## Another Lottery Ticket

As you can see in the graphic below the numbers that were picked by the computer when I purchased the ticket did not match the numbers that were selected that evening.

So a question that occurred to me was: “How close did I get?” And one way to think about distances between two numbers is to subtract their values and take the absolute value of the result. So the distance between the number $2$ and the number $7$ would be $5$. Here is the formula:

$\lvert 2-7 \rvert = \sqrt{(2-7)^2} = \sqrt{(-5)^2} = \sqrt{25} = 5$

This is very easy to extend into higher dimensions. For example, if you go to two dimensions this becomes the pythagorean theorem as applied to two points in the plane. Here is the formula using the example of the points $(a,b) and (c,d)$

$dist( (a,b), (c,d) ) = \sqrt{(a-c)^2 + (b-d)^2}$

In the case of the lottery tickets we can think of a lottery ticket as a point in a six dimensional space, since the ticket has six numbers. This means that the distance between the ticket I purchased and the winning ticket is:

${\sqrt{(10-6)^2 + (13-8)^2 + (14-31)^2 + (22-46)^2 + (52-52)^2 + (11-29)^2 }}$

This expression simplifies to be $\sqrt{1230}$ which is about $35.1$, this means that I was pretty close when you consider the maximum distance between two sets of numbers in the lottery is ($125$). The maximum distance is computed by considering the distance between the two points $(1,2,3,4,5,1)$ and $(55,56,57,58,59,35)$ This type of distance is called the Euclidean Distance between two points..

## E to the ‘i’ Pie

I’ve seen the following before but it is so cool I have to write it down:

The derivation of the formula below is really cool but this post is not about the derivation but an application of the formula:

[1]  $e^{ix} =$ $\cos x + i\sin x$

If we let $x = \pi$ we get

[2] $e^{i\pi} =$ $\cos \pi + i\sin \pi$

Simplify this by recalling that $\cos \pi = -1$ and $\sin \pi = 0$ we get:

[3] $e^{i\pi} = -1$

This is all really straight forward.  Here is the cool part.  What if we let $x = \pi/2$ then we get:

[4]  $e^{i\pi/2} =$ $\cos \pi/2 + i\sin \pi/2$

Since $\cos \pi/2 = 0$ and $\sin \pi/2 = i$ then this simplifies to be

[5]  $e^{i\pi/2} = i$

Finally if we raise both sides of the equation to the $i$ power we get:

[6] ${(e^{i\pi/2}})^i = i^i$

Combining the exponents on the left and noting that $(i)(i) =$ $i^2$ and $i^2 = -1$ we get

[7] $e^{-\pi/2} = i^i$

or

[8] $\dfrac{1}{e^{\pi/2}} = i^i$

or

[9] $\dfrac{1}{\sqrt{e^\pi}} = i^i$

Since the left hand side no longer has $i = \sqrt{-1}$ then if $e^{\pi}$ is a real number then $i^i$ is also a real number!   And, it is a real number see, for example, the articles “http://en.wikipedia.org/wiki/Gelfond%27s_constant” and “http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem

If you go to WolframAlpha.COM and enter the sting “evaluate $i^i$” it actually gives you a  number, marked as transcendental and it’s value is:

[10] $i^i =$ $0.207879576350761908546955 ...$

Finally I wrote this post after reading today “Saturday Morning Breakfast Cereal” cartoon.  WARNING: This day’s cartoon is most definitely N.S.F. (Not suitable for work!)

## A Mathematicians Lottery

I follow the Spiked Math Blog and I especially liked the ‘Mathematicians Lottery‘ post which relies on the divergence of the harmonic series for the payout of the prize to insure that the payout can be managed.

