NationStates Jolt Archive

Here's something for all you math people out there.

Any number over infinity is so close to 0 that it's usually called 0 because that's the limit. Any number over itself is 1. But what's infinity over infinity? Is it 1 or 0? I know infinity is a concept, but is there mathematical consensus on this?

Also, why is 0! 1?

eep.. does not compute

This is really only hypothetical, as you could never get infinity to divide by since it is contiuous and has no limits. But since the laws of mathmatics states that, any number divided by itself is one, I suppose if you could, then infinity over infinity would HAVE to equal 1.

Infinity over infinity is not defined.
It can be any number between minus infinity and plus infinity.

However, in certain mathematical exercises, it's possible that you have to divide, and you get to a certain number.

The rules of mathematics are slightly different when it's about infinity.

infinity divided by infinity? Show me an equation where you would have infinity divided by infinity....

I may be wrong on this, but infinity is a concept, not a number. Dividing a concept by a concept is too wierd, so lets move on to infinite numbers.

If you divide an infinite number by an infinite number, the answer depends on which infinite numbers you choose. (There's an lot of different infinite numbers.)

So the answer could be 2. Which would be neat.

Infinity over infinity is not defined.
It can be any number between minus infinity and plus infinity.

However, in certain mathematical exercises, it's possible that you have to divide, and you get to a certain number.

The rules of mathematics are slightly different when it's about infinity.

How can the rules be different when the number doesn't exist, There isn't a number to donate infinity, it is greater than all numbers, and has no boundry, therefore you can't do any kind of mathmatics on it, it is mearly a concept and a hypothetical situation.

Yes, well, you can do hypothetical mathematic exercises on it.

You can still use it, even though it's hypothetical.

Please explain to me how you can do calculations with a number which has no limits.

Yuck! Not only Maffs, but Maffs that holds no purpose in everyday life! HATE IT!!!

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

Any number to the power of 0 equals 1, which is what I said to start with.

lim as x->0 of 1/x on the positive side = +infinity.

lim as x->0 of 1/x on the negative side = -infinity.

lim as x->0 of 1/x from both sides = does not exist.

Just FYI

eep.. does not compute



Translation = DUH!!!

Well, lets look on the bright side. Even though there is no defined answer of infinity/infinity at least the non existant answer is real and not imaginary.

How about (infinity x i)/infinity?

-GL

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

Any number to the power of 0 equals 1, which is what I said to start with.

0^0 != 1

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

Any number to the power of 0 equals 1, which is what I said to start with.

0^0 != 1

And thats what I said.

technically, 0/0 should be 1, if you take the limit.

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

Any number to the power of 0 equals 1, which is what I said to start with.

0^0 != 1

And thats what I said.

You said that 0^0 = 1, it does not. 0^0 is an indeterminate form. L'Hopital's Rule (or possibly a rule I'm not familiar with) must be used.

Kanteletar: Take the limit as x->infinity of x/x... it's 1 at all real numbers except for 0. I really don't know what to do about that, you might be right that it does not exist :)

Well according to the rules of mathmatics, any number to the power of 0, which I might add is what the ^ symbol means, equals 1.

Well according to the rules of mathmatics, any number to the power of 0, which I might add is what the ^ symbol means, equals 1.except 0.

to say x^0 means x/x

Look:

x^3=xxx/1
x^2=xx/1
x^1=x/1
x^0=x/x
x^-1=1/x
x^-2=1/xx
x^-3=1/xxx

If you plug 0 in for x, it stops working at <x^0.

Kanteletar: Take the limit as x->infinity of x/x... it's 1 at all real numbers except for 0. I really don't know what to do about that, you might be right that it does not exist :)

It is 1, but if you do the normal procedure for limits (ie direct substitution) you end up with 0/0. Using L'Hopital's rule (basically taking the derivative of the top and bottom functions separately) you end up with lim x-> infinity of 1/1 which is 1.

Well according to the rules of mathmatics, any number to the power of 0, which I might add is what the ^ symbol means, equals 1.

If you put 0^0 into a calculator you'll get an error.

Kanteletar: Take the limit as x->infinity of x/x... it's 1 at all real numbers except for 0. I really don't know what to do about that, you might be right that it does not exist :)

It is 1, but if you do the normal procedure for limits (ie direct substitution) you end up with 0/0. Using L'Hopital's rule (basically taking the derivative of the top and bottom functions separately) you end up with lim x-> infinity of 1/1 which is 1.with a removable discontinuity at x=0....

CAN you remove that discontinuity? Are we allowed to? :P

Well according to the rules of mathmatics, any number to the power of 0, which I might add is what the ^ symbol means, equals 1.

If you put 0^0 into a calculator you'll get an error.

I know I've tried. Probably because it can't be done, it is entirly hypothetical

But if you were to be able to do it, by the elementary laws of mathmatics, you would get an answer of 1.

