Mathematics [Forum Game]

[Carbon]

okok, one more forum game

it's easy:
poster 1: write a task (like 80x5-90=?)

poster 2: solve it and write a new task.

Rules:
You have to solve it without calculator!
Don't write it down, or use paper/pen. -Use your head
The task has to be solve able without calculator.
If noone can solve it in 48h, it's ok to use calculator, solve it, and write a new one.

ok, i'll start with an easy one:

5(12/2)+54=?

[JaneBennet]

84.

If are positive real numbers, prove that .

[11rdc11]

Not sure if this right but I think it has to with the Denominator. The left hand side gets closer and closer to 2 because the denominator is always larger than the numerator and as it tends to approach infinity it converges to 0. Same arguement for the right side, as i grows to infinity it converges to 0. So 2 > 0 Hope this right

[JaneBennet]

Hope this right

Not quite. I did not say anything about x, y or z tending to infinity. What if x, y and z tend to 0 instead? All cases must be taken into account.

x, y and z are just positive real numbers. Prove the inequality for all positive real numbers x, y and z.

[river_rat]

There probabily is an easier way of doing this but it's late down south and i'm tired.

So we start off by noting that we have that .

Some algebra shows that this is equivalent to which as we can write in the nice compact form

.

So whats the point of all that? Well, let and recalling the AM-GM inequality we see that

.

Flip both sides and multiple by 3 to see that





So my question (if you like my solution) is this:

In Bulgrandia, it is customary and law for couples to have children until they have their first son, after which they stop having children all together. What is the proportion of boys to girls in Bulgrandia right now?

[JaneBennet]

So we start off by noting that we have that .

Some algebra shows that this is equivalent to which as we can write in the nice compact form

.

So whats the point of all that? Well, let and recalling the AM-GM inequality we see that

.

Flip both sides and multiple by 3 to see that

My solution was this (of course I have a solution because I made the problem up starting from the “solution” ):

By AM–GM,

Similarly

And

Adding up.

i.e.

In Bulgrandia, it is customary and law for couples to have children until they have their first son, after which they stop having children all together. What is the proportion of boys to girls in Bulgrandia right now?

[DivideByZero]

So my question (if you like my solution) is this:

In Bulgrandia, it is customary and law for couples to have children until they have their first son, after which they stop having children all together. What is the proportion of boys to girls in Bulgrandia right now?

If a chance of boy or girl was 50% then I think that the ratio of boys to girls in Bulgrandia will be 1/2 or 50%.


out of 1000 couples:

round 1) 500 boys are made, 500 girls are made.

round 2) after the families get their first girl they want to get another baby. A total of 250 more girls are made and 250 more boys are made in Bulgrandia. Thus grad totaling 750 boys and 750 girls.

round 3) after the 250 families get a girl from round 1, a total of 125 boys are made and 125 girls are made. Thus grand totaling 875 girls and 875 boys.

round 4, 5, 6, etc...) You see where this is going.

So the answer is 50-50 ratio of boys to girls.

I hope I'm right >.<
EDIT: I'll submit my question once I get know whether my solution was satisfactory.

[William McCormick]

So my question (if you like my solution) is this:

In Bulgrandia, it is customary and law for couples to have children until they have their first son, after which they stop having children all together. What is the proportion of boys to girls in Bulgrandia right now?

If a chance of boy or girl was 50% then I think that the ratio of boys to girls in Bulgrandia will be 1/2 or 50%.


out of 1000 couples:

round 1) 500 boys are made, 500 girls are made.

round 2) after the families get their first girl they want to get another baby. A total of 250 more girls are made and 250 more boys are made in Bulgrandia. Thus grad totaling 750 boys and 750 girls.

round 3) after the 250 families get a girl from round 1, a total of 125 boys are made and 125 girls are made. Thus grand totaling 875 girls and 875 boys.

round 4, 5, 6, etc...) You see where this is going.

So the answer is 50-50 ratio of boys to girls.

