Exponentiation of real numbers to real exponents #
in progress
Statement #
For zero, if \(n>0\), then \(0^{n}=0\). If \(n<0\), then \(0^{n}\) is undefined.
For a positive base, exponentiation to a real power cannot make the number negative or introduce any imaginary component to the number.
\[ b^{m}\cdot b^{n}=b^{m+n} \]
\[\left(b^{m}\right)^{n}=b^{m\cdot n}\] \[b^{n}\cdot c^{n}=\left(b\cdot c\right)^{n}\]This means that for a rational number \(\frac{p}{q}\),
\[x^\frac pq= \left(x^p\right)^\frac 1q=(x^\frac 1q)^p\]\(x^{\frac{1}{n}}\) or \(\sqrt[n]x\) denotes the unique nonnegative real \(n\)th root of \(x\). Even if there are multiple \(n\)th roots of \(x\), there is only one value of \(x^{\frac{1}{n}}\).
Irrational exponents can be defined using limits of rational exponents (this is intuitive). A formal description of this is
\[b^x = \lim_{r (\in \mathbb{Q}) \to x} b^r \quad (b \in \mathbb{R}^+,\, x \in \mathbb{R})\] \[b^{-n}=\frac{1}{b^{n}}\]For a negative base, exponentiation to a real power can introduce imaginary components.
Exponent rules hold for integer exponents.
\[\left(\left(-1\right)^{\frac{1}{2}}\right)^{2}=-1\] \[\left(\left(-1\right)^{2}\right)^{\frac{1}{2}}=1\] \[\left(-1\right)^{2\cdot\frac{1}{2}}=-1\]\(\left(b^{m}\right)^{n}=b^{m\cdot n}\) does not hold. Instead, \(\left(b^{m}\right)^{n}\) must be calculated by first calculating \(b^{m}\), and then taking that result to the \(n\)th power.
\[b^{x}=\left(-b\right)^{x}\left(\cos\left(\pi x\right)+i\sin\left(\pi x\right)\right)\] \[b^{x}=\left(-b\right)^{x}\left(\cos\left(\pi x+2\pi kx\right)+i\sin\left(\pi x+2\pi kx\right)\right)\]Since \(-b\) is positive (\(-b=\left|b\right|\)), the rules above can be used for \(\left(-b\right)^{x}\).
Note that \(b\in\mathbb{R}\) if \(x\in\mathbb{Z}\), regardless of the choice of \(k\). \(\pi x+2\pi kx=\left(x+2kx\right)\pi\), and \(\left(x+2kx\right)\in\mathbb{Z}\). Therefore, \(\sin\left(\pi x+2\pi kx\right)=0\). We can reduce the expression to \(b^{x}=\left(-b\right)^{x}\cos\left(\pi x\right)\). This is what you would expect from repeated multiplication.
\(\left(b^{m}\right)^{n}=b^{m\cdot n}\) holds for complex base and real exp?
\(b^{n}\cdot c^{n}=\left(b\cdot c\right)^{n}\) holds for complex base and real exp?
\(b^{m}\cdot b^{n}=b^{m+n}\) does not hold?
principal root = greatest real part, then greatest imaginary if real tied