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Prove or disprove that

$$ f(x)=\left(\sum_i a_i x^i\right)\left(\sum_j b_j e^{-\lambda_j x}\right) $$ where $\forall i, a_i>0$, $\forall j, b_j>0,\lambda_j>0$, and both sums are finite,

has a unique maximum on $[0,+\infty)$.

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2 Answers

up vote 2 down vote accepted

@DejanGovc has already succinctly disproved the statement by counterexample. This little exposition concerns itself with some concepts, tips and techniques for treating any particular case you might encounter, and how to know whether it has extrema and how to find them with the least pain. Also, I consider the more general case that $p(x)$ is any real polynomial, and I provide an explicit example of a function $f$ with any number $k$ of distinct points having maxima (in fact, the same global maximum).

If there are several exponential terms $b_j\,e^{-\lambda_j\,x}$, the term with the lowest value of $\lambda_j$ will dominate the other terms as $x\rightarrow\infty$: the other terms will be negligable by comparison. So let us analyze the case of just a single exponential term $be^{-\lambda x}$, the dominant term, in depth, and assume for the moment that the dominated terms are left out. Since we thus only have one exponential term, let us assume without loss of generality that its coefficient is $b=1$ (this loses no generality because we could multiply it into $p(x)$ as a factor.)

If $f(x)=p(x)e^{-\lambda x}$ for any polynomial $p(x)\in\mathbb{R}[x]$ with real coefficients (positive or negative), then $$ f\,'(x)=\left[p'(x)-\lambda p(x)\right]e^{-\lambda x} $$ so that $$ \text{sign}\left(f\,'(x)\right)= \text{sign}\left[p'(x)-\lambda p(x)\right]. $$ For $\lambda > 0$, we know the asymptotic behavior as $x\rightarrow\infty$: $f(x)\rightarrow0$. This is because the exponential function dominates (grows more quickly than) any polynomial. On the graph, this is a horizontal asymptote along the positive $x$ axis. The quantity $e^{-\lambda x}$ in the expression for $f(x)$ is also called a decay factor. Without this factor, we would just take the global maximum of the polynomial $p(x)$ on the positive real number line, which we could determine from the critical points of $p$ and a determination of the sign changes of $p'(x)$. The local maxima (and minima) will be at sign changes from $+$ to $-$ ($-$ to $+$), with the endpoint $x=0$ also needing to be checked. Then, evaluating the function $p(x)$ at each change in sign in $p'(x)$, we could determine the global extrema on $[0,\infty)$. For $f(x)$, the decay factor not only changes where the critical points lie (as described above), but it also discounts the values of $f(x)$, relative to $p(x)$, by the ever diminishing decay factor, making it harder and harder for (nonzero) local extrema to be global extrema as $x$ increases, since both will tend closer toward the asyptotic value of zero.

Some other observations that might be of use are that:

  • the zeros of $f\,'$ will include any roots of $p$ of multiplicity greater than unity, but will also include new roots lying between those of $p$ and those of $p'$ (does anyone know of a good citation for this?)
  • $f$ need not have any global maximum; for example, take $p(x)=-(x^2+1)$
  • $f$ has a global maximum iff it has a nonnegative value on $[0,\infty)$
  • $f$ need not have a unique global maximum; for example, take $p(x)=-\prod_{i=1}^k(x-x_i)^2$ for $k > 1$ distinct roots $\{x_i\}_{i=1}^k$. If you graph this, it lies in the fourth quadrant, with $k$ equal global maxima touching (and an asymptote along) the positive $x$-axis.

