You do Do in calculator, too.
Sequence formulas that give you the exact definition of a term are explicit formulas. Formula: a n = 5n Explicit, every term is defined by this formula. Recursive formulas define terms based on preceding terms. Formula: a a 1 2 = 1 = 1 a = a + a, for_ n n n 1 n 2 3 Recursive, after the second term and the two previous terms to get the next term.
Formula: a a 1 2 = 1 = 1 a = a + a, for_ n n n 1 n 2 3 Find the first 6 terms of the sequence. a a 1 2 = 1 = 1 a = a + a = 1+ 1 = 2 3 2 1 a = a + a = 2 + 1 = 3 4 3 2 a = a + a = 3+ 2 = 5 5 4 3 a = a + a = 5+ 3 = 8 6 5 4
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Problems for you.. Page 587-588 1,5,9,13,17,21,25,27,31,35,37, 45,51,53
More problems to do. Page 588 57,59,61,63,65,67,69,71,73,77
Notation for multiple sums: sigma Definition_ of _ Sigma_ Notation: The_ sum_ of _ n_ terms_ a, a, a,..., a is_ written_ as n ai = a1 + a2 + a3+... + a i= 1 i = index_ of _ summation a = ith_ term_ of _ the_ sum i lower_ bound = 1 upper_ bound = n n 1 2 3 n
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Summation Formulas n i= 1 c = cn 3 5 = 5+ 5+ 5 = 5( 3) = 15 i= 1 n i= 1 i 2 = n n n + i = ( 1) 2 i= 1 n( n + 1)( 2n + 1) 6 4! i 2 = 4(5)(9) = 30 " 1 2 + 2 2 + 3 2 + 4 2 = 30 6 i=1 ( ) 3 3(3 +1)! i = = 6 " 1+ 2 + 3 = 6 2 i=1 ( ) n i= 1 i 3 = n 2 ( n + 1) i 3 = 22 (2 +1) 2 i=1 4 4 2 2 ( )! = 9 " 1 3 + 2 3 = 9
Evaluate_ the_ sum n i + 1 2 i= 1 n for_ n = 10_ & _ n = 1000 n i +1! = 1 n!( i +1) " by _Theorem n 2 i=1 n 2 i=1 n 1!( i +1) = 1 n n " %! i +! 1 n 2 i=1 n 2 $ ' # i=1 i=1 & ( by _Theorem 1 n 2 n n " % $! i +! 1' # i=1 i=1 & = 1 " n(n +1) n 2 # $ n % + n & ' ( by _Theorem
Evaluate_ the_ sum n i + 1 2 i= 1 n for_ n = 10_ & _ n = 1000 1 n 2 n n " % $! i +! 1' # i=1 i=1 & = 1 " n(n +1) n 2 # $ n % + n & ' ( by _Theorem 1! n(n +1) $ + n n 2 " # n % & = n + 3 2n when _ n = 10 ' 13 20 =.65 when _ n = 1000 ' 1003 2000 =.50015
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Problems for you. Page 588-589 79-109 odds
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Problems for you From page 598 #1-9 odds, 13, 15, 19, 23, 27, 29, 31, 35, 37
Problems for you. Page 598-599 39-69 odds
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Calculator List Math 5 List Ops 5
You do. (by hand and calculator) Problems for you. Page 607 1,5,9,13,17,19,23,27,31,33,35
We need to do a cumulative sum for an infinite geometric Cumsum under LIST OPS Scroll right
An example of inductive reasoning Looking for patterns
Some of these patterns hold for several early terms but break down afterwards. One of the most famous of these is Fermat s conjecture. These are all primes. One of the troubles with evaluating Fermat s conjecture was that the numbers grow very large very quickly
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Problems for you.. Pg 617 1-11 all
Part 2
You do.
Let s do the first one together.
Showing it works for when 1 i= 1 i = 1 n(n + 1) = 2 (1)(2) 2 n = 1: = 1 check. Showing if it works for nth case it works for n + 1case : Formula for n + 1case : (n + 1)((n + 1) + 1) 2 ( n + 1)( n + 2) = 2 n + i= 1 1 1 + 2 + 3+ + n + ( n + 1) = n( n + 1) + ( n + 1) 2 = n( n + 1) 2( n + 1) + 2 2 = n 2 + n + 2n + 2 = 2 n 2 + 3n + 2 2 = n 2 + 3n + 2 2 = ( n + 1)( n + 2) 2 check
If you are using Mathematical induction to show that something holds for n greater than some number, say for 4, then.. 1. Show it holds for the case when n = 4 2. Show if it holds for the case n then that it holds for the case n+1..(exactly like before)
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Problems for you.. From pg 617 17-25 odds
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Note:
You do.. 1 4 4 4 3 4 6 2 4 4 1 4 1 0 4 = = = = = ( ) ) (1)(2 ) (2 (4) ) (2 (6) ) (2 (4) (1) 2 4 3 2 2 1 3 4 4 + + + + = + x x x x x
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Problems for you.. Page 624 1-9 odds, 17-29 odds
Part 2 Finding Binomial Coefficients in the Calculator
An example.
