BR Maths Corner-1

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Amber G.
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Re: BR Maths Corner-1

Postby Amber G. » 22 Jul 2008 20:10

wasn't there another theory somewhere that the number of prime numbers decrease as you go along to higher numbers (obvious) and that may lead to a final drying up of primes as you crossed a certain number ? this is from childhood memory, so please forgive me if I the details aren't right !

No. As I posted simple proof (Eculid's). They just become less dense -- on average you find about (x/ln x) prime numbers between 1 and x . (ln= log to the base e) The function is called pi(x), Actually Ramanujan's fame was that he claimed (and actually wrote to Hardy) about neat appox of this function)
This may be interesting to you: (how the prime numbers "decrease" etc..)
http://primes.utm.edu/howmany.shtml#pi_def

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Re: BR Maths Corner-1

Postby Amber G. » 22 Jul 2008 20:49

A little OT - Just realized, checking the link of pepin's test, I gave in my post, the wiki article, also mentions AKS test -
this is, of course, famous Agrawal-Kayal-Saxena primality test. (two were undergraduates - and their prof were/are in IIT Kanpur). that caused a major sensation in math world in 2002.. getting coverage in WSJ and NY times..

Beauty was that their work, was so elegant it could be understood by a math college student.

Another btw (and that may be related to our nuclear thread wrt to cooperation between us/india) - was Kayal/Saxena were refused visa when Harvard/Princeton/Duke etc invited them for a lecture tour..(and to receive some honors here in US) only the prof got the visa from US

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Re: BR Maths Corner-1

Postby ramana » 22 Jul 2008 20:53

Looks like the corner is a hit. Already into page two and over 400 views. Another winner.

Now only if Nayakuddin decides to step out of the jirga! 8)

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Re: BR Maths Corner-1

Postby Rahul M » 22 Jul 2008 21:21

Amber G. wrote:Another btw (and that may be related to our nuclear thread wrt to cooperation between us/india) - was Kayal/Saxena were refused visa when Harvard/Princeton/Duke etc invited them for a lecture tour..(and to receive some honors here in US) only the prof got the visa from US

speaking of such issues, I know a very respected prof working in Stat mech from the IACS (Indian Assoc. for the Cultivation of Sciences, kolkata, India's first research center and same place where Raman did his work in) who had an amusing experience while appearing for a visa interview to go to the US.
The time was post 11th May 1998 and the US consulate summarily refused to issue visas to anybody remotely connected with physics.

Interviewer : Occupation ?
Said Prof. : Scientist.
Intrvwr (visibly turning stern faced) : Which Institute ?
Said Prof. : mmm.. Indian Assoc. for Cultivation...
Intrvwr (heaves a sigh of relief) : OK. (thinks the person is from some agriculture related field !!) :lol:

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Re: BR Maths Corner-1

Postby Anujan » 22 Jul 2008 21:29

Okay here is another pooch (famous greek problem)

Let us suppose that there is a car and a bullock cart. The speed of the car is 40Kmph and Bullock cart is 20Kmph.

Warm up question:
Q1. If they race, will the car win the race ? ( Hint: yes :P )

Real question:
Q2. If the bullock cart is given a 10Km headstart (Bullock starts at a point 10Km ahead of the car), will the car overtake the bullock cart at all ? (assuming bullock never gets tired and infinite petrol tank capacity :mrgreen: )

Answer (A) Yes - after 1 hour for example, car would have covered 40Km, Bullock would have covered 10+20 = 30Km, so car has overtaken the bullock

Answer (B) We notice that the car is twice as fast as the bullock cart. So by the time the car covers the 10Km, Bullock would have covered half that distance (5Km) and would be ahead. By the time the car covers the 5Km, bullock would have covered 2.5Km and would be ahead by that distance. By the time the car covers the 2.5Km, bullock would have covered 1.25Km and would ahead by that distance....So no matter how small the distance between the car and the bullock, by the time the car covers that distance, the bullock would have covered half that and would be ahead by a teeny weeny amount. Reasoning that way, the car can never catch up :oops:

Which answer is correct and why ? (Hint: It is related to Amber G-saar's Question 1.)

ramana wrote:Looks like the corner is a hit. Already into page two and over 400 views. Another winner.
Now only if Nayakuddin decides to step out of the jirga! 8)

Atleast for me, my participation here is directly because the BENIS dhaaga was shaheedized :oops: If you restart the BENIS dhaaga, I promise to issue my pooches and fatwas there and not participate here :mrgreen:
Last edited by Anujan on 22 Jul 2008 21:41, edited 1 time in total.

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Re: BR Maths Corner-1

Postby Rahul M » 22 Jul 2008 21:40

lakshmic wrote:Answer (B) We notice that the car is twice as fast as the bullock cart. So by the time the car covers the 10Km, Bullock would have covered half that distance (5Km) and would be ahead. By the time the car covers the 5Km, bullock would have covered 2.5Km and would be ahead by that distance. By the time the car covers the 2.5Km, bullock would have covered 1.25Km and would ahead by that distance....So no matter how small the distance between the car and the bullock, by the time the car covers that distance, the bullock would have covered half that and would be ahead by a teeny weeny amount. Reasoning that way, the car can never catch up :oops:

another hint :
only the last line is wrong !

amber g, thanks for the link !

