Timm Ahrendt, Schnelle Berechnung der Exponentialfunktion auf hohe Genauigkeit, Berlin: Logos 1999, section 3.2.3 "Gekoppelte Newton-Iteration für Quadratwurzeln" has an elegant division-free coupled iteration for the computation of square roots, for which an unpublished manuscript by A. Schönhage is cited as the reference.

Given $r_{approx}$, a low-precision approximation to $a^{-\frac{1}{2}}$, compute $s_{0}=a*r_{approx}$ and $r_{0}=r_{approx}/2$ then iterate

$$s_{i+1} = s_{i} + r_{i} \cdot (a - s_{i} \cdot s_{i}) \\ \space \space \space \space \space \space \space \space r_{i+1} = r_{i} + r_{i} \cdot (1 - r_{i} \cdot 2 \cdot s_{i+1})$$

where the $s_{i}$ are approximations to $\sqrt{a}$ and the $r_{i}$ are approximations to $\frac{1}{2}a^{-\frac{1}{2}}$. The convergence is quadratic, as the method is related to the Newton iteration. For arbitrary-precision computation this coupled iteration has the advantage that to compute an $m$-bit square root, one only needs the reciprocal square root accurate to about $m/2$ bits. When using binary floating-point computation, this scheme allow for the efficient use of fused multiply-add, with the $i$-th step requiring four FMA operations plus a doubling, which is exact.

I have searched the literature for a similar division-free coupled scheme for real-valued cube roots, but have come up empty-handed. I looked for papers addressing cube roots in particular, and also for the more general case of $n$-th root.

Are any division-free coupled iterations for cube roots available in the literature? The convergence should be at least quadratic. Particularly useful would be schemes that lend themselves naturally to the use of fused multiply-add (FMA), for maximum performance on modern hardware platforms.

  • $\begingroup$ Solutions to polynomials can be found with companion matrices expressed as fractions of two numbers (how they may be represented) and since cube root of $k$ is a solution to $x^3-k=0$ it could be useful. And yes it can utilize FMA. $\endgroup$ – mathreadler Mar 10 '17 at 20:10
  • $\begingroup$ @mathreadler I am not familiar with companion matrices. If you could provide an answer that shows how they can be used to construct a division-free iteration for the cube root (starting with a low-precision approximation to either $a^{\frac{1}{3}}$ or $a^{-\frac{1}{3}}$), that would be helpful. $\endgroup$ – njuffa Mar 10 '17 at 20:14
  • $\begingroup$ Translation of the title of the article: "Quick computation of exponential function with high precision" and of the praragraph : "coupled Newton iteration for square roots" $\endgroup$ – Jean Marie Mar 11 '17 at 13:38

For third order roots solving $f(x)=x^3-a$ you would have to approximate the inverse of the derivative $f'(x)^{-1}=\frac13x^{-2}$ with the sequence $r_i$.

So in analogy let $r_{approx}$ be a coarse approximation of $a^{-\frac23}$, set $s_0=a·r_{approx}$, $r_0=\frac13·r_{approx}$ and iterate $$ s_{i+1}=s_i+r_i·(a-s_i^3) $$ and compute $r_{i+1}$ as approximate solution to $$1=3r_{i+1}s_{i+1}^2=3(r_i+Δr_i)s_{i+1}^2$$ so that $3r_is_{i+1}^2=1-3Δr_is_{i+1}^2$ where the right side can be approximativly inverted to $$ 1=(1+3Δr_is_{i+1}^2)3r_is_{i+1}^2 $$ so that an approximate solution of quadratic error in similar form to the quadratic situation is \begin{align} r_{i+1}&=r_i(1+3Δr_is_{i+1}^2) \\ &=r_i+r_i·(1-3r_is_{i+1}^2) \end{align}

