Here is an interesting excursion to try; take a quadratic equation that you know has rational roots, and then permute the values of A, B, and C through all possible variations. What is the probability that they would ALL have rational roots?
For example, you might try 2x2+7x + 3, which has roots of -1/2 and -3. If you switch to 2x2+3x + 7 there are no real roots at all. In fact, four of the six permutations have imaginary roots, and only 3x2+7x +2 also has rational roots.
If instead, you had used x2+3x - 10 (roots at 2 and -5) you would find three other permutations had rational roots, 3x2+x -10, -10x2+3x + 1, and -10x2+1x + 3 . The other two permutations have real irrational roots.
What happened with the choices you made for A, B, and C (you DID do it didn't you???).
Several questions pop up...
Is it possible for All six of the permutations to have rational roots?
Both my examples have an even number of the permutations that have rational roots. Is it possible that the number of permutations with rational roots could be odd?
Is it possible that a choice of A, B, and C would have two permutations with rational roots, two with irrational roots, and two with imaginary roots? (sort of the ultimate root trifecta)
The answer to the first question is yes, in fact, I can prove (and you probably can too) that if we select A, B and C so that A+B+C = 0, then all six permutations will have rational roots. The roots will be rational if b^2-4*A*C is a square of a rational, so we can substitute -(A+B) for C and see that (B,2-4A(-A+B)=B2-4AB+4A2=(B-2A)2... An unresolved problem at the moment, which I will have to play with some, is whether this works both ways. Is it possible that all six permutations have rational roots when A+B+C is not Zero??? (please advise if you get this before I get back to it..)
I like the second question because it can be solved with a simple appeal to symmetry. Since B2-4AC is equal to B^2-4CA, exchanging A and C will always produce the same value for the discriminant (B2-4AC). So the number of permutations which have rational roots will always be even.
For the Third, I resorted to a computer search and the first I found was with the values of 1, -4, and -5. x2-4x-5 has rational roots, x2-5x-4 has irrational roots, and -4x2+x-5 has imaginary roots.
Thursday, 7 January 2010
Tuesday, 5 January 2010
Lotteries and Math
Strange how often problems pop up in two or three different ways at once.. recently I was asked by another teacher how often consecutive numbers showed up in the lottery... He had noticed that it seemed to happen pretty often and he thought it might indicate a problem with the way they were being drawn. He was assuming a lottery with 45 numbers, (other places use other numbers)... Fortunately, I had come across the solution only a few days before in two different journal articles dated about 50 years apart.
I have often heard (yes and even said) that the lottery is a tax on the mathematically ignorant. Lately I have been changing my mind a little. I think they are still a VERY bad gamble. The probability of winning, as one pundit put it, is about the same whether you buy a ticket or not. My new appreciation for them is in the way they have contributed to math, and perhaps along the way, to some good social causes also.
My interest was rekindled recently when I came across a letter in the Library of St. Johns College in Cambridge from Brook Taylor, the guy in the calculus theorem, to his friend, the Reverend Newcome in November of 1711 (almost 400 years ago now). What surprised me about the letter was not what he wrote, but what he included... a lottery ticket. I had no idea that the "money prize" kind of lotteries went back so far. Ok, I'm not totally unread. I knew about the soldiers drawing lots from Agamemnon’s helmet During the Persian Wars, but that was not a pay-to-win-a-prize thing... That was a draw-the-lucky-straw and get to fight against the champion kind of lottery. It turns out that except for rare exceptions, most lotteries were of this sort until around the 14-15 th century. The Chinese apparently had one of some form to build their Great Wall, and Augustus Caesar held one to fund public improvements in Rome, but regular lotteries of the modern type began to appear in the middle ages.... the first recorded official lottery was chartered by Queen Elizabeth I, in the year 1566, and was drawn in 1569. Benjamin Franklin organized one to fund the purchase a cannon to defend the state of Philadelphia. THEN?? They got so corrupt and unsavory that they were literally banned by almost every government in the world. Around 1892 the US passed a federal law that lotteries could not use the public mail, and by 1900 they seemed to be totally gone...... until recently. I was simply not aware of their hundreds of years of popularity in the US and England.
Any game that becomes that popular, will find its way into math, and Lotteries did too. In Genoa in the 1500's they used a lottery to replace members of the ninety man Republic Council. Every six months the ninety names would be placed in a bowl and five names drawn would be replaced with new members. Later the process of buying insurance against their loss of position by members, and others who might have depended on their presence on the council, led to the idea of a lottery with a cash pay-off.
The Genoese lottery problem came to the attention of Leonhard Euler, most probably, at the interest of Frederick the Great who implemented a lottery to rebuild the state revenues after the Seven Years War. Euler wrote ," On the Probability of Sequences in the Genoese Lottery" and calculated the probability that if five numbers were selected at random from the interval [1, 90], there would be two or more consecutive numbers. With the kind of magical inductive work he is known for, he came up with $\frac{\dbinom{n-m+1}{m}}{\dbinom{n}{m}}$.. Ok, If Euler felt lotteries were worth his mathematical time... I'll consider them worthy of study... but I won't be buying a ticket this week....
