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For natural numbers $a,b$ the operation $a^b$ is defined as $a * a * a * a \cdots b$ times. However when you get to negative powers, complex powers, complex bases, polynomial expressions etc then this definition falls flat. So what exactly is exponentiation?

sai-kartik
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2 Answers2

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The basic idea is to take the concept we have, then keep extending it so as to generalize an identity. To wit:

  • The definition $a^0:=1,\,a^{b+1}:=a^ba$ obtains $a^b$ for integers $b\ge0$. It implies $a^{b+c}=a^ba^c$ for $b,\,c\ge0$, but we can extend this to any integers $b,\,c$ provided $a\ne0$ by saying $a^{-b}=1/a^b$.
  • We now have $(a^b)^c=a^{bc}$ for integers $b,\,c$, so to extend this to rationals we say $a^{p/q}=\sqrt[q]{a^p}$. However, if $a<0$ this only gets a real result if $q$ is odd when $p/q$ to be in its lowest terms. If on the other hand $a\ge0$, we've now defined $a^b$ for all $b\in\Bbb Q$.
  • We can handle arbitrary real $b$ by requiring $a^b$ to be a continuous function of $b$.
  • Defining the exponential function $\exp x$ as the solution of $\frac{dy}{dx}=y$ for which $y=1$ at $x=0$, and $\ln x$ as its inverse, we can verify the above all satisfy $a^b=\exp(b\ln a)$ if $a>0$. We can use this generalization to define $a^b$ for any nonzero complex $a$ and any complex $b$, but it becomes multi-valued.
J.G.
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Well the very basic definition of exponentiation defined by you is valid only for positive powers. Different exponents have different definitions.

When we talk about negative powers , the way negative powers is defined is basically like so : $$a^x\cdot a^y =1\ \ (\forall a>0)$$ $$a^{x+y}=a^0\ \ \text{ (law of exponents)}$$ $$x+y=0 $$ $$\Rightarrow y=-x$$ So basically what we can infer from here is that to define negative powers we define it as the reciprocal of positive exponents i.e. $$a^{-x}= \frac{1}{a^x}$$ Therefore to define negative powers we basically multiply the reciprocal of the positive power $x$ times.

sai-kartik
  • 2,090