Bit of Fun. HAPPY NEW YEAR $30\,016$ !! Re-calibration of the Year! What year is it really in $2016$?

Bit of Fun. Re-calibration of the Year! What year is it really? What year is it really in $2016$?
Our Calendar starts with the birth of a religious figure. What if Jesus had never existed. What year would it be now?
There are four options as I see it and I prefer the third one. The problem is really that there is no one date that can be used as an absolute date back in time as a start date for humanity. For a bit of fun though we can think of the year $2016$ as being re-calibrated in either of four ways. If you can think of any others let me know!

Any re-calibration of the Year will be an approximation, and we will simply have to make a choice and stick with it. The important thing about the re-calibrated years below is the order of magnitude of the number of years, which reflects more about what humanity has gone through.

THE YEAR $5516$ - Since Historical records
The earliest written historical records seem to go back $3500$ years BC to the first Egyptian Dynasty. Some people think we will eventually find written historical record back to $4000$ BC, for example in the archaeological digs in Egypt, which are ongoing. If you take the first Egyptian Dynasty as the oldest written historical record which dates to $3500$ BC and hence as the start point of our time. Then $2016 + 3500 = 5516$. This one is for those who demand actual symbolic writing as the start of the modern human race. 

THE YEAR $41\,016$ - Since Earliest Rock Painting
This is the one if you accept that the earliest rock paintings as the start point of the re-calibration. This may change with new confirmed discoveries. Rock art and rock paintings have been discovered which date to at least as far back as $41\,000$ years (El Castillo Spain). There is some talk of rock art in Australia going further back to $50\,000+$ but these have not been accurately dated in a scientific way yet, so inconclusive. So use $41\,000$ and then add the $16$  years. That way we still relate to the year $16$ as in $2016$. 

THE YEAR $30\,016$ - Since Agriculture
This one is for you if you accept that the switch to agriculture as the start of the time. Scientists tell us that humans have evolved rapidly in the past $30,000$ years as we have switched from hunting and gathering to agriculture (Scientific American, September 2014 Volume 311, Number 3), I can send you the article) and so it has basically taken us this long to get our act together.
This is my favorite and I decided to go back to the year $2000$ and backdate the $30\,000$ from there and then add the $16$  years since then. That way we still relate to the year $16$ as in $2016$. The order of magnitude of the year is now correct with regard to when it all began!


THE YEAR $7\,000\,016$ - Since Gorillas/Chimps
This is for those who take the start point from where evolution turned to create humans. Scientists say that it was about $7\,000\,000$ years ago, Start of the human lineage, following a split with the lineage containing chimpanzees and gorillas.

HAPPY NEW YEAR $30\,016$!!

Bit of Fun. PRIME DECOMPOSITION OF CALENDAR YEARS 2000-2100

PRIME DECOMPOSITION OF CALENDAR YEARS  2000-2100

Are you like me and always wondering what prime factorisation the next calendar year will bring? LoL. Well look no further, here are the prime decompositions of the whole century for your nerdy convenience!

$2016$ is composed of all single digit primes, while I'm really looking forward to $2017$, prime!, we haven't had a prime year since $2011$. This century there are a total of $14$ prime calendar years and the decade of highest density of primes is $2080-2090$ within which there are four prime years. Fascinating isn't it! $\Large{ {o\;o\atop |} \atop\smile}$

