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I want to minimize the function $$f(x) = e^{ax^2} + e^{b(1-x)}$$ with respect to $x$ (where $a$ and $b$ are constants), subject to the constraint that $0 \leq x \leq 1$.

I know that $x = 0$ and $x = 1$ are both critical points, but I am interested in finding the potential minimum which lies between the boundary values.

Surely the first step is to set the derivative equal to $0$, yielding $$0 = f'(x) = 2axe^{ax^2} - be^{b(1-x)}$$ The minimizing $x$ must solve $2axe^{ax^2} = be^{b(1-x)}$, but I'm not sure how to "use" the constraint $0 \leq x \leq 1$ to do so...

Note: In my specific problem, I have $a = -\frac{N}{16}$ and $b = -\frac{N}{4} (\frac{1}{2} - p)^2$, where $N$ is a constant positive integer, and $p$ is a constant with $0 \leq p < \frac{1}{2}$. Ideally, we will be able to express a minimizing $x$ in terms of $N$ and $p$.

jamaicanworm
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    The constraint will mean that the minimum might not be at a point where the derivative is $0$; it could be at one of the endpoints. Did the problem really say you should use the constraint to solve the equation that sets the derivative to $0$? – Michael Hardy Apr 01 '12 at 01:15
  • @MichaelHardy I know the endpoints can be minima/maxima, but I'm interested in finding the other candidate between the endpoints. – jamaicanworm Apr 01 '12 at 01:16
  • What am I supposed to do with the bounty if no one has yet answered the question? :) – jamaicanworm Apr 08 '12 at 19:10

2 Answers2

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If $b \neq 0$ and $2ax \geq b$ for all $0 \leq x \leq 1$ then $ax^2-bx+b$ is increasing on $0 \leq x \leq 1$ and $\frac{2a}{b} e^a \leq 1$. These statements imply that

$$ \frac{2ax}{b}e^{ax^2 - bx + b} < 1 $$

for $0 \leq x < 1$, which means that $f'(x) \neq 0$ there. Thus you may conclude that your minimum occurs at one of the endpoints.

What restrictions are there on $a$ and $b$? Perhaps similar arguments exist for other specific cases.

Edit: As I will probably not have enough time to work on this in the next 7 days I will sketch here what I was thinking about in the hopes it will benefit others. My goal was to put a lower bound on the minimum of the function $f(x) = e^{ax^2} + e^{b(1-x)}$ using the constraints for $a$ and $b$ in the note at the end of the question.

For $y > 0$, replacing

$$ e^{ax^2} \hspace{0.5cm} \text{with} \hspace{0.5cm} \frac{1}{y+e^{-ax^2}} $$

or

$$ e^{b(1-x)} \hspace{0.5cm} \text{with} \hspace{0.5cm} \frac{1}{y+e^{-b(1-x)}} $$

in $f(x)$ will yield a function $f_y(x)$ which is less than $f(x)$. Depending on which choice was made, $f_y'(x)$ will be either nonnegative or nonpositive for $y$ large enough.

To determine how large it needs to be, we start by solving $f_y'(x) = 0$ for $y$, which I believe will always reduce to a quadratic equation. These two roots of $f_y'(x) = 0$ are functions of $x$ which I think can always be made less than a sum of two exponentials. This sum will be bounded (since x is bounded), so we would like to find an upper bound (to elminate the $x$ dependence). Here I would have tried to use the inequality $e^x \leq 1/(1-x)$, which holds for $x<1$, to bound this sum of exponentials above by a sum of rational functions. Thus we end up wanting to solve something like

$$ y > \frac{1}{1-p(x)} + \frac{1}{1-q(x)} $$

where $p(x)$ and $q(x)$ are quartic (or less) and $0 \leq x \leq 1$. An upper bound of the right-hand side is theoretically possible to write down. Taking $y$ larger than this will make $f_y(x)$ monotonic, and hance the minimum of $f(x)$ will be greater than the lowest endpoint of $f_y(x)$, whose height can (theoretically) be calculated.

Antonio Vargas
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  • I do know $b \neq 0$, but I cannot guarantee that $2ax \geq b$. In this specific problem, I have $a = -\frac{N}{16}$ and $b = -\frac{N}{8} (1-p)^2$, where $N$ is a constant natural number and $p$ is a constant with $0 \leq p < \frac{1}{2}$. – jamaicanworm Apr 01 '12 at 16:20
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    @jamaicanworm There is definitely some tricky stuff happening there. I am able to bound $f(x)$ from above with those constants but not explicitly maximize or minimize it so far. – Antonio Vargas Apr 01 '12 at 17:23
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Alway plug in the boundary values. Optima occur among critical points; the oundary values are, by definition critical. Here $f(0)= e^b$, while $f(1)=e^a.$ To find your other critical points, try dividing one side by the other.

Scott Carter
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    Thanks! As I mentioned in a comment, I'm interested in the non-boundary value solution. When I divide, I get $\frac{2ax}{b}e^{ax^2 - bx + b} = 1$, but I'm not sure how that helps.. – jamaicanworm Apr 01 '12 at 02:35
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    I would try completing the square in the exponent, so you want to solve $(py+q)\exp(y^2) = 1$. Still a mess, but a simpler one. Maybe these is a generalization of the W function with $x \exp(x^2) = 1$. – marty cohen Apr 01 '12 at 03:24