The Wikipedia page on the harmonic series is very well done and the demonstation that this series diverges using an improper integral reminded me of the proof that I first saw when I was taking third semester calculs back in 1972 (forty years ago). Here is how the payout goes and the proof that you can award an arbitrary large sum of money as prizes in the lottery and get away with it even if you suppose that every ticket is a winner. Here is how all that goes:

[1] $\sum_{n=1}^{\infty}\frac{1}{n}$

The idea is that payout works like this: You get $1.00 the first week then (1/2)($1.00) the second week then (1/3)($1.00) the third week and because this series is divergent its partial sum is eventually greater than any finite number (even$750,000,000,000,000 “750 trillion dollars”).

The key is that the harmonic series diverges very, very slowly. To see that the series diverges at all it is enough to notice that it is bounded (from below) by something else that you can show ‘goes-to-infinity’. The choice is the following integral:

[2] ${\int^{\infty}_1\frac{1}{n}\,dn}$

To see why the integral bounds the series from below this picture is a great help:

Since the area under the curve from 1 to infinity is strictly less than the sum of the areas of the rectangles then if we can show that the integral diverges then that will show that the series diverges. This is because each of the rectangles has base equal to 1 and height equal to 1/n and thus area equal to 1/n and the sum of them all is exactly the value of the infinite series.

When you actually perform the integration in [2] you get:

[3] ${\int^x_1\frac{1}{n}\,dn = ln(x)-ln(1)}$

To see why this integrates to ln(x) you think about the inverse function theorem and the fact that the derivative of exponential is itself and the inverse of the exponential is the lograthim.

[4] ${\frac{d}{dx}\left(e^x\right) = e^x}$

and

[5] if ${f(x) = e^x}$ then ${f^{-1}(x) = ln(x)}$

This also explains why $ln(x)$ is an increasing function and why it is a very, very slowly increasing function (since its inverse is a very very quickly increasing function). And all this says that:

[6] $\sum_{n=1}^{\infty}\frac{1}{n} > {\lim_{x\to\infty}\int^x_1\frac{1}{n}\,dn = \lim_{x\to\infty}ln(x) = \infty}$

So this settles it, the harmonic series is divergent and you can always exceed an arbitrary large by just picking a greater and greater partial sum. The question then comes up just how many years do you need to wait to collect 750 trillion dollars. This comes out to be:
[7] $ln(n) = \frac{(7.5)(10^{14})}{52}$ because there are 52 weeks in a year.

Now we just solve for n getting:

[8] $n = e^{\frac{(7.5)(10^{14})}{52}}$

I did visit Wolfram Alpha web site and entered the exponential and, of course, it is way beyond computation, but interestingly Wolfram Alpha displayed the following:

[9] $e^{\frac{(7.5)(10^{14})}{52}} = 10^{10^{10^{1.119224798480347}}}$

And this totally settles the matter as to whether or not it is OK award 750 trillion dollars to each ticket holder.

Original strip http://spikedmath.com/477.html

The “Shoes and Socks Rule” is all about ‘un-doing’ things. If you want to ‘un-do’ putting on your socks and shoes then you have to do things backward, first take your shoes off then take your socks off, which is just the opposite of how you got them on. Well this much is all clear.

The question is what does this have to do with the following formula:

[1]  (XY)-1 = Y-1 X-1

What it is trying to say is when you do that “Minus-One-Exponent-Thingy” you have to reverse the order.  And the “Minus-One-Exponent-Thingy”, well that really means “INVERSE FUNCTION”. In other words the function that ‘un-does’ the original one (it’s inverse).

The normal way to write this is to do it this way:

[2]  (f o g)-1(x) = (g-1 o f-1)(x)

Another way to write (f o g)(x) is to say f(g(x)) so the above becomes:

[3]  f(g(x))-1 = g-1(f-1(x))

And the proof is really cool see:

The basis of the proof is that you only need to show that one function undoes a second function and that second function undoes the first function to prove that the two functions are inverses of each other.  The rest of the proof relies on the fact that function composition is associative (you can re-group it) and that a function composed with its own inverse gives the ‘identity-function’ (i.e. the do-nothing function) and the identity function composed with anything just gives back that anything since it does nothing.