$\begin{array}{ll}
2000 &= 2^4\times 5^3 \\
2001 &= 3\times 23\times 29\\
2002 &= 2\times 7\times 11\times 13\\
2003 &= \mbox{prime} \\
2004 &= 2^2\times 3\times 167\\
2005 &= 5\times 401 \\
2006 &= 2\times 17\times 59\\
2007 &= 3^2\times 223\\
2008 &= 2^3\times 251\\
2009 &= 7^2\times 41\\
2010 &= 2\times 3\times 5\times 67\\
2011 &= \mbox{prime} \\
2012 &= 2^2\times 503 \\
2013 &= 3\times 11\times 61\\
2014 &= 3\times 19\times 53\\
2015 &= 5\times 13\times 31\\
2016 &= 2^5\times 3^2\times 7\\
2017 &= \mbox{prime}\\
2018 &= 2\times 1009\\
2019 &= 3\times 673\\
2020 &= 2^2\times 5\times 101\\
2021 &= 43\times 47\\
2022 &= 2\times 3\times 337\\
2023 &= 7\times 17^2\\
2024 &= 2^3\times 11\times 23\\
2025 &= 3^4\times 5^2\\
2026 &= 2\times 1013\\
2027 &= \mbox{prime}\\
2028 &= 2^2\times 3\times 13^2\\
2029 &= \mbox{prime}\\
2030 &= 2\times 5\times 7\times 29\\
2031 &= 3\times 677\\
2032 &= 2^4\times 127\\
2033 &= 19\times 107\\
2034 &= 2\times 3^2\times 113\\
2035 &= 5\times 11\times 37\\
2036 &= 2^2\times 509\\
2037 &= 3\times 7\times 97\\
2038 &= 2\times 1019\\
2039 &= \mbox{prime}\\
2040 &= 2^3\times 3\times 5\times 17\\
2041 &= 13\times 157\\
2042 &= 2\times 1021\\
2043 &= 3^2\times 227\\
2044 &= 2^2\times 7 \times 73\\
2045 &= 5\times 409\\
2046 &= 2\times 3 \times 11\times 31\\
2047 &= 23\times 89\\
2048 &= 2^{11}\\
2049 &= 3\times 683\\
2050 &= 2\times 5^2\times 41\\
2051 &= 7\times 293\\
2052 &= 2^2\times 3^3\times 19\\
2053 &= \mbox{prime}\\
2054 &= 2\times 13\times 79\\
2055 &= 3\times 5\times 137\\
2056 &= 2^3\times 257\\
2057 &= 11^2\times 17\\
2058 &= 2\times 3\times 7^3\\
2059 &= 29\times 71\\
2060 &= 2^2\times 5\times 103\\
2061 &= 3^2\times 229\\
2062 &= 2\times 1031\\
2063 &= \mbox{prime}\\
2064 &= 2^4\times 3\times 43\\
2065 &= 5\times 7\times 59\\
2066 &= 2\times 1033\\
2067 &= 3\times 13\times 53\\
2068 &= 2^2\times 11\times 47\\
2069 &= \mbox{prime}\\
2070 &= 2\times 3^2\times 5\times 23\\
2071 &= 19\times 109\\
2072 &= 2^3\times 7\times 37\\
2073 &= 3\times 691\\
2074 &= 2\times 17 \times 61\\
2075 &= 5^2\times 83\\
2076 &= 2^2\times 3\times 173\\
2077 &= 31\times 67\\
2078 &= 2\times 1039\\
2079 &= 3^3\times 7\times 11\\
2080 &= 2^5\times 5\times 13\\
2081 &= \mbox{prime}\\
2082 &= 2\times 3\times 347\\
2083 &= \mbox{prime}\\
2084 &= 2^2\times 521\\
2085 &= 3\times 5\times 139\\
2086 &= 2\times 7\times 149\\
2087 &= \mbox{prime}\\
2088 &= 2^3\times 3^2\times 29\\
2089 &= \mbox{prime}\\
2090 &= 2\times 5\times 11\times 19\\
2091 &= 3\times 17\times 41\\
2092 &= 2^2\times 523\\
2093 &= 7\times 13\times 23\\
2094 &= 2\times 3\times 349\\
2095 &= 5\times 419\\
2096 &= 2^4\times 131\\
2097 &= 3^2\times 233\\
2098 &= 2\times 1049\\
2099 &= \mbox{prime}\\
2100 &= 2^2\times 3\times 5^2\times 7
\end{array}$

Bit of Fun. The proof that 1=2 !!

And if $1=2$ then $2=3$, and $3=4$ so

$$0=1=2=3=4=5=6=7=...$$

All numbers are equal!

