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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 23 1 2 3 2 3 4 1 2 3 6 2 45 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 31 11 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