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Definition of yx for irrational x
By Olof Sisask (P3033) on Thursday, April 26, 2001 - 05:54 pm:

I was just thinking, how is yx defined when x is irrational (i.e. you can't say it's the q-th root of y to the power of p, if x = p/q). Also, why is it that x1/q = the q-th root of x? It's probably defined that way, but you could define it however you wanted - what's the reasoning behind it? If that makes sense.

Regards,
Olof

By Dan Goodman (Dfmg2) on Thursday, April 26, 2001 - 06:03 pm:

Hi Olof, there's various ways of doing it (as always).

You can define yx to be ex log(y) which deals with rationals, irrationals, complex numbers, or whatever you like. If you define it that way you have to show that this corresponds to your definitions of yx when x is rational, but this isn't too difficult.

The other way of doing it is to define it as a limit. So, if the sequence of rationals pn/qn tends to x as n tends to infinity (and every irrational has such a sequence) then we define yx to be the limit of ypn/qn as n tends to infinity. The reason for defining x1/q to be the q-th root of x is because we want to preserve the idea that (xa)b=xab for all a and b. If we want this relationship to hold for all a and b, then it a=1/p and b=p we want (x1/p)p=x, which just says x1/p is the p-th root of x. Does that help?

By Olof Sisask (P3033) on Thursday, April 26, 2001 - 06:12 pm:

I see Dan, thanks a lot. I suppose that the relationship (xa)b=xab comes from logs? How would one show that log ab = log a + log b? Am I right in thinking this can be used to show that log an = n log a, which would then show (xa)b=xab?

Thanks again,
Olof

By Dan Goodman (Dfmg2) on Thursday, April 26, 2001 - 08:52 pm:

Well, (xa)b=xab can be proved for integer a,b without using logs, and this could be a reason for thinking that it should also be true if a and b are nonintegers.

Alternatively you can do it usings logs. You can prove that log ab = log a + log b by saying that ab=elog(ab), a=elog(a), b=elog(b), so elog(ab)=elog(a)elog(b)=elog(a)+log(b), taking logs we get log(ab)=log(a)+log(b).

There are various starting points for all of these things. If you've defined xy to be eylog(x) then obviously log(xy)=log(eylog(x))=ylog(x). However, if you've defined xy in another way you have to show that xy=eylog(x), and so on. We can prove (xa)b=xab for integer a and b just by noting that (xa)b=xaxa...xa (b times), and so =xa+a+...+a=xab.

By Ali Korotana (P4195) on Friday, April 27, 2001 - 12:09 am:

There is another way of proving that lnab = lna +lnb, which should show that log ab= loga +logb, though I haven't given this much thought.

It uses integration of the curve of y=1/x.
If we integrated between the limits of 1 and a, where a is greater than 1.

ò1 a (1/x)dx = lna = A1

Call this A1. b is another x-value greater than 1. We integrate between 1 and ab

ò1 ab (1/x)dx = lnab = A3

We call this A3 since the area between a and ab is A2. It can be seen that A1 + A2 = A3

We cannot simply integrate y between a and ab as our answer won't be in terms of b only.

If we use the substitution x=au, where u is a function of x then
dx= adu

The only way that x=au is if
a=1 and
ab=b

To find A2

òa ab(1/x)dx

Using the substitutions
a=1
ab=b
x=au
dx=adu

A2= ò1 b (1/au)adu = ò1 b (1/u)du =lnb

Since A1 + A2 = A3
then lna + lnb = lnab

By Olof Sisask (P3033) on Friday, April 27, 2001 - 04:31 pm:

Thanks Ali and Dan. I hadn't thought of it that way before Ali - thanks for pointing it out.

One last thing - is it as easy to show that xaxb = xa+b, for real a & b? I'll have a go in a moment.

/Olof

By Kerwin Hui (Kwkh2) on Friday, April 27, 2001 - 05:02 pm:

Just use the definition Dan gave.

Kerwin

By Olof Sisask (P3033) on Saturday, April 28, 2001 - 02:32 pm:

Do you do more rigorous proofs of this kind of thing at university?

By Dan Goodman (Dfmg2) on Saturday, April 28, 2001 - 03:05 pm:

Olof, yes, when proving it rigorously you have to worry about convergence, that's the main problem (which is true if you define yx as the limit of ypn/qn, or if you define yx as exlog(y), since you have to prove that exp and log are well defined functions, lots of work on power series and stuff).