Rahul.

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Re: BR Maths Corner-1

Postby Amber G. » 22 Jul 2008 21:48

Which answer is correct and why ? (Hint: It is related to Amber G-saar's Question 1.)

Sorry for being too dense, but do not get the hint/understand how it is related to my first problem. I know about zeno's paradox etc.. just that don't see how it relates to Q1.
Perhaps, put the detail hint/answer just keep the fonts tiny so those who do not want to peek can still ignore.

Thanks.

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Re: BR Maths Corner-1

Postby ramana » 22 Jul 2008 22:12

That Greek question is the famous Zeno's problem.

BTW it separates the pre-calculus age afrom the post calculus age. The pre-calculus will argue that the bullock cart will be ahead infitesmally while the post calculus will say as delta t tends to zero the car will overtake the cart and race over.

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Re: BR Maths Corner-1

Postby Anujan » 22 Jul 2008 22:16

Amber G. wrote:Sorry for being too dense, but do not get the hint/understand how it is related to my first problem. I know about zeno's paradox etc.. just that don't see how it relates to Q1.


Strong hints to Amber G-saar's Question 1 and Car-Bullock question ! Do not Read !
Amber G-saar

(1/2)(3/4)(5/6)...(99/100) can be re-written as
(1)(3/2)(5/4)..(99/98)(1/100). And hence
(1)(1+1/2)(1+1/4)(1+1/6)....(1+1/98)(1/100).
This is approx one hundredths of
1+(1/2)+(1/4)+...+(1/98) + (1/2) [(1/4)+(1/6)+...+(1/98)]
and hence one hundredths of
1+(1/2)[1+(1/2)+(1/3)+(1/4)...+(1/49)] + (1/2)(1/2) [(1/2)+(1/3)+...+(1/49)]
the series in the square bracket is a divergent series and the sum is integral 1/x to appropriate limits (ln 49= approx 3.5) I meant my pooch was related to the concept of convergent-divergent series
The bullock covers 10( 1/2 + 1/4 + 1/8 + ...) which tends to 10 when summed to infinty because the series is a convergent series. So total distance covered by the bullock = 10 (headstart) + 10(summation of convergent series)=20 and they cross over when they are 20Km from the start point


:mrgreen:

Perhaps, put the detail hint/answer just keep the fonts tiny so those who do not want to peek can still ignore. Thanks.


Please also include "Answer post" in bold font ! I have set up firefox option to minimum font size of 12 and could read all answers because they didnt show up as tiny :cry: Took me a while to figure out the answer to the pooch of "why are fonts not tiny" :cry:

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Re: BR Maths Corner-1

Postby Amber G. » 22 Jul 2008 23:05

lakshmic wrote:
Amber G. wrote:Sorry for being too dense, but do not get the hint/understand how it is related to my first problem. I know about zeno's paradox etc.. just that don't see how it relates to Q1.


Strong hints to Amber G-saar's Question 1 and Car-Bullock question ! Do not Read !
Amber G-saar

(1/2)(3/4)(5/6)...(99/100) can be re-written as
(1)(3/2)(5/4)..(99/98)(1/100). And hence
(1)(1+1/2)(1+1/4)(1+1/6)....(1+1/98)(1/100).
This is approx one hundredths of
1+(1/2)+(1/4)+...+(1/98) + (1/2) [(1/4)+(1/6)+...+(1/98)]
and hence one hundredths of
1+(1/2)[1+(1/2)+(1/3)+(1/4)...+(1/49)] + (1/2)(1/2) [(1/2)+(1/3)+...+(1/49)]
the series in the square bracket is a divergent series and the sum is integral 1/x to appropriate limits (ln 49= approx 3.5) I meant my pooch was related to the concept of convergent-divergent series
The bullock covers 10( 1/2 + 1/4 + 1/8 + ...) which tends to 10 when summed to infinty because the series is a convergent series. So total distance covered by the bullock = 10 (headstart) + 10(summation of convergent series)=20 and they cross over when they are 20Km from the start point


:mrgreen:

Perhaps, put the detail hint/answer just keep the fonts tiny so those who do not want to peek can still ignore. Thanks.


Please also include "Answer post" in bold font ! I have set up firefox option to minimum font size of 12 and could read all answers because they didnt show up as tiny :cry: Took me a while to figure out the answer to the pooch of "why are fonts not tiny" :cry:

Sorry I may be missing something but to me it makes no sense ..
For example granted, (1+1/2+1/3... 1/n) is approx (ln n) (actually even if n tends to infinity the error between ln (n) and the sum is about gamma (=.53..) so not sure, how you can take approx.. (because gamma itself is 0.53 much larger than 1/10 or 1/15 .. and the error in approx is of the order of (1/n) even if you exclude gamma which itself is about half.

Also (1+1/2) (1+1/4) (1+1/6) ... (1+1/98) expansion will have about 2^50 terms not just two set of series you have shown ...

Perhaps I can not follow the steps you are taking ... can you expand it further so that they are easy to follow (so say even a high schooler can understand?.)
Or take the series to , say n (instead of 100) and give a close form of the result/appox so that one can follow what you are saying.

thanks in advance.

