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Kurt Robert Rudolph edited this page Jun 11, 2012 · 13 revisions

Mon Jun 11 14:00:34 CDT 2012

For Wed

pg.16 # 3, 7, 8, 14 pg.18 # 8, 9, 12, 13

Read 1.1 - 1.5

Binomial Theorem

[{\left( {1 + x} \right)^h} = \sum\limits_{k = 0}^n {n \choose k} x^k ]

[ {n \choose k} = \frac{ n!}{ k!( n - k)!}]

[{\left( {1 + x} \right)^h} = (1 + x)(1 + 3x + 3x^2 + x^3)]

[=] [1] [3x] [3x^2] [x^3]

Prop.

[ {n \choose k} + { n \choose k+1} = {n+1 \choose k+1}]

[ \frac{ n!}{ k!( n-k)!} + \frac{ n!}{ (k+1)!(n-k-1)!} = \frac{ n!}{k!(n-k-1)!} \left( \frac{ 1}{ n-k} + \frac{ 1}{ k+1} \right)]

[ \frac{ (n+1)!}{ (k+1)!(n-k)! } = {n+1 \choose k+1}]

History

Geometry

2600 years ago: many facts

"theorems from axioms"

2300 years ago (Euclid).

Probability

Accurate facts about odds in gambling known around 17th centry

"theorems from axioms" about 1930 (kolmogovov)

ex

Chevalrer de Mere c.1654

his bet: I can get a 6 on a six sided dice in four tosses of a die.

His calculation: [ 4 \times \frac{ 1}{ 6} = \frac{ 2}{ 3}]

[prob( Win) + prob( Loose) = 1 ]

[p + \frac{ 5}{ 6}^4 = 1]

[ p = 1 - \frac{ 5}{ 6}^4 = 1 - \cdots ]

ex

An immortal plays a game in which his chance of winning is [ \frac{ 1}{ 3} ]. If he plays once per day what is the probability he eventually wins?

[ p( win) = \sum\limits_{k = 0}^\infty \frac{ 1}{ 3} {2 \choose 3}^k = 1 ]

something else

[ (1 + x)^{-1} = 1 - x + x^2 - x^3 +x^4 - x^5 \cdots ]

[ (1 - x)^{-1} = \frac{ 1}{ 1 - x} = 1 + x + x^2 + x^3 + x^4 + \cdots ] Memorize

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