I am looking for an expansion of the exponential function (using a negative argument value) where each of the terms is an algebraic expression, and none of the terms are negative. That is, I would like an expansion of the form:

$$\exp(-x) = \sum_{n=0}^\infty f_n(x) \quad \quad \quad \text{for all } x>0,$$

where each term $f_n(x)$ is a non-negative algebraic expression. Obviously the Taylor expansion is no good because it has negative terms, so I am wondering if there is an alternative (preferably with a simple form). Is there an expression for the exponential function of this form?

  • $\begingroup$ Not sure about global convergence, but you could look into Pade approximation (not that it offers any guarantee about sign, but you have a shot). $\endgroup$ – Ian Mar 23 at 11:50
  • $\begingroup$ Why do you want $f_n$ to be non-negative algebraic expressions? It seems like an oddly specific restriction where some context might be useful. $\endgroup$ – Simply Beautiful Art Mar 23 at 15:12

This is not possible.

For $$ \mathrm{e}^{-x} = \sum_{n=0}^\infty f_n(x) \text{,} $$ where $f_n(x)$ is an algebraic expression in $x$ for each $n$ that is positive for all $x > 0$ requires that $f_n(x) \leq \mathrm{e}^{-x}$ for each $n$ and all $x > 0$. This is equivalent to $$ \frac{1}{f_n(x)} > \mathrm{e}^x \text{.} $$ Note that $\frac{1}{f_n(x)}$ is an algebraic expression. We will show that no algebraic expression in $x$ can grow as fast or faster than $\mathrm{e}^x$ on $x > 0$.

We define a function, $O$ on the set of algebraic expressions.

  • For any term, $a x^n$, with $a \in \Bbb{R} \smallsetminus \{0\}$ and $n \in \Bbb{Q}$, $O(a x^n) = n$. Note that this includes nonzero constants.
  • For any sum of algebraic expressions, $s+t$, $O(s+t) = \max\{O(s), O(t)\}$.
  • For any difference of algebraic expressions, $s-t$, $O(s-t) = \max\{O(s), O(t)\}$.
  • For any product of algebraic expressions, $s \cdot t$, $O(s \cdot t) = O(s) + O(t)$.
  • For any quotient of algebraic expressions, $s/t$, $O(s/t) = O(s) - O(t)$.
  • For any exponentiation by a rational number, $s^t$, $O(s^t) = t O(s)$.

(We do not concern ourselves with $O(0)$ because adding, subtracting, subtracting from, multiplying by, dividing by, and taking rational powers of zero are all trivial operations, so need not concern us in this problem.)

We claim that for any rational expression, $s(x)$, $O(s(x)) \in \Bbb{Q}$ and, there exists a positive real number, $M$, and a real number $x_0$ such that, for all $x \geq x_0$, $$ |s(x)| \leq M x^{O(s(x))} \text{.} $$ Why is this?

  • For any term, $a x^n$, take $M = a$ and observe that the claim follows for that term. In fact, we may write $a x^n = x^{O(a x^n)} \left( a x^0 \right)$. Notice that the power of $x$ inside the pair of parentheses is nonpositive.
  • Henceforth, we use "$(\dots)$" to represent a subexpression containing only nonpositive powers of $x$. For any sum or difference, $s \pm t$, suppose we have already written $s = x^{O(s)}(\dots)$ and $t = x^{O(t)}(\dots)$, then $$ s \pm t = x^{O(s \pm t)}\left( x^{O(s) - \max\{O(s), O(t)}(\dots) \pm x^{O(t) - \max\{O(s), O(t)}(\dots) \right) \text{.} $$ Notice that all powers of $x$ inside the outermost pair of parentheses are nonpositive.
  • For a product, $s \cdot t = x^{O(s)}(\dots) \cdot x^{O(t)}(\dots)$, $$ s \cdot t = x^{O(s \cdot t)}(\dots)(\dots) \text{.} $$
  • For a quotient, the analogous process as we just observed for multiplication occurs.
  • For a rational power, $s^t = \left( x^{O(s)}(\dots) \right)^t = x^{O(s^t)}(\dots)$.