So here is how that would work: the original function is:

[4]  (f o g)(x)

And, the inverse function of that — we want it to turn out to be:

[5]  (g-1 o f-1)(x)

So, to prove this then we think what does a function do?  It takes elements from a set (let’s call it set X) to another set (let’s call it set Y).  The inverse of the function will take elements of the set Y back to the original set X.  Also if you take a function and then apply its inverse the inverse function will UN-DO what the function does. Another way to say this is to say that it leaves things alone.

Or you could say if the combination of a function and another function LEAVES the input set ALONE and the combination of the second function with the first function also leaves the the input set ALONE then the two functions are INVERSES of each other. In symbolic notation this is:

if
[6]  (p o q)(x) = x
and
[7]  (q o p)(x) = x
then
[8]  p(x) and q(x) are inverse functions of each other.

For the proof we are working on we have to show the following two things (here we know f(x)1 and f(x) are already inverses of each other and that g1(x) and g(x) are inverses of each other:

[9]    ((f o g) o (g-1 o f-1))(x) = x
[10]  ((g-1 o f-1) o (f o g))(x) = x

Here is where the fact that function composition being associative comes in.  Notice that if we re-group the parenthesis in [9] the middle two functions turn in to the ‘do-nothing’ function call it “I” for “identity” so we get:

[10]    (f o I o f-1)(x) = x

And if we notice that composing a function with the “I” or “identity” function just gives us the same function then that “I” in the middle just drops out giving us:

[11]    (f o  f-1)(x) = x

And for sure statement [11] is true because that is just the definition of the action of a function and it’s inverse.

Since you can do the same thing to statement [10] this shows that both statement [9] and statement [10] are true so it is the case that statement [2] is true.

Showing that function composition is associative would be the final thing to show.  (Well that and showing that a function composed with the “identity” function is the same as just the function itself.)

## Another Visit To Char-Grill Burgers

Here is my number today:

See this post to understand why I think that figuring out why the prime factorization of my Char-Grill number is cool.

Also, since this was not too hard to figure out. I started thinking what else can I do with this number.

A good idea is to compute the value of Euler’s Totient Function at that number.

Here is the definition of $\varphi(n)$

The number of positive integers less than or equal to n that are co-prime to n.

Two positive integers are co-prime if they share no common factors other than one.

So it turns out that $\varphi(n)$ is multiplicative which means $\varphi(mn) = \varphi(m)\varphi(n)$

And it also turns out that:

$\varphi(p^k) = p^k(1-1/p) = p^{k-1}(p-1)$

If you put this together with $k=1$ and note that

$1-1 = 0$  and $x^{0}=1$

then $\varphi(p) = p-1$

And applying the totient function to the prime factorization of a number yields:

$\displaystyle\varphi(n) = \displaystyle\varphi(\displaystyle\prod_{i=1}^m(p_i^{k_i}))$ where $m$ is the number of prime factors of $n$

And now since $1781 = 43^1 \times 167^1$ we get

$\varphi(1781) = \varphi(43) \times \varphi(167) = (43-1)(167-1) = (42)(168) = 7056$

# Now, this is a watch!

And what follows is my first post to my wordpress.com blog using $LaTeX \displaystyle\ddot\smile$

(12 O’Clock) The Cube Root Of 1728

$\sqrt[3]{1728}$

$\sqrt[3]{3*576}$ (Dividing by three)

$\sqrt[3]{3*4*144}$ (Dividing by four)

$\sqrt[3]{3*4*12*12}$ (I know what twelve times twelve is)

$\sqrt[3]{12*12*12}$ (That’s three of them $12$ and the cube root is easy now)

(1 O’Clock) $B'_L = 1$ or Legendre’s Constant or The Prime Number Theorem

$\displaystyle\lim_{n\to\infty} \left( ln(n) - \frac{n}{\pi(n)} \right) = 1$

where ln(n) is the natural logrithm of the number n, and $\pi(n)$ is equal to the number of prime numbers less than the number n.