Kanteletar: Take the limit as x->infinity of x/x... it's 1 at all real numbers except for 0. I really don't know what to do about that, you might be right that it does not exist :)

It is 1, but if you do the normal procedure for limits (ie direct substitution) you end up with 0/0. Using L'Hopital's rule (basically taking the derivative of the top and bottom functions separately) you end up with lim x-> infinity of 1/1 which is 1.with a removable discontinuity at x=0....

CAN you remove that discontinuity? Are we allowed to? :P

I really don't know. I might be able to ask my prof if he's still around the college next Monday (might not be though cuz he may not have any finals to give next week).

Kanteletar: Take the limit as x->infinity of x/x... it's 1 at all real numbers except for 0. I really don't know what to do about that, you might be right that it does not exist :)

It is 1, but if you do the normal procedure for limits (ie direct substitution) you end up with 0/0. Using L'Hopital's rule (basically taking the derivative of the top and bottom functions separately) you end up with lim x-> infinity of 1/1 which is 1.with a removable discontinuity at x=0....

CAN you remove that discontinuity? Are we allowed to? :P

I really don't know. I might be able to ask my prof if he's still around the college next Monday (might not be though cuz he may not have any finals to give next week).I'm sure you can.

Just like (x^2-1)/(x-1) has a big ol hole in it, you can still graph it if you factor out the bottom.

So if you do the same with x/x, and just make it 1, I suppose the limit of x/x would be 1 everywhere! Including 0!

Kanteletar: Take the limit as x->infinity of x/x... it's 1 at all real numbers except for 0. I really don't know what to do about that, you might be right that it does not exist :)

It is 1, but if you do the normal procedure for limits (ie direct substitution) you end up with 0/0. Using L'Hopital's rule (basically taking the derivative of the top and bottom functions separately) you end up with lim x-> infinity of 1/1 which is 1.with a removable discontinuity at x=0....

CAN you remove that discontinuity? Are we allowed to? :P

I really don't know. I might be able to ask my prof if he's still around the college next Monday (might not be though cuz he may not have any finals to give next week).I'm sure you can.

Just like (x^2-1)/(x-1) has a big ol hole in it, you can still graph it if you factor out the bottom.

So if you do the same with x/x, and just make it 1, I suppose the limit of x/x would be 1 everywhere! Including 0!

I believe you're right.

Kanteletar: Take the limit as x->infinity of x/x... it's 1 at all real numbers except for 0. I really don't know what to do about that, you might be right that it does not exist :)

It is 1, but if you do the normal procedure for limits (ie direct substitution) you end up with 0/0. Using L'Hopital's rule (basically taking the derivative of the top and bottom functions separately) you end up with lim x-> infinity of 1/1 which is 1.with a removable discontinuity at x=0....

CAN you remove that discontinuity? Are we allowed to? :P

I really don't know. I might be able to ask my prof if he's still around the college next Monday (might not be though cuz he may not have any finals to give next week).I'm sure you can.

Just like (x^2-1)/(x-1) has a big ol hole in it, you can still graph it if you factor out the bottom.

So if you do the same with x/x, and just make it 1, I suppose the limit of x/x would be 1 everywhere! Including 0!

I believe you're right.

Yes

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)

I said it couldn't be done, purely hypothetical. Good night, where are you old chap, its good morning over here.

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)

I said it couldn't be done, purely hypothetical. Good night, where are you old chap, its good morning over here.It's... 2:40?! aww crap man, I need some sleep!

Mix :shock: with :evil: and you're have my sad blood-shot eyes icon :P

Oh yeah, well solve this!

2x-4y=0

It's a doozy!

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)

When I said take the derivative of x/x (not quite what I said), I meant take the derivative of x, and put it over the derivative of x (you'd end up with 1/1 and the limit of a constant is a constant.

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)GAH

I messed up YET AGAIN!

The derivative of x/x is 0.

Oh yeah, well solve this!

2x-4y=0

It's a doozy!umm

x=2
y=1

Go away :)

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)

When I said take the derivative of x/x (not quite what I said), I meant take the derivative of x, and put it over the derivative of x (you'd end up with 1/1 and the limit of a constant is a constant.derivative of x over derivative of x? That's not how it works lol

Oh yeah, well solve this!

2x-4y=0

It's a doozy!umm

x=2
y=1

Go away :)

Wrong!

x=0, y=0

Hahahahahaha! Got you!

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)

When I said take the derivative of x/x (not quite what I said), I meant take the derivative of x, and put it over the derivative of x (you'd end up with 1/1 and the limit of a constant is a constant.derivative of x over derivative of x? That's not how it works lol

That's how L'Hopital's rule for indeterminate limits works.

Here's something for all you math people out there.