Prove by integration Volume Sphere is $V=\displaystyle {4\pi R^3\over 3}$

Volume of a Sphere by integration

Prove by integration Volume Sphere of radius $R$ is $$V=\displaystyle {4\pi R^3\over 3}$$

Volume of revolution of the semicircle about the $x$ axis gives the volume

$$\begin{array}{ll}
V&=\displaystyle \pi \int^R_0 y^2\,dx\\
&=\displaystyle 2\pi \int^R_0 y^2\,dx\\
&=\displaystyle 2\pi \int^R_0 R^2-x^2\,dx\\
&=\displaystyle 2\pi \left.\left(R^2x-{x^3\over 3} \right)\right|^R_0\\
&=\displaystyle 2\pi \left(R^2-{R^3\over 3} \right)\\
&=\displaystyle 2\pi\times {2R^3\over 3}\\
&=\displaystyle {4\pi R^3\over 3}
\end{array}$$

Area of a Circle $A=\pi r^2$ using integration

Area of a Circle $A=\pi r^2$ using integration

To find the area of the circle we simply integrate to find the area between $x=0$ and $x=r$ and multiply by $4$.

Area $$A=4\int^r_0 \sqrt{r^2-x^2}\,dx$$

Using the substitution $x=r \sin\theta$ gives $$dx=r  \cos\theta$$ and the integral converts to

$\displaystyle\begin{array}{ll}
A&=\displaystyle 4\int^{\pi/2}_0 \sqrt{r^2-r^2\sin^2\theta}.r\cos\theta \,d\theta\\
&=\displaystyle 4\int^{\pi/2}_0 \sqrt{r^2(1-\sin^2\theta)}.r\cos\theta \,d\theta\\
&=\displaystyle4\int^{\pi/2}_0 \sqrt{r^2\cos^2\theta}.r\cos\theta \,d\theta\\
&=\displaystyle 4\int^{\pi/2}_0 r\cos\theta.r\cos\theta \,d\theta\\
&=\displaystyle 4\int^{\pi/2}_0 r^2\cos^2\theta \,d\theta\\
&=\displaystyle 4r^2\int^{\pi/2}_0 {1+\cos 2\theta\over 2}\, d\theta\\
&=\displaystyle 2r^2\left. \left( \theta+{\sin 2\theta\over 2}\right)\right|^{\pi/2}_0\\
&=\displaystyle 2r^2\left( {\pi\over 2} +{\sin\pi\over 2}-0- {\sin 0\over 2}\right)\\
&=\displaystyle\pi r^2
\end{array}$

Bit of Fun. How Card Counting in Blackjack works and when to bet!

Ever played blackjack in a casino?
I found this on another website but think it is worthy of blogging about.


Card Counting
Most casino players think of card counting as an esoteric skill requiring feats of superhuman memory and computer-like mathematical prowess. The reality is that thousands of players know how to count cards in blackjack, and most of them have average intelligence. Counting cards is one of the only ways to get a consistent mathematical edge over a casino, but it does require significant amounts of dedicated practice and study before you can reliably count on that edge.

Entire books and websites have been written about this particular method of advantage gambling. The purpose of this page isn’t to provide a comprehensive guide to card counting, but it does aim to provide a detailed discussion of the basics.

Why Does Counting Cards Work
In almost all casino games, each wager is made on an independent event, so there’s no way to overcome the game’s insurmountable mathematical edge. Blackjack is an exception, though—the odds change with every card that’s dealt in a blackjack game. For example, if you spin the wheel on a roulette game, and you wager on black, then 18 of the possible outcomes are winners. On your second spin, that same wager on black has the same number of potential winning outcome: 18. In blackjack, though, the cards that have been dealt change the number of potential outcomes. For example, if all four aces have been dealt, the probability of being dealt a natural 21 have been reduced to 0%. (You can’t get a natural blackjack without an ace in the deck.)

As a general rule, a deck with a lot of high cards (tens and aces) left in it is better for the player, while a deck with a lot of low cards (twos, threes, fours, fives, and sixes) is better for the dealer. When the deck is rich in high cards, the player has a better chance of being dealt a natural, which pays out at 3 to 2. Perceptive readers might notice that the dealer also has a better chance of getting a blackjack, too, but the dealer doesn’t get a 3 to 2 payout—only the player does.

Some of the advantage from counting cards is had from making some simple basic strategy changes. For example, in a deck that’s rich in tens and aces, it makes sense to avoid hitting stiff hands. The dealer doesn’t get to make this decision; the dealer has to hit a certain total and stand on a certain total, regardless of whether or not it’s the right play given the composition of the deck.

Another strategy change with a deck that’s composed more than 2/3 of high cards is to take insurance. Normally that’s a sucker bet with a huge negative expectation, but not when you’re counting cards and know the score.