In general, $$ f'(s_{i+1})^{-1}=\frac{r_i}{1-[1-r_if'(s_{i+1})]} =r_i·(1+[1-r_if'(s_{i+1})]+O([1-r_if'(s_{i+1})]^2) $$ suggests the update of the approximate inverse derivative as $$r_{i+1}=2r_i-r_i^2f'(s_{i+1}).$$

  • $\begingroup$ I tried the cube root iteration and it seems to work just fine, with some minor issues caused by additional rounding error (as compared to the square root case), because the computation does not map perfectly to FMAs. If no better answer comes along in the next couple of days, I will happily accept this one. $\endgroup$ – njuffa Mar 10 '17 at 21:29
  • $\begingroup$ You will always in some way have to include the current error, i.e., the function value, in the computation of the next step, i.e., you can not avoid the general form $x_{k+1}=x_k-g(x_k)f(x_k)$ of the fixed point iteration. Which implies third powers or at least 3 term products for the cubic root. $\endgroup$ – Lutz Lehmann Mar 11 '17 at 10:05

You can use companion matrices of polynomials. Here is one for 3rd degree:

$${\bf C} =\left[\begin{array}{ccc}0&0&-c_0\\1&0&-c_1\\0&1&-c_2\end{array}\right]$$

It's a monic polynomial so $c_3$ is supposed be 1 and we are interested in solving the equation $$\sum_k c_kx^k = 0$$

If we set $c_0=a,c_1=0,c_2=0$ we see that we get to solve $x^3 - a = 0$ and $x=\sqrt[3]{a}$ is the real root.

Now eigenvalues of this matrix are the roots. The largest modulus eigenvalue is the one which will do best under multiplication so we will need to encourage the real eigenvalue a bit more than the complex conjugate ones, we can do this by adding $\lambda\bf I$ to the matrix. This will move all eigenvalues $+\lambda$ along the real line and ensure our real eigenvalue will be largest in modulus.

$${\bf C}+\lambda{\bf I} = \left[\begin{array}{ccc}\lambda&0&666\\1&\lambda&0\\0&1&\lambda\end{array}\right]$$

If we start with vector $[1,1,1]^T$ and calculate $({\bf C}+9{\bf I})^{16}[1,1,1]^T$ the mean value of the quotient is 8.73289174558397, which powered by three is 666.000004672982. One correct decimal digit every two iterations.

Row 1 is two multiplications and one addition. Other rows can be fused multiply+add or if we select $\lambda = 2^k$ even a bit-shift+add will do. Rather parallellizable and the only time division could come into play is last step when we need to compare two iterates to measure the eigenvalue.

Convergence plot when we use $+9{\bf I}$:

  • x is iteration,
  • y is $\log_2\left(\epsilon + \frac{|est-true|}{true}\right)$

enter image description here So convergence is a line on a log scale. Averaging about 2 new bits of precision per iteration. 32 bits of precision after 16 iterations starting with vector full of ones.

  • $\begingroup$ I am not following. So the start vector is ${[1,1,1]}^{T}$ regardless of $a$ and one iterates by continuing to square the matrix? What's the rate of convergence of that process? What does "mean value of the quotient" mean in this context? Your original $C$ shows $-c_{0}$ in the upper right corner, if $c_{0} = a = 233$, shouldn't $C + I$ show $-233$ in the upper right corner? $\endgroup$ – njuffa Mar 10 '17 at 23:28
  • 1
    $\begingroup$ The convergence is linear with rate $\frac{233^{1/3}}{1+233^{1/3}}$ or around $\frac67$, which is rather slow, one digit every 15 iterations. $\endgroup$ – Lutz Lehmann Mar 11 '17 at 0:19
  • $\begingroup$ Convergence is a line on a log-scale, averaging 2 new bits of precision per iteration for the example above. But there probably exist smart things we can combine it with... $\endgroup$ – mathreadler Mar 11 '17 at 12:10
  • $\begingroup$ [+1] Very thorough explanation. $\endgroup$ – Jean Marie Mar 11 '17 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.