One more example You do
Problems for you. Page 624 11-16 all (use calculator) 49-67 odds
Problem: With a simple counting problem, it usually easiest to list the possible ways the result can be found. (3,3), (2,4), (4,2), (1,5), (5,1): 5 ways distinct means different, no doubles of the same number (1,9), (9,1), (2,8), (8,2), (3,7), (7,3), (4,6), (6,4): 8 ways
You try For a problem like this it would be unwise to count all the possibilities. Use Reason! (30 possibilities for 1 st number) x (30 possibilities for 2 nd number) x (30 possibilities for 3 rd number) 27,000 possibilities!
How many different local telephone numbers are possible (disregard the area code)? Note: The first two numbers of a local number cannot be 0 or 1.
Permutations are the number of ways to order n elements, an important subset of the Fundamental Counting Principle
You do. Remember, with Permutations order matters!
Last problem reworked using the above formula
Distinguishable permutations mean they are unique Question: How many ways can you arrange the letters A,B,C,D? All permutations are different. 4 x 3 x 2 x 1 = 24 ways A as 1 st letter: B as 1 st letter: C as 1 st letter: D as 1 st letter: ABCD BACD CBAD DBCA ABDC BADC CBDA DBAC ACDB BCDA CADB DCAB ACBD BCAD CABD DCBA ADBC BDAC CDBA DABC ADCB BDCA CDAB DACB All these arrangements are unique: no repeats
Distinguishable permutations mean they are unique Question: How many ways can you arrange the letters A,A,B,D? AABD where the first A is used and AABD where the second A is used are NOT Distinguishable! AABD BAAD DAAB ABAD BADA DABA ABDA BDAA DBAA AADB ADAB ADBA 12 distinguishable ways or 24 undistinguishable ways
You do. Page 634-635 1-27 odds
Part II Recall that with Permutations, the order of the chosen items was crucial When you choose a subset of items from a set, where order does not matter, then you want Combinations. 10 ways
AB, AC, AD, AE, AF, AG, BC, BD, BE, BF, BG, CD, CE, CF, CG, DE, DF, DG, EF, EG, FG: 21 ways Note: this is the same formula as the binomial coefficient
How many 5-card poker hands are there? There are 52 cards and you choose 5 (order does not matter) You do. For both problems order does not matter. You should start thinking of that first (Does order matter?) when you look at counting problems.
Choosing seven boys
You do. Page 634-635 33-55 odds
Let s start by learning the terms. An experiment is something that happens whose result is uncertain. The set of all results or outcomes is called the Sample Space of an experiment. Any subset of the Sample Space is called an Event.
You do. The Probability of an event (where all the outcomes are equally likely) equals the ratio of the outcomes of the event to all the possible outcomes. ½ because there are two ways for this to happen (TH or HT) out of a total of Sample Space of 4 items (TT,HH,TH,HT), so 2/4.
¼ because there are 13 hearts in a deck of 52 total cards and 13/52 equals ¼.
Use the Fundamental Counting Principle to Determine the Sample Space Size 6 x 6 = 36 Figure how many ways to make 7
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You do. page 645-646 1-29 odds
More Probability Choosing an item at random, means all items are equally likely to be selected. Since the graduate is chosen at random, each graduate is equally likely to be selected.
Number of Colleges in Various Regions
Mutually Exclusive Events have no outcomes in common. Rolling a 6 on a die and choosing a spade from a deck of cards are mutually exclusive. Rolling a 6 on one die and having the sum of the two dice you rolled equally 4 are not mutually exclusive events.
Mutually exclusive? No, a card can be both a face card and a heart.
You do. Mutually exclusive because an employee cannot belong to more than one row. You cannot have both 0-4 and 5-9 years of service.
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Problems for you.. Page 646-47 #31-39 odds
Probability Part 3 If the computer is really generating random numbers, then each number should be chosen independent of the others.
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The complement of event A, labeled A, is the collection of all outcomes in the sample space that are NOT in A.
Sometimes it is easier to find the probability of the complement than the probability of the event you are interested in. The complement would be that none of the units produced are faulty. Probability that the unit is produced correctly. Probability that all 200 are produced correctly.
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Problems for you Pages 647-648 #41-53 odds