I found another article that spelled out a recursive way to create a table to find the number of ways to NOT get two adjacent.
If we define the number of ways of picking m objects from n objects without getting adjacent values as f(n,m) then we can state that f(n,1) = n and f(1,p)=0 for any p>1..
[It is also easy to show that f(n,2)=$\dbinom{n}{2}$-n+1).. all the ways of picking two except the n-1 ways that have two adjacent numbers]
All the members of f(n,m) are of two kinds, then. One kind is those which include item one, but not item two, and there must be f(n-2,m-1) of these.. (if you have a pattern that doesn't include adjacent values when picking 5 things out of 12 for example, then you can make one that works picking 6 things out of 14 by picking the first and not the second and then the remaining pattern from any successful f(12,5) solution).
There are others which do not include ball 1 but do include ball 2; f(n-1,m) of them... SOO the number of solutions for f(n,m)= f(n-2,m-1) + f(n-1,m) Now we can start to make a table,
N ...1...2...3...4...5...6...7...8...9
m
1........1...2...3...4...5....6...7...8...9
2........0...0...1...3...6...10..15..21..28.
3 ......0.....0...0...0...1...4...10..20..35..
4.......0................0....0....1....5...15..
Note that each new entry is the entry to its left, added to the entry two to the left in the previous row...
For example, the 4 found for f(6,3) is the one to its left f(5,3) plus the 3 in the location for f(4,2)...
Now we can quickly extend the table for any number of values...
So by either method, the number of ways to NOT get two consecutive numbers (or more) in drawing five balls from a sequence of 45 is given by $\dbinom{45-5+1}{5}$ To find the number of draws that do include adjacent values we subtract this from $\dbinom{45}{5}$ which gives 472,361 ways to get at least one adjacent pair out of the 1,221,759 possible sets of five, a probability of about 38%... so you should expect it pretty often.
I have often heard (yes and even said) that the lottery is a tax on the mathematically ignorant. Lately I have been changing my mind a little. I think they are still a VERY bad gamble. The probability of winning, as one pundit put it, is about the same whether you buy a ticket or not. My new appreciation for them is in the way they have contributed to math, and perhaps along the way, to some good social causes also.
My interest was rekindled recently when I came across a letter in the Library of St. Johns College in Cambridge from Brook Taylor, the guy in the calculus theorem, to his friend, the Reverend Newcome in November of 1711 (almost 400 years ago now). What surprised me about the letter was not what he wrote, but what he included... a lottery ticket. I had no idea that the "money prize" kind of lotteries went back so far. Ok, I'm not totally unread. I knew about the soldiers drawing lots from Agamemnon’s helmet During the Persian Wars, but that was not a pay-to-win-a-prize thing... That was a draw-the-lucky-straw and get to fight against the champion kind of lottery. It turns out that except for rare exceptions, most lotteries were of this sort until around the 14-15 th century. The Chinese apparently had one of some form to build their Great Wall, and Augustus Caesar held one to fund public improvements in Rome, but regular lotteries of the modern type began to appear in the middle ages.... the first recorded official lottery was chartered by Queen Elizabeth I, in the year 1566, and was drawn in 1569. Benjamin Franklin organized one to fund the purchase a cannon to defend the state of Philadelphia. THEN?? They got so corrupt and unsavory that they were literally banned by almost every government in the world. Around 1892 the US passed a federal law that lotteries could not use the public mail, and by 1900 they seemed to be totally gone...... until recently. I was simply not aware of their hundreds of years of popularity in the US and England.
Any game that becomes that popular, will find its way into math, and Lotteries did too. In Genoa in the 1500's they used a lottery to replace members of the ninety man Republic Council. Every six months the ninety names would be placed in a bowl and five names drawn would be replaced with new members. Later the process of buying insurance against their loss of position by members, and others who might have depended on their presence on the council, led to the idea of a lottery with a cash pay-off.
The Genoese lottery problem came to the attention of Leonhard Euler, most probably, at the interest of Frederick the Great who implemented a lottery to rebuild the state revenues after the Seven Years War. Euler wrote ," On the Probability of Sequences in the Genoese Lottery" and calculated the probability that if five numbers were selected at random from the interval [1, 90], there would be two or more consecutive numbers. With the kind of magical inductive work he is known for, he came up with $\frac{\dbinom{n-m+1}{m}}{\dbinom{n}{m}}$.. Ok, If Euler felt lotteries were worth his mathematical time... I'll consider them worthy of study... but I won't be buying a ticket this week....
I found another article that spelled out a recursive way to create a table to find the number of ways to NOT get two adjacent.
If we define the number of ways of picking m objects from n objects without getting adjacent values as f(n,m) then we can state that f(n,1) = n and f(1,p)=0 for any p>1..
[It is also easy to show that f(n,2)=$\dbinom{n}{2}$-n+1).. all the ways of picking two except the n-1 ways that have two adjacent numbers]
All the members of f(n,m) are of two kinds, then. One kind is those which include item one, but not item two, and there must be f(n-2,m-1) of these.. (if you have a pattern that doesn't include adjacent values when picking 5 things out of 12 for example, then you can make one that works picking 6 things out of 14 by picking the first and not the second and then the remaining pattern from any successful f(12,5) solution).