$\pi(n)$ ~ $\dfrac{ln(n)}{n}$ And the difference between these two numbers is some really deep math. Not the least of which (to me) is how to get that limit to go to 1 $\displaystyle\ddot\smile$ given the approximation that follows it.

(2 O’Clock) The sum $\displaystyle\sum_{i=0}^{\infty}\frac{1}{2^i}$ comes out to be equal to 2.

$\displaystyle\sum_{i=0}^{\infty}\frac{1}{2^i} = \frac{1}{2^0} + \frac{1}{2^1} + \frac{1}{2^2} + ... = 1 + \frac{1}{2} + \frac{1}{4} + ...$

If you think about adding one to one half to a quarter to an eighth to a sixteenth and so on that does come out to be precisely two (in the limit).

(3 O’Clock) –

The hexadecimal (base 16) value of the symbol for the number three in the Unicode Symbol Set is $33_{16}$ . To get a feel for this the hexadecimal value of the symbol for the numbers zero to nine would be $30_{16}, 31_{16}, 32_{16}, 33_{16}, ... 38_{16}, 39_{16}$ etc… for all of the the ten possible digits $0,1,2,3 ... 8,9$

(4 O’Clock) $2^{-1} (mod 7)$ This one is a bit tricky. because $2^{-1} = \frac{1}{2}$ or one half modulo seven does not seem to make any sense since modular arithmetic is traditionally thought of as involving only the natural numbers. However once the definition of $\frac{1}{x} mod y$ is made you get the following:

$a = \frac{1}{x} (mod y)$

$ax = 1 (mod y)$

or in our case

$2a = 1 (mod 7)$ which comes out to be $a = 4$ since $2*4 = 8 = 1 (mod 7)$

(5 O’Clock) This formula $( 2\varphi - 1)^2$ contains the greek letter phi (in italics) $\varphi$. This is the common symbol for the ‘Golden Ratio‘. The golden ration is that proportion that is formed when you divide a line segment in to two parts and so that the ratio of the total to the longer segment is equal to the ratio of the longer segment to the shorter segment. So, now, suppose you have a line segment divided into two parts the longer is length $a$ and the shorter is length $b$. And here is that equality:

$\frac{a + b}{a} = \frac{a}{b} = \varphi$

$\frac{a}{a} + \frac{b}{a} = \frac{a}{b}$

$1 + \frac{b}{a} = \frac{a}{b}$

$1 + \frac{1}{\frac{a}{b}} = \frac{a}{b}$

$1 + \frac{1}{\varphi} = \varphi$

$\varphi + 1 = \varphi^2$

$\varphi^2 - \varphi - 1 = 0$

$\varphi = \frac{-(-1) \pm \sqrt[2]{ (-1)^2 - 4(1)(-1)}}{2(1)}$

$\varphi = \frac{1 + \sqrt[2]{1 + 4}}{2}$ We must take the positive root since $\varphi$ is positive.

$\varphi = \frac{1 + \sqrt[2]{5}}{2}$

So now what does $( 2\varphi - 1)^2$ come out to be?

$\left( 2 \left(\frac{1 + \sqrt[2]{5}}{2} \right) -1 \right)^2$

$\left(1 + \sqrt[2]{5} - 1\right)^2$

$\left( \sqrt[2]{5} \right)^2$ and that is just $5$

(6 O’Clock) $\displaystyle 3!$ or three factorial which is $\displaystyle 3*2*1 = 6$

(7 O’Clock) $6.\bar{9}$ or $6.999 \cdots$ There are lots of folks who say this is not really seven at all but instead just something very close to seven. Here is a great article on repeating decimals and how and why they represent rational numbers. And here is a nice heuristic process for converting $6.\bar{9}$ to $7$

$x = 6.\bar{9}$

$x = 6.999 \cdots$

$10x = 69.999 \cdots$

$10x-x = 63$

$9x = 63$

$x = 7$

(8 O’Clock) “$\displaystyle\bullet \displaystyle\circ \displaystyle\circ \displaystyle\circ$” This notation can be thought of as being binary where the symbol “$\bullet$” is a one and the symbol “$\circ$” is a zero so we have for the number eight the binary number: “$1000_2$“. This is converted to base ten the way all numbers are converted from one base to another.

