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

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.

• Short answer, because that's the definition of an algebraic expression. You could define it to be something else, but then the same results would not hold for these "new algebraic expressions" without modifications. If that's your doubt, ask about the specific results, not the definition. – Federico Poloni Jul 17 '17 at 7:38

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.

• I edited exponent to 'base'. That is what i meant. – yoyo_fun Jul 16 '17 at 23:08
• That makes more sense. I edited my answer accordingly. – G Tony Jacobs Jul 16 '17 at 23:09
• "The roots of polynomials are algebraic numbers"... uhm, this seems to go directly against the Abel-Ruffini theorem, which "states that there is no algebraic solution to the general polynomial equations of degree five or higher with arbitrary coefficients"... maybe you meant something else... – Mehrdad Jul 17 '17 at 6:29
• @Mehrdad That use of the term "algebraic" is incorrect and should be changed. I notice that the article now reads that it should be understood as "solution by radicals", which is correct. – Ryan Reich Jul 17 '17 at 6:43
• @RyanReich: Ohh, I didn't realize an algebraic solution meant something different than a solution that is an algebraic number. Might be worth adding that to your answer in case someone else has the same confusion. – Mehrdad Jul 17 '17 at 6:47

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.

• what if the exponent of x would be a non-integer? The the expression would still be considered to be repetitive multiplication, even if we do not have an integer number at the exponent? – yoyo_fun Jul 16 '17 at 23:38
• @yoyo_fun, it would still be algebraic for rational exponent since $x^{p/q} = \sqrt[q] {x^p}$. – Ennar Jul 17 '17 at 0:01
• But, then your explanation is meaningless. $$\sqrt{x}=\sum_{n\geq0}{1/2\choose n}(x-1)^n$$ is analytic. – David Hill Jul 17 '17 at 0:38
• @David Hill, well you could have said the same for $1/(1-x)$, right? Algebraic function being analytic is not a bad thing, in fact it is the best kind of analytic functions. But, coming back to roots. They can be defined in purely algebraic terms, completely independent of the Taylor series. Exponential on the other hand cannot, since $e$ is transcedental. – Ennar Jul 17 '17 at 7:03

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 $f' = f$.

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".