Any number over infinity is so close to 0 that it's usually called 0 because that's the limit. Any number over itself is 1. But what's infinity over infinity? Is it 1 or 0? I know infinity is a concept, but is there mathematical consensus on this?

It depends on how you get there. L'Hospital says, differentiate the top and bottom, and try the division again.

Also, why is 0! 1?

How many orders can you arrange 3 objects in?
{1,2,3}
{1,3,2}
{2,1,3}
{2,3,1}
{3,1,2}
{3,2,1}
(3!) = 6

How many orders can you arrange 2 objects in?
{1,2}
{2,1}
(2!) = 2

How many orders can you arrange 1 object in?
{1}
(1!) = 1

How many orders can you arrange 0 objects in?
{}
(0!) = 1

---
The Republic of This Name will DEAT Me
http://www.nationstates.net/forum/viewtopic.php?t=140286

[quote=Sdaeriji]Oh yeah, well solve this!

2x-4y=0

It's a doozy!umm

x=2
y=1

Go away :)

Wrong!

x=0, y=0


How? I was under the impression that to be called x and y, the two quantities had to be different. To my understanding, that would make.

2x-4x=0

Oh yeah, well solve this!

2x-4y=0

It's a doozy!umm

x=2
y=1

Go away :)

Wrong!

x=0, y=0
How? I was under the impression that to be called x and y, the two quantities had to be different. To my understanding, that would make.

2x-4x=0

Hahahahahaha! Got you!

I know. I'm just being a jerk because the stuff you guys are talking about in this thread is miles over my head.

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

Any number to the power of 0 equals 1, which is what I said to start with.

0^0 != 1

And thats what I said.

f(x) = e^(-1/x^2)
g(x) = x^2

lim x->0 f(x) = 0
lim x->0 g(x) = 0

lim x->0 [f(x)]^[g(x)] is left as an exercise to the reader.

---
The Republic of This Name will DEAT Me
http://www.nationstates.net/forum/viewtopic.php?t=140286

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

Any number to the power of 0 equals 1, which is what I said to start with.

0^0 != 1

And thats what I said.

f(x) = e^(-1/x^2)
g(x) = x^2

lim x->0 f(x) = 0
lim x->0 g(x) = 0

lim x->0 [f(x)]^[g(x)] is left as an exercise to the reader.

---
The Republic of This Name will DEAT Me
http://www.nationstates.net/forum/viewtopic.php?t=140286

It's e isn't it? I tried it in my head but I never was very good at limits. :)

edit: sorry e^-1

infinity over infinity (plus or minus) only happens when taking limits. In which case L'Hopital's rule is used. Same thing if a limit produces 0 over 0, or 0^0.

Any number to the power of 0 equals 1, which is what I said to start with.

0^0 != 1

And thats what I said.

f(x) = e^(-1/x^2)
g(x) = x^2

lim x->0 f(x) = 0
lim x->0 g(x) = 0

lim x->0 [f(x)]^[g(x)] is left as an exercise to the reader.

---
The Republic of This Name will DEAT Me
http://www.nationstates.net/forum/viewtopic.php?t=140286

It's e isn't it? I tried it in my head but I never was very good at limits. :)

edit: sorry e^-1


Thats what I thought, but when I tried to work it through, I got some strange things so gave up.

Here's something for all you math people out there.

Any number over infinity is so close to 0 that it's usually called 0 because that's the limit. Any number over itself is 1. But what's infinity over infinity? Is it 1 or 0? I know infinity is a concept, but is there mathematical consensus on this?

Also, why is 0! 1?

1/0 is nothing, not infinity
1/x with x going to 0 has as a limit infinity

if you have f(x)/g(x) with g(a)=f(a)=0 then the limit for x going to a can be calculated using l'Hopital:

in this case, limit f(x)/g(x), x->a equals limit (df(x)/dx)/(dg(x)/dx)
as a result, a 0/0 can be infinity, or just an number.

example: sin(x)/(e^x-1), with x going to 0 gives 0/0
use l'Hopital:
*=cos(x)/e^x, which is 1 for x going to 0
so sin(x)/(e^x-1) = 1

note: it get's harder once you start using function of more variables, as a limit of a function f(x) is taken by letting x 'get closer' to a certain value a, and there are only 2 ways to get closer to a. but if we have a function f(x,y) then getting closer to, let's say (a,b), is a lot harder as there are many ways to get closer to it, and if a limit exists, it has to be the same for all possible ways to get closer.

an example:

g(x,y) = (xy)/(x^2+y^2), for (x,y) going to (0,0)

first we get closer by a path ((1/n,1/n)), n going to infinity:
g(1/n,1/n) = (1/n^2)/(2/n^2) = 1/2

then use the path ((1/n,2/n)) :
g(1/n,2/n) = 2/5

so g(x,y) has no limit for (0,0)

but, nevertheless, it is still discontinuous.