When the deck is rich in lower ranked cards, the dealer is less likely to bust, making it harder for the player to win that way. It’s also less likely that a player will be dealt a natural when there are still a lot of low cards in the deck.

The bulk of the player’s advantage comes from betting more when the deck is rich in high cards and betting less when the deck is rich in low cards. The 3 to 2 payout for a blackjack is where the counter gets most of his edge. Most counters range their bets from 1 to 5 units based on how favorable the count is, but some even range their bets from 1 to 10 units. The casinos watch players who ranger their bets, though, so don’t be surprised if you don’t get some “heat” from the casino. Professional blackjack players use various means of “camouflage” to disguise the fact that they’re counting, and they even sometimes work in teams. You can see examples of camouflage in action in the movie 21, which shows one way in which a team takes advantage of a hot deck.

How Card Counting Works
You don’t have to memorize all the cards that have been played in order to count cards. Card counting uses a heuristic method of tracking the ratio of high cards to low cards. The Hi-Lo system is one of the most common ways of doing this. It assigns the following values to the following cards:
2,3,4,5,6 = +1
10, J, Q, K, A = -1
The other cards (7,8,9) =0

You start your count at 0, and then as each card is dealt, you adjust the account by either adding 1 or subtracting 1. That running total is called the “running count”, but the running count alone isn’t enough to make you profitable. Most casinos use multiple decks, so you have to convert the running count into what’s called the “true count” in order to make the correct decisions.

In order to calculate the true count, you first estimate how many decks are still left in the shoe (the device which holds the cards for the blackjack dealer). Then you divide the running count by the number of decks left. For example, if your running count is +4, and you estimate that 4 decks remain in the shoe, then the true count is +1.

The higher the count, the more you bet. Competent counters can reduce the house edge, in which the player has a disadvantage of 1% or 2% to the casino, to an edge of 0.5% to 1.5% over the casino. If you’re playing for $\$ $20 per hand on average, and you play 100 hands per hour, then you’re putting $\$ $2000 per hour into action. With a 1% advantage, that means you can expect, in the long run, to earn $\$ $20 per hour for your efforts.

Keep in mind that this number isn’t guaranteed, either. In the short run, anything can happen, and even card counters with large bankrolls can go broke quickly if luck turns against them. Besides short term fluctuations in the mathematics of the game, casinos will often refuse to let you play blackjack. In Atlantic City, it’s illegal for a casino to bar a player from blackjack, but they’ve made the rules there so tight that it’s practically impossible to get an edge over the casino.

Other card counting strategies offer varying degrees of complexity in exchange for varying degrees of accuracy. Most players look for a balance of practicality and playability when choosing a system. Other pages on our site look at the specifics of these other counting techniques.

References: 
http://www.gamblingonline.com/casino/blackjack/strategy/card-counting/

http://www.slideshare.net/BryanPenfound/game-theory-8497888 image

Sum of an Arithmetic Series. How easy is it to add up the numbers $1$ to $100$? $1$ to $1000$? $1$ to $1\,000\,000$?

How easy is it to add up the numbers $1$ to $100$? $1$ to $1000$? $1$ to $1\,000\,000$?

Example: Add up the numbers from $1$ to $100$.
There is a nice way to do this by writing the sum in two ways, adding them up and dividing by $2$.

$$\begin{array}{llllll}
1&+2&+3&+4&+\cdots &+100\\
100&+99&+98&+97&+\cdots &+ 1\\\hline
101&+101&+101&+101&+\cdots &+ 101
\end{array}$$

Noting that $101$ occurs $100$ times these add up to $100\times 101$.

Hence the sum

$$1+2+3+4+\cdots +100=\displaystyle {100\times 101 \over 2}$$

Did you notice that the first and last terms of the series add up to $101$. This is how the general case works too.

Sum of an Arithmetic Series

The sum of the arithmetic series with $n$ terms

$$A+ (A+d) + (A+2d)+\cdots + L$$

is

$$S_n={n(A+L)\over 2}$$

where

$A=$ the first term

$L=$ the last term

$n=$ the number of terms in the series

$L=$ the last term

For the example above, 
$A=1$, $L=100$ and $n=100$ (and $d=1$ but we didn't need it here).

The Birthday Problem :-)

The Birthday Problem :-) 

Question: What are the chances or what is the probability that two people in a room of say $30$ people have their birthday on the same day of the year?