BTW - A Hint: the problem is taken from an old International contest ( bright students of middle school ), (similar problem has also appeared in one national MO for high schoolers - ((of course the series and its its limit for n tends to infinity is fairy well known) The calculus/advance math may make the problem easy but it is not required) :)

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Re: BR Maths Corner-1

Postby Sriram » 22 Jul 2008 23:22

Amber G. wrote:A little OT -

I remember reading the article by Lee Gomes (WSJ), subsequently, there were angry letters for refusing visas to those students.

Added later:
Math discovery rattles Net security
Last edited by Sriram on 22 Jul 2008 23:27, edited 1 time in total.

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Re: BR Maths Corner-1

Postby vina » 22 Jul 2008 23:26

Amber G. wrote:Vina et all -
The problems is fairly well discussed, at:
http://en.wikipedia.org/wiki/Monty_Hall_problem
As I said before, Your initial chances of success was (1/3) opening of the door by the host gives absolutely NO other information and the probability so it still remains same (1/3) that it was in the original door. Hence the other remaining door must have prob = 1-(1/3) = 2/3


Sorry Amber G. Your explanation is wrong. Please read the link you posted carefully ( I had no idea about the background of this Monty Hall thing). To say that the host gives NO other information is wrong . He actually gives tremendous amount of information , over stages. Initially he tells you that there are 2 goats and 1 car and 3 doors. When at a later stage, you touch a door and he opens another door and shows a goat, what he tells you is that the car is either in either the door you are touching or another one, he eliminates one possibility. That is tremendous value and information. That is exactly what lets you make that switch. If he didn't make open the door and show the goat, you wouldn't switch, because you have NO further information.

The 2/3 and 1/3 that you are referring to and the explanation that the link gives are overall probabilities, which of course don't change . But the CONDITIONAL probabilities are different , when better information becomes available and just like I calculated are 1/6 and 1/3 . They too have shown the decision tree with the conditional probabilities and and shown it on the page in the link you posted.

Dileep, you are right, this problem is all about "timing"/"stages" and remembering across "stages" . If you lose that memory, you end up choosing between the 2 doors after the goat is shown, with equal probabilities of 1/2 and hence the wrong answer of " don't switch ".

The entire thing is quite intuitive. The Wikipedia link despite all the pictures does a bad job of explaining it. Let me take a stab at it in simple english.

At the beginning of the game , there are 2 goats and 1 car and 3 doors. So the chance if you touch a door, what lies behind it is a goat is twice the probability that what is behind it is a car.. ie. 2/3 vs 1/3 .

So you touch a door and don't open it. Now the friendly goat dealer /car dealer opens one of the other two doors and shows you a goat (gives you info) !. Now if you have no knowledge of the starting conditions, you would say.. oh well, 2 doors, 1 goat , 1 car. so equal probability , so let me stick with my choice.

But if you as, wait a minute.. Either the unopened door I am touching has a goat or the other unopened door has a goat . Now I now that initially there were twice as many goats as cars . So when I initially touched the door, the possibility that it was a goat is TWICE that of that of a car. So oh my god, the kind dealer has exposed one more goat. Now if like a schmuck, I open the door I am touching, it is twice as probable, that i will get a goat and not the car. Since you want the car, you switch and open the other door , raising your chances of getting a car.

See , you could switch only because you remembered the intial state of the problem. If you didn't know that there were twice as many goats as cars when you touched , you couldnt make the decision on whether to open or switch.

This kind of thing /decision trees is used a heck of a lot in finance/strat planning . That is the reason I still can do this math with any amount of clarity. There is tremendous value in information. For eg, in a venture investing situation,you would structure the deal in stages ( round 1, round 2 round 3 etc) , with specific metrics in each stage, because you can use better inforamation as time goes by and information becomes avialable to increase your expected pay offs.

Same thing with M&A transactions. You structure many deals with earn outs, based on certain conditions because of availabilty of information over time and as a risk shifting measure.

In project planning /investing in new projects etc, you use Real Options valuations and do this kind of thing to see whether investing makes sense and to get the full value of the project. In fact, if you are not careful and dont value the real options you end up leaving tremendous amount of money on the table or end up making skewed decisison.

Ok, if all that sounds esoteric, let me give you an example we all know.

If in 1989 when the LCA project was launched, say, the DRDO gave out a timeline of 10 years and said, the investment is $10b and the expected payoff is $12 b ,you probably would not put the money right away. You would tell them, okay prove that the stuff works by building a TD program in 3 years and take $2b now. Lets make a go no go decision in 3 years ,depending on what progress gets made and how the TD shapes up. Now if you do that staging, the expected value is probably going to be much higher than the $12b you would have got if you plonked the $10b right away , because, you gain options to kill the project after 3 years if it is lemon with minimal losses, and also the option to invest in it fully and make it a success if it comes good after 3 years. Deferring the $7b investment to 3 years later also has tremendous value (time value of money).

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Re: BR Maths Corner-1

Postby Amber G. » 23 Jul 2008 00:04

Vina Possible I may have misspoken (or you have misunderstood me) but that may be besides the point, as the wiki article, hope clears the problem and the solution. May be original original question could have been more explicit/clear about what host knows and the fact (s)he always does his/her business of showing a goat door ( showing of goat door could and would be done - if the problem was stated a little more clearly -- no matter what door you choose).
In any case wiki article may be to your satisfaction,
My point about 'NO INFORMATION' (perhaps I should have put 'NO NEW INFORMATION which is useful' etc) was that opening of a goat door is irrelevant ( just if some one said host's name is Ram Narayan -- Information but irrelevant to change the odds) ... because host is going to do that no matter what (that is open a goat door). Hence the original prob of 1/3 for your original choice does not change.

