Math Problems

[Janny]

Eep, no can do. It was part of a larger and more evil problem, that was set with the same 1950's air of can't-you-get-that-right.

Arc length needs to be found for a surface area of revolution. So that requires two processes at which I suck. Firstly the y component of the CoG of the arc has to be found, which I can't do. And then the length of arc has to be found. I was hoping to use the answer and the second soultion to get a figure for y bar and decrypt the book's explanation of how to find it.
SA = 2 pi . arc length . y(bar)

arc length is = integral of (1 + [dy/dx]^2 ) ^1/2
The line: y^2 = 4x
2y dy/dx = 4
dy/dx = 4 / 2y
= 2 / 2 root x
(dy/dx)^2 = 1 / x

And so the integral is required betwix 0 and 1.

If you want to solve the whole thing:
Find the surface area of revolution generated by the parabola y^2 = 4x when the part between (0,0) and (1,2) revolves about the x-axis.

Solution:

8 pi over 3 times (2 root 2 - 1)

[Beren3000]

Sorry, can't help much with that: our syllabus didn't cover SA of revolution.
But I can get 8 pi over 3 as the final answer (no roots, though)
I don't know, but I have a feeling that the method you're using (even though I can't begin to understand it) is a bit too complicated for the problem...

[Janny]

I perpetually have that feeling. How else would you go about it though? That particular formula is the literal solution: the length of the line times the distance moved by the centre of gravity (which represents the 'average point'?)
It also says that you can integrate y w.r.t. the arc length, and multiply that by 2pi. But i have no idea how to do that.

[Mercutio]

I've officially started AP ("Advanced Placement") Calculus. So I may start posting in this thread more...

[Janny]

So... what's the sum of the infinite series:
1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128...

How do you deal with our unspellable Italian friend's series? :S

[Mercutio]

Ha. We haven't gotten there yet...we did do a good bit on series last year in pre-calc though. If we learned that then I sure don't remember!

I have a feeling we didn't learn it last year since it needs limits, right? That's what we've started in calculus now.

EDIT: It took me a bit to realize by "unspellable Italian friend's series" you were referring to the Fibonnacci (?) Sequence!

I'm still not back in the school mode.

[Janny]

You don't know how to spell Fibonacci (?) either! It's not the powers on the bottom that are the problem, it's just the question of how to express the numerator series. The problem is that I think you can only express it by using previous terms. Hmm...

[Janny]

Is the answer:

2

?

[Mercutio]

I have no idea

I'll come back to it once we get to that section.

[Elanor]

Here's the infamous 3x+1 problem:

First pick a positive integer.

if it's even, divide it by 2.
if it's odd, multiply it by 3 and add 1.

Repeat this process until you get 1, then stop.

For example: 10 5 16 8 4 2 1
But will you always get 1, or will some numbers just keep getting bigger and never terminate?
Actually, I've tested this (by writing a simple program) on every positive integer from 1 to about 200000000, and it's terminated every time. But can you prove that it always will?

[Beren3000]

Originally Posted by Elanor

But will you always get 1, or will some numbers just keep getting bigger and never terminate?

This problem is very interesting! I've (literally) filled sheets trying to figure it out; and even though I still can't, I like to think that I'm getting closer...
Do you have the answer to it? I mean, if I reach a proof would you be able to tell me if it's wrong or right?

[Elanor]

My Computer Science teacher said it's been tested up to something like 10^16 (though my computer couldn't go that high without overflowing), and it's terminated every time.

But no one has ever been able to prove that it always will... even really freaky genius mathematicians. In class we figured out that if 3x+1 will always get to a power of 2 at some point, it will terminate by the x/2 rule, but will 3x+1 always reach a power of 2?

Someone in the class proved that 4^n = 3x+1 by induction:

Base: 4^1 = 3(1) + 1

Inductive Hypothesis: 4^k = 3x+1
Prove: 4^(k+1) = 3y+1

4^(k+1) = 4 * 4^k
= 4(3x+1) by Inductive Hypothesis
= 12x+4
= 12x + 3 + 1
= 3(4x+1) +1
= 3y+1

...

This doesn't prove that the 3x+1 problem always terminates, but it might help as a step along the way. But apparently Paul Erdos said, "Mathematics is not yet ready for these kinds of questions," or something like that.

[Curubethion]

a=b+c
b+c=a
5a=5b+5c
4b+4c=4a
Add 4b+4c=4a to 5a=5b+5c:
5a+4b+4c=4a+5b+5c
Subtract 9a:
4b+4c-4a=5b+5c-5a
4(b+c-a)=5(b+c-a)
4=5
Tell me what's wrong. let's see how long it takes you .

[Beren3000]

Originally Posted by Elanor

This doesn't prove that the 3x+1 problem always terminates

I don't see why... it doesn't prove that every number of the form 3x+1 is a power of 4. But it does prove that any power of 4 can be expressed in the form 3x+1. And if you notice in the proof, the next power of 4 (4^k+1) is of the form 3y+1 where y= 4x+1 but this last equation has an infinity of solutions, so that y and x can take any value. The way I see it, this means that, no matter how long it takes, one of the 3x+1 numbers will become a power of 4. Powers of 4 are automatically powers of 2 and will eventually reach 1. Please correct me if I'm wrong... (I know the answer can't be that easy; I've read up on the problem )

Originally Posted by Curubethion

Tell me what's wrong.

Simple:

b+c-a = 0 which means that the last equation is correct without 4 being equal to 5

and I think it has been posted before...

[Elanor]

Originally Posted by Beren3000

I don't see why... it doesn't prove that every number of the form 3x+1 is a power of 4. But it does prove that any power of 4 can be expressed in the form 3x+1. And if you notice in the proof, the next power of 4 (4^k+1) is of the form 3y+1 where y= 4x+1 but this last equation has an infinity of solutions, so that y and x can take any value. The way I see it, this means that, no matter how long it takes, one of the 3x+1 numbers will become a power of 4. Powers of 4 are automatically powers of 2 and will eventually reach 1. Please correct me if I'm wrong... (I know the answer can't be that easy; I've read up on the problem )

I know-- it feels right, and I certainly believe that it will always terminate, but I guess for mathematicians it's not good enough. I guess the point is that you can't prove everything with the tools of mathematics we use today.

[Curubethion]

Originally Posted by Beren 3000

Simple:


and I think it has been posted before...

Darn...you got it

[Janny]

It's one of those tricks maths people have to learn. If someone is proving something that doesn't work, look for the algebra that equals zero and find where they divided using it.

[Curubethion]

And don't forget it's dangerous to divide/multiply by a variable. This creates extraneous roots. For example:
x=1-->x^2=1
Where there was one solution, now there are two (-1, +1)

[Beren3000]

Here's a nice one (don't know if this has been posted before):
A certain room has no windows whatsoever. It only has a light bulb on the inside. When you're outside the room, the door is closed so that you can't see whether the light bulb is on or not. Outside the room, there are 3 switches; you know that 1 of these switches turns the lamp on: the other two are not connected to anything. The trick is, you only get to enter the room once. How can you find out for certain which switch controls the lamp?

[Curubethion]

Oh, that's a classic

Flip on switch 1 and switch 2. After a minute, turn off switch 2. Go to the other side of the room. If the light's on, then switch 1 controls the light. If it's off but warm, then switch 2 is the control. If it's off and cold, it's switch 3.

Here's a twist: there are 3 switches and 3 lights. Can you match them up with the same parameters? (I saw this on The Scholar)