# Approximating $\log x$ by a sum of power functions $a x^b$

Let's approximate $$\log x$$ on the interval $$(0,1)$$ by a power function $$a x^b$$ to minimize the integral of the squared difference $$\delta_0(a,b)=\int_0^1\left(\log x-a x^b\right)^2dx.\tag1$$ It's easy to verify that the minimum is attained at $$a_0=-\frac34,\,b_0=-\frac13$$ that gives the approximation $$\log x=-\tfrac34x^{-1/3}+\mathcal R(x),\tag2$$ where $$\mathcal R(x)$$ is the error term. Now, let's again approximate $$\mathcal R(x)$$ by a power function $$a x^b$$ to minimize $$\delta_1(a,b)=\int_0^1\left(\mathcal R(x)-a x^b\right)^2dx=\int_0^1\left(\log x-\left(-\tfrac34x^{-1/3}+a x^b\right)\right)^2dx.\tag3$$ The minimum is attained at \begin{align}a_1&=\frac{17}4-\sqrt{58} \sin \left[\frac13 \arctan \left(\frac{433}{33\sqrt7}\right)\right]\approx0.88760008404...,\\b_1&=\frac{1}{3}+\frac{4}{3} \sqrt{2} \cos \left[\frac{1}{3} \arctan\left(\frac{\sqrt{7}}{11}\right)\right]\approx2.21311796239...,\end{align}\tag4 which are algebraic numbers of degree $$3^\dagger$$. If we repeat this process once again, we will get the next term $$a_2x^{b_2}$$, where $$a_2\approx-0.1406322691...,\, b_2\approx-0.2430593194...\tag5$$ are algebraic numbers of degree $$15^\ddagger$$, for which I don't know any closed form. The following steps will similarly produce pairs of algebraic numbers of higher degrees, resulting in an approximation of $$\log x$$ on the interval $$(0,1)$$ by a generalized power series $$\log x\approx-\tfrac34x^{-1/3}+a_1x^{b_1}+a_2x^{b_2}+\dots,\tag6$$ where each next term causes the integral of the squared error term to progressively decrease. The powers $$b_n$$ and coefficients $$a_n$$ are not monotone and do not exhibit any clear pattern (although the coefficients generally tend to decrease in absolute value, with some sporadic spikes).

Question: What does the series $$(6)$$ converge to? Does it converge to $$\log x$$ on any interval?

If it does converge to $$\log x$$, then empirically the convergence appears to be quite slow and erratic.

$${^\dagger}$$ The corresponding minimal polynomials are $$\small64 z^3-816 z^2+684 z-9$$ and $$\small9 z^3-9 z^2-21 z-7.$$
$${^\ddagger}$$ The corresponding minimal polynomials are $$\small5035261952 z^{15}+180729937920 z^{14}+19190513664 z^{13}-60948402536448 z^{12}-383744783499264 z^{11}+6281308897579008 z^{10}+50474690060451840 z^9-155303784466089984 z^8-1906255797863421024 z^7+805421030545306296 z^6+670389754270702752 z^5+127003127714790264 z^4+8514399973766202 z^3+130643635592430 z^2-127629387774 z-79827687$$
and
$$\small118098 z^{15}-1299078 z^{14}-15628302 z^{13}-52936335 z^{12}-55068660 z^{11}+119832291 z^{10}+512627130 z^9+898647291 z^8+984822786 z^7+742152591 z^6+396538632 z^5+150470676 z^4+39697272 z^3+6920496 z^2+715716 z+33172.$$
Although the polynomials look scary, they are quite nice in some sense, e.g. $$\small5035261952=2^{21}\cdot7^4$$ and $$\small79827687=3^8\cdot23^3,$$ and they also can be factored into quintics over some $$\mathbb Q[q]$$ with $$q$$ expressible in radicals.

• I guess it converges to $\log(x)$ in $L^2((0, 1))$ since even just the polynomials are dense in $L^2((0, 1))$. But to prove this you would need to connect your series expansion with orthogonal projections onto subspaces of the span of the power functions. Commented Jun 2 at 1:12
• Although a series of this form can converge to $\log x$, it doesn't obviously mean that this particular procedure will produce a series that converges to it, right? Commented Jun 2 at 19:30
• It could be interesting to post this problem in stats.stackexchange.com to see how they would react Commented Jun 3 at 9:37
• @ClaudeLeibovici How is it related to statistics? Commented Jun 3 at 19:08
• Frankly speaking, fitting residuals after each step does not make me very comfortable. Commented Jun 4 at 6:05