Hope this is a little more clearer.

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Re: BR Maths Corner-1

Postby Abhijit » 23 Jul 2008 00:10

here is a nice puzzle (i know i am extending the boundary of this thread - which is supposed to be math, but what the heck)
The math whizzes here should find it pretty easy.

a paki jarnail and his begum, a RAPE and his RAPETTE and a common abdul and his aayesha are standing on the bank of a river. The river can be crossed by a boat that can hold up to 2 people but can be rowed by a single person too (doesn't need 2 to go across, but needs at least 1 to make a ferry obviously). all 6 folks want to cross but
- the 4 upper crust people will not travel in the boat with either of the lower crust
- the men are incredibly horny. they will ravish any woman who is not in close proximity of her husband. In this matter there are no 'class' distinctions. any man can molest any woman if he even approaches her while she is not accompanied by her husband. and no, the molesting man has no compunctions doing it in presence of his own wife (in other words, they are perfect pakis ( :D )). And they can't even approach any bank that has a woman without her husband (you can't say that only begum and aayesha are on one bank and abdul approaches that bank alone - because he would simply jump off the boat and proceed to issue canadian vija to the begum, right in front of his own motorham).
how do you take them across without any canadian vijas being issued in the process or any of the RAPE fours not sullying their selves with the aam abdul or ayesha in a boat ?
And the answer is not trivial like - let all 6 of them drown, good riddance - if it were, i would have posted it in the towels thread :D

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Re: BR Maths Corner-1

Postby Anujan » 23 Jul 2008 00:18

vina wrote:Now the friendly goat dealer /car dealer opens one of the other two doors and shows you a goat.


I never thought I would see this many "goat" discussions in a BRF non-BENIS dhaaga :D

Here is another pooch.

There is a deck of cards with jokers removed (total of 52 cards). There are two mathematicians M1, M2 sitting across each other. The host shuffles the cards, draws 5 cards (A,B,C,D,E) and places them face down in two groups (A) - (B,C,D,E).

M1 looks at A, then looks at B,C,D,E.
M1 then shows one of the four cards B,C,D,E to M2.
M1 then takes all the 4 cards and gives it to M2.

M2 looks at the four cards and guesses card A correctly.

How is this possible ?

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Re: BR Maths Corner-1

Postby Amber G. » 23 Jul 2008 00:37

M1 then takes all the 4 cards and gives it to M2.

I think critical information not clearly stated here.. is M1 orders the card in certain way and then passes it to M2.

There was actually a magician who used to perform a similar type of trick.

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Re: BR Maths Corner-1

Postby vina » 23 Jul 2008 00:49

Amber G. wrote:My point about 'NO INFORMATION' (perhaps I should have put 'NO NEW INFORMATION which is useful' etc) was that opening of a goat door is irrelevant ( just if some one said host's name is Ram Narayan -- Information but irrelevant to change the odds) ... because host is going to do that no matter what (that is open a goat door). Hence the original prob of 1/3 for your original choice does not change.

Hope this is a little more clearer.


No.. no Amber G. jee. The goat door opening is very relevant as it reveals information . The goat dealer opens the door based on the door you touch first. So it is a conditional thing. Let me put it this way. If he opened a goat door before you put your hand on a particular door and selected it,you will wait for the door to open and then make the choice. It then it becomes equal probability , so no switching.

Let me post a paragraph from the wiki link you set under the title "Sources of Confusion" that addresses this point.

A competing deeply rooted intuition at work in the Monty Hall problem is the belief that exposing information that is already known does not affect probabilities (Falk 1992:207). This intuition is the basis of solutions to the problem that assert the host's action of opening a door does not change the player's initial 1/3 chance of selecting the car. For the fully explicit problem this intuition leads to the correct numerical answer, 2/3 chance of winning the car by switching, but leads to the same solution for other variants where this answer is not correct (Falk 1992:207).

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Re: BR Maths Corner-1

Postby Amber G. » 23 Jul 2008 01:19

Size reduced to tiny so it does not disturbs others ..:)
[quote]No.. no Amber G. jee. The goat door opening is very relevant as it reveals information . The goat dealer opens the door based on the door you touch first. So it is a conditional thing. Let me put it this way. If he opened a goat door before you put your hand on a particular door and selected it,you will wait for the door to open and then make the choice. It then it becomes equal probability , so no switching.[quote]
May be we are going in circle. Main point here, is opening of the goat door give you No new information about how the probability of your original choice.-which is already has been selected.. (Sure now the prob. of finding a car in the door which is reveled to be a goat is zero - but that is besides the point)

For example: if some body said (after you choose a door) .. " :Hey I know that one of the remaining door has a goat" so what ... and the host knows which one it is .. so what?
Main point here, the the prob. (1/3) or your original choice, remains the same as (1/3) even after host shows one goat door because he can always do it.

You get the *same* answer, if you do the whole math as is shown in wiki article.


