Irrational Indices

I was just thinking about powers, when suddenly I thought about the concept of irrational indices, which I found pretty confusing :s-smilie:

Does anyone know if/how something like this would work.

For example, take

\left( ( 2 )^{\sqrt 2} \right)^{\sqrt 2} = 2^{2}= 4

, so what would

2^{\sqrt 2}

be?

I got it as

\sqrt[\sqrt 2]{4} = 4

, which makes no sense to me :confused:

(edited 10 years ago)

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Reply 1

ThePerfectScore

What's sqrt(2)? If you understand what means, then 2^(sqrt(2)) is 2 raised to sqrt(2).

Reply 2

davros

Original post by majmuh24

I was just thinking about powers, when suddenly I thought about the concept of irrational indices, which I found pretty confusing :s-smilie:

Does anyone know if/how something like this would work.

For example, take

\left( ( 2 )^{\sqrt 2} \right)^{\sqrt 2} = 2^{2}= 4

, so what would

2^(\sqrt 2)

be?

I got it as

\sqrt[\sqrt 2]{4} = 4

, which makes no sense to me :confused:

There isn't an "intuitive" way of looking at irrational indices as far as I'm aware.

I can think of 2 ways of looking at something like this:

One way, if you know the decimal representation of your irrational, is to look at the sequence of numbers

2^{1.4}, 2^{1.41}, 2^{1.414},...

and see if it tends to a limit. If this limit exists, then it will be the number you are looking for.

Another way is to think about how we define

z^a

in general for a complex number z and real exponent a.

If we first define the exponential function exp(z) by its standard power series and also define what we mean by the logarithm log(z) of any complex number, then we can define:

z^a = exp(a log z)

which takes care of irrational indices as well as rational ones.

Reply 3

interstitial

Original post by ThePerfectScore

What's sqrt(2)? If you understand what means, then 2^(sqrt(2)) is 2 raised to sqrt(2).

What?

\sqrt 2 = 2^{\frac{1}{2}}

, but I fail to see how that helps :s-smilie:

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Reply 4

firegalley246

Original post by davros

2^{1.4}, 2^{1.41}, 2^{1.414},...

and see if it tends to a limit. If this limit exists, then it will be the number you are looking for.

Another way is to think about how we define

z^a

z^a = exp(a log z)

which takes care of irrational indices as well as rational ones.

This method with the exponentials and logarithms is really the way to go; though approximating the power with rationals does work (it does tend to a limit, by continuity).

Reply 5

ThePerfectScore

Original post by majmuh24

What?

\sqrt 2 = 2^{\frac{1}{2}}

, but I fail to see how that helps :s-smilie:

Posted from TSR Mobile

Excuse me, I was trying to help you there. :rolleyes:

Reply 6

interstitial

Original post by ThePerfectScore

Excuse me, I was trying to help you there. :rolleyes:

I can't see how that would help, I knew that bit already.

Posted from TSR Mobile

Reply 7

ThePerfectScore

Original post by majmuh24

I can't see how that would help, I knew that bit already.

Posted from TSR Mobile

So you know that

\sqrt{2}

2^{1/2}

. Okay, then what exactly is

2^{1/2}

Reply 8

interstitial

Original post by ThePerfectScore

So you know that

\sqrt{2}

2^{1/2}

. Okay, then what exactly is

2^{1/2}

As a decimal? 1.414....

Posted from TSR Mobile

Reply 9

FireGarden

Irrational powers don't 'make sense' as you might be trying to think of them. They are essentially a generalisation of rational powers that is sensible because the function

a^x

for fixed

a\in\mathbb{R}

is continuous.

The generalisation is then the definition

a^c = \displaystyle\lim_{x\to c} (a^x)

for any

c\in\mathbb{R}

This may not seem particularly nice and a bit abstract, but that's life in analysis; some generalisations of concepts need these kinds of definitions. Imagine if someone said to you "

2^{\pi}

is just 2 multiplied by itself pi times".

Reply 10

interstitial

Original post by FireGarden

Irrational powers don't 'make sense' as you might be trying to think of them. They are essentially a generalisation of rational powers that is sensible because the function

a^x

for fixed

a\in\mathbb{R}

is continuous.

