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

Question

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).

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

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

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_{${^\dagger}$ 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.}

Answer 1

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.

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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 $\{-1/3, -0.735.., -0.478.., 2.21.. \}$.