Find the value of e
It's approximately 2.7182818 bla bla bla You can use Wikipedia if you want more digits.

Now, get a calculator and do the following:
(1+1/x)^x
And have x start at something small, but increase the x value as you go on.
The higher your x, the closer and closer you get to representing e, but you can not represent e fully because it would require you use an infinitesimal number and infinity.
Infinity and infinitesimal have their uses, but not many.
But yeah, .9 repeating still is not a valid number as it can not be represented with a fraction.
.9999999999999999999999999999999999999999999999999 for instance is 1-0.0000000000...1
.9 repeating would require 1-.0000000000000000000000000000000000000000000000000 and continue on forever with the end that doesn't exist to be 1.

It doesn't work, but at the same time it does.
Math likes to **** with you like that.

(x^(x-1))x
How do I simplify this into (x^x) ?
I mean, I know if you plug in a number, say three, you end up with (3^2)3 = 3^3 and it checks, but how do I mathematically move it around so that both sides are the same?
I wish this migraine would go away so I could think clearly again.

Also, Felt, answer this.
What would you get if you subtracted an infinitesimal from one? Would it be anything else besides .9 repeating? :l
Because .9 repeating is a repeating decimal, it will never fall on any point, ever. It will not ever reach 1, nor will it fall on any .999999999...9 spots. It continues on forever.
:s

(x^(x-1))x
How do I simplify this into (x^x) ?
I mean, I know if you plug in a number, say three, you end up with (3^2)3 = 3^3 and it checks, but how do I mathematically move it around so that both sides are the same?
I wish this migraine would go away so I could think clearly again.

(X^(Y)) * X = X^(Y+1)

I forget exactly which property that is, but it's a property that applies to all real numbers. That's all you really need to do: Work your way to an identity using only properties that are known to apply to all real numbers.

I don't think this is the place to debate about .999...but I'll do it anyways, except with logic and definitions.
So you're in love with the number e. It is defined by lim(x→inf) (1 + 1/x)^x . Hold up. Did we just use a limit? Looks like we used it for another number.
lim(x→inf) (1 - 1/x) = 0.999...
According to you, there's an infinitesimal that makes 0.999... and 1 distinct numbers, because we took the limit.
Now, because we used a limit, there must be an infinitesimal separating lim(x→inf) (1 + 1/x)^x from e. e is the true number, and the definition approaches but never reaches it since there's an infinitesimal in the way.
In fact, when you take the limit of a function, you approach a value arbitrarily close, but never exactly on the value. That means there's an infinitesimal in between the limit and the actual value.
Therefore, the validity of 0.999... = 1 depends if you accept that lim(x→inf) (1 + 1/x)^x = e, and if you accept the concept of limits in general.

Anyways, a number x can be represented as x^1, and using laws of exponent of added indexes together with the same base being multiplied together):
x^x * x = x^x * x^1 = x^(x+1)

Also
(x^x)/x = x ^ (x-1)
IS NOT TRUE FOR ALL REAL NUMBERS
simply because of how 0 screws it over

Also, don't mess with infinity, especially if it's inside a function. Make sure you know calculus before doing stuff like that.

Rollback Post to RevisionRollBack

My Youtube channel.
Contains Pachebel's Canon made with noteblocks, a working Rubik's cube made with pistons, and the ultimate TNT cannon.

In fact, when you take the limit of a function, you approach a value arbitrarily close, but never exactly on the value. That means there's an infinitesimal in between the limit and the actual value.

Your understanding of limits is flawed. The limit is the actual value. That's why it's called a limit. The value of a limit is the limiting value. If a limit exists, there is one, and exactly one, value for it. This gets into the whole epsilon-delta thing with limits, but that's key to understanding why the limit doesn't just "approach" some value.

So in your examples, for any finite x, the function is different from the limit. That doesn't change the value of the limit to some infinitesimal difference. To further clarify why this is an erroneous way of thinking, consider the following:

lim(x->infinity) sin(x)/x

We know that the limit is 0. According to you it never actually "becomes" 0, but is, instead, infinitesimally displaced from it. Now, tell me, on which side of 0 is it infinitesimally displaced? Since you claim it can't be exactly 0, it must be either above or below it. Which one is it? This example should illuminate exactly why this infinitesimal discussion is nonsense.

It's almost as if this topic was started just so people could have another excuse to ignore the fact that they never bothered to address the proofs I raised there. In fact, at this point people are repeating themselves to the point that I can just link to one of my replies in the original topic and it'll actually be relevant to the mistakes being made here:

The limit DOES approach it, that's why we use it and not just say "0", because that doesn't work. Also, you can tell which side it's on, because you have negative and positive limits, and they yield different results too.