The generalisation is then the definition

a^c = \displaystyle\lim_{x\to c} (a^x)

for any

c\in\mathbb{R}

This may not seem particularly nice and a bit abstract, but that's life in analysis; some generalisations of concepts need these kinds of definitions. Imagine if someone said to you "

2^{\pi}

is just 2 multiplied by itself pi times".

This whole question stems from something I recently saw, saying that if you raise an irrational number to an irrational power, the answer can also be rational.

Wouldn't the complex exponent definition that involved taking a power series for the general exponential function be a better way to go about this?

Also, would this definition apply for all

x \in \mathbb R

, for example negative numbers as well as positive ones? (I have a feeling that this would give either imaginary or complex values)

To expand on this, what about numbers in

a \in \mathbb C

, would this definition still be valid for imaginary or complex exponents?

(edited 10 years ago)

Reply 11

FireGarden

Original post by majmuh24

This whole question stems from something I recently saw, saying that if you raise an irrational number to an irrational power, the answer can also be rational.

Haha, I'd imagine that was my post in the favourite proofs thread!

Wouldn't the complex exponent definition that involved taking a power series for the general exponential function be a better way to go about this?

If you want to consider negative reals, then yeah; from the perspective of complex numbers, they have a non-zero argument, which means for exponents between integers the result is going to get stuck somewhere which has an imaginary part.

Also, would this definition apply for all

x \in \mathbb R

, for example negative numbers as well as positive ones? (I have a feeling that this would give either imaginary or complex values)

You'll get complex values for negative base numbers, and it won't work as a function

\mathbb{R}\mapsto\mathbb{R}

anymore.

To expand on this, what about numbers in

a \in \mathbb C

, would this definition still be valid for imaginary or complex exponents?

Hmm.. here I'm not so sure off the top of my head. There may be problems with the arguments of the complex numbers when we try to take limits. It is a better idea to use the power series for the complex exponential in this case even if it could work out, 'cause it'll be a lot simpler.

Reply 12

davros

Original post by firegalley246

This method with the exponentials and logarithms is really the way to go; though approximating the power with rationals does work (it does tend to a limit, by continuity).

Agreed.

The power series technique and sorting out a consistent definition of the logarithm takes more "machinery" to set up, but gives you lots of powerful results in return.

The limit idea was more my way of trying to motivate the result by suggesting that the original calculation still makes sense if you consider it as the limit of a series of rational powers that home in on the irrational.

Reply 13

firegalley246

Original post by davros

Out of interest, when/how did you come about using the limit idea (it is a good, low-level way of explaining it)? I was trying to tell some y9 students last summer about the number

e^{\pi \sqrt{163}}

and one of them asked about irrational powers - I told him to take the limit of rational powers, but I honestly can't remember when/how I learnt about this because I'm sure it never got taught when I was in university.

Reply 14

davros

Original post by firegalley246

Out of interest, when/how did you come about using the limit idea (it is a good, low-level way of explaining it)?

I honestly don't remember either.

I don't think it was something I was formally taught - probably something I absorbed from reading a lot of separate textbooks when I was a lot younger :smile:

Reply 15

Yung_ramanujan

Original post by firegalley246

Out of interest, when/how did you come about using the limit idea (it is a good, low-level way of explaining it)? I was trying to tell some y9 students last summer about the number

e^{\pi \sqrt{163}}

whats so special about

e^{\pi \sqrt{163}}

Reply 16

firegalley246

Original post by Yung_ramanujan

whats so special about

e^{\pi \sqrt{163}}

This link explains it a bit: http://mathoverflow.net/questions/4775/why-are-powers-of-exp-pi-sqrt163-almost-integers

Reply 17

Yung_ramanujan

Original post by firegalley246

This link explains it a bit: http://mathoverflow.net/questions/4775/why-are-powers-of-exp-pi-sqrt163-almost-integers

what does the function j(t) mean. i know tau is ramanujans tau constant, i am aware of it but dont really know anything about it. What is the meaning of j(t)?

Reply 18

firegalley246

Original post by Yung_ramanujan

what does the function j(t) mean. i know tau is ramanujans tau constant, i am aware of it but dont really know anything about it. What is the meaning of j(t)?