We have shown that $O$ migrates out a power of $x$ to the front of an algebraic expression where the remainder of the expression contains only nonpositive powers of $x$. If we inspect the argument for the power that was moved out of subexpressions at each step, we discover that we have moved the smallest rational power of $x$ that leaves nonpositive powers of $x$.

The consequence is that any rational expression is equivalent to the product of a rational power of $x$ times an algebraic expression containing only nonpositive powers of $x$, where at least one of the powers of $x$ is zero.

Now we analyze the rational expression containing only nonpositive powers of $x$. In the interest of brevity, I summarize.

  • For a term, $a x^n$, keep the $|a|$.
  • For a sum or difference, $a \pm b$, keep $|a| + |b|$. By the triangle inequality, the sum of subexpressions having nonpositive powers of $x$ is bounded by this kept number on $x \geq 1$ (so we add the constraint $x_0 \geq 1$).
  • For a product, $a \cdot b$, keep $|a||b|$, a bound for the product of the subexpressions.
  • For a quotient, $a / b$, keep $|a/c|$, where $c$ is the minimum of set of kept numbers in the subexpression in the denominator of the quotient. (Note that $c \neq 0$.)
  • For a rational power, $a^b$, keep $|a|^b$.

The kept number from the subexpression containing only nonpositive powers of $x$ is a suitable choice for $M$.

But now we have a problem, $\mathrm{e}^x$ grows faster than any constant multiple of a power of $x$ and we have shown that any $\frac{1}{f_n}$ only grows as fast as a constant times a rational power of $x$. Consequently, there is an $x_1 \in (0,\infty)$ such that for all $x > \max \{1, x_0, x_1\}$, $\frac{1}{f_n(x)} < \mathrm{e}^x$, which means there is no algebraic expression, $f_n(x)$ satisfying $$ f_n(x) < \mathrm{e}^{-x} $$ for all $x \in (0,\infty)$. Relevant to the problem, there can be no sum of algebraic expressions, all positive on $(0, \infty)$, whose sum is as small as $\mathrm{e}^{-x}$, because no one of them can be that small.

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You can take the usual Taylor expression $e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!}$and group the terms. If $0 < x < 1$, the terms are decreasing in absolute value, so $$ e^{-x} = (1 - x ) + \left(\frac{x^2}{2} - \frac{x^3}{3!}\right) + ..$$ is an expression of all positive terms. If $x < -1$ you can do essentially the same thing but put more terms in the first bracket. Starting at $k > |x|$, the terms are decreasing, so if you group the first $k$ terms together, and then pair up the rest, you still get only positive terms.

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  • $\begingroup$ I believe that quarague is writing out a simplification of $\exp(-x)$, where $x > 0$, rather than a simplification of $\exp(x)$, where $x < 0$. $\endgroup$ – John Hughes Mar 23 at 13:14
  • $\begingroup$ @user It is, got confused with the negative x, should be correct now. $\endgroup$ – quarague Mar 23 at 13:15
  • $\begingroup$ -1: This doesn't answer the question, as your$f_n$ depends on $x$. $\endgroup$ – Martin Argerami Mar 24 at 7:08

Not possible. What you write is a power series, and if a power series converges to a function , it is identical to Maclaurin's series.

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  • $\begingroup$ I would assume that not all algebraic expressions can be written as a power series. $\endgroup$ – user Mar 23 at 19:34
  • $\begingroup$ @user, all "decent" functions (continuous, with derivatives of all orders over $\mathbb{C}$ at some point in $\mathbb{R}$) have Maclaurin (or Taylor) series, valid at least over an interval of the real numbers. $\endgroup$ – vonbrand Jul 22 at 16:25

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