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math man

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1 to the power of say n (n is some finite natural number) is 1 of course, and we all are taught 1 to the power of anything is 1.
So it would be obvious to think that 1^(infinity) is 1. Is this correct or not... What do you guys believe?
 

seanieg89

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No, 1^inf is undefined...you cannot just extrapolate something like that and claim it to be true of the 'object' infinity. I stress the use of the word object here as you have gone from considering a real number with a real exponent to a real number with infinity (not a real number) as a superscript.

Of course you could DEFINE this to be anything you like, but I don't think its particularly useful to do so. You would run into big problems when you tried to define things like (-1)^inf.
 

seanieg89

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At least thats how I interpret what you are asking, if you are instead referring to some limiting process you might need to be a bit more specific.

Eg If x->1 and y->inf does x^y->1?

The answer is an emphatic no here, as can be seen from the identity (1+1/n)^n->e.
 

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Daniel Daners made special emphasis on that, taking care to highlight this common error:

 

math man

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Yes I was more referring to the limiting
Process. However this question was just
Designed to see how many people think
It is one as it is commonly mistaken to
Be and I feel as though it is a beautiful
Phenomenon how 1 to the infinity can tak
On any value
 

seanieg89

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Whenever you do something (which you suspect may be illegal) in mathematics you should be asking yourself why you CAN do it. Steps in your working are invalid until proven valid, not the other way around!

(So if you want a more specific answer on why we cannot say (1+1/n)^n->1 or something similar, you must first play devils advocate and try to "justify" why it IS 1.)
 
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seanieg89

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idgi why wouldnt 1^inf not equal 1
Why wouldnt 1 to the power of a chair equal 1? Infinity is not a real number, the current definition of exponentiation does not apply to this situation and we cannot extrapolate any information from the fact that 1 to the power of a real number is 1.

In any case the OP meant something different, read my earlier post on limits.
 

seanieg89

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Who would make the though? It's obvious you can't do that lol.
Its not THAT obvious...

If x->a and y-> b, then xy->ab.

This is true for real a,b. It is even true for infinite a,b given suitable definitions.
From this it is natural to 'guess' that:

If x->a and y->b then x^y-> a^b might hold for positive/infinite a,b given suitable definitions. This is not the case.
 

RealiseNothing

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Its not THAT obvious...

If x->a and y-> b, then xy->ab.

This is true for real a,b. It is even true for infinite a,b given suitable definitions.
From this it is natural to 'guess' that:

If x->a and y->b then x^y-> a^b might hold for positive/infinite a,b given suitable definitions. This is not the case.
I thought it was obvious though that since it is approaching infinity (and hence not a real value), that you couldn't use basic methods that you would use on 'real' values.
 

mirakon

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I remember my maths teacher once said that some people think parallel lines meet at the point of infinity

The point being that infinity is a very annoying concept to work with mathematically, philosophically or in any way.
 

seanieg89

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I thought it was obvious though that since it is approaching infinity (and hence not a real value), that you couldn't use basic methods that you would use on 'real' values.
The product law for limits works even if infinity is allowed though. And we could try to define:

x^inf=0 if 0=<x<1
x^inf=inf if x>1
x^inf=? if x=1.

However, no choice of '?' would make our desired exponentiation limit law true.

Many times infinity can be adjoined to the real numbers usefully, but this is not one of those times.</x<1
 

seanieg89

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I remember my maths teacher once said that some people think parallel lines meet at the point of infinity

The point being that infinity is a very annoying concept to work with mathematically, philosophically or in any way.
They do in projective geometry :).
 

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