You didn't answer the question. Which side of 0 is the following on:

lim(x->infinity) sin(x)/x

The limit is exactly equal to 0 because, given a neighborhood around 0, there is some neighborhood "around infinity" which results in the expression sin(x)/x being in the given neighborhood around 0 (this is the plain-words explanation of a delta-epsilon proof). I like this example because there are even values of x for which this is exactly 0.

So in this example, using 0 would fail, becuase something/0 = undefined, but when approaching 0 (infinitely close, but without becoming 0), we can get the result we want.

Read the expression that I wrote. It's the limit as x goes to infinity, not the limit as x goes to 0. If we took the limit as x goes to 0, we'd get 1. If we take the limit as x goes to infinity, we get 0. You don't know what you're talking about and your understanding of limits sucks.

Rollback Post to RevisionRollBack

Never attribute to malice what can adequately be explained by incompetence.

Did I accidentally wander into /sci/ instead of the Minecraft forums?

Rollback Post to RevisionRollBack

The inquisitors were torturing Harry.

First, Ignatius used the rock.

Then Billy asked Harry if he wanted to read his BDSM blog. Harry was so surprised that his pants flew right off. He was wearing women's underpants. The inquisitors were wearing them, too.

In fact, when you take the limit of a function, you approach a value arbitrarily close, but never exactly on the value. That means there's an infinitesimal in between the limit and the actual value.

Your understanding of limits is flawed. The limit is the actual value. That's why it's called a limit. The value of a limit is the limiting value. If a limit exists, there is one, and exactly one, value for it. This gets into the whole epsilon-delta thing with limits, but that's key to understanding why the limit doesn't just "approach" some value.

So in your examples, for any finite x, the function is different from the limit. That doesn't change the value of the limit to some infinitesimal difference. To further clarify why this is an erroneous way of thinking, consider the following:

lim(x->infinity) sin(x)/x

We know that the limit is 0. According to you it never actually "becomes" 0, but is, instead, infinitesimally displaced from it. Now, tell me, on which side of 0 is it infinitesimally displaced? Since you claim it can't be exactly 0, it must be either above or below it. Which one is it? This example should illuminate exactly why this infinitesimal discussion is nonsense.

It's almost as if this topic was started just so people could have another excuse to ignore the fact that they never bothered to address the proofs I raised there. In fact, at this point people are repeating themselves to the point that I can just link to one of my replies in the original topic and it'll actually be relevant to the mistakes being made here:

(x^x)/x = x ^ (x-1)

I don't know what causes my mind to even bother thinking about things like this, but it happens, and this one is driving me batshit.

(Also, lololol .9 repeating does NOT equal 1 because then what is 1 - (1 / infinity)? It is .9 repeating. .9 repeating ^ infinity = 1/e )

(I know you can't really use infinity in the technical sense, but still, .9 repeating does not equal 1.)

(Also, (1 + 1/infinity)^infinity = e )

:3

Edit: I'm beginning to think this works for all complex numbers, not just the real ones, too. I'm checking this now.

:l

.9 repeating would essentially be the result of 1 minus an infinitesimal.

Do you know what euler's constant is and how it's calculated?

Whenever you use infinity to calculate something, instead just use a very large number. The larger the number, the more accurate your result will be.

.9 repeating is not 1

:l

(Really, .9 repeating is just as invalid as infinity.)

Find the value of e

It's approximately 2.7182818 bla bla bla You can use Wikipedia if you want more digits.

Now, get a calculator and do the following:

(1+1/x)^x

And have x start at something small, but increase the x value as you go on.

The higher your x, the closer and closer you get to representing e, but you can not represent e fully because it would require you use an infinitesimal number and infinity.

Infinity and infinitesimal have their uses, but not many.

But yeah, .9 repeating still is not a valid number as it can not be represented with a fraction.

.9999999999999999999999999999999999999999999999999 for instance is 1-0.0000000000...1

.9 repeating would require 1-.0000000000000000000000000000000000000000000000000 and continue on forever with the end that doesn't exist to be 1.

It doesn't work, but at the same time it does.

Math likes to **** with you like that.

I still prefer trig, to be honest.

Yes, you do basic algebra to it. Multiply both sides by X and see what happens.

How do I simplify this into (x^x) ?

I mean, I know if you plug in a number, say three, you end up with (3^2)3 = 3^3 and it checks, but how do I mathematically move it around so that both sides are the same?

I wish this migraine would go away so I could think clearly again.

Also, Felt, answer this.

