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April 2nd, 2018, 05:58 AM   #1
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Help on exponents.

How are non-natural exponents are defined?

example : 2^(sqrt5)
2.2.2.2.....(sqrt5 times) makes no sense.

(please give a detailed solution.)
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April 2nd, 2018, 06:11 AM   #2
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Originally Posted by ThreXys View Post
How are non-natural exponents are defined?

example : 2^(sqrt5)
2.2.2.2.....(sqrt5 times) makes no sense.

(please give a detailed solution.)
$\displaystyle \sqrt{5} \approx 2.236$

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
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April 2nd, 2018, 06:25 AM   #3
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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.
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Last edited by JeffM1; April 2nd, 2018 at 06:36 AM.
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April 2nd, 2018, 09:13 PM   #4
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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 $\mathbb{C}$.
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April 3rd, 2018, 01:44 PM   #5
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Originally Posted by SDK View Post
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 $\mathbb{C}$.
I'm perhaps caught in a notation problem. Shouldn't we be using something on the order of
$\displaystyle \prod _{n \to \infty} 2^{a_n/10^n}$ 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.
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April 3rd, 2018, 02:36 PM   #6
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$2^{\sqrt5} = e^{\sqrt5\ln2}$
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April 3rd, 2018, 03:21 PM   #7
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Originally Posted by skipjack View Post
$2^{\sqrt5} = e^{\sqrt5\ln2}$
I had a brain fart with the whole Cauchy sequence thing. I understand the notation now.

Thanks!

-Dan
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