Week of Feb 21

mathwithbaddrawings.com/2017/02/08/how-to-tell-a-mathematician-you-love-them

Homework 4:

3a: stars-and-bars for picking six jellybeans from among five flavors. What is indistinguishable? The flavors are distinguishable.

Here's another way of looking at it: let's
distribute six **identical** pebbles in front of the five
jars of **distinguishable** jelly beans. This is exactly the
same as distributing six identical cookies to five children, and is done
with six stars and 5-1 = 4 bars.

Now just trade in each pebble for one of that flavor of jelly bean, and we've converted this to the original problem!

2bc: starting with a0 vs starting with a1

2.1.6: more summation

p 148: *Investigate!*

Arithmetic sequences: constant difference

Geometric sequences: constant ratio

Example 2.2.1: arithmetic sequences

Example 2.2.2: geometric sequences

Sums of arithmetic and geometric sequences

1 + 2 + 3 + ... + 100

Proof by induction: 1 + 2 + ... + N = N(N+1)/2

Example 2.2.4: reverse-and-add, often easier than reducing to triangle numbers

Example 2.2.5

Example 2.2.6 on Levin p 154

Let b_{k} = 1+3k. Then the sum

a_{n} = 2 + 1 + 4 + 7 + 10 + · · · +
(1 + 3(n−1))

can be rewritten as

a_{n} =2 + b_{0} + b_{1}
+ ... + b_{n-1}.

Written this way it is clearer that there are n terms of the form b_{k}
here, ranging from 0 to (n-1).

Evaluate 1+2+4+...+512

Example 2.2.7

Example 2.2.8

Example 2.2.9

S = 0.464646.... When we subtract from 100S, we get +46 on the right, but there is no minus term. Why?

Let b_{N} = 1^{2} + 2^{2} + 3^{2} + ... +
N^{2}

Is b_{N} an arithmetic sequence? No, but look at the **second
differences**.

Find a closed formula for b_{N}.

What about 1^{3} + 2^{3} + 3^{3} + ... + N^{3}?

If the n-th differences are constant, then a_{k} can be expressed
as a polynomial in k of degree n, and the sums can be expressed as a
polynomial of degree n+1.

*Investigate!*, page 167

Start here Wednesday

def a(n):

if n== 0: return 1

if n==1: return 2

return 5*a(n-1) - 6*a(n-2)

def b(n):

if n== 0: return 1

if n==1: return 3

return 5*b(n-1) - 6*b(n-2)

def c(n):

if n== 0: return 1

if n==1: return 4

return 5*c(n-1) - 6*c(n-2)

Example 2.4.2, p 168; proof by induction

Example 2.4.3: a_{n} = a_{n-1} + n. This is a **constant
2nd-difference** sequence. Use **telescoping**.

Example 2.4.4: same problem, alternative approach

Example 2.4.5: introduce a factor: a_{n} = 3*a_{n-1} + 2.
Note Levin's algebra restart!

def a(n):

if n== 0: return 1

return 3*a(n-1) + 2

**Characteristic roots**: works when a_{n+1} is the
sum of a linear combination of the two previous terms a_{n} and a_{n-1}.

Example 2.4.6

Example at start of chapter: a_{n} = 5*a_{n-1} - 6*a_{n-2}

Fibonacci example, Levin p 173

import math

phi=(1+math.sqrt(5))/2

pho=(1-math.sqrt(5))/2

def s(n):

return (pow(phi,n)-pow(pho,n))/math.sqrt(5)

for i in range(20):

print(i,'\t',s(i))

Postage in *Investigate!*

Example 2.5.1: triangle numbers

Example 2.5.2: 6^{n}-1 is divisible by 5

Example 2.5.3: n^{2} < 2^{n} for n>=5

Warning: Canadians

Euclidean algorithm theorem: for integers a and b<a, we can find q and r so a=qb+r