Math: two exponent questions

[zev_steinhardt]

Two quick questions concerning exponents:

1) We all know the following to be true:


21 = 2
22 = 4
23 = 8
24 = 16
25 = 32

What is 24.5 and how is that calculated?

2) Why is X0 = 1 where X = any number?

Zev Steinhardt

[AWB]

Quick answer: 24.5
= 29/2
= (29)1/2
= sqrt(29)
= sqrt(512)
= 22.6274

[zev_steinhardt]

Originally posted by AWB
Quick answer: 24.5
= 29/2
= (29)1/2
= sqrt(29)
= sqrt(512)
= 22.6274

Thank you AWB. I never would have thought to figure it out that way. Do you know the answer to my second question?

Zev Steinhardt

[CalMeacham]

Zev:

You ought to be able to find the answers to your questions in any good advanced algebra text, or an introductory calculus book. I know it's in my copy of George B. Thomas' "Calculus".

For the first question, recall that squaring x to the n is the same as x to the 2 n. In other words, (x^n)^2 = x^2n. Taking the square root ought thus to be the same as dividing the exponent by 2: SQRT(x^n) = x^(n/2) In your case x^4.5 = SQRT(x^9)

Since (x^n)/(x^m) = x^(n-m), you can see that 1 = (x^n)/(x^n) = x^0

This should hold true for any x, with the possible exception of x = 0. However, you find that a lot of sequences and series can be written succinctly if you define 0^0 = 1. There is, in fact, a reference in a footnote in Thomas' book about this.

[KneadToKnow]

22=2*2*1
21=2*1
20=1

[Cantrip]

Originally posted by zev_steinhardt
Two quick questions concerning exponents:

1) We all know the following to be true:

21 = 2
22 = 4
23 = 8
24 = 16
25 = 32

What is 24.5 and how is that calculated?
Zev Steinhardt

Well, we know that xn*xm=xn+m. For example, 22*23=25=32, which also = 4*8.

So, 24.5=24*21/2, which equals 16*square root of 2.

I think the other one is definitional, although I'm sure there's a mathematical explanation. Try graphing it. (I always hated when my teachers told me that.) :)

[AWB]

1. Fractional exponents
If the exponent on a symbol x is a fraction p/q, then xp/q is defined as (x1/q)p, where x1/q is the positive qth root of x if x is positive, and the (negative) qth root if x is negative and q is odd. It follows that xp/q = (xp)1/q.

[AWB]

2) Zero Exponent
According to the 2nd law of exponents:

am/an = am-n

If both m and n are equal, say to p, then:

ap/ap = ap-p

The left side of this equation is 1, since a number divided by itself is 1. The right side becomes a0. So,

a0 = 1

[AWB]

Another way to figure non-integer exponential expressions:

Say you have an expresstion ay, where y is a non-integer expression (and perhaps not an easy decimal expression such as the OP's 4.5).

Let V = ay
ln V = ln ay
ln V = y ln a
eln V = ey ln a
V = ey ln a

[sailor]

If the limit of X^Y as Y approaches 0 from both sides is 1 then I think it makes sense to define X^0=1 (unless the electoral college has a better idea).

[scr4]

25 = 32
24 = 16
23 = 8
22 = 4
21 = 2

[/B]


20 = 1
2-1 = 1/21 = 1/2
2-2 = 1/22 = 1/4
2-3 = 1/23 = 1/8


This is not a proof, but I think it helps to show that 20=1 is natural, or at least makes sense.

[ZenBeam]

Just keep in mind that X^0 = 1 only as long as X is not zero. For example, in AWB's last two posts, if a = 0, he either divides by zero, or takes the log of zero, both of which are undefined.

[Chronos]

00 depends on the context in which you're asking. In most situations where it comes up, it makes things simpler to define it as being 1. For instance, lim(x --> +0) xx = 1 (I'm not sure about the limit from the left). However, there's no consistent way to define it for all cases.

[Joe_Cool]

Originally posted by AWB
Another way to figure non-integer exponential expressions:

Say you have an expresstion ay, where y is a non-integer expression (and perhaps not an easy decimal expression such as the OP's 4.5).

Let V = ay
ln V = ln ay
ln V = y ln a
eln V = ey ln a
V = ey ln a



AWB:
The problem with that explanation is that you're still left with raising a number (e) to a fractional power (a trancendental one, no less!). So basically you're stuck with a harder version of the same problem you started out with.