Wiki part, you quote, is not really valid here, as long as one is careful in applying the logic to only valid cases. (because, if you really think about it, in other cases, new information can not be ignored - One has to be scientifc there)

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Re: BR Maths Corner-1

Postby Amber G. » 23 Jul 2008 01:32

Tanaji wrote:
3 prove
(cos 20 ) (cos 40) (cos 80) = 1/8


Bwahaaha, I actually managed something...

cos 20 = cos x

(cos x)(cos [60-x])(cos[60+x])
Expanding this out, we get
(1/4)(cosx){cos^2 x - 3sin^2x}
But sin^2 x = 1 - cos^2 x
So above reduces to cos^3 x - (3/4)cosx
which can be further simplified to cos3x / 4
cos3x = cos 60 = 1/2
so above is 1/8


Proved

.

Nice work, Here is another method, you may like:
We use sin (2a) = 2 (sin a (cos a)
let x = cos 20 cos 40 cos 80
Multiply both sides with sin 20 and we get
so (sin 20)(x) = sin 20 cos 20 cos 40 cos 80
= (1/2) sin 40 cos 40 cos 80
=(1/4) sin 80 cos 80
= (1/8) sin 160 = (1/8) sin 20
Hence x=1/8 QED. :)

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Re: BR Maths Corner-1

Postby JwalaMukhi » 23 Jul 2008 01:41

Abhijit wrote:here is a nice puzzle (i know i am extending the boundary of this thread - which is supposed to be math, but what the heck)
The math whizzes here should find it pretty easy.

a paki jarnail and his begum, a RAPE and his RAPETTE and a common abdul and his aayesha are standing on the bank of a river. The river can be crossed by a boat that can hold up to 2 people but can be rowed by a single person too (doesn't need 2 to go across, but needs at least 1 to make a ferry obviously). all 6 folks want to cross but
- the 4 upper crust people will not travel in the boat with either of the lower crust
- the men are incredibly horny. they will ravish any woman who is not in close proximity of her husband. In this matter there are no 'class' distinctions. any man can molest any woman if he even approaches her while she is not accompanied by her husband. and no, the molesting man has no compunctions doing it in presence of his own wife (in other words, they are perfect pakis ( :D )). And they can't even approach any bank that has a woman without her husband (you can't say that only begum and aayesha are on one bank and abdul approaches that bank alone - because he would simply jump off the boat and proceed to issue canadian vija to the begum, right in front of his own motorham).
how do you take them across without any canadian vijas being issued in the process or any of the RAPE fours not sullying their selves with the aam abdul or ayesha in a boat ?
And the answer is not trivial like - let all 6 of them drown, good riddance - if it were, i would have posted it in the towels thread :D


Step 1: Ayesha and Abdul take the boat and go to the other bank. Ayesha dropped there and abdul returns with boat.
Step2: Rapette and Begum take the boat and go to the other bank. Ayesha takes the boat joins abdul.
Step 3: Jernail and RAPE take the Boat and go to the other bank. Jernail and begum return to original bank.
Step 4: Abdul and ayesha take the boat and reach other bank. RAPE and RAPette take it back to original bank.
Step 5: Jernail and RAPE take the boat to other bank. Ayesha takes the boat back to original bank.
Step6: Rapette and Begum take the boat to the other bank. Abdul takes the boat back to original bank.
step7: Abdul and ayesha take the boat to the other bank.
Seven step onlee because fortunately Jernail and RAPE do not issue Candian vijas during their journey midway.

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Re: BR Maths Corner-1

Postby Nandu » 23 Jul 2008 02:10

Amber G. wrote:
1. Prove that (1/2)(3/4)(5/6)………..(99/100)
is less than (1/10) and greater than (1/15)
For those who have calculators change the last term to say (99999999/100000000) and then show the result to say 4 significant figures.


I feel like this might be a "cheat" since a bit of "higher" math is involved, though anyone with high school math will understand it. I will still be curious to see what the more elementary soln. is.

Stirling's Formula says n! = sqrt(2pi*n)*(n/e)^n * exp(e). exp(e) is the error factor, and 1/(12n) > e > 1/(12n+1)

The product of even numbers up to 100 (call it d), is 50! * 2^50 (mulitply each factor in 50! by 2).
This is the denominator in the above problem is d. The numerator is 100! /d. So the expression is 100!/(50!^2 * 2^100).

Substitute in Stirling's formula (sans the error factor) and simplify to get sqrt(2pi*100)/(2pi*50) = 1/sqrt(pi*50), which is clearly between 1/12 and 1/13. I won't go into the details of the error factor, but it is too close to 1.0 to make any difference to this solution.

I am guessing the fact that the expression is 100C50/2^100, with 100C50 the middle binomial coefficient for (1+x)^100, has something to do with a more elementary soln, but can't figure out what yet.

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Re: BR Maths Corner-1

Postby Rahul M » 23 Jul 2008 02:16

nandu, I too couldn't find a school level solution to the problem. My approach has been identical to yours.

Amber G, plz help !

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Re: BR Maths Corner-1

Postby negi » 23 Jul 2008 02:31

lakshmic wrote:
There is a deck of cards with jokers removed (total of 52 cards). There are two mathematicians M1, M2 sitting across each other. The host shuffles the cards, draws 5 cards (A,B,C,D,E) and places them face down in two groups (A) - (B,C,D,E).

