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October 26th, 2016, 03:23 PM   #11
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Quote:
 Originally Posted by topsquark But $\displaystyle \infty$ is not a number so you can't say that $\displaystyle 1^{\infty}$ is equal to anything. In fact, $\displaystyle 1^{\infty}$ is undefined. -Dan
What would you say 1*1*1*1*... is equal to as an infinite sequence. Yes, technically infinity is not a number but 1^infinity is a technically false way of saying 1, multiplied by itself infinitely.

And why would it be undefined? 1*1*1*1... It's just repeating itself. It doesn't matter how many times I state it.

October 26th, 2016, 03:49 PM   #12
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Quote:
 Originally Posted by VisionaryLen What would you say 1*1*1*1*... is equal to as an infinite sequence. Yes, technically infinity is not a number but 1^infinity is a technically false way of saying 1, multiplied by itself infinitely. And why would it be undefined? 1*1*1*1... It's just repeating itself. It doesn't matter how many times I state it.
If you have a finite number of 1's to multiply then yes, of course the product is equal to 1. So alright, a bit more accuracy: $\displaystyle \prod_{n = 1}^{\infty} 1 = \lim_{n \to \infty} 1^n$ is undefined.

-Dan

October 26th, 2016, 04:21 PM   #13
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 Originally Posted by topsquark $\displaystyle \prod_{n = 1}^{\infty} 1 = \lim_{n \to \infty} 1^n$ is undefined.
No, $\lim \limits_{n \to \infty} 1^n = 1$ because every term in the sequence is equal to 1.

$$\lim _{n \to \infty} \left(1+\frac{1}{n^2}\right)^n=1$$ because the base heads to 1 "more quickly" than the exponent grows.

$$\lim _{n \to \infty} \left(1+\frac{1}{n}\right)^{n^2}=+\infty$$ because the exponent grows "more quickly" than the base heads to 1.

$$\lim _{n \to \infty} \left(1+\frac{1}{n}\right)^{n}=e$$ because the base and the exponent remain in balance causing the expression to neither expand nor contract.

Each of the indeterminate forms has this conflict between two terms pulling in opposite directions. Like tossing a coin, most of the time one wins over the other: heads or tails. But just occasionally the coin balances on it's edge and the system remains in a sort of unstable equilibrium.

Last edited by v8archie; October 26th, 2016 at 04:48 PM.

October 26th, 2016, 04:31 PM   #14
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 Originally Posted by v8archie No, $\lim \limits_{n \to \infty} 1^n = 1$ because every term in the sequence is equal to 1.
I think I've been reading too many of zylo's posts.

Thanks for the catch.

-Dan

 October 27th, 2016, 07:38 AM #15 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 a$\displaystyle _{\infty}$=1 if a$\displaystyle _{n}$= n/(1+n) but a$\displaystyle _{n}$ never equals 1. This isn't a paradox. a$\displaystyle _{\infty}$ is a limit, not a$\displaystyle _{\infty}$ evaluated at n=$\displaystyle \infty$. Last edited by zylo; October 27th, 2016 at 07:52 AM.
 October 27th, 2016, 09:23 AM #16 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra $a_\infty$ isn't standard notation. It is misleading too. The concept of evaluating $a_n$ at $n=\infty$ makes no sense because $\infty$ is not a number.
 October 27th, 2016, 05:43 PM #17 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 I would start with the function \displaystyle \begin{align*} f(x) = a^x \!\end{align*}, then \displaystyle \begin{align*} f'(x) &= \lim_{h \to 0} \frac{a^{x + h} - a^x}{h} \\ &= \lim_{h \to 0} \frac{a^x\,a^h - a^x}{h} \\ &= a^x\lim_{h \to 0} \frac{a^h - 1}{h} \end{align*} Now if we want \displaystyle \begin{align*} f(x) = f'(x) \end{align*}, in other words, for \displaystyle \begin{align*} a = \mathrm{e} \end{align*} that would require \displaystyle \begin{align*} \lim_{h \to 0} \frac{a^h - 1}{h} &= 1 \\ \lim_{h \to 0} \left( a^h - 1 \right) &= \lim_{h \to 0} h \\ \lim_{h \to 0} a^h &= \lim_{h \to 0} \left( 1 + h \right) \\ a &= \lim_{h \to 0} \left( 1 + h \right) ^{\frac{1}{h}} \end{align*} and if we define \displaystyle \begin{align*} n = \frac{1}{h} \end{align*} and note that as \displaystyle \begin{align*} h \to 0 , \, n \to \infty \end{align*}, this limit is equivalent to \displaystyle \begin{align*} \mathrm{e} = a = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) ^n \end{align*} Thanks from greg1313 Last edited by skipjack; October 28th, 2016 at 05:43 AM.
October 28th, 2016, 06:05 AM   #18
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Quote:
 Originally Posted by Prove It that would require \displaystyle \begin{align*} \lim_{h \to 0} \frac{a^h - 1}{h} &= 1 \\ \lim_{h \to 0} \left( a^h - 1 \right) &= \lim_{h \to 0} h \\ \lim_{h \to 0} a^h &= \lim_{h \to 0} \left( 1 + h \right) \\ a &= \lim_{h \to 0} \left( 1 + h \right) ^{\frac{1}{h}} \end{align*}
$\displaystyle lim\frac{f(x)}{g(x)} = \frac{limf(x)}{limg(x))}$ if lim g(x) $\displaystyle \neq$ 0

 October 28th, 2016, 07:30 AM #19 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra What are you trying to say? Or are you just partially quoting random results from the theory of limits?
 October 28th, 2016, 11:29 AM #20 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Ah! I see what you are trying to say... you think the derivation is incorrect. \begin{align*} \lim_{h \to 0} \frac{a^h - 1}{h} &= 1 \\ \left(\lim_{h \to 0} h\right) \left(\lim_{h \to 0} \frac{a^h - 1}{h}\right) = \lim_{h \to 0} \left(\cancel{h} \frac{a^h - 1}{\cancel{h}}\right) &= \lim_{h \to 0} h \\ \lim_{h \to 0} \left(a^h - 1\right) &= \lim_{h \to 0} h \\ \lim_{h \to 0} \left(1\right) + \lim_{h \to 0} \left(a^h - 1\right) &= \lim_{h \to 0} \left(1\right) + \lim_{h \to 0} \left(h\right) \\ \lim_{h \to 0} \left(1 + a^h - 1\right) = \lim_{h \to 0} a^h &= \lim_{h \to 0} \left(1+h\right) \\ \left(\lim_{h \to 0} a^h\right)^{\frac1h} &= \left(\lim_{h \to 0} 1+h\right)^{\frac1h} \\ \lim_{h \to 0} \left(a^h\right)^{\frac1h} = \lim_{h \to 0} a = a &= \lim_{h \to 0} \left(1+h\right)^{\frac1h} \\ \end{align*}

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