Fix real numbers $a_1$, $a_2$, $a_3$ and $b_1$, $b_2$, $b_3$ and $c_1$, $c_2$, $c_3$. Consider the equation $$ a_1\exp(b_1 (x-c_1)) + a_2\exp(b_2 (x-c_2)) = a_3\exp(b_3 (x-c_3)) $$ in $x$. My question is this: For what values of the constants $a_1$, $a_2$, $a_3$ and $b_1$, $b_2$ and $b_3$ and $c_1$, $c_2$ and $c_3$ does the above hold for all $x\in\mathbb{R}$? My guess would be that this is precisely when $c_1 = c_2 = c_3$ and $b_1 = b_2 = b_3$ and $a_1 + a_2 = a_3$. I would probably be able to prove this result (or whatever the true result is) by just looking at different cases, differentiating or the like and finding properties that differ for the function on the left-hand side and right-hand side.

However, it seems like a somewhat silly and clumsy way to proceed. Surely there is some "intrinsic", perhaps "algebraic" feature of the exponential function (and many other functions as well, presumably) which ensures that sums of exponentials of linear transforms of $x$ again are exponentials only in very particular cases?

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    $\begingroup$ Please see under Wronskian, and (for this type of situation) Vandermonde determinant. But that is for more general situations. In this case you evaluate at $0$, and first and second derivatives at $0$, reach conclusion. $\endgroup$ – André Nicolas Jul 22 '13 at 16:25
  • $\begingroup$ If for example $a_2=0$, we don't care what $b_2$ is. So assume all the $a_i$ are non-zero. Then indeed all the $b_i$ are equal, say $b$. That only forces $a_1 \exp(-bc_1)+a_2\exp(-bc_2)=a_3\exp(-bc_3)$, which can happen in lots of ways, $\endgroup$ – André Nicolas Jul 22 '13 at 17:27

Here's the way I would do it, using only algebra and basic properties of the exponential function, with no (explicit) use of calculus: start by setting $d_i = a_i \exp(-b_i c_i)$ for $i = 1, 2, 3$. Then your equation becomes

$d_1 \exp(b_1 x) + d_2 \exp(b_2 x) = d_3 \exp(b_3 x)$.

In this equation we can safely assume that $b_1 \ge b_2 \ge b_3$; since the $a_i$ may have any sign, we can re-arrange/re-name them any way we like. Now assume not all the $b_i$ are equal. Then we may write

$d_1 + d_2 \exp((b_2 - b_1)x) = d_3 \exp((b_3 - b_1)x)$.

Now if $b_1 > b_2, b_3$, so that $b_2 - b_1, b_3 - b_1 < 0$, we let $x \to \infty$ and conclude that $d_1 = 0$, hence $a_1 = 0$ and we obtain

$d_2 \exp((b_2 - b_1)x) = d_3 \exp((b_3 - b_1)x)$,

and multiplying through by $\exp(b_1x)$,

$d_2 \exp(b_2x) = d_3 \exp(b_3x)$.

Next we observe that either $d_2 = d_3 = 0$, or $d_2 \ne 0 \ne d_3$, since neither exponential factor can ever vanish. The former case corresponds to the trivial "solution" $a_1 = a_2 = a_3 = 0$; in the latter case we have

$d_2 = d_3\exp((b_3 - b_2)x)$,

which in turn forces $b_2 = b_3$ since otherwise the left-hand side is constant and the right-hand side is not! Of course then $d_2 = d_3$ as well.

Having dealt with the case $b_1 > b_2 \ge b_3$, and remembering our hypothesis that $b_1 \ge b_2 \ge b_3$, it only remains to deal with the case $b_1 = b_2 > b_3$. Then we find that

$d_1 \exp(b_1 x) + d_2 \exp(b_1 x) = d_3 \exp(b_3 x)$,


$d_1 + d_2 = d_3 \exp((b_3 - b_1)x)$,

and again the fact that $d_1 + d_2$ is constant but $d_3 \exp((b_3 - b_1)x)$ isn't creates a contradiction with the assertion $b_1 = b_2 > b_3$ unless of course $d_1 + d_2 = d_3 = 0$.

A brief summary of what has been accomplished so far, remembering that $b_1 \ge b_2 \ge b_3$, and excluding the case $a_1 = a_2 = a_3 = d_1 = d_2 = d_3 = 0$:

Case I: $b_1 > b_2 \ge b_3$: $d_1 = a_1 = 0$, $b_2 = b_3$, $d_2 = d_3$;

Case II: $b_1 = b_2 > b_3$: $d_1 + d_2 = d_3 = 0$;

Case III: $b_1 = b_2 = b_3$: $d_1 + d_2 = d_3$.

