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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

I made the statement in another topic that:

"I believe it is commonly accepted that 1/infinity = 0, due to ..."

I invite discussion

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

means

Thats how its defined in standard real analysis. In other words, given any real positive number epsilon, you can always find a big enough positive number delta such that if you plug in any value of *x* greater than delta into (*x*), (*x*) will be within epsilon of 0. This avoids any mention of infinity, see?

*Last edited by JaneFairfax (2007-03-02 16:08:19)*

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Yes, exactly. It carefully avoids mentioning Real Infinity. Or what happens on infinity.

*Last edited by George,Y (2007-03-02 16:09:33)*

**X'(y-Xβ)=0**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

It's OK, I chose the words "commonly accepted" carefully, and think a discussion would be enlightening ...

In fact my recent work on compound interest highlights an interesting thing.

A formula for "e" is

A naive interpretation of this would be

(1+1/∞)[sup]∞[/sup] => (1+0)[sup]∞[/sup] => 1

(*assuming (!)* 1/∞ = 0, and also that 1[sup]∞[/sup] = 1 which is interesting in its own right)

But plugging in large values of n gives us e=2.7182... getting consistently more accurate.

Anyway, I will be writing this up somehow.

In the meantime feel free to tear this apart.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

George,Y wrote:

Yes, exactly. It carefully avoids mentioning Real Infinity. Or what happens on infinity.

There is no such thing as "on infinity". Functions are defined on the real numbers. Infinity is not a real number, and thus, not in the domain of functions.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

MathsIsFun wrote:

It's OK, I chose the words "commonly accepted" carefully, and think a discussion would be enlightening ...

In fact my recent work on compound interest highlights an interesting thing.

A formula for "e" is

A naive interpretation of this would be

(1+1/∞)[sup]∞[/sup] => (1+0)[sup]∞[/sup] => 1

(

assuming (!)1/∞ = 0, and also that 1[sup]∞[/sup] = 1 which is interesting in its own right)But plugging in large values of n gives us e=2.7182... getting consistently more accurate.

Anyway, I will be writing this up somehow.

In the meantime feel free to tear this apart.

*Last edited by JaneFairfax (2007-03-03 11:27:54)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

So I propose we give a well-defined definition to operations on infinity. Please note that any definition we give to operations on infinity will not be a binary operator (though that would be nice) as there are things such as indeterminate forms. We will leave these out and ignore them since there isn't really much we can do with them anyhow.

Without further ado, here is Shadrach's Extension of Binary Operators to Infinity (SEBO for short):

So with this, we may remark that 1/infinity = 0 and 1^infinity = 1, but we can't say anything on what infinity / infinity is.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

This is the page I am currently working on: Limits (An Introduction)

You can get a rare insight into my scattered thought processes as I put a page together

I am currently pushing my brain to show the formal definition in an easy way, through words, diagrams or animation. Inspiration will arrive I hope.

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**Toast****Real Member**- Registered: 2006-10-08
- Posts: 1,321

Cool, I'm looking forward to a limits page.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Nice half-page there.

Where you have the example of (x²+1)/(x-1), it might be worth noting that as well as trying values that get progressively closer to 1 and looking for a pattern, you can also find the limit algebraically.

(x²+1)/(x-1) = (x+1)(x-1)/(x-1) = x+1 for all x ≠ 1. Therefore, as x approaches 1, then the function will approach 2.

If you want to get really advanced, you could even perhaps throw in a mention of L'Hopital's rule.

Also, at one point you have (1+1/n)^2, where the 2 should be an n.

It looks like it's going to be a very useful and interesting page when it's finished though, so keep up the good work!

(The method is certainly talkative. )

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Thanks, mathsy ... umm talkative in a bad way?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Don't worry, I was just joking.

Limits Page wrote:

The method basically says

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Ah yes good point, inanimate animation. Poetic license?

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

There is no such thing as "on infinity". Functions are defined on the real numbers. Infinity is not a real number, and thus, not in the domain of functions.

Sorry Sir! I don't know you recently begin to censor the phrase " on infinity".

But I suppose as MIF has put infinity into the function 1/x, I can interpret he intended to evaluate the function of 1/x on infinity.

*Last edited by George,Y (2007-03-04 20:30:55)*

**X'(y-Xβ)=0**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

George,Y wrote:

I can interpret he intended to evaluate the function of 1/x on infinity.

It may have been my intention, but ... did I succeed?

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

You didn't. Neither did I stated you did. I meant "It (the definition of limit) avoids (discussing) what happens on infinity)".

However, you can check a similiar problem posted by Heldensheld in the Helpme section. Very Tricky!

*Last edited by George,Y (2007-03-04 22:24:37)*

**X'(y-Xβ)=0**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

I just noticed that ... you subtracted 0.9 times the series from the series and were left with just a few terms. Neat work.

URL: http://www.mathsisfun.com/forum/viewtopic.php?pid=61986

I can see your point of view: the "n"th term is very small and so can be dismissed. But could you not eliminate it all together (somehow)?

And back to 1/infinity, I may have trouble actually evaluating the function, but I can evaluate it's limit.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Yes, that is exactly what I mean. Thank you for understanding.

**X'(y-Xβ)=0**

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