I hope I'm right >.<
EDIT: I'll submit my question once I get know whether my solution was satisfactory.

My neighbor was a Postmaster in the United States Postal system. And he was funny. He said that if the odds were 50/50 it meant nothing really. It is still a total gamble. Because in actuality you can get 100 straight similar occurrences of a 50/50 gamble. By actual experience I believe.

I know from day to day life, that usually things run around 3-7 in a row of the same occurrence, and then even out over time in most cases when the odds are 50/50. You could get some pretty unbalanced first round runs though with only 1000 babies.

My thought here is that there is an initial gamble. Because I do not see 500/500 boys to girls being born. To me the odds are against that.

But I believe you are right if the odds are 50/50 that says it all.

Sincerely,


William McCormick

[river_rat]

So my question (if you like my solution) is this:

In Bulgrandia, it is customary and law for couples to have children until they have their first son, after which they stop having children all together. What is the proportion of boys to girls in Bulgrandia right now?

If a chance of boy or girl was 50% then I think that the ratio of boys to girls in Bulgrandia will be 1/2 or 50%.


out of 1000 couples:

round 1) 500 boys are made, 500 girls are made.

round 2) after the families get their first girl they want to get another baby. A total of 250 more girls are made and 250 more boys are made in Bulgrandia. Thus grad totaling 750 boys and 750 girls.

round 3) after the 250 families get a girl from round 1, a total of 125 boys are made and 125 girls are made. Thus grand totaling 875 girls and 875 boys.

round 4, 5, 6, etc...) You see where this is going.

So the answer is 50-50 ratio of boys to girls.

I hope I'm right >.<
EDIT: I'll submit my question once I get know whether my solution was satisfactory.

Looks good enough to give you a go at asking the next question

[DivideByZero]

ok then. Here is my question:

What is the cube root of the number 1729 to the nearest hundredth?

(Remember that you can only use your head, i.e. no calculator. Its pretty easy if you know what you're doing.)

[JaneBennet]

[DivideByZero]

Good! :-D
Did you do 1/432 = 0.002 in your head?
Thats pretty intense.

Anyways,
its your turn.

[JaneBennet]

Did you do 1/432 = 0.002 in your head?
Thats pretty intense.

Not really. ; in other words .

Give me some time to think of the next problem.

[JaneBennet]

Okay, try this.

Let be a real number. Prove that for all positive integers , .

[thyristor]

[river_rat]

Okay, try this.

Let be a real number. Prove that for all positive integers , .

How's this, firstly if then .

So suppose true for then then if we have that



But

since

Thus



And we are done by induction on

[thyristor]

And what is your problem?

[JaneBennet]

Great work once again, river_rat! :-D

My own solution was

By AM–GM (again!) we have



but your proof by induction is much better because it also proves a few extra results, e.g. . :P

Next problem, please!

[thyristor]

What's AM-GM? Couldn't you post one while we wait?

[DrRocket]

Okay, try this.

Let be a real number. Prove that for all positive integers , .

Fix n.
Let .

Then f(0)= n. f'(x) =nx^(n-1)-n. So f'(x) is negative between 0 and 1, zero at 1 and positive for x>1. f(1) = 1, which is a minimum for f.

BTW how do you make (n-1) show up as the exponent in TEX ? I had it in parentheses and it came out garbled.

[JaneBennet]

DrRocket: Nice! I completely forgot that you could also use calculus for this kind of problem.

To do in TeX, you must use curly brackets, not round brackets: x^{n-1}


thyristor: http://en.wikipedia.org/wiki/Inequal...eometric_means


river_rat: Don’t keep us waiting too long for the next problem, will you?

[thyristor]

I'll post one while we wait.

What's wrong

[MagiMaster]

In the last couple of steps, you divided by 0, or (a-b). (Though, I think you also have a typo at the end there too.) I'll wait for river_rat's problem though.

[thyristor]

You're right. If we divide by a-b then a-b can not be zero.
Which is false.
What would my typo be, though?

[river_rat]

That's where the dragon is lying:

Here is my problem.