Lastly, note that, over over $\mathbb{C}$, every polynomial factors into linear factors, and that if $p(x)$ has real coefficients, then its complex roots must come in conjugate pairs of equal multiplicity (corresponding to powers of quadratic factors irreducible over $R$). Now if the factorization over $\mathbb{C}$ is $$ p(x)=\prod_{i=1}^k(x-x_i)^{n_i} $$ then $$ \frac{p'(x)}{p(x)}=\sum_{i=1}^k \frac{n_i}{x-x_i} $$ so that $$ \frac{f'(x)}{f(x)}= \frac{p'(x)}{p(x)}-\lambda= -\lambda+\sum_{i=1}^k \frac{n_i}{x-x_i}. $$ Expanding this partial fraction representation and finding the roots of the denominator will find the "new" roots of $f\,'$, or critical points of $f$, mentioned above. This should not be such a great surprise if one has done enough of these types of problems, and demonstrates the utility of the logarithmic derivative.

Going back to the general case of $d > 1$ exponential (decay) terms, let us suppose that $$ f(x)=p(x)g(x) \qquad\text{for}\qquad g(x)=\sum_{j=1}^d b_j\,e^{-\lambda_j\,x} $$ where $\lambda_1 < \cdots < \lambda_d$ (so that the first term of $g$ dominates for $x$ large enough). Then $$ g(x)=e^{-\lambda_1\,x}\sum_{j=1}^d b_j\,e^{\lambda_1-\lambda_j\,x} $$ where the first summation term is constant, and each subsequent term decays at a faster rate. Similarly, its derivative is $$ g'(x)=e^{-\lambda_1\,x}\sum_{j=1}^d b_j\,\lambda_j\,e^{\lambda_1-\lambda_j\,x} $$ and its logarithmic derivative is $$ \frac{g'}{g}= \frac{ \sum_{j=1}^d b_j\,\lambda_j\,e^{\lambda_1-\lambda_j\,x} }{ \sum_{j=1}^d b_j\, e^{\lambda_1-\lambda_j\,x} } \qquad\text{(which }\rightarrow\lambda_1\text{ as }x\rightarrow\infty\text{)}, $$ both of which are never zero when $\forall j~b_j > 0$. Finally, then, using the product rule $$f\,'=\left(pg\right)'=pg'+p'g\,,$$ the logarithmic derivative of $f$ in this general case is $$ \frac{f'}{f}= \frac{p'}{p}+ \frac{g'}{g}, $$ for whose roots we would require an iterative rootfinding technique such as Newton's method or whatever our computer has implemented. But we can say that these "new" roots will occur somewhere in the vicinity of wherever (i.e., whatever $x$ values make) $$ \frac{p'}{p}<-G \qquad\text{or}\qquad p'(x)+G\,p(x) < 0 \qquad\text{for}\qquad G=\inf\frac{g'}{g} > 0, $$ which can be found procedurally and graphically.

In the exceptional case that all $\lambda_i$ were integral multiples of $\lambda_1$, we would be able to considerably simplify this by observing that $g(x)=h\left(e^{-\lambda_1\,x}\right)$ for $h(t)=\sum_{j=1}^d b_j\,t^{\lambda_j/\lambda_1}$ a polynomial in $\mathbb{R}^+[t]$.

If all of this seems terribly capricious, please note that similar reasoning and techniques are routinely found in the context of Laplace transforms in signal processing and likelihood calculations in probability.

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Some experimenting with Wolfram|Alpha disproves the claim. For example, the function $f(x)=(1+3x^2+\frac{1}{2000}x^8)\mathrm{e}^{-x}$ has more than one local maximum. (As a simple calculation should show.)

Intuition suggests that a function of the form $f(x)=p(x)e^{-x}$ (where $p$ is a polynomial with nonnegative coefficients) may have any finite number of local maxima. (Except for $0$, as it always does have a maximum.) This is because $x^k$ will at first grow faster than the exponential function, so by multiplying it with a small positive number, we should be able to construct a function $x\mapsto a_k x^ke^{-x}$ that is very close to $0$ on some interval of the form $[0,A]$, then achieves a maximum and then returns close to zero again. So by summing such functions we should be able to achieve any positive number of maxima we like. (I haven't written the proof of this though, this is just an idea.)

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