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I'm studying Calculus and I've stumbled across a concept I have some difficulties in fully grasping. That is "real powers".

I don't understand the theory behind it and I think I don't understand very well how to practically UTILIZE such concepts.

THEORY

My book (Bertsch's Analisi matematica - Seconda edizione by McGraw Hill - it's an Italian book) claims that real powers such as: $$ y=a^b $$ are well-defined only when $a\ge0$.

I can see this is because if you're operating in the Reals then you shouldn't do stuff like $(-3)^\frac24$ or $(-5)^\frac12$. But it still makes little sense to me. I mean, there are $a$'s smaller than zero for which $y=a^x$ is defined. Just think of $(-7)^3$, for example, which equals $-343$. Doesn't this make the definition inadequate?

If anyone could clarify on that, I'd be extremely thankful to him/her. If you can, PLEASE, also provide references and such. It's not that I don't trust you or something, I just want to know which books/websites are the good ones.

PRACTICE

Let's say I'm doing my homework and there's an exercise which asks me to graph some function like this: $$y=\ln(x^2+3)$$ the first thing I'd usually do is finding the domain. So I consider all the restrictions typical of the function I'm handling (in this case, being a logarithm, I impose that the argument must be greater than zero) and what I get is: $\mathscr{D}_f : x<-\sqrt3 \cup x>\sqrt3$.

Now if, hypothetically, my function was: $$y=\frac{x^3 + x^2}{9x}$$ shouldn't I also consider $x>0$ as a restriction for my domain? Because, after all, $x^3$, $x^2$ and even $x$ are all forms of $y=a^b$ where $a=x$ and, respectively, $b=3$, $b=2$ and $b=1$.

What if I had a function like this then? $$ y=x^{\ln{(x^2+3)}} $$ is this one only defined when $x\ge0$? (aside from all the other conditions...)

And this? $$ y=x^x $$ is it only defined when $x\ge0$? but what if $x$ was $-5$? wouldn't then $y$ be defined and simply be $-0.00032$?

I could go on for hours with countless examples. But I suspect you might have gotten my point. Please excuse my poor English, I'm not a native speaker.

Thank you for your time.

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user2397616
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    You are absolutely right. It is not true that the power $a^b$ is only well-defined when $a\geq 0$. You can ascribe a well-defined value to $a^b$ for any $a$ when $b$ is a positive integer, for example; in fact, you can extend the function far more than that. But you lose a lot of its nice properties (e.g. $a^{b_1} a^{b_2} = a^{b_1+b_2}$) when you do so. It's fairly safe to say that the power $a^b$ is only defined for all $b$ and in a nice way when $a\geq 0$. – Billy Jul 27 '13 at 16:44
  • Your example with $\log(x^2+3)$ doesn't quite work. You probably meant $\log(x^2-3)$. As to the definition of exponentiation, it depends on context. If we are working purely in the integers, there is no problem in assigning meaning to such things as $(-3)^{11}$. However, if we are working in the reals, we would like $(-3)^{11}$ to be real close to $(-3)^{11.00001}$, and then we run into trouble. – André Nicolas Jul 27 '13 at 16:45
  • In mathematics, it is acceptable for one definition to restrict itself to a particular context. We can simply provide additional definitions for other contexts (e.g. when $a\ne 0$ and $b$ is an integer), though we are duty-bound to ensure the definitions agree where they overlap. What would not be acceptable is an overextended definition that fails to define what happens in every case that $a<0$. – Erick Wong Jul 27 '13 at 16:49

3 Answers3

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The expression $a^b$ is well-defined if $b$ is an integer (with the exception of $a = 0$ and $ b \le 0$). Indeed, the whole concept of polynomial functions depends on this fact.

However, the more generalized definition of $a^b = e^{b \ln a}$ depends on $\ln a$ being well-defined, which is only the case for positive real numbers. The logarithm of negative (or complex) numbers requires the use of complex numbers and “branch cuts”.

Response to asker's comments: No, using powers of $e$ to define other powers isn't a circular definition. In calculus, $e^x$ (perhaps less-confusingly written here as $\exp(x)$) is typically defined by either:

  • The inverse of the natural logarithm function, which itself can be defined as $\ln{x} = \int_1^x \frac{1}{t} \mathrm{d}t$
  • The solution to the differential equation $f'(x) = f(x)$, $f(0) = 1$.
  • The Taylor series $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$

(Proof that the above are equivalent is left as an exercise for the reader.)

It happens that $\exp(0) = 1$, $\exp(1) = e$, $\exp(2) = e^2$, and in general $\exp(n) = e^n$ for $n \in \mathbb{Z}$ as defined by the familiar “repeated multiplication” definition of exponentiation. Using $\exp$ and $\ln$ to define powers allows exponentiation to be generalized to irrational exponents.

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Dan
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  • That's extremely interesting. Could you please point me to some place where I could find more info on this "generalized definition" you mention?

    On a side note. That poses another question. Why are we defining powers by using logarithms? I mean logarithms are, at least according to what I learned from high school, introduced as "some kind of opposite to powers". So it's like, we're defining powers through logarithms, which are defined through powers.

    I know I might seem like a bit of a drag but I really can't seem to wrap my head around this thing!

     – user2397616 Jul 29 '13 at 12:00
  • Also, if we define $a^b:=e^{b\ln{a}}$ we're using a power ($e^{b\ln{a}}$) to define another power! – user2397616 Jul 29 '13 at 12:21
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For the exponential function $\,f(x)=a^x\;$ to be properly defined in the whole real line we must have $\,a\ge 0\,$ ...

About the (real) logarithmic function: this is defined for positive reals, and since $\,x^2+3>0\;\;\forall\,x\in\Bbb R\,$ we have no problem here.

About $\,x^{\log(x^2+3)}\;$ : we must have here, again, $\,x>0\,$ for the same reason as above.

DonAntonio
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At some point, you must have defined $\exp$, either as $x\mapsto \sum\limits_{n=0}^{+\infty}\cfrac{x^n}{n!}$ or as the unique solution of $\left\{\begin{array}{l}y'=y\\y(0)=1\end{array}\right.$.

Then you have defined $\ln$ as the inverse function of $\exp$ or as $x\mapsto \int\limits_1^x\cfrac{1}{t}dt$.

Anyway all those definitions are equivalent.

Then you define $a^b := \exp (b\ln a)$. Then you let $e:= \exp(1)$ and you get that $e^x=\exp(x\ln e)=\exp x$. Then, the cool thing is that for integer values of $a$ and $b$, this new $a^b$ still is the one we knew. For example $2^3=\exp(3\ln 2)=\exp(\ln 2^3)=\exp(\ln 8)=8$ where the first $2^3$ is the "new" $2^3$.

So this is why we impose $a > 0$: We need $\ln a$ to be defined. But we had already defined $a^b$ for some $a$s and $b$s so we keep those too since there can't be a conflict. For example $(-1)^2=1$. We simply extended the domain of of $(a,b)\mapsto a^b$. It is possible to extend it again (with complex numbers) but you lose many properties, get multivalued functions etc.

xavierm02
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