The graph of x/x is not y=1 at all x's. At 0 there is a hole. And there ain't nothing you can do about it!

And the derivative? Well, the derivative does not exist either! Go ahead! Try it! Take the derivative of x/x! It is 0/0! LOL AND THERE AGAIN WE ARE BACK TO THE SAME PROBLEM! LOL

It does not exist :)

There, solved the problem, good night :)

actually there is something you can do about it :p

if a function f(x) is continuous in a certain interval ]a,b[ and the limits in a (only right limit required) and b (only left limit) exist and are finite,

then we define the extended function f*(x) as:
f*(x) =
limit f(t), t->a, if x = a
f(x), if x is in ]a,b[
limit f(t), t->b, if x = b
(obviously only when both limits exist and are finit)
and this function f*(x) is then uniformally continuous in [a,b]

what's even better is that mathematicians normally mix up f*(x) and f(x), and basicly, we use f*(x) all the time instead of f(x) :p

if you have f(x)/g(x) with g(a)=f(a)=0 then the limit for x going to a can be calculated using l'Hopital:

in this case, limit f(x)/g(x), x->a equals limit (df(x)/dx)/(dg(x)/dx)
as a result, a 0/0 can be infinity, or just an number.

example: sin(x)/(e^x-1), with x going to 0 gives 0/0
use l'Hopital:
*=cos(x)/e^x which is 1 for x going to 0
so sin(x)/(e^x-1) = 1

You differentiated e^(x-1) wrong, if that's what you meant. Or is it (e^x) - 1?

Here's something for all you math people out there.

Any number over infinity is so close to 0 that it's usually called 0 because that's the limit. Any number over itself is 1. But what's infinity over infinity? Is it 1 or 0? I know infinity is a concept, but is there mathematical consensus on this?

It depends on how you get there. L'Hospital says, differentiate the top and bottom, and try the division again.

Also, why is 0! 1?

How many orders can you arrange 3 objects in?
{1,2,3}
{1,3,2}
{2,1,3}
{2,3,1}
{3,1,2}
{3,2,1}
(3!) = 6

How many orders can you arrange 2 objects in?
{1,2}
{2,1}
(2!) = 2

How many orders can you arrange 1 object in?
{1}
(1!) = 1

How many orders can you arrange 0 objects in?
{}
(0!) = 1

---
The Republic of This Name will DEAT Me
http://www.nationstates.net/forum/viewtopic.php?t=140286

this can be explained with the use of the gamma function (note: i am going to use G instead of 'gamma'):
definition:
G(p) = integral(exp(-x)*x^(p-1)dx, x=0..infinity)

note that this isn't a Riemann integral, it's an, i don't know how you call it in english, unreal integral, which is defined by taking the limit of a Riemann integral. also the resulting function is only function of p.

we can show that G(p) exists for p>0
another thing we can show is:
G(p+1) = p*G(p), p>0
and we can also find that G(1) = 1
which brings us to a definition of n!:
n! = G(n+1)

now take 0!:
0! = G(0+1) = G(1) = 1

here you have it :p

if you have f(x)/g(x) with g(a)=f(a)=0 then the limit for x going to a can be calculated using l'Hopital:

in this case, limit f(x)/g(x), x->a equals limit (df(x)/dx)/(dg(x)/dx)
as a result, a 0/0 can be infinity, or just an number.

example: sin(x)/(e^x-1), with x going to 0 gives 0/0
use l'Hopital:
*=cos(x)/e^x which is 1 for x going to 0
so sin(x)/(e^x-1) = 1

You differentiated e^(x-1) wrong, if that's what you meant or is it (e^x) - 1?

of course it is (e^x) -1 :p

as the power operator ^ precedes the substraction operator - this is the only way you are allowed to interprete e^x-1

if you have f(x)/g(x) with g(a)=f(a)=0 then the limit for x going to a can be calculated using l'Hopital:

in this case, limit f(x)/g(x), x->a equals limit (df(x)/dx)/(dg(x)/dx)
as a result, a 0/0 can be infinity, or just an number.

example: sin(x)/(e^x-1), with x going to 0 gives 0/0
use l'Hopital:
*=cos(x)/e^x which is 1 for x going to 0
so sin(x)/(e^x-1) = 1

You differentiated e^(x-1) wrong, if that's what you meant or is it (e^x) - 1?

of course it is (e^x) -1 :p

as the power operator ^ precedes the substraction operator - this is the only way you are allowed to interprete e^x-1

Fair enough, I'm just used to seeing my own typing with math equations and I tend to overuse brackets to avoid any possiblity of confusing myself. :mrgreen:

Infinity over infinity is undefined, since infinity itself is undefined.

But if a function at a certain point yields infinity over infinity, you can take the derivative of the top and the bottom to get teh limit of the function at that point, which usually will turn out to be an actual number.