Under these circumstances the case $b_1 > b_2 > b_3$ is ruled out under Case I, since it forces $b_2 = b_3$. And of course in Case III we still have $a_1\exp(b_1c_1) + a_2\exp(b_1c_2) = a_3\exp(b_1c_3)$ which allows the values of the $a_i$ and $c_i$ to be traded off against each other to a certain extent. Similar trade-offs apply in some of the other cases as well; I leave it to the reader to discover these.

The key point here is that, unless the coefficients are very specifically constrained, the $b_i$ must all be the same.

I believe these results generalize to sums of more than two exponentials. The key to the analysis, without calculus, I believe, is to divide out by the factor $\exp(b_ix)$ with the largest (or smallest) $b_i$, and use properties of the $\exp(b_ix)$ as $x \to \pm \infty$ to force the values of certain of the other constants, as we did with the $a_i$, $d_i$.

The reason I said, at the beginning of this answer, is that there is "no (explicit) use of calculus" is that there is no differentiation, integration, etc., deployed in this answer. Of course, it could reasonably be argued that the definition of $\exp(b(x - c))$ itself involves arguments based on continuity, limits, and other topological properties of the real number system; but these are not quite calculus in my book.

To make more general progress, and/or to analyze similar statements for other functions, say $\sin(bx)$ etc., may very well require the machinery of differentiation, Wronksians and Vandermonde determinants, as Andre Nicolas mentioned in his comment.

  • $\begingroup$ You could probably prove a similar result for $2^{b(x-c)}$ with $x$ restricted to the rationals, and then there's not even any continuity involved in the definitions. $\endgroup$ – Gerry Myerson Jul 23 '13 at 9:05

Thanks for the answers so far. After looking at Lewis' post, I came up with the following (hopefully correct) argument, which is not really a solution to the original problem, but at least a solution to a related problem.

Consider a family $(p_i)$ of $n$ distinct polynomials and define $f_i(x)=\exp(p_i(x))$. We may then ask: Is the family $(f_i)$ linearly independent?

An answer to this would give a partial answer to the original question (it would not specify what linear independence implies about the polynomials, but it would specify one sufficient criterion for linear independence of exponentials of general polynomials, in particular the linear functions of the original question). As Andre Nicolas pointed out, Wronskians may be of use here. However, I'll try to make an argument which explicitly uses some of the properties of the exponential.

In order to prove that $(f_i)_{i\le n}$ is linearly independent, we must show that there exists no set of coefficients $\lambda_1,\ldots,\lambda_n$ with at least one nonzero coefficient such that $$ \sum_{i=1}^n \lambda_i f_i = 0. $$ As we need to prove linear independence for all sets of exponents of distinct polynomials, it suffices to consider the case where all the coefficients $\lambda_1,\ldots,\lambda_n$ are nonzero and obtain a contradiction in this case. Furthermore, note that for any polynomial $p(x)=\sum_{k=0}^m a_kx^k$, it holds that $$ \exp(p(x)) = \exp(a_0)\exp(p(x)-a_0), $$ so it suffices to consider the case where all the polynomials $p_i$ has no constant term.

Now letting $p(x)=\sum_{k=1}^m a_kx^k$ and $q(x)=\sum_{k=1}^m b_kx^k$ be two polynomials with no constant terms, define $\alpha(p,q)=\max\{1\le k\le m\mid a_k\neq b_k\}$. If the coefficient of the $\alpha(p,q)$'th term of $q$ is strictly less than that of $p$, we say that $q$ is strictly weaker than $p$, or that $p$ is strictly stronger than $q$, and write $q \prec p$. Furthermore, we write $q\preceq p$ when $q\prec p$ or $q=p$. It then holds that $q\prec p$ if and only if $\lim_{x\to\infty} p(x)-q(x)=\infty$. The relation $\preceq$ is a linear ordering. Therefore, among a family of distinct polynomials with no constant terms, there is always a unique polynomial which is strictly stronger than all the other polynomials. For the family $(p_i)$, let $i^*$ be the index of the unique strongest polynomial. We then have $$ \lambda_{i^*}+\sum_{i\neq i^*} \lambda_i \exp(p_i(x)-p_{i^*}(x))=0. $$ For any $i\neq i^*$, it then holds that $\lim_{x\to\infty}p_i(x)-p_{i^*}(x)=-\infty$. Therefore, the above implies $\lambda^*=0$, a contradiction.


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