[DrMatrix]

Joe_Cool makes a excellent point. Let me see if I can simplify. (Is simplify the word I'm looking for?)

Exponentiation to a positive power is a compact way to represent repeated multiplication. Using this as a definition, only positive integral powers are defined. We notice that exponentiation follows the following rules:[list=1] a1 = a am * an = am+n (am)m = am*n am/an = am-n (for m > n, a not 0)[/list=1]The proofs are left as exercises for the reader.

We would like to extend the definition of exponentiation to allow for exponents that are not positive integers while still following the above rules. We can define exponentiation as an operation that follows the above rules (removing the restriction m>n in rule 4). It is convenient to require the base to be a positive number when using an exponent that is not a positive number. By applying rule 4 with m=n we have:
am/am = am-m
1 = a0

a1/2 = sqrt( (a1/2)2)
= sqrt(a1/2 * 2)
= sqrt(a)

Using this result;
24.5 = 29*1/2
= sqrt(29)
= sqrt(28*2)
= 24*sqrt(2)
= 16*sqrt(2)

This answers the OP.

The astute reader will notice that this definition of exponentiation (an operation that obeys rules 1-4) only defines rational powers. Another rule must be added to extend to irrational powers.

I've never used Preview Reply more times than for this post.

[egkelly]

What is i (square root of -1) to the ith power?

[ZenBeam]

What is i (square root of -1) to the ith power?
e^(-pi/2) = 0.20787957635076

[DrMatrix]

[b]ZenBeam[/i]

I'm not sure how one would define a non-positive real raised to a non-integral exponent, let alone an imaginary raised to an imaginary. A complex base may be raised to a positive integral (real) exponent, and a positive (real) number may be raised to a complex exponent.

How does one define a complex raised to a complex? I'm not saying you are incorrect, I just want to know where your answer came from since you didn't show your work. :)

[panamajack]

I think the point of AWB's analysis is that natural logs have been tabulated and are thus more easily computable than arbitrary roots (which is what you'd have to work out otherwise (see AWB's first post)). This is what calculators & log tables are for; if you want more precise answers, do the series.

And ZenBeam, I'm not sure on your answer; this is what I get :
Find X = jj
logjX = j (rewrite)
ln X / ln j = j (change of base)
ln X = j ln j (multiply by ln j)
ln X = j (pi/2) (ln j = pi/2 by Euler's Formula)
X = exp(j (pi/2)) (take exp() of both sides)

Am I missing something?

[DrMatrix]

Posted by panamajack
ln j = pi/2 by Euler's FormulaI don't know Euler's Formula off the top of my head, but the above implies (by raising both sides to e):
i = epi/2

:confused:

[ZenBeam]

I just want to know where your answer came from since you didn't show your work.

That's because I cheated. I plugged j^j into matlab, and backed out the e^(-pi/2) bit. Here's a derivation.

C = j^j
ln(C) = ln(j^j) = j * ln(j)
but
ln(j) = j*pi/2
so
ln(C) = -pi/2
C = exp(-pi/2)

panamajack, when I first derived it myself, I got the same answer as you. Maybe it's a branch cut issue?

[jcgmoi]

Take special case of Euler's formula:

ei*pi = -1

Take square root:

ei*pi/2 = i

Raise to power of i:

e(i*pi/2)*i = ii

So e-pi/2 = ii

That is, the imaginary unit i, raised to the power of i is a real number.

[panamajack]

I wrote :

Am I missing something?


Why yes, I am, one more j to be precise. I figured the angle for ln j but failed to use Euler's Formula correctly.

so ln j = j* pi/2 which yields the correct answer,
exp(-pi/2) for jj for the method I showed. Though jcgmoi did it cleanly (and right the first time).

Dr. Matrix, Euler's formula is :

ej*theta = cos(theta) + j sin(theta)

It actually is a pretty simple formula if you understand what's behind it, though I probably shouldn't try and explain it right now considering how I mishandled it earlier.

[RickG]

What is it with you engineers using j for the square root of -1? As all good mathematicians and physicists know, the correct notation is i.

Now off to Great Debates to reside with the Java/C Curly Bracket Position War. :)

Rick