What would you get if you subtracted an infinitesimal from one? Would it be anything else besides .9 repeating? :l

Because .9 repeating is a

repeatingdecimal, it will never fall on any point, ever. It will not ever reach 1, nor will it fall on any .999999999...9 spots. It continues on forever.:s

(X^(Y)) * X = X^(Y+1)

I forget exactly which property that is, but it's a property that applies to all real numbers. That's all you really need to do: Work your way to an identity using only properties that are known to apply to all real numbers.

So you're in love with the number e. It is defined by lim(x→inf) (1 + 1/x)^x . Hold up. Did we just use a limit? Looks like we used it for another number.

lim(x→inf) (1 - 1/x) = 0.999...

According to you, there's an infinitesimal that makes 0.999... and 1 distinct numbers, because we took the limit.

Now, because we used a limit, there must be an infinitesimal separating lim(x→inf) (1 + 1/x)^x from e. e is the true number, and the definition approaches but never reaches it since

there's an infinitesimal in the way.In fact, when you take the limit of a function, you approach a value arbitrarily close, but never exactly on the value. That means there's an infinitesimal in between the limit and the actual value.

Therefore, the validity of 0.999... = 1 depends if you accept that lim(x→inf) (1 + 1/x)^x = e, and if you accept the concept of limits in general.

Anyways, a number x can be represented as x^1, and using laws of exponent of added indexes together with the same base being multiplied together):

x^x * x = x^x * x^1 = x^(x+1)

Also

(x^x)/x = x ^ (x-1)

IS NOT TRUE FOR ALL REAL NUMBERS

simply because of how 0 screws it over

Also, don't mess with infinity, especially if it's inside a function. Make sure you know calculus before doing stuff like that.

Contains Pachebel's Canon made with noteblocks, a working Rubik's cube made with pistons, and the ultimate TNT cannon.

That's right, I hadn't thought of that. Both 0/0 and 0^0 are indeterminate. It does work for any non-zero real number, though.

Actually, it works for any non-zero complex number.

:3

Oh wow! If I knew I could frustrate people this hard with math, I would of taken calculus.

Your understanding of limits is flawed. The limit

isthe actual value. That's why it's called a limit. The value of a limit is the limiting value. If a limit exists, there is one, and exactly one, value for it. This gets into the whole epsilon-delta thing with limits, but that's key to understanding why the limit doesn't just "approach" some value.http://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit

So in your examples, for any

finitex, the function is different from the limit. That doesn't change the value of the limit to some infinitesimal difference. To further clarify why this is an erroneous way of thinking, consider the following:lim(x->infinity) sin(x)/x

We know that the limit is 0. According to you it never actually "becomes" 0, but is, instead, infinitesimally displaced from it. Now, tell me, on which side of 0 is it infinitesimally displaced? Since you claim it can't be exactly 0, it must be either above or below it. Which one is it? This example should illuminate exactly why this infinitesimal discussion is nonsense.

It's almost as if this topic was started just so people could have another excuse to ignore the fact that they never bothered to address the proofs I raised there. In fact, at this point people are repeating themselves to the point that I can just link to one of my replies in the original topic and it'll actually be relevant to the mistakes being made here:

viewtopic.php?f=6&t=13282&start=120#p224032

If you want to start a new math topic, can we actually start it with something interesting that isn't a completely trivial property of exponents?

Yeah, sorry 'bout that one. I just could remember the goddamned property at the time.

It's both since they both represent the same real number.

I'm not.

You didn't answer the question. Which side of 0 is the following on:

lim(x->infinity) sin(x)/x

The limit is exactly equal to 0 because, given a neighborhood around 0, there is some neighborhood "around infinity" which results in the expression sin(x)/x being in the given neighborhood around 0 (this is the plain-words explanation of a delta-epsilon proof). I like this example because there are even values of x for which this is exactly 0.

Read the expression that I wrote. It's the limit as x goes to infinity, not the limit as x goes to 0. If we took the limit as x goes to 0, we'd get 1. If we take the limit as x goes to infinity, we get 0. You don't know what you're talking about and your understanding of limits sucks.

That it's misleading to consider that a limit doesn't reach its value or that it's somehow infinitesimally displaced from it.

Let x=.999...

Therefore, 9x=9, so x=1

Did I accidentally wander into /sci/ instead of the Minecraft forums?

First, Ignatius used the rock.

Then Billy asked Harry if he wanted to read his BDSM blog. Harry was so surprised that his pants flew right off. He was wearing women's underpants. The inquisitors were wearing them, too.

They realized that they were all men of the lord.

- 30 Hs

I was actually using the logic of Towel's, for proof via contradiction.

Contains Pachebel's Canon made with noteblocks, a working Rubik's cube made with pistons, and the ultimate TNT cannon.

0^0 is an indeterminate, form, it is not 1. It is not anything.

You don't need to apologize for anything anyway.