# Approximate the square root of (1-X) efficiently through (nested) products

Currently, I encountered a problem of approximating the following series:

$$(I-X)^{-\frac{1}{2}}=I+\frac{1}{2}X+\frac{1\cdot3}{2\cdot4}X^{2}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}X^{3}+\ldots$$

where $X$ is a diagonalizable matrix and the largest (in absolute value) eigenvalue is less than $1-\frac{1}{\kappa},\kappa>1$. What I am looking for is something like the following

$$(1-X)^{-1}=I+X+X^{2}+X^{3}+\dots\approx\prod_{k=0}^{d}(1+X^{2^{k}})$$

With $O(\log(\kappa\log(1/\epsilon)))$ multiplications and summations, we can achieve accuracy of $O(\epsilon)$.

We can assume the following representations are cheap to get (we assume that their computational cost is $1$):

1. $1+aX^{n},a\in(-1,1),n\in[-1,0,1,\dots)$

2. Multiplication between any two cheap representations.

3. Summation between any two cheap representations.

So how can we approximate $(I-X)^{-\frac{1}{2}}$ efficiently?

My initial idea is

$$(I-X)^{-\frac{1}{2}}=I+\frac{1}{2}X(1+\frac{3}{4}X(1+\dots(1+\frac{2n-1}{2n}X(1-X)^{-1}))\dots)$$

But it is not efficient enough.

Not exactly the same thing, but you can use Newton's method easily in your framework. As you can see for instance here, Newton's method for the inverse square root can be computed with additions and multiplications only, and the same algebraic expression that works for scalars can be used for diagonalizable matrices without modifications (proof: in the eigenvector basis, doing the iteration on the matrix level is equivalent to running $n$ scalar iterations inside a diagonal matrix).
• Thank you very much for your answer. It partially solved my problem. Although the quadratic convergence of Newton's method is exactly what I am looking for, but expanding the Newton's update ($d$ steps,e.g., from $y_0=I$ to $y_1$ to $y_d$ ) will result in $2^d$ times matrix operations. The reason I don't want to store $y_1$ to $y_d$ is that the matrix multiplication is usually as expensive as calculating its SVD and sorry for not making this clear in the question. Your answer definitely gives me something to work on. – cdh Apr 4 '14 at 19:58
• @DehuaCheng I see. Don't you have the same problem in your example with $(I-X)^{-1}$, then? If you expand out all the products, it becomes $2^d$ operations again. – Federico Poloni Apr 5 '14 at 6:57
• Yes, you are right. In that inverse case, by applying sparsifier on $X^2,X^4,\dots,X^{2^d}$, we can actually afford these multiplications, so it takes $d$ multiplications to get $X^2,X^4,\dots,X^{2^d}$. In the inverse square root case, I am now trying to do the similar thing by sparsifying $S_n$ (assuming the update rule is $y_{n+1} = y_n S_n$). There are still difficulties, but it looks promising. – cdh Apr 6 '14 at 7:34