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April 2nd, 2018, 05:58 AM  #1 
Newbie Joined: Apr 2018 From: India Posts: 4 Thanks: 0  Help on exponents.
How are nonnatural exponents are defined? example : 2^(sqrt5) 2.2.2.2.....(sqrt5 times) makes no sense. (please give a detailed solution.) 
April 2nd, 2018, 06:11 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,888 Thanks: 765 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
So $\displaystyle 2^{\sqrt{5}} \approx 2^2 \cdot 2^{2/10} \cdot 2^{3/100} \cdot 2^{6/1000}$ $\displaystyle = 2^2 \cdot \left ( 2^{1/10} \right )^2 \cdot \left ( 2^{1/100} \right )^3 \cdot \left ( 2^{1/1000} \right )^6$ So let's look at $\displaystyle 2^{1/10}$. This is a solution to the equation $\displaystyle x^{10} = 2$, which is $\displaystyle 2^{1/10} \approx 1.0718$. Thus $\displaystyle 2^{\sqrt{5}} \approx 2^2 \cdot (1.0718 )^2 \cdot \text{ ... }$ etc. Dan  
April 2nd, 2018, 06:25 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,127 Thanks: 468 
I have a decidedly minority view on irrational numbers. There is no operational meaning to irrational numbers in any physical sense. They are instruments of pure thought. They allow us to work without messy approximations. We posit a number x such that x * x = 5. Can we express that number in decimal form? No. But what we can do is to give successively better approximations in terms of rational numbers. $\dfrac{2}{1} < x < \dfrac{3}{1} \implies x \approx 2.5 \text { and } 2.5^2 = 6.25.$ Poor approximation. $\dfrac{22}{10} < x < \dfrac{23}{10} \implies x \approx 2.25 \text { and } 2.25^2 = 5.0625.$ Approximation off by about 1%. $\dfrac{223}{100} < x < \dfrac{224}{100} \implies x \approx 2.235 \text { and } 2.235^2 = 4.995225 \approx 5.$ A decent approximation. And we can keep doing that to get closer and closer to x. So basically, in my minority opinion, $\sqrt{5}$ is how we say a good enough approximation to x such that x > 0 and x * x = 5. What is amazing is that using these idealized numbers permits doing practical math much more easily than using good approximations. Last edited by JeffM1; April 2nd, 2018 at 06:36 AM. 
April 2nd, 2018, 09:13 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 444 Thanks: 254 Math Focus: Dynamical systems, analytic function theory, numerics 
The exponential function is the analytic continuation of the function $2^x$ for $x \in \mathbb{Q}^+$. A rigorous definition of $2^{\sqrt{5}}$ is \[2^{\sqrt{5}} = \lim_{n \to \infty} 2^{a_n} \] where $a_n \to \sqrt{5}$ is a Cauchy sequence of rationals. This limit can easily be shown to be well defined (i.e. it doesn't depend which Cauchy sequence you pick) and exists for all $x \in \mathbb{R}$. In fact, it extends as a complex analytic function on all of $\mathbb{C}$. 
April 3rd, 2018, 01:44 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,888 Thanks: 765 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \prod _{n \to \infty} 2^{a_n/10^n}$ rather than $\displaystyle \lim_{n \to \infty} 2^{a_n}$ or do they mean the same thing? Your notation would lead me to say that $\displaystyle \lim_{n \to \infty} 2^{a_n} \to 2^{a_{\infty}}$ which is, of course, a ridiculous statement. Dan Last edited by topsquark; April 3rd, 2018 at 01:48 PM.  
April 3rd, 2018, 02:36 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,546 Thanks: 1754 
$2^{\sqrt5} = e^{\sqrt5\ln2}$

April 3rd, 2018, 03:21 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,888 Thanks: 765 Math Focus: Wibbly wobbly timeywimey stuff.  

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