And that method is to notice that when writing down digits to represent numbers we use a system that gives the digits different values depending on their position.

For example the number ninety nine in base ten is just two “9” digits side by side, written “99”. The first of these digits, the one on the right, does represent the number $9$ or to be more precise $10^0 * 9$ (or one times nine since ten to the zero power is one) while the second of these digits represents the number $10^1 * 9$ (or ten times nine which is ninety).

So the arithmetic “$10^1 * 9 + 10^0 * 9$” gives ninety plus nine which comes out to be ninety nine, as we expect to get with two nines stuck together.

If there were three nines side by side the left most would be the number $10^2 * 9$ or one hundred times nine, the resulting number would be nine hundred ninety nine.

When working with a different base (other than ten) we do the same thing except to convert from that base to base ten we use that number base instead of ten.

For example what would $1000_2$ (or one thousand base two) be if converted to base ten (hopefully the number eight).

$1000_2 =$

$(2^3)(1) + (2^2)(0) + (2^1)(8) + (2^0)(0) =$

$(8)(1) + (4)(0) + (2)(0) + (1))(0) =$

$8 + 0 + 0 + 0 =$

$8$ as expected.

(9 O’Clock) $21_4$ – This is interesting because on a 24 hour clock 2100 is 9PM. And if you refer back to 8 O’Clock you can see how to convert twenty one base four to a base ten number, and get nine.

$21_4 =$

$(4^1)(2) + (4^0)(1) =$

$(4)(2) + (1)(1) =$

$8 + 1$

Which gives $9$ as expected

(10 O’Clock) $\binom{5}{2}$ the binomial coefficient “Five Choose Two” or the number of two element subsets of a set of five objects.

Another way of saying this is to say how many ways are there of choosing two things from a group of five things without accounting for order and without replacement. The bit about not accounting for order means if you choose object three and four that is the same as choosing object four and three.

So when you think about, how to choose two things from among five you first choose one thing and there are five possibilities. Next you choose a second thing and there are four objects remaining. So for each of the five choices of the first thing there are four choices for the second thing.

This gives twenty ways of choosing. If you do not account for order then there are one half as many since choosing thing three then thing four is the same as choosing thing four then thing three.

So the final result is ten, as expected. There is also a formula for computing this and goes like this:

$\binom{n}{k} = \frac{n!}{(n-k)!n)}$ and in our case we get the following:

$\binom{5}{2} = \frac{5!}{(5-2)!2!} =$

$\frac{120}{(3!)(2!)} =$

$\frac{120}{(6)(2)} =$

$\frac{120}{12} = 10$ as expected.

(11 O’Clock) $0x0B$ – This is a hexadecimal number (base 16). When you count by a different number base you always reach the number ten when you get to the base. For example in base $10$ the numer that follows nine is ten. And this number is always written as the number $1$ stuck to the number $0$, or $10$ (see how to convert from one base to another and positional value above for why a one and a zero are chosen for the numer ten.

So, for example if you count in base three you get the following, $0,1,2,10,11,12,20, ...$

If you count in some number base that is greater than ten then you have extra digits before reaching $10$. It is customary to give these extra digits letter values. So for example the digits for base sixteen are $0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F$ and following $F$ is the number base or $10$.

To distinguish base sixteen from other bases the prefix $'0x'$ is traditionally given indicating base sixtee.

So the counting numbers for this base are:

$0x00, 0x01, 0x02, 0x03$

$0x04, 0x05, 0x06, 0x07$

$0x08, 0x09, 0x0A, 0x0B$

$0x0C, 0x0D, 0x0E, 0x0F$

$0x10$.

Going through the list we see that eleven base sixteen is written $0x0B$ as expected.

## Sadly I have two kids in their 20’s And This Does Apply

From the Monday 30 July 2007 http://indexed.blogspot.com/ – Making Do