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1^∞

Name: Anonymous 2006-12-11 19:10

hay /sci/ what's 1^∞

well it's indeterminate but shouldn't it be 1?

Name: Anonymous 2006-12-11 19:29 (sage)

It is 1.

Name: Anonymous 2006-12-11 19:51

Unfortunately, you can't put infinity as an exponent. If you look at exponents as an operation (a valid approach), then infinity (in this sense) is not part of any particularly useful number system.

What you're looking for is the limit of 1^n as n->∞

That, my friend, is 1.

Name: 4tran 2006-12-13 0:35

e

[1+(1/x)]^x = e by definition (as x->∞)

Name: Anonymous 2006-12-13 1:41

>>4
Fine, then get a calculator, start with 1*1, then keep multiplying the answer by 1.  Stop when it equals e.

Name: 4tran 2006-12-13 5:04

I'll get back to you after an infinite quantity of time

Name: Anonymous 2006-12-13 16:02

>>5
His point (or what his point should have been, at least) is that "1^∞" is indeterminate. Since lim x->∞ 1^x = 1, and lim x->∞ (1 + 1/x)^x = e, but both would equal 1^∞ if it had a numeric value, and thus would be equal to each other.

Name: Anonymous 2006-12-13 18:20

>>7
If you add infinitesimals to the fun, you could say (1 + 1/∞)^∞ = (1 + ε)^∞ = indeterminate.

Or, more seriously, does (1 + 1/∞)^∞ = 1 necessarily contradict lim x->∞ (1 + 1/x)^x = e?

Name: Anonymous 2006-12-13 20:06

>>8
does (2-2)/(2-2) = 1 necessarily contradict lim x->∞ (x-2)/(2-x) = i?

Name: Anonymous 2006-12-13 20:28

1^∞ is not indeterminate you fools, it's not even a well-formed expression. ∞ IS NOT A NUMBER.

>>3 has the correct answer. >>4 fails. We're looking for the limit as n->∞ of 1^n, not (1+1/n)^n. The former is 1, the latter is e.

Name: Anonymous 2006-12-13 21:45 (sage)

>>10
Welcome to the extended real number line.

>>9
I don't see how that's relevant. Explain.

Name: Anonymous 2006-12-13 21:48

>>10
Saying 1^∞ can only 'mean' lim x->∞ of 1^x is like saying 0/0 can only 'mean' lim x->0 of x/x.

Name: Anonymous 2006-12-16 20:17

>>12 is almost right, you can have lim f(x)^g(x) as x->a where f(a)=1 and g(a)=∞.  lim f(x)^g(x)=e^(lim g(x)*ln(f(x)))=e^(lim ln(f(x))/(1/g(x)))   ln(f(x))=0 and 1/g(x)=0 so you can evaluate the limit with L'Hopital's rule

Name: Anonymous 2006-12-16 21:32

This is an awesome thread.

Name: Anonymous 2006-12-16 23:31

>>12
No, you're completely, 100% wrong. The difference between saying 1^∞ must mean lim x->∞ of 1^x, and saying that 0/0 must mean lim x->0 of x/x is the fact that ∞ is not a number. 1^∞, as has been said before, is not a valid expression, whereas 0/0 *is* a valid expression.

>>11
You point out that 1^∞ is a valid expression on the extended number line. Ok, that's nice. Define what an exponent is in the extended number line in terms of a function. Finding it a little.. difficult? There's a reason for that.

Name: Anonymous 2006-12-16 23:51

>>15
x^∞ =
   0 if -1 < x < 1,
   1 if x == 1,
   ∞ if x > 1,
   undefined otherwise.

Works for me.

Name: Anonymous 2006-12-17 15:20

x^∞ = x^(y/0) = (x ^ y) ^ (1/0) = (x ^ y) ^ (1 / 0)

Name: Anonymous 2006-12-17 15:23

>>17
Oh wait, I found something!

x^(y/0) = (x ^ (1/0)) ^ y = (x ^ ∞) ^ y

By which we can draw the conclusion that x ^ ∞ is 1, which is the same no matter what y is.

Name: Anonymous 2006-12-17 15:30 (sage)

>>17,18
I can't tell what it is you're trying to proof, but even the extended real line doesn't allow division by zero.

Name: Anonymous 2006-12-17 15:32

>>19
This is /sci/.

Name: Anonymous 2006-12-19 7:19

>>20

... and if your observent self would notice, the title for these boards is "Science & Math", emphasis on the latter.

Name: Anonymous 2006-12-19 16:44

>>21
I don't think you understand. This is /sci/.

Name: Anonymous 2006-12-19 19:00

http://dis.4chan.org/read/sci/1165881984/20
Your point being?
Don't change these.
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