Choose a, b and c uniformly and independently from [-1, 1]. What is the probability that has real roots

[DrRocket]

That's where the dragon is lying:

Here is my problem.

Choose a, b and c uniformly and independently from [-1, 1]. What is the probability that has real roots

13/24 details to follow

[DrRocket]

Choose a, b and c uniformly and independently from [-1, 1]. What is the probability that has real roots

Forgive the lack of TEX but I haven't quite figured out how to get something quite this involved into TEX in my lifetime.

First, note that by a scaling argument you can assume that a=1 unless a=0 and the case a=0 is of probability zero so we can ignore it. So assume henceforth that a=1. Then we are looking for the probability that b^2-4c is at least 0 where b and c are uniformly distributed on [-1,1].

You can look at this as requiring the probability density or probability measure for the random variables b^2 on [0,1] and 4c on [-1,1] or for what follows on the entire real line, with support on those intervals.

A relatively straightforward analysis shows the probability density for 4c is simply 1/8 supported on [-4,4] and for b^2 it is 1/(2 sqrt(x)) supported on [0,1].

The density for b^2 - 4c is then the convolution product of the two density functions, which involves evaluating some integrals. If you evaluate those integrals you get the density function F for b^2-4c as

F(x) = 0 for x less than 4
1/[8sqrt(x+4)] for x between -4 and -3
1/8 for x between -3 and 4
1/8[1-sqrt(x-4)] for x between 4 and 5
0 for x greater than 5

Then you integrate F between 0 and infinity, to get the probability that b^2 -4c is non-negative and the answer is 13/24.

Assuming that I have not screwed up the arithmetic somewhwere this seems to check out. Everything that is a probability density integrates to 1 and is supported where it ought to be so I think I got the arithmetic right.

[river_rat]

First, note that by a scaling argument you can assume that a=1 unless a=0 and the case a=0 is of probability zero so we can ignore it.

if a, b and c are iid uniform variables, are and iid?

Ok, slight problem. I'm off to conference this week so i'm not sure how easy it will be for me to check up on here (we are in the middle of the bush) so if you don't here from me (and seeing that doc rocket has almost got it right) he can treat his solution as "correct" for the purposes of continuing the game in my absence.

[DrRocket]

First, note that by a scaling argument you can assume that a=1 unless a=0 and the case a=0 is of probability zero so we can ignore it.

if a, b and c are iid uniform variables, are and iid?

Ok, slight problem. I'm off to conference this week so i'm not sure how easy it will be for me to check up on here (we are in the middle of the bush) so if you don't here from me (and seeing that doc rocket has almost got it right) he can treat his solution as "correct" for the purposes of continuing the game in my absence.

You are correct. That scaling argument does not hold water. But the correct solution is truly hideous. Butt ugly !

The probability density function for 4ac turns out to be supported on [-4,4] and on that interval is 1/2[ln(abs x/4)] which you can then convolve with the density for b^2. This is a godawful integral that can be done with perserverence and a good integral table, but is so ugly that I don't want to spend the time. It is is neither pretty nor instructive. I don't know if the resulting convolution can be integrated in closed form from 0 to 5 (the nasty convolution integral is supported on [-4,5]) to get the required probability of a non-negative discriminant. Yech!!

So, if somebody want to slog through this mess and come up with a final numerical answer they are welcome to do so. If not and you want me to come up with another problem I can do that -- I have something involving convex functions in mind that would result in a proof of the AM-GM inequality that has gotten some play lately. Or if somebody else has a problem that they think ought to be posed then that is OK too.

[JaneBennet]

Poke.

[Guitarist]

Try this, in a light-hearted manner.

Suppose .

Suppose that, for all , induces the partition on such that and .

Suggest a possible description of for

a) n = 0;
b) n = 1;
c) n = 2;
d) n = prime
e) n = card(N)
f) any other n that take your fancy

[river_rat]

a) The empty set (ill defined?)
b)
c)
d) where
e) The empty set (ill defined again?)