
New Users Post up here and introduce yourself! 
 LinkBack  Thread Tools  Display Modes 
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: 1,763 Thanks: 706 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: 993 Thanks: 407 
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: 356 Thanks: 194 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: 1,763 Thanks: 706 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: 18,844 Thanks: 1566 
$2^{\sqrt5} = e^{\sqrt5\ln2}$

April 3rd, 2018, 03:21 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,763 Thanks: 706 Math Focus: Wibbly wobbly timeywimey stuff.  

Tags 
exponents 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
exponents  jeho  Algebra  2  October 13th, 2016 08:01 PM 
Exponents Law  Meezus  Algebra  1  October 25th, 2015 09:09 PM 
exponents  euphmorning  Algebra  5  April 29th, 2015 01:13 AM 
Exponents  Aman Verma  Algebra  6  December 28th, 2013 09:55 PM 
exponents  johnny  Algebra  2  August 1st, 2007 03:54 AM 