M1 looks at A, then looks at B,C,D,E.
M1 then shows one of the four cards B,C,D,E to M2.
M1 then takes all the 4 cards and gives it to M2.

M2 looks at the four cards and guesses card A correctly.

How is this possible ?



What is the use of the bolded part ? I mean M1 anyways gives all four cards to M2.

My solution , nothing mathematical but specualtion onlee.

The pack of cards comes arranged according to suites.

When hosts shuffles them and divides into group A vs B,C,D,E ( I assume he picks these cards in order; for if he picks these five from midst of the pack in random fashion I guess
one cannot guess the card A )

Their arrangement depends on this shuffling pattern.

Now when M2 gets to look at the four cards B,C,D and E
He can actually extrapolate and can guess the card 'A'
from the shuffling pattern.

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Re: BR Maths Corner-1

Postby Nandu » 23 Jul 2008 02:47

lakshmic wrote:
Here is another pooch.

There is a deck of cards with jokers removed (total of 52 cards). There are two mathematicians M1, M2 sitting across each other. The host shuffles the cards, draws 5 cards (A,B,C,D,E) and places them face down in two groups (A) - (B,C,D,E).

M1 looks at A, then looks at B,C,D,E.
M1 then shows one of the four cards B,C,D,E to M2.
M1 then takes all the 4 cards and gives it to M2.

M2 looks at the four cards and guesses card A correctly.

How is this possible ?



A prearranged code between M1 and M2.
Assign an arbitrary order to all the cards.

The suit of A can be signalled by which of BCDE is shown first. Lowest = Clubs, Next = Spades, Highest = Hearts etc..

The number of A can be signalled by how M1 shuffled BCDE before handing them to M2. There are four cards, so 4! = 24 different shuffles, enough to convey information about one of 13 numbers.


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Re: BR Maths Corner-1

Postby Amber G. » 23 Jul 2008 03:00

Nanu - Yep. Actually the way the problem was worded here, I can guess if the card deck had 100 (instead of 52) cards to start with.

The other variation of this trick is interesting. and is MUCH MORE harder.(where first card is not shown) and is less well known.

For who do like to work on this go ahead.
A good analysis for the harder problem is here: Best card Trick
(Don't peek if you don't want too)

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Re: BR Maths Corner-1

Postby Anujan » 23 Jul 2008 03:43

negi wrote:What is the use of the bolded part ? I mean M1 anyways gives all four cards to M2.

My solution , nothing mathematical but specualtion onlee.

The pack of cards comes arranged according to suites.

When hosts shuffles them and divides into group A vs B,C,D,E ( I assume he picks these cards in order; for if he picks these five from midst of the pack in random fashion I guess
one cannot guess the card A )

Their arrangement depends on this shuffling pattern.

Now when M2 gets to look at the four cards B,C,D and E
He can actually extrapolate and can guess the card 'A'
from the shuffling pattern.


Negi saar, Nandu-saar is right. Answer post, do not read !
The deck is shuffled and cards are picked randomly. There is no cheating about that.

Let us start numbering the cards from 1 to 52. Ace Spade gets 1, King spade gets 13, Ace of Hearts gets 14, King of hearts 26, Ace Clubs gets 27, King clubs gets 39, Ace Diamond gets 40, King diamond gets 52.

Given cards BCDE, the cards can be arranged from lowest to highest, because they are uniquely numbered. So without loss of generality, let us assume B=2, C=5, D=7, E=9.

M1 shows me D, I cant figure out the suit yet, but I simply write 7 down on paper.

M1 then arranges the cards in some fashion. There are 4!=24 ways of arranging 2,5,7,9. So he can indicate any number between 1 and 13. Let us assume 2,5,7,9 indicates 1, and 2,5,9,7 indicates 2 and so on and he gave me the cards arranged as 2,5,9,7. So I know that the hidden card is a "2". Since he had also shown me a "D" which is the second highest number, the hidden card is a hearts.

So the hidden card is 2 of hearts.

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Re: BR Maths Corner-1

Postby Nandu » 23 Jul 2008 19:42

Nandu wrote:A seemingly simple probability problem with application to real life.

Doctors have developed an early test for a fatal illness X, which affects one in 1,000 people. The test is fairly accurate, giving only 5% false positives.

You took the test. It came back positive. What is the probability you are afflicted with X?



Nobody wants to take a crack?

Before the answer, an assumption: The test has no false negatives, i.e. if you have the disease, the test will be 400% positive onlee.

Answer:
Let us say one lakh people at random takes the test. 100 of them have the disease and will get positive test result. 99,900 of them do not have the disease, but 5% of those, i.e. 4995 will get a positive test result. Thus if you have a positive test result, your chances of having the disease is 100/(4995+100) = 1.96%.

The general probability formula behind this is called Bayes Theorem.
If you want a long, but very readable discussion of this: http://yudkowsky.net/bayes/bayes.html

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Re: BR Maths Corner-1

Postby SK Mody » 23 Jul 2008 23:49

A cute problem I remember mulling over many years ago:

y(1,x) = x
y(n+1,x) = x^y(n,x) (n = 1, 2, 3, ... )

Q1: If y(n,x) --> 2 as n --> inf, what is the value of x?