Firstly let's assume 365 days in a year (ignoring leap years) and that birthdays are equally likely on all days (which isn't actually entirely true)

Let's start by adding people to the room. 

Suppose there are two people in the room, how many possible combinations are there for their two birthdays. 
There are $365 \times 365$ possible combinations. Specifically, $$1\, 1, 1\, 2, 1 \,3, \ldots1 \,365, 2 \,1, 2 \,2, \ldots, 2\, 365, \ldots,365\,1, 365\,2 \ldots 365\, 365$$
How many of these combinations have different birthdays? There are $365\times 364$ possible combinations where the two birthdays are different.

Therefore the probability that two birthdays are different is $$\displaystyle{365\times 364 \over  365^2}$$
Hence the probability that the two birthdays are the same is thus $$P=\displaystyle 1-{365\times 364 \over 365^2} = 0.002739... \approx 0.3\%$$ 
There is an easier way to look at it for 2 people, but I am setting up the general case! So now let's assume there are $n$ people in the room!, where $n$ is a whole number. 

The number of possible combinations where everyone in the room have different birthdays is
$$365\times 364\times 363\times \ldots $$ 
where there are $n$ terms in the product.
In mathematics this is the number of combinations where order counts of $n$ objects from $365$, which has symbol 
$$\displaystyle ^{365}P_n$$
The probability of no two people in the room having the same birthday is then
$$\displaystyle{^{365}P_n\over 365^n}$$Therefore the probability of at least two people in the room having the same birthday is
$$P(n)=\displaystyle 1 - {^{365}P_n \over 365^n}$$

The graph of the plot of probability $P$ against $n$ follows. 

From the graph I can approximately read that the probability of two people having the same birthday in a room of $15$ people is about $25\%$.

Using the equation directly,
$$P(15)=\displaystyle 1 - {^{365}P_{15} \over 365^{15}}=0.2529=0.25\,\mbox{(2 d.p.)}$$
A popular result occurs when the number of the people in the room is $23$ as then the probability of two people with the same birthday is over $50\%$! So if you're a betting man, let the betting begin! Its a good party trick as this sort of probability is counter intuitive, the probability of two people having the same birthdays is much higher than we think.
From the graph you can see if $n=23$ the probability is about $50\%$ and verifying the formula we have,
$$P=\displaystyle 1 - {^{365}P_{23} \over 365^{23}}=0.50729...=0.51\,\mbox{(2 d.p.)}$$

My favorite is that when $n=60$ or so, the probability of two people having the same birthday is about $99.4\%$ or as a decimal $0.994$ which incidentally rounds to $0.99$ correct to $2$ decimal places,$$P=\displaystyle 1 - {^{365}P_{60} \over 365^{60}}=0.994122...=0.99\,\mbox{(2 d.p.)}$$
What does that mean? Well it means in a room with $60$ guests its pretty much a dead certainty that at least two people will have the same birthday! It's time to rope your friends into a bet you can't really lose, but they don't know it!

And finally I almost forgot to answer my initial question. Just put $n=30$,
$$P=\displaystyle 1 - {^{365}P_{30} \over 365^{30}}=0.7063...=0.71\,\mbox{(2 d.p.)}$$

The chance of two people having the same birthday in a room with $30$ people is about $71\%$.

Have a great day!
Jan Hansen 2015 :-)

Example of Chain Rule for differentiation. Find $y'$ if $y=\left( \left( 3x^2+2\right)^4+1\right)^5$.

Example of Chain Rule for differentiation. 
Find $y'$ if $$y=\left( \left( 3x^2+2\right)^4+1\right)^5$$
The chain rule for differentiation is really quite simple. You just differentiate the 'outside function' first. Then multiply by the derivative of the 'inside function'. Easy peasy :-)

$\displaystyle y'=5\left( \left( 3x^2+2\right)^4+1\right)^4\times 4\left( 3x^2+2\right)^3\times 6x$

$\displaystyle y'=120x\left( 3x^2+2\right)^3\left( \left( 3x^2+2\right)^4+1\right)^4$

Bit of Fun. Japanese Stick Multiplication :-)

Bit of Fun! Japanese Stick Multiplication.
You're trapped on a desert island and you need to do some multiplications for crops and stuff, but don't have a calculator. What to do? Don't worry, there is always Japanese Stick Multiplication!


