Why is the power function considered an algebraic function, but the exponential function is NOT algebraic?

Question

I am curious to know why if a mathematical expression contains an exponential function that expression may NOT be considered an algebraic expression, but if it contains a power function (if the variable is the base of a power expression) then that expression as a whole can be considered an algebraic expression.

Answer 1

An algebraic expression is made up of polynomials. In these expressions, no operations are used other than the ring operations of multiplication and addition. Thus $x^6$ is just $x\times x\times x\times x\times x\times x$. In order to express $6^x$ on the other hand, for a real variable $x$, we really need transcendental functions such as $\exp$ and $\log$.

The roots of polynomials are algebraic numbers; roots of transcendental functions may, in general, be transcendental numbers. As soon as we use transcendental functions we really are playing a different game. We are no longer simply working in a ring, but we are dealing with functions that we need convergence to describe. Thus, topology is now present, and it's no longer a purely algebraic system.

Answer 2

It is not clear what you had in mind.

But if $K = \mathbb{Q}, \mathbb{R}$ or $ \mathbb{C}$ then $K[x]$ is the ring of polynomial functions $P(x) = \sum_{n=0}^N a_n x^n$ and $K(x)$ is the field of rational functions $\frac{P(x)}{Q(x)}, P,Q \in K[x]$.

Then you can add more elements (such as $\sqrt[l]{x}$) to obtain the algebraic closure $\overline{K(x)}$, the field of functions $f$ satisfying a polynomial equation $\forall x,\sum_{n=0}^d P_n(x) f(x)^n=0, P_n \in K(x)$.

Also note $K(x)$ is closed under differentiation, but it doesn't contain any (non-zero) solution for $f' = cf , c \in K^*$ and more generally $\sum_{k=0}^M a_k f^{(k)} = 0$. Adding those yields a new set of functions, containing the trigonometric functions, the exponential being a solution of $f' = f$.

Answer 3

Algebraic expressions are those that only include algebraic operations, i.e. addition and multiplication, and when it makes sense, taking multiplicative inverses. Also, taking roots is still algebraic.

Thus, $x^n = x\cdot x\cdots x$ is algebraic expression.

Any polynomial $p(x) = \sum_{k=0}^n a_kx^k$ is algebraic expression.

However, $e^x$ for some real number $x$ is different kind of beast and there are many ways to define it, but I think the most illustrative way in this case would be to define it by series:

$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = \lim_{n\to\infty}\sum_{k=0}^{n}\frac{x^k}{k!}$$

which is no longer algebraic expression, but a limit of algebraic expressions. Limits are no longer algebraic constructs, but belong to topology instead and hence, exponential function is no longer algebraic, but analytic.

Answer 4

To add to the other answers, $x^\sqrt{2}$ where $x$ is a positive real is typically not considered algebraic because it is not obtainable by arithmetic and taking roots of polynomial equations. Note that $x^\sqrt{2} = \exp(\sqrt{2}\ln(x))$. So it is typically not true that "if the variable is the base of a power expression then that expression can be considered algebraic".