Math thread! Come in and have some pi...

[Muckle_Ewe]

http://forum.bodybuilding.com/showth...#post551947533

Are these guys serious? Is (x^2-1)/(x+1) really discontinuous? Wtf man?

Ok...

I'm having a mind melt down trying to sort this all out.

I'll just rip this off wiki. My knowledge of continuous functions isn't solid so hopefully MiscMath can sort this out.

If two functions f and g are continuous, then f + g, fg, and f/g are continuous. (Note. The only possible points x of discontinuity of f/g are the solutions of the equation g(x) = 0; but then any such x does not belong to the domain of the function f/g. Hence f/g is continuous on its entire domain, or - in other words - is continuous.)


But what I'm thinking is...

Say you have some function. f(x) = x. Pretty obvious this is continuous.

But do we change this into a discontinuous function by multiplying by .

It doesn't seem to make much sense to me... Are x and different functions?

[Greg1983]

It doesn't seem to make much sense to me... Are x and different functions?

Yes. Both are straight lines with a slope of 1, but x^2 / x has a removable discontinuity at the origin.

[charity4thepoor]

I know that x/0 = undefined, but I always thought that 0/0 was just 0.

I learned in statistics today that 0! is 1. That's pretty weird too. lol.

yea man if 0! didn't equal 1 then no probabilities would make sense. i think it's just the way 0! is defined

[DangerDan]

yea man if 0! didn't equal 1 then no probabilities would make sense. i think it's just the way 0! is defined

Yeah, that's what I was reading on the internetz. Pretty interesting.

[DangerDan]

So then is every rational function with the denominator able to equal to zero discontinuous at the point where the denominator is 0?

[charity4thepoor]

So then is every rational function with the denominator able to equal to zero discontinuous at the point where the denominator is 0?

yes sir

[Muckle_Ewe]

Yeah, that's what I was reading on the internetz. Pretty interesting.

A quick not very rigorous proof.

n! = n(n-1)!

=> (n-1)! = n!/n

Set n=1

0! = 1!/1 = 1

Hence 0! = 1

[DangerDan]

yes sir

That's so weird. I've never learned that. Ridiculous.

[DangerDan]

A quick not very rigorous proof.

n! = n(n-1)!

=> (n-1)! = n!/n

Set n=1

0! = 1!/1 = 1

Hence 0! = 1

Very nice. That's pretty cool.

[Errorist]

Got a 65% on mine and the class average was a 67%. No one got an A, 3 got a B, 3 got a C, 3 got a D, and 6 got Fs.

I think we had 2 perfect As, the rest were Bs-Fs, lots of Ds and Fs. I think like 1/3rd the class was missing today. Guessing they dropped.

Wtf man?

How come I didn't learn this in my calc class?


What if 0/0 = one? **** would be crazy.

http://www.google.com/images?hl=en&q...w=1280&bih=619

[DangerDan]

I think we had 2 perfect As, the rest were Bs-Fs, lots of Ds and Fs. I think like 1/3rd the class was missing today. Guessing they dropped.


http://www.google.com/images?hl=en&q...w=1280&bih=619

Hell yeah, a lot of my class probably won't show up on Tuesday. Drizzopped.


And yeah, I'm aware of the whole dividing by zero thing. It just seems weird that 0/0 would even be dividing by zero because 0 isn't anything. So you're not really dividing anything by zero.

[Errorist]

And yeah, I'm aware of the whole dividing by zero thing. It just seems weird that 0/0 would even be dividing by zero because 0 isn't anything. So you're not really dividing anything by zero.

Yea I don't really get it either.

1/0 = Should still be 1 because you're dividing by nothing, so you still have the whole thing that you're dividing.
0/2 = Shouldn't be possible because how can you divide nothing by something? Doesn't make sense.

Dunno, weird logic.

[MiscMathematician]

Yea I don't really get it either.

1/0 = Should still be 1 because you're dividing by nothing, so you still have the whole thing that you're dividing.
0/2 = Shouldn't be possible because how can you divide nothing by something? Doesn't make sense.

Dunno, weird logic.

Something wrong with the way youre thinking about this. If you split a dollar between no people, how much money to they get? Doesn't make sense, there is no "they".

Mathematically, if 1/0 is not 0, then all numbers are equal. Proof: Exercise . You can define a system of numbers with this property, but it isn't very interesting.

0/2? If john and Alice make x dollars and split it, they each get x/2 dollars. They end up getting 0/2 dollars a piece. But since they made a total of 0 dollars, 0/2=0.

[Errorist]

Something wrong with the way youre thinking about this. If you split a dollar between no people, how much money to they get? Doesn't make sense, there is no "they".

Mathematically, if 1/0 is not 0, then all numbers are equal. Proof: Exercise . You can define a system of numbers with this property, but it isn't very interesting.

0/2? If john and Alice make x dollars and split it, they each get x/2 dollars. They end up getting 0/2 dollars a piece. But since they made a total of 0 dollars, 0/2=0.

What do you get when you divide something by nothing?
1/0 = 1
2/0 = 2
3/0 = 3
You get exactly what you started out with, because you haven't divided it by anything. There's nothing to divide the original number by, so it's still a whole number.

What do you get when you divide 0 by something? You can't because there wasn't anything to divide in the first place. It's impossible.

These 2 rules really need to be switched.

[Greg1983]

What do you get when you divide something by nothing?
1/0 = 1
2/0 = 2
3/0 = 3
You get exactly what you started out with, because you haven't divided it by anything. There's nothing to divide the original number by, so it's still a whole number.

What do you get when you divide 0 by something? You can't because there wasn't anything to divide in the first place. It's impossible.

These 2 rules really need to be switched.