If we suppose we have already computed the first $$a_0, \ldots, a_{n-1}$$ and $$b_0, \ldots, b_{n-1} > -1$$, then we can derive the formula for $$a_n$$ and $$b_n$$ in terms of the preceding values. To that end, we first we compute the distance function with the additional term \begin{align} \delta &= \int_0^1 \left(\log x - \sum_{i=0}^n a_i x^{b_i} \right)^2 \text dx \\ &= \int_0^1 (\log x)^2 \text dx -2 \int_0^1 \log x \sum_{i=0}^n a_i x^{b_i} \text dx + \int_0^1\left(\sum_{i=0}^n a_i x^{b_i} \right)^2 \text dx \\ &= 2 -2 \sum_{i=0}^n a_i \int_0^1 \log(x) x^{b_i} \text dx + \sum_{i,j=0}^n a_i a_j \int_0^1x^{b_i + b_j} \text dx \\ &= 2 -2 \sum_{i=0}^n a_i \left(\underbrace{\left[\log(x) \frac{x^{b_i+1}}{b_i+1}\right]_0^1}_{=0} - \int_0^1 x^{-1} \frac{x^{b_i+1}}{b_i+1} \text dx \right) + \sum_{i,j=0}^n \frac{a_i a_j}{b_i+b_j+1} \\ &= 2 +2 \sum_{i=0}^n \frac{a_i}{(b_i+1)^2} + 2\sum_{i=0}^n \sum_{j=0}^{i-1} \frac{a_i a_j}{b_i+b_j+1} + \sum_{i=0}^n \frac{a_i^2}{2b_i+1}. \end{align} Then we compute the gradient \begin{align} \frac{\text d\delta}{\text da_n} = \frac{2}{(b_n+1)^2} + 2\sum_{j=0}^{n-1} \frac{a_j}{b_n+b_j+1} + 2 \frac{a_n}{2b_n+1}, \end{align} \begin{align} \frac{\text d\delta}{\text db_n} = -4\frac{a_n}{(b_n+1)^3} - 2a_n\sum_{j=0}^{n-1} \frac{a_j}{(b_n+b_j+1)^2} - 2 \frac{a_n^2}{(2b_n+1)^2}. \end{align} Equating the second equation with $$0$$ gives $$a_n = (2b_n+1)^2 \left( -\frac{2}{(b_n+1)^3} - \sum_{j=0}^{n-1} \frac{a_j}{(b_n+b_j+1)^2} \right)$$ Which we can plug in the first equation equated with $$0$$ (and divided by $$2$$) \begin{align} 0 &=\frac{1}{(b_n+1)^2} + \sum_{j=0}^{n-1} \frac{a_j}{b_n+b_j+1} + (2b_n+1) \left( -\frac{2}{(b_n+1)^3} - \sum_{j=0}^{n-1} \frac{a_j}{(b_n+b_j+1)^2} \right) \\ &= \frac{b_n+1 - 2(2b_n+1)}{(b_n+1)^3} + \sum_{j=0}^{n-1} \frac{a_j(b_n+b_j+1) - (2b_n+1)a_j}{(b_n+b_j+1)^2} \\ &= -\frac{3b_n+1}{(b_n+1)^3} + \sum_{j=0}^{n-1} \frac{a_j(b_j-b_n)}{(b_n+b_j+1)^2}. \end{align} This equation can of course be transformed into an equivalent polynomial equation in $$b_n$$. To find $$b_n$$, we need to solve the polynomial equation and then find the actual real root $$> -1$$ that minimizes the function, with the determinant of the hessian matrix criterion, for example.
For $$n=0$$, the condition $$\frac{3b_0+1}{(b_0+1)^3} = 0$$ gives $$b_0 = -1/3$$. Then $$a_0 = -2\frac{(2b_0 + 1)^2}{(b_0+1)^3} = -\frac{3}{4}.$$ For $$n=1$$, $$b_1$$ satisfies $$\frac{3b_1+1}{(b_1+1)^3} = \frac{a_0(b_0 - b_1)}{(b_0 + b_1 + 1)^2} \iff 27 b_1^4 - 18 b_1^3 - 72 b_1^2 - 42 b_1 - 7 = 0.$$ This gives four possible candidates for $$b_1$$ among $$\{-1/3, -0.735.., -0.478.., 2.21.. \}$$.
• I did not go intro thé détails of pour answer but this is not what thé OP wants to do. Could you show the résults of pour approach for $n=2$ or $n=3$ ? Thanks Commented Jun 8 at 13:48
• I think that I misexplained my point. What you do is exactly what I had in mind that is to say minimize the norm for all parameters at the time. What the OP seems to have in mind is to find the parameters $(a_i,,b_i)$ one at the time. Commented Jun 9 at 5:15