There are others which do not include ball 1 but do include ball 2; f(n-1,m) of them... SOO the number of solutions for f(n,m)= f(n-2,m-1) + f(n-1,m) Now we can start to make a table,
N ...1...2...3...4...5...6...7...8...9
m
1........1...2...3...4...5....6...7...8...9
2........0...0...1...3...6...10..15..21..28.
3 ......0.....0...0...0...1...4...10..20..35..
4.......0................0....0....1....5...15..
Note that each new entry is the entry to its left, added to the entry two to the left in the previous row...
For example, the 4 found for f(6,3) is the one to its left f(5,3) plus the 3 in the location for f(4,2)...
Now we can quickly extend the table for any number of values...
So by either method, the number of ways to NOT get two consecutive numbers (or more) in drawing five balls from a sequence of 45 is given by $\dbinom{45-5+1}{5}$ To find the number of draws that do include adjacent values we subtract this from $\dbinom{45}{5}$ which gives 472,361 ways to get at least one adjacent pair out of the 1,221,759 possible sets of five, a probability of about 38%... so you should expect it pretty often.
Sunday, 3 January 2010
Weird Al, doing the Palindrome song..
Thanks to Ron Dirkse, A great Stats teacher who still resides in his adopted homeland of Japan..(huge envy here) This is great.. Thanks Ron...
Which leads me to my next question.... Who can come up with the best math palindrome... If you are in my math class, you can get extra credit for this one.... (don't be greedy, take the challenge, then we'll discuss what it is worth)..They don't have to be numbers, but they can... 12(63)= (36)21...
Which leads me to my next question.... Who can come up with the best math palindrome... If you are in my math class, you can get extra credit for this one.... (don't be greedy, take the challenge, then we'll discuss what it is worth)..They don't have to be numbers, but they can... 12(63)= (36)21...
Saturday, 2 January 2010
A Palindromic Date, and A Problem
01/02/2010.... or 01022010 Ok, technically that is not a Palindrome, or else numbers like 110 would be palindromes. But still,close enough to be special as pointed out by my lovely sweetheart.
How unusual is that??? Well, there are only 19,998 numbers less than 100,000,000 that are palindromes...I just came across this at the Wolfram Mathworld page.. The sum of the reciprocals of the palindromic numbers ( 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212...) converges to a constant...... approx 3.37018..
Strange, but as numbers get bigger, prime palindromes get very rare it seems...2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311, 11411, 12421, 12721, 12821, 13331, 13831, 13931, 14341, and it seems that after eleven, there are no palindromic primes that have an even number of digits (Every palindrome with an even number of digits is divisible by 11).
Next year on Nov 20, we get a real palindromic date, 11/02/2011.... Now that IS unusual.. Ok, and the next one after that is?????
Addendum, Professor Charles Wells of Case Western Reserve suggested a follow up, "Count the number of dates in the American system that are palindromes by something cleverer than brute force. " Well, I didn't actually do that, but I did take a moment to figure out that, if my math doodles are right, there was not such an event from 12/31/1321 (oops, think that should be 09/31/1390... "mia culpa") until 10/02/2001... then after the Jan 2nd this year, and Nov of next year... (and the one you have to figure out that comes after that)... well, there just are not very many of them...
also... I just read somewhere that we who are alive today (those of us over 20) are kind of blessed because we have lived through two palindromic years... 1991 and 2002, and that won't happen to anyone again for about a thousand years unless someone does discover a way not to die in anything like what we consider a normal lifetime now-a-days... lucky me.. lucky you???
How unusual is that??? Well, there are only 19,998 numbers less than 100,000,000 that are palindromes...I just came across this at the Wolfram Mathworld page.. The sum of the reciprocals of the palindromic numbers ( 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212...) converges to a constant...... approx 3.37018..
Strange, but as numbers get bigger, prime palindromes get very rare it seems...2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311, 11411, 12421, 12721, 12821, 13331, 13831, 13931, 14341, and it seems that after eleven, there are no palindromic primes that have an even number of digits (Every palindrome with an even number of digits is divisible by 11).
Next year on Nov 20, we get a real palindromic date, 11/02/2011.... Now that IS unusual.. Ok, and the next one after that is?????
Addendum, Professor Charles Wells of Case Western Reserve suggested a follow up, "Count the number of dates in the American system that are palindromes by something cleverer than brute force. " Well, I didn't actually do that, but I did take a moment to figure out that, if my math doodles are right, there was not such an event from 12/31/1321 (oops, think that should be 09/31/1390... "mia culpa") until 10/02/2001... then after the Jan 2nd this year, and Nov of next year... (and the one you have to figure out that comes after that)... well, there just are not very many of them...
also... I just read somewhere that we who are alive today (those of us over 20) are kind of blessed because we have lived through two palindromic years... 1991 and 2002, and that won't happen to anyone again for about a thousand years unless someone does discover a way not to die in anything like what we consider a normal lifetime now-a-days... lucky me.. lucky you???
Subscribe to:
Posts (Atom)