Q2 : If y(n,x) --> 3 as n --> inf, what is the value of x? :twisted:

Notation: a^b is a to the power of b.

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Re: BR Maths Corner-1

Postby Vriksh » 24 Jul 2008 00:21

For SK Mody's problem

A cute problem I remember mulling over many years ago:

y(1,x) = x
y(n+1,x) = x^y(n,x) (n = 1, 2, 3, ... )

Q1: If y(n,x) --> 2 as n --> inf, what is the value of x?

Q2 : If y(n,x) --> 3 as n --> inf, what is the value of x? :twisted:

Notation: a^b is a to the power of b.


for y(n,x)-->2; x = 2^(1/2), and for y(n,x)-->k; x= k^(1/k)

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Re: BR Maths Corner-1

Postby Amber G. » 24 Jul 2008 00:48

cshankar wrote:For SK Mody's problem

A cute problem I remember mulling over many years ago:

y(1,x) = x
y(n+1,x) = x^y(n,x) (n = 1, 2, 3, ... )

Q1: If y(n,x) --> 2 as n --> inf, what is the value of x?

Q2 : If y(n,x) --> 3 as n --> inf, what is the value of x? :twisted:

Notation: a^b is a to the power of b.


for y(n,x)-->2; x = 2^(1/2), and for y(n,x)-->k; x= k^(1/k)


Your answer is correct, for 2 but not for 3 ... problem is not trivial for values less grater than e = (2.718.... ) (or less than 1/e) (Big hint why????)

The answer is not very trivial but really very nice. (and sort of hard)
Hint: take the case when y(n,x) --> 4, from your logic, x = 4^(1/4) = 2^(1/2) ...
but then the limit tends to 2 and NOT 4.

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Re: BR Maths Corner-1

Postby SK Mody » 24 Jul 2008 00:57

cshankar wrote:
for y(n,x)-->2; x = 2^(1/2)

400% correct 8)

and for y(n,x)-->k; x= k^(1/k)

Muhahahaha ... :mrgreen:

Amber G - you have it.
It is a tricky problem.

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Re: BR Maths Corner-1

Postby SK Mody » 24 Jul 2008 01:16

A heuristic argument as to why e is the max value for which n--> inf, y(n,x)=e can be solved:

Consider x(a)^a = a (think of x as a function of a)
It is clear that the solution to n--> inf, y(n,x)=a is a non-decreasing function of a.
Assume that x(a) is continuous and differentiable.
take natural log of both sides and differentiate w.r.t.a.
you get dx/da = (1/a - ln(x))(x/a)
This shows that dx/da is 0 at a = e.
The second derivative, d2x/da2, also turns out to be negative.
Therefore x(a) has a maximum at a=e.
But the solution to n--> inf, y(n,x)=a is non-decreasing function of a.
So it cannot exist beyond a=e.

Note that the corresponding value for x is x = e^(1/e).

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Re: BR Maths Corner-1

Postby Nandu » 24 Jul 2008 01:48

Wasn't Amber G's post suggesting that there is a solution for k > e, but that it is "hard".

I don't really see how there can be a solution, at least among reals.

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Re: BR Maths Corner-1

Postby Amber G. » 24 Jul 2008 02:12

Nandu - (No.. that is you are correct, as I read my earlier post I am not clear.... range of the limit is between (1/e and e) and (considering only real numbers etc..) (n^n^n ...) in convergent only for (1/e^e<=n<=e^(1/e)) ...
Just check out the reference given below for example.
http://planetmath.org/encyclopedia/Powe ... uence.html

Tricky, and rather hard part could be , give the problem that n = 3^(1/3) and ask to find the limit of the sequence ... if they guess the answer 3, it is wrong.. (because there is another real number... you can find its value numerically (not 3 but less than e) .. say a where a^(1/a)=3^(1/3), and the limit is actually a of the sequence.
Last edited by Amber G. on 24 Jul 2008 02:29, edited 2 times in total.

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Re: BR Maths Corner-1

Postby Anujan » 24 Jul 2008 02:14

Rahul M wrote:there was a similar problem (well not quite !) that may be interesting in this case.
you toss an unbiased coin 6 times and it comes up head 4times. what is the probability that the 7th toss will be a head ?


Okay here is a related pooch.

Three people, A,B,C, participate in this game. The host has 2 kind of caps, white and black. He chooses uniformly at random and places a cap on each of the peoples heads. The trick is, each person can see the other two people's cap but cannot see her own cap.

They are strictly prohibited from communicating with each other (ofcourse, before the game, they can agree on a scheme). The host gives them each a piece of paper. Each participant has to guess her own cap. They can write "White", "Black" or "Pass" on their paper. They cannot see each other's papers.

The group wins the game if at least one participant guesses correctly and no other participant guesses it wrong. For example if A guesses it correctly, and B,C write "Pass" they win. But if A guesses correctly and B guesses her own cap wrong, they lose.

What are the best odds of winning this game ?

