Very Tough Problem

[lesch2k]

looks like i got off to the right start

what you would do if you want to solve it is compare the difference in increasing terms and the differences of those terms to the nth degree until you find a constant increase.

then you can apply those numbers to a polynomial to solve the average number of tosses, but i dont care to do all of that.

g'night

 

[chickenhead]

<< looks like i got off to the right start

what you would do if you want to solve it is compare the difference in increasing terms and the differences of those terms to the nth degree until you find a constant increase.

then you can apply those numbers to a polynomial to solve the average number of tosses, but i dont care to do all of that.

g'night >>



That is a painful way to do it, kind of like pulling a tooth with a monkey wrench.

There are much more elegant ways to solve it.

I will demonstrate one at the end of this thread.

 

[Mephistopheles]

Tell me if I'm on the right track:

1(1/2) + 2(1/2^2) + 3(1/2^3) + ...

= 1/2 [ 1 + 2(1/2) + 3(1/4) + ... ]

 

[Moonbeam]

"If engineers built bridges the way you do math, millions of people would die every day under collapsed bridges."

Not hardly, all the bridges would have fallen long ago.

 

[silverpig]

1(1/2) + 2(1/4) + 3(1/8) + 4(1/16)... = Sum i=1->inf [i/2^i]

 

[chickenhead]

<< Tell me if I'm on the right track:

1(1/2) + 2(1/2^2) + 3(1/2^3) + ...

= 1/2 [ 1 + 2(1/2) + 3(1/4) + ... ] >>



Looks good so far.

Now you need to figure out whether this series converges or diverges, and if it converges, which number it converges to.

 

[EpsiIon]

<<

<< Hmmm, interesting. I'd like to see what it is. Because I just did a tiny, tiny sample with a quarter, and I seemed to average about two flips... >>



I took a tiny sample with women, and it seemed like they were all bisexual.

(I live in a college town.)

The lesson to learn from this is to never trust tiny samples. >>



The point was not that it was a proof. Nor that it showed what would happen in a larger sample. The point was to show why I think it's two until I see the math. My experience = 2. Forgive me for generalizing; I understand that a tiny sample does not necessarily a correct answer give.

 

[silverpig]

It's easy to see the series converges as 2^i grows much much faster than i. Don't make me pull out good old epsilon-delta now... (I hated those proofs anyways). Now let's see if someone can find what it converges to.


edit: spelling sucks

 

[chickenhead]

Yes, the interesting thing about this problem is that all uneducated attempts at solving it result in 2, but the real answer is something else.

 

[rgwalt]

The series doesn't converge (my math book is at school and I can't remember everything about convergence, but I don't think it converges). Anyway, by writing a program to test coin tosses using random numbers, the average number of tosses is 2.

Good enough for engineering work.

Ryan

 

[rgwalt]

OK, what does the series converge to?

Ryan

 

[chickenhead]

I will let some more people try before I spoil it.

 

[Mephistopheles]

I give up. I'll just guess. 4?

 

[chickenhead]

<< I give up. I'll just guess. 4? >>



No guessing. Let someone solve it.

If nobody solves it, I'll post the solution.

 

[calbear2000]

the solution is 2.

 

[silverpig]

bump

 

[Ionizer86]

Chickenhead, I believe that's what limits in calculus are for...Limits are useful, and yes, it'll come out two if you designed such an equation to the nth term.

 

[calbear2000]

I just reread this whole thread.

To Epsilon and everyone else that said 2, you're correct.

This guy's either:
a) bs'ing us and being pretty rude about it
b) pulling stuff out of his ass
c) needs to go to community college to learn basic calculus

 

[dullard]

We want to know the value of the series: X = .5*1+.25*2 +.125*3 etc.

This can be written in a simpler method: X = sum,i [i/2^i].

Or another way of writting it is: X = 0 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + ...

There is a well known infinite series: 1 = sum, [1/2^i] = 0 + 1/2 + 1/4 + 1/8 + 1/16 + 1/ 32 + ...
This is the same as the classic problem where you step half way to the goal, then step another half way, repeat...

Now what is X - 1?

X - 1 = (1/2 - 1/2) + (2/4 - 1/4) + (3/8 - 1/8) + (4/16 - 1/16) + (5/32 - 1/32) + ...
X - 1 = 0 + 1/4 + 2/8 + 3/16 + 4/32 + 5/64 + 6/128

Now compare the bold text lines (Compare X to X - 1). It is obvious that X - 1 = 1/2 * X.

Rearranging:

1/2 X = 1

or

X = 2

 

[silverpig]

Or you could use maple:

sum('x/(2^x)','x'=1..infinity);

2

 

[chickenhead]

<< We want to know the value of the series: X = .5*1+.25*2 +.125*3 etc.

This can be written in a simpler method: X = sum,i [i/2^i].

Or another way of writting it is: X = 0 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + ...

There is a well know infinite series: 1 = sum, [1/2^i] = 0 + 1/2 + 1/4 + 1/8 + 1/16 + 1/ 32

Now what is X - 1?

X - 1 = (1/2 - 1/2) + (2/4 - 1/4) + (3/8 - 1/8) + (4/16 - 1/16) + (5/32 - 1/32) + ...
X - 1 = 0 + 1/4 + 2/8 + 3/16 + 4/32 + 5/64 + 6/128


Now compare the bold text lines (Compare X to X - 1). It is obvious that X - 1 = 1/2 * X.

Rearranging:

1/2 X = 1

or

X = 2 >>



Ok, this solution is good. I must have made a mistake in my calculations. I got 3.

 

[TuffGirl]

<< Ok, this solution is good. I must have made a mistake in my calculations. I got 3. >>


Oh my, after insulting just about everyone in this thread, he finally finds a solution acceptable enough for him.

Also you never answered if you were ever part of these forums under the name greenfirs. If not, you certainly share the same charm as him. :disgust:

 

[thelanx]

Just integrate the function y=(1/2)^x from 0 to infinity. (I think)

 

[gopunk]

Ok, this solution is good. I must have made a mistake in my calculations. I got 3.

3?! you rejected all those 2's in favor of a 3? maybe there really is an education crisis in america...

 

[dullard]

<< Ok, this solution is good. I must have made a mistake in my calculations. I got 3. >>



Don't worry, Dullard can usually solve any math problem. And if I can't do it, my friend Dimwit can.