# Why is the exponential function not in the subspace of all polynomials?

The exponential function can be written as

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots.$$

The subspace of all polynomials is $$\text{span}\{1, x,x^2, \dots \}$$

Sure $e^x$ is in this set?

• Any function in the subspace must be a linear combination of a finite number of basis elements. Commented Oct 6, 2016 at 11:53
• What Parcly wrote, and adding: what is true is that $\;e^x\;$ is in the closure (with respect to uniform convergence) of the span of the polynomials (in fact, any continuous function in a closed bounded interval) is. This is the famous Weierstrass approximation theorem. Commented Oct 6, 2016 at 11:56
• @DonAntonio The restriction of $e^x$ to a closed bounded interval is in the closure of $\mathbb{R}[x]$, but not $e^x$ itself. An experienced reader can deduce that from your comment in parentheses, but I'm not sure a beginner would get the correct idea from your comment. Commented Oct 6, 2016 at 13:43
• With your definition, for any $n$, i'ts not hard to deduce $\lim_{x\rightarrow\infty }{e^x\over x^n}=\infty$. This is not true for any given polynomial. Commented Oct 7, 2016 at 9:26
• Similarly, $\sqrt2 = 1 + 1/(2 + 1/(2 + 1/(2 +\cdots$, but we don't expect to find $\sqrt2$ in the rationals. Commented Oct 7, 2016 at 13:13

If $p$ is a polynomial of degree $n$, then the $n$th derivative of $p$ is constant. Note that the $n$th derivative of $e^x$ is $e^x$. Now all you have to do is prove that $e^x$ is not constant.

• I find it more natural to consider the $n+1$st, not only but also for avoiding to consider "the $0$th derivative" which might confuse somebody.
– quid
Commented Oct 6, 2016 at 14:30
• I think this is what the OP wanted, whereas the accepted answer, albeit true, doesn't really address the reason why $e^x$ isn't a polynomial.
– Vim
Commented Oct 6, 2016 at 14:52
• @Vim if you consider what a finite linear combination of the basis vectors looks like, then this answer is equivalent to the accepted one. Commented Oct 6, 2016 at 20:25
• (+1) In the same vein: the only polynomial equal to its own derivative is 0. Honestly, I don't think "this answer is equivalent to the accepted one". Realizing this doesn't require much "reading competence". (+ linear combinations are " finite" by definition) Commented Feb 15, 2023 at 5:23

The function $e^x$ is not in $\text{span}\{1, x,x^2, \dots \}$ because it is no finite linear combination of basis elements (but a countable one). What is true is that $$e^x \in \overline{\text{span}\{1, x,x^2, \dots \}}$$ is in the closure because you can find a sequence in $\text{span}\{1, x,x^2, \dots \}$ which converges to $e^x$. I hope it helps you :)

• What you mention about the closure is a bit unclear. You need to put a topology on the space of continuous functions to even speak about "closure" and "convergence". For example the restriction of $\exp$ to $[0,1]$ is in the closure of $\mathbb{R}[x]$ for the uniform convergence topology, as also mentioned by DonAntonio, but you really have to restrict. If you want you can also say that the sequence of polynomials converges pointwise, but that's not quite satisfactory either. Commented Oct 6, 2016 at 13:45
• @NajibIdrissi You are right. Thanks for pointing that out, I just tried to keep it brief. The closure I meant is respecting the space $(C(K), \Vert \cdot \Vert_\infty)$ for some compact $K \subset \mathbb R$ and that works just fine! Commented Oct 6, 2016 at 14:09
• or closure in formal power series ... :) Commented Oct 6, 2016 at 22:20
• Anyway, " it is no finite linear combination of basis elements" is a rewording of the statement, not a proof of it. Commented Feb 14, 2023 at 8:23
• @AnneBauval Awkward to answer a comment of a 7 year old question. Anywas, this question was about why $\mathrm e^x$ is not contained in the span of the monomials and that was just an explanation. I never claimed that my explanation was a proof, to begin with. Reading competence is also a valuable skill. Commented Feb 15, 2023 at 2:01

$\mathrm{span}(A)$ is the set of finite linear combinations of terms from $A$. Infinite sums require notions of limits and bring up issues of convergence radii (there are plenty of infinite polynomial that converge only at a single point).