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Re: BR Maths Corner-1

Postby Rahul M » 24 Jul 2008 02:27

at least 2 can surely win. the third one has a 50% chance.
added later :assuming they can communicate somehow. if they can't see the papers, I don't understand how they will be able to do that. :?:

amber g, I'm still waiting for the high school level solution to the 1/10 1/15 problem. :(

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Re: BR Maths Corner-1

Postby Anujan » 24 Jul 2008 02:39

Rahul M wrote:at least 2 can surely win. the third one has a 50% chance.
added later :assuming they can communicate somehow. if they can't see the papers, I don't understand how they will be able to do that. :?:

amber g, I'm still waiting for the high school level solution to the 1/10 1/15 problem. :(


Rahul-M-saar,

Maybe I should have stated more clearly. The *entire* group is deemed a winner or a loser. Meaning if A guesses correctly, B and C pass, the *entire* group wins. Similarly if A and B pass and C guesses correctly, the *entire* group wins. If on the other hand, A gueses correctly, B writes "pass" and C guesses wrong, the *entire* group loses. So there is no case where one person wins and one loses.

Hint: Highlight for the hint, the text below is wearing camo onleee :mrgreen:
The objective is to let the best placed person win. Who is the best placed person ?

However, they cannot see each other's papers or communicate in any way. There is no trick involved here.
Last edited by Anujan on 24 Jul 2008 02:42, edited 1 time in total.

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Re: BR Maths Corner-1

Postby Amber G. » 24 Jul 2008 02:42

Nandu wrote:
Amber G. wrote:
1. Prove that (1/2)(3/4)(5/6)………..(99/100)
is less than (1/10) and greater than (1/15)
For those who have calculators change the last term to say (99999999/100000000) and then show the result to say 4 significant figures.


I feel like this might be a "cheat" since a bit of "higher" math is involved, though anyone with high school math will understand it. I will still be curious to see what the more elementary soln. is.

Stirling's Formula says n! = sqrt(2pi*n)*(n/e)^n * exp(e). exp(e) is the error factor, and 1/(12n) > e > 1/(12n+1)

The product of even numbers up to 100 (call it d), is 50! * 2^50 (mulitply each factor in 50! by 2).
This is the denominator in the above problem is d. The numerator is 100! /d. So the expression is 100!/(50!^2 * 2^100).

Substitute in Stirling's formula (sans the error factor) and simplify to get sqrt(2pi*100)/(2pi*50) = 1/sqrt(pi*50), which is clearly between 1/12 and 1/13. I won't go into the details of the error factor, but it is too close to 1.0 to make any difference to this solution.

I am guessing the fact that the expression is 100C50/2^100, with 100C50 the middle binomial coefficient for (1+x)^100, has something to do with a more elementary soln, but can't figure out what yet.
[/quote
Nandu,
Answer is correct, of course. Actually I was "recycling" this in an contest (few years earlier), I did have the number 1/12 and 1/13 but jury (others who judge the "hardness" of Q) thought it was too hard :) so I changed them to (1/10), (1/15) ...:)

Of course, I could have asked the problem .. of throwing 100 coins and getting (exactly) 50 heads.. (the answer is same) Or Vina's gamma functions.. (eg integral (0 to pi/2) of (sin x)^n) :)

Beauty for me, is out of nowhere, (you have 1,2,3,4 .. simple numbers only) one gets "pi" (as in your formula $1/sqrt(pi*50)$) in the solution..)

- BTW did you check out the link I gave (best card game)..
RahulM -
I will wait for a few days more, just in case, some one puts some elegant solution. before I post my solution.

Also : This leaves with one problem (find (3) proper factors of 2^58+1 from the original 3 problems.

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Re: BR Maths Corner-1

Postby Anujan » 24 Jul 2008 03:08

Amber G. wrote:Also : This leaves with one problem (find (3) proper factors of 2^58+1 from the original 3 problems.


Amber G-saar,

I found one, I suspect the method for others is similar.

Answer post, do not read
Take powers of 2: 2,4,8,16,32,64....
Notice that their remainders when divided by 5 is: 2,4,3,1,2,4,3,1
Notice that the remainder of 2^2 is 4, 2^6 is 4 and so on
Notice that 58-2 is divisible by 4 and hence 2^58 when divided by 5 should have a remainder of 4
So 2^58+1 should be divisible by 5

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Re: BR Maths Corner-1

Postby Amber G. » 24 Jul 2008 03:34

lakshmic wrote:
Amber G. wrote:Also : This leaves with one problem (find (3) proper factors of 2^58+1 from the original 3 problems.


Amber G-saar,

I found one, I suspect the method for others is similar.

Answer post, do not read
Take powers of 2: 2,4,8,16,32,64....
Notice that their remainders when divided by 5 is: 2,4,3,1,2,4,3,1
Notice that the remainder of 2^2 is 4, 2^6 is 4 and so on
Notice that 58-2 is divisible by 4 and hence 2^58 when divided by 5 should have a remainder of 4
So 2^58+1 should be divisible by 5


lakshmic: The 5 is obviously a trivial solution, that is the reason I asked 3 factors. This is becauase (a+1) ) divides (a^(any odd power)+1) so (4+1) divides (4^29+1).. In fact, RahulM has already pointed that out/ hinted that .. in Fermat_primes post ... (16,32,..etc has no odd factors .. 58 does... :)

Any way for your 3 caps solution .. one can easily get 75% of saving by:
ANSWER:
Simple strategy, if you see other two of the same color, guess yours opposite, else pass. One looses only when all three caps are the same color.


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