Download Survey of Mathematical Ideas Math 100 Chapter 1

Survey of Mathematical Ideas
Math 100
Chapter 1
John Rosson
Tuesday January 23, 2007
Chapter 1
The Art of Problem Solving
1.
2.
3.
4.
Solving Problems by Inductive Reasoning
Number Patterns
Strategies for Problem Solving
Calculating, Estimating and Reading
Graphs
Number Patterns:
Successive Differences
The method of successive difference tries to find a
pattern in a sequence by taking successive differences
until a pattern is found and then working backwards.
2
57 220 575 1230 2317 ….
55 163 355 655 1087 ….
108 192 300 432
….
84 108 132
….
24 24 ….
Number Patterns:
Successive Differences
Fill in the obvious pattern and work backwards by
adding.
2
57 220 575 1230 2317 3992
55 163 355 655 1087 1675
108 192 300 432 588
84 108 132
156
24 24
24
The method of successive differences predicts
3992 to be the next number in the sequence.
Number Patterns:
Successive Differences
The method of successive differences is not always
helpful. Consider
1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 ….
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ….
0 1 1 2 3 5 8 13 21 34 55 89 144 233 ….
1 0 1 1 2 3 5
8 13 21 34 55 89 ….
Since the sequence reproduces itself after
applying successive differences, the method
can give us no simplification.
Number Patterns; Sums
We can use patterns to conjecture formula for extended sums.
1 2
1
2
23
1 2 
3
2
3 4
1 2  3 
6
2
45
1 2  3  4 
 10
2
1
red is n
blue is n+1
Conjecture: The sum of the first n numbers
n (n  1)
1  2  3  ...  n 
2
True
Number Patterns; Sums
A little more interesting is the sum of squares…
1 (1  1)  (1  2 1)
1
6
2  (1  2)  (1  2  2)
5  12  2 2 
5
6
3  (1  3)  (1  2  3)
14  12  2 2  32 
 14
6
4  (1  4)  (1  2  4)
30  12  2 2  32  4 2 
 30
6
1  12 
Conjecture: The sum of the squares of the first n numbers
n (1  n)(1  2n)
2
2
2
2
True
1  2  3  ...  n 
6
Number Patterns; Sums
Consider the patterns that we have seen….
n
1  2  3  ... n 
1
n 1 n
1
1
1
1
1  2  3  ... n  
1 2
n 1 n 1 2n
12  2 2  32  ... n 2  

1 2
3
0
0
0
0
Conjecture: The sum of the cubes of the first n numbers
n 1 n 1 2n 1 3n
3
3
3
3
 1  2  3  ... n  


1 2
3
4
Number Patterns; Sums
Check some cases.
1 1 1 1 2 1 1 31
11  


1
1 2
3
4
2 1 2 1 2  2 1 3 2 35
3
3
9 1  2  



1 2
3
4
4
3 1 3 1 2  3 1 3 3
3
3
3
36  1  2  3  


 35
1 2
3
4
3

So 2 is a counterexample (as is 3) and the conjecture is
false. In this case close is not enough
Number Patterns; Sums
Looking at the sums of cubes, the true conjecture is ….
Conjecture: The sum of the cubes of the first n numbers
2


n
1
n
True
13  2 3  33  ... n 3   

1
2 
2


1
1
1
1  13   
  1
1 2 
2

2 1 2 
9  13  2 3   
  9
1 2 
2


3
1
3
36  13  2 3  33   
  36
1 2 
2


4
1
4
100  13  2 3  33  4 3   
  100
1 2 
The sums of higher powers
have no simple formula.
They are formulated in terms
of a new idea: the Bernoulli
polynomials.
Number Patterns;
Figurate Numbers
Number Patterns;
Figurate Numbers
The figurate numbers are a classical source of number sequences.
Triangular numbers: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55…
Square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100…
Pentagonal numbers: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145…
Hexagonal numbers: 1, 6, 15, 28, 45, 66, 91, 120, 153, 190…
Heptagonal numbers: 1, 7, 18, 34, 55, 81, 112, 148, 189, 235…
Octagonal numbers: 1, 8, 21, 40, 65, 96, 133, 176, 225, 280…
Nonagonal numbers: 1, 9, 24, 46, 75, 111, 154, 204, 261, 325…
Number Patterns;
Figurate Numbers
We can calculate the figurate numbers using successive differences.
Consider the nonagonal numbers.
1 9 24 46 75 111 154 204 261 325 396
8 15 22 29 36 43 50
57 64 71
7 7 7 7 7 7
7
7
7
Number Patterns;
Figurate Numbers
Formulas for the figurate numbers:
n(n  1)
2
Sn  n2
Tn 
Pn 
Hn 
Hp n 
On 
Non 
n(3n  1)
2
n ( 4 n  2)
2
n(5n  3)
2
n ( 6 n  4)
2
n(7 n  5)
2
Considering these formulas leads
us to conjecture a formula for a
general N-agonal number:
Nn 
n(( N  2)n  ( N  4))
2
True
Note that this formula works for N=3 and N=4.
It even works for N=2 (biagonal numbers).
Strategies;
Polya’s Four Step Process
1.
2.
3.
4.
Understand the problem
Devise a plan
Carry out the plan
Look back and check
Other Strategies
• Make a table or chart.
• Look for a pattern.
• Solve a similar
simpler problem.
• Draw a sketch.
• Use inductive
reasoning
• Solve an equation.
•
•
•
•
•
Use a formula.
Work backward.
Guess and check
Common sense (?)
Look for a “catch” if
the problem seems too
easy or impossible.
Example
How must one place the integers from 1 to 15 in each of the spaces
below in such a way that no number is repeated and the sum of the
numbers in any two consecutive squares is a perfect square?
Example
Maybe the problem is easy. Let us just try to fill in the squares
following the “consecutives sum” to square rule. Start, say, with 1.
This “playing with the problem” is part of understanding it.
1
15
10
6
3
1
3
6
10
15
13
12
4
5
11
14
2
7
9
Nothing left to
add to 9.
Nothing left to
add to 15.
Let us construct a table to see how things “add up”.
Example
1+3=4
1+8=9
1+15=16
6+3=9
6+10=16
11+5=16
11+14=25
2+7=9
2+14=16
7+2=9
7+9=16
12+4=16
12+13=25
3+6=9
3+13=16
8+1=9
13+3=16
13+12=25
4+5=9
4+12=16
9+7=16
14+2=16
14+11=25
5+4=9
5+11=16
10+6=16
10+15=25
15+1=16
15+10=25
The table
tells us that
we have to
start with
either an 8
or a 9 since
these two
numbers can
only be
paired with
one other
number.
Example
The plan is now to start with either 8 or 9 see if we can fill in
the table. Most numbers have only two choices for neighbors
and one choice will eliminate the next.
9
16
7
2
9
14
16
25
11
16
5
4
9
12 13
16
25
3
16
6
9
10 15
16
25
1
16
8
9
Answer
Sum Check
Assignments 2.1, 2.2, 2.3, 2.4
Read Section 2.1
Due January 25
Exercises p. 54
1-8, 21-28, 33-40, 41-50, and
67-76
Read Section 2.2
Due January 25
Exercises p. 61
1-6, 23-42, 44, 49-54
Read Section 2.3
Due January 30
Exercises p. 73
1-6, 7-27 odd, 47, 51, 52, 71, 75,
97, 115, 127, 129, 131, 133.
Read Section 2.4
Due February 1
Exercises p. 79
1, 3, 5, 7, 9, 17, 19, 25, and 27