Example 1: Using Japanese stick multiplication, show 
                              $$23\times 14=322$$
Draw the sticks according to the numbers as shown and separate into different groups as shown with the lines, working your way outside in, so start with the two end groups.

Add up the intersection points in each group. The are 3 separate areas here, separated by the 2 lines.

Write down these numbers along one line as shown below.

Starting with the number on the far right, 12. 

Use the left digit in each number. 

The right digit carries over to the next one on the left.

Example 2: Using Japanese stick multiplication, show 
                              $$412\times 1462=602 \,344$$
Draw the sticks according to the numbers as shown and separate the groups as shown, working your way outside in, so start with the two end groups.

Add up the intersection points in each group. The are 6 separate areas, separated by the lines.

Write down these numbers along one line as shown below..

Starting with the number on the far right, 4 

Use the left digit in each number. 

The right digit carries over to the next one on the left.

Enjoy your crops!

Grove Prelim 3U EX 6.17 - Q1e, 2b, Trigonometric Equations

 QUESTIONS: Q1e,f,2b
Q1e =================

 
Q1f =================

Q2b =================

Further Trigonometry, t-method, BOSTES - 3U, example t-method

Grove Prelim 2U EX 7.6, Q2bc

 QUESTIONS: Q2bc

Grove Prelim 2U EX8.6 - Q8,9

QUESTIONS: 8,9

 

Grove Prelim 2U EX 6.7 - Q11 Trigonometry,

Grove Prelim 2U Test Yourself 6 - Q8,9, 15 - Trigonometry,

QUESTIONS: Q8,9, 15

The Monty Hall Game Show Problem - in brief.

The Monty Hall Game Show Problem.

In a game show there are 3 doors.
Behind each of two doors is a Goat!
Behind one of the doors is a brand new Car! 

The host asks you to select one door, but not open  it yet.
The host then goes over to the other two doors and opens one of them to reveal a Goat.

He then asks you if you want to swap your door with the other unopened door.

Should you Swap or Stick with your original choice?

Many people will intuitively say that it doesn't make any difference and that the chance either way is 50-50!

However, the answer is that you should swap, as it gives you twice the chance of winning the car!

Think about it this way.

On your first choice you have 1/3 chance to win the Car, and you have 2/3 chance of getting a goat.

If you're sitting on the goat and the host choses the other goat then only the car remains so when you swap you win the car!

It's really very simple, but quite counter intuitive. Even the great number theorist Paul Erdős didn't believe this result at first along with many other academics.
The diagram below does the formal math probabilities which seems more confusing than the simple explanation above! (simple is always better)

The tree diagram assumes that the Player has initially chosen Door 1.

More in depth explanations with good diagrams:
http://mathforum.org/mathimages/index.php/The_Monty_Hall_Problem

https://en.wikipedia.org/wiki/Monty_Hall_problem

Grove HSC 3U EX2.9, Grove HSC 2U EX2.9, - Q10 - Maximum and Minimum values

Grove Prelim 3U EX 6.15 Q3h, 5, 9 - Trigonometry,

Q3h===========================

Q5===========================

Q9===========================

Grove Prelim 3U EX 6.16 - Q1cd -, Trigonometry,

Grove Prelim 3U EX3.10, Further Inequalities, Q30

Grove HSC 3U Test Yourself 6 Q1,2,3

Grove HSC 3U EX10.3, - QUESTIONS Q5a, 5g, 5j, 9, 10, 11, 16, - Binomial Theorem

QUESTONS:  Q5agj,  9, 10, 11, 16, 

Q5a ====================== 

Q5g ======================


Q5j ======================


Q9 ======================


Q10 ======================



Q11 ======================

Thx POPPY :-)

Q16 ======================

GROVE PRELIM 3U TY6 - Q2abc, 7b - Trigonometric identities

QUESTIONS: Q2abc, 7b
 

Past Prelim Exam 2012 Math Ext1 Vinnies - Recurring decimals to Fractions

GROVE PRELIM 2U/3U TEST YOURSELF 2 - Q4ab, 26 - algebra, factorising, surds

Q4ab, 26

  

Algebra type questions from previous exams (BOSTES 2U)

Rationalising the denominator questions: