP Calculus AB, SJHS 2013-2014

At first, you should verbalize the pattern, then figure out a
way to express it mathematically
Always mentally check the first few terms by substituting
in
n = 1, 2, etc to see if they match up with your sequence
Some common patterns:
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ARITHMETIC SEQUENCE EXAMPLES
For each sequence, determine the common difference d, then write
an expression for the sequence. Assume each sequence starts at
n=0.
a)
an  7,15,23,31, 
b)
an  77,54,31,8,   
d  23
an  77  23n
Notice that you can plug in
n = 0, 1, 2, 3,... And the
result is the given
sequence
AP Calculus AB, SJHS 2013-2014
d 8
an  7  8n
Pattern: each number is 8 more than
the previous! That’s the common
difference, and each term adds 8.
27
GEOMETRIC SEQUENCES
Find the common ratio r to each geometric sequence. Then write an
expression for the sequence. Assume each sequence begins at n = 0.
a)
an  81,54,36,24,   
r
2
3
b)
an  64, 32,16, 8,...
1
r
2
n
 1
an      64
 2
c)
AP Calculus AB, SJHS 2013-2014
n
2
an     81
3
Pattern: each number is 2/3 of the
previous! That’s the common ratio, and
each term is multiplied by 2/3. This is
expressed as an exponent.
an  8,20,50,125,   
5
r
2
n
5
an     8
2
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CONSTRUCTING SEQUENCES
Alternating:
factor of (-1)n
- as n goes from even to odd, so will your sequence
Odd numbers:
use (2n+1)
- This is always odd
Recursive:
include an-1, an-2, etc
- These indicate “the term before n”, “two terms before n”,
etc
AP Calculus AB, SJHS 2013-2014
Even numbers:
use 2n
- This is always even
29
DESCRIBE EACH SEQUENCE, AND EXPRESS EACH IN SEQUENCE NOTATION.
1 1 1
a)
*the denominators are all perfect odd
an  1, , , , 
9 25 49
squares
1
an 
2
 2n  1
an  3,7,19,39, 
an   2n   3
*each term is an even square, plus three
2
c)
an  0, 7,14, 21,28,   
an   1  7n
n
d)
an  0,1,4,9,16,   
an  n 2

an  an1   2n  1
e)
3 1
1
, ,0,  , 
2 2
2
 n 
an  cos 

 6 
an  1,
*these are the factors of seven, alternating
positive and negative
*this is clearly the sequence of squares
*we can also look at it as recursive, with each
term being the previous plus the odd numbers
in sequence
- Start at zero, add 1, add 3, add 5, add 7...
*these are exact values of cosine, on
intervals of pi/6
AP Calculus AB, SJHS 2013-2014
b)
30
NUMERICAL METHODS ASSIGNMENT
This quiz opens July 29 and closes Aug 12
 This is the last quiz and assignment, so please
enjoy the last part of your summer before we
begin the class

AP Calculus AB, SJHS 2013-2014
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END
AP Calculus AB, SJHS 2013-2014
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