Minimizing a function (with boundary conditions)

Question

I tried the following thing just for fun (so it does not have a deeper sense), but unfortunately i failed to solve it.

Assume this function is given: $$ f\left(x\right)=\sum_{k=1}^{n}{a_kx^{2k}}=a_1x^2+a_2x^4+a_3x^6+...+a_nx^{2n} $$

With these bounardy conditions: $$ f(0)=0\\ f(n)=n $$

Of course the first condition is always true by definition for the given function / polynomial.

$a_k$ should be chosen to minimize this expression (which equals to the length of the function from $0$ to $n$): $$ \int\limits_0^n \sqrt{1+\left(f'(x)\right)^2}\mathrm{d}x $$

Is there a way to do this?

Thank you very much

Best regards

Kevin

Answer 1

Edit 20151230

I've been studying the root structure of the polynomials we're creating. (Perhaps it will be easier to bound the roots and hence find bounds on the arc length? Alternatively, it may be faster/easier to find the roots numerically than find the coefficients numerically.) What I've seen:

• The roots of $f_n$ are not bad starting points for the roots of $f_{n+2}$.
• There tends to be a root at each of $\pm n$. This is strongly true for odd $n$. For even $n$, sometimes there is a conjugate pair very near each of $\pm n$. (An example of conjugate pairs is $n=10$.)
• If $\alpha \in \Bbb{R} \setminus\{0\}$ is a root, so is $-\alpha$. Similar reflection holds for purely imaginary roots. (Only example of the latter seems to be $n=10$ having roots $\pm 45.230...\,\mathrm{i}$.)
• If $\alpha + \mathrm{i}\beta \in \Bbb{C} \setminus \Bbb{R}$ is a root, so are $\pm \alpha \pm \mathrm{i}\beta$. (This is a good thing... Searching over roots should involve searching over a space of half the dimension. (There are two coordinates in each complex root, so we go from four coefficients to two coordinates instead of to one coordinate, so the reduction is only a factor of two.))
• For square $n$, the roots of $f_n$ tend to lie very near a circle (except for the real roots). A near-circle of roots appears sporadically for other $n$. I don't recognize a pattern in the radii of the circles.

From the theory of (complex) holomorphic functions, we should never expect the arc length to be particularly good. We are attempting to approximate $x$ with $x^2 f(x^2)$ for some truncated power series $f$. That is, we wish to find $f$ such that $f(x^2) = \frac{1}{x}$. (The way the problem is set, we would expand the RHS in powers around $n/2$ so that the disk of convergence includes all of $[0,n]$.) Unfortunately, this gives coefficients for every degree, all of the same general magnitude. We then deform this by suppressing the terms of odd degree, but this deformation is so large that almost anything can happen.

A table of small arclengths:$$ \begin{align} \mathbf{n} && \mathbf{n \sqrt{2}} && \textbf{arclength} && \textbf{excess}\\ 1 && 1.41421... && 1.47894... && 0.0647293... \\ 2 && 2.82843... && 2.90038... && 0.0719484... \\ 3 && 4.24264... && 4.31624... && 0.0735977... \\ 4 && 5.65685... && 5.73166... && 0.0748082... \\ 5 && 7.07107... && 7.14639... && 0.0753231... \\ 6 && 8.48528... && 8.56107... && 0.0757936... \\ 7 && 9.89949... && 9.97557... && 0.0760791... \\ 8 && 11.31370... && 11.38999... && 0.0762907... \\ \end{align}$$ We can see the excess slowly increasing as $n$ increases. I haven't seen any evidence that this reverses. (Although numerically integrating the arc length for large $n$ is slow, so I haven't spent the time for $n>10$ and I'm not convinced I have even one significant digit in the $n \in \{9,10\}$ arclengths I have.)

Original Answer

This seems hopeless analytically. Let's define $f_n(x) = \sum_{k=1}^n a_k x^{2k}$, so all the functions of interest have distinct names. Require $f_n(0) = 0$ (trivial) and $f_n(n) = n$. Then...

• $f_1(1) = 1 = a_1 (1)^2$, so $a_1 = 1$. That was easy...
• $f_2(2) = 2 = a_1 (4) + a_2 (16)$, so $a_2 = \frac{1}{8} - \frac{a_1}{4}$. Then we want to minimize $\int_0^2 \frac{1}{2} \sqrt{4 a_1^2 x^6-4 a_1 x^6-16 a_1^2 x^4+8 a_1 x^4+16 a_1^2 x^2+x^6+4} \,\mathrm{d} x$. This integral seems to resist Mathematica. If we try to minimize by differentiating and setting equal to zero, we might try (without bothering to justify) interchanging the order of integration and differentiation, yielding the much more tractable (heh) $$\int_0^2 \frac{8 a_1 x^6-32 a_1 x^4+32 a_1 x^2-4 x^6+8 x^4}{4 \sqrt{4 a_1^2 x^6-4 a_1 x^6-16 a_1^2 x^4+8 a_1 x^4+16 a_1^2 x^2+x^6+4}} \,\mathrm{d}x, $$ which also resists Mathematica. However, numerically, we find $a_1 = 0.8365528...$ and $a_2 = -0.08413821...$. Further, this curve stays within $0.3115852...$ of the line $y=x$.
• For $f_3$, after eliminating $a_3$, the integral we wish we could attack is $$\int_0^3 \frac{1}{81} \left(36 a_1^2 x^{10}+2916 a_2^2 x^{10}-24 a_1 \ x^{10}+648 a_1 a_2 x^{10}-216 a_2 x^{10}-34992 a_2^2 x^8-3888 a_1 a_2 \ x^8+1296 a_2 x^8-1944 a_1^2 x^6+104976 a_2^2 x^6+648 a_1 x^6-17496 \ a_1 a_2 x^6+104976 a_1 a_2 x^4+26244 a_1^2 x^2+4 x^{10}+6561 \right)^{1/2} \,\mathrm{d}x,$$ which is also hopeless. Numerically, we get $$\begin{align} a_1 &= 0.7923981... \\ a_2 &= -0.09279764... \\ a_3 &= 0.004643382... \end{align}$$ with maximum deviation $0.3335419...$
• For $f_4$: $$\begin{align} a_1 &= 0.7641241... \\ a_2 &= -0.09148441... \\ a_3 &= 0.006028571... \\ a_4 &= -0.0001449432... \end{align}$$ with maximum deviation $0.3480680...$.

I did look at applying the Euler-Lagrange equations (random reference), but did not find a tractable way to attack the resulting system. I'd be happy to see someone else make some headway by this method.