You can't be serious. Division is the inverse of multiplication. 0/2 should return what you have to multiply 2 by to return 0. That would be 0. 2/0 should return what you have to multiply 0 by to return 2. There's nothing big enough, so it's infinite.

[Errorist]

You can't be serious. Division is the inverse of multiplication. 0/2 should return what you have to multiply 2 by to return 0. That would be 0. 2/0 should return what you have to multiply 0 by to return 2. There's nothing big enough, so it's infinite.

If I cut 1 pizza into 0 slices, and eat them, how much pizza do I have left? I end up with 1 whole pizza and I'm still hungry.

Also, I can't cut a pizza into slices that I don't have, it's impossible to do.

[Greg1983]

If I cut 1 pizza into 0 slices, and eat them, how much pizza do I have left? I end up with 1 whole pizza and I'm still hungry.

Also, I can't cut a pizza into slices that I don't have, it's impossible to do.

When you're dividing by numbers smaller than 1, you can't think of it as dividing up the pizza. I wish I could take a pizza, decide I'm cutting it into half a slice, and wind up with 2 pizzas. That would rule. Think of it like this.

How many slices of size 0.5 can you pull out of the pizza? 2. 1/0.5 = 2.
How many slices of size 0.1 can you pull out of the pizza? 10. 1/0.1 = 10.
How many slices of size 0 can you pull out of the pizza? As many as you want. Go for the rest of your life if you want. And that's why it's infinite.

[charity4thepoor]

the **** is going on in here?

[Errorist]

How many slices of size 0 can you pull out of the pizza? As many as you want. Go for the rest of your life if you want. And that's why it's infinite.

So dividing by 0 is possible, 1/0 = infinite.

the **** is going on in here?

lol

[Greg1983]

So dividing by 0 is possible, 1/0 = infinite.

As possible as infinite is, anyways.

[MiscMathematician]

If I cut 1 pizza into 0 slices, and eat them, how much pizza do I have left? I end up with 1 whole pizza and I'm still hungry.

Also, I can't cut a pizza into slices that I don't have, it's impossible to do.

If you have a pizza and cut it into zero slices? A pizza already is "one slice." So then, you couldn't have it to being with if is cut into 0 slices. Secondly when you add up all your parts of the "pizza" (mathematicians would call this a partition) you need to get back the whole "pizza". If 1 pizza is divided into 0 slices, this means that you now HAVE 0 slices of pizza. So adding up all 0 slices should give you what you started with... Again, it makes no sense.

So dividing by 0 is possible, 1/0 = infinite.

No, but the limit as x approaches 0 of 1/x^2 is infinity. "1/0 = infinity" is okay for calculus I suppose, but I wouldn't write it, just think it. Here's one example: what is the limit as x goes to 0 of 1/x? Its not infinity.

[DangerDan]

http://www.wolframalpha.com/input/?i...2C+-Pi%2C+Pi}]

[Iapetus]

http://www.wolframalpha.com/input/?i...2C+-Pi%2C+Pi}]

Win!

[Errorist]

I got but it's not right. What am I doing wrong? I took the derivative of both sides, used quotient rule for dy/dx on the right and simplified, then divided both sides by the right side.

[Greg1983]

I got but it's not right. What am I doing wrong? I took the derivative of both sides, used quotient rule for dy/dx on the right and simplified, then divided both sides by the right side.

(1/3)x^(-2/3) = 13(1/2)y^(-1/2) (dy/dx)
dy/dx = [ 2y^(1/2) ] / [ 39x^(2/3) ]

Too lazy to tex it up for ya but I think that's right. Been a while...

[charity4thepoor]

No, but the limit as x approaches 0 of 1/x^2 is infinity. "1/0 = infinity" is okay for calculus I suppose, but I wouldn't write it, just think it. Here's one example: what is the limit as x goes to 0 of 1/x? Its not infinity.

dat **** aint even exist brah


(yes genius)

[MiscMathematician]

I got but it's not right. What am I doing wrong? I took the derivative of both sides, used quotient rule for dy/dx on the right and simplified, then divided both sides by the right side.

If you do this the easy way, you can just differentiate y= x^(2/3)/169 with the power rule. You've made an arithmetic mistake it seems. Notice there should be no minus sign (positive power of x).

But to differentiate implicitly wrt x, on the RHS you should have 1/3*x^(-2/3), on the LHS 13/2*y^(-1/2) y'.

Now multiply both sides by 2/13*y^(1/2) to isolate y'.

[Errorist]

(1/3)x^(-2/3) = 13(1/2)y^(-1/2) (dy/dx)
dy/dx = [ 2y^(1/2) ] / [ 39x^(2/3) ]

Too lazy to tex it up for ya but I think that's right. Been a while...

If you do this the easy way, you can just differentiate y= x^(2/3)/169 with the power rule. You've made an arithmetic mistake it seems. Notice there should be no minus sign (positive power of x).

But to differentiate implicitly wrt x, on the RHS you should have 1/3*x^(-2/3), on the LHS 13/2*y^(-1/2) y'.

Now multiply both sides by 2/13*y^(1/2) to isolate y'.

You guys get two different answers. I reworked it and got which is wrong.

[MiscMathematician]

You guys get two different answers. I reworked it and got which is wrong.

They aren;t different. Rewrite with positive powers and replace y^1/2 with x^(1/3)/13. And for gods sake, simplify!

edit: not sure how you got 26...

[Errorist]

They aren;t different. Rewrite with positive powers and replace y^1/2 with x^(1/3)/13. And for gods sake, simplify!

edit: not sure how you got 26...

Either do I, it should have been 39. Runnin on 3 hours of sleep. Also, I didn't know I could just swap out the signs of powers. What exactly do you mean replace y^1/2 with x^(1/3)/13. Do I keep the 39 in the denominator still?