By defining these 3 matrices:

$\displaystyle M_1 = \begin{bmatrix} 1&0&0&0&0 \\ 1&1&0&0&0 \\ 1&1+x&1&0&0 \\ 1&1+x(1+x)&1+2x&1&0 \\ 1&1+x(1+x(1+x))&1+x(1+x)+x(1+2x)&1+3x&1 \end{bmatrix}$

$\displaystyle M_2 = \begin{bmatrix} 1&0&0&0&0 \\ x&1&0&0&0 \\ 1&x&1&0&0 \\ 0&1&x&1&0 \\ 0&0&1&x&1 \end{bmatrix}$

$\displaystyle M_3 = \begin{bmatrix} 1&1&1&1&1 \\ \frac{1}{2}(-1+x)x&1&1&1&1 \\ \frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1&1&1 \\ 1&\frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1&1 \\ 1&1&\frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1 \end{bmatrix}$

and multiplying and dividing elementwise $M_1\cdot M_2 / M_3$ , we get the following matrix $A$:

$\displaystyle A = \begin{bmatrix} 1&0&0&0&0 \\ \frac{2}{-1+x}&1&0&0&0 \\ \frac{2}{(-1+x)x}&\frac{2(1+x)}{-1+x}&1&0&0 \\ 0&\frac{2(1+x(1+x))}{(-1+x)x}&\frac{2(1+2x)}{-1+x}&1&0 \\ 0&0&\frac{(2(1+x(1+x)+x(1+2x)))}{((-1+x)x)}&\frac{(2(1+3x))}{(-1 + x)}&1 \end{bmatrix}$

Then the matrix inverse of $A$ is a new matrix $B$ starting:

$\displaystyle B = \begin{bmatrix} 1&0&0 \\ \frac{-2}{-1 + x}&1&0 \\ (\frac{4}{(-1+x)^2} - \frac{2}{(-1+x)x} + \frac{4x}{(-1 + x)^2})&(\frac{-2}{-1+x} - \frac{2x}{-1+x})&1 \end{bmatrix}$

If we then calculate the ratios of the first column in matrix $B$ we get a sequence $c\;$:

$\displaystyle c = \frac {B(n,1)}{B(n+1,1)}, \; n=1,2,3\;...$

and if we define a sequence $b$:

$\displaystyle b = 1,\; 1+x,\; 1+2x,\; 1+3x \;...$

and multiply element-wise $c$ with $b$ and add $x$, we get a sequence $d\;$:

$\displaystyle d = x + b*c$

$\displaystyle d = \frac{1+x}{2}, \frac{x(2+x+x^2)}{1+x+2x^2}, \frac{1+x(6+x(7+2x(4+x)))}{2(1+x)(2+x+3x^2)}\;...$

My question is: Do the terms in this sequence $d$ tend to $\displaystyle \sqrt x\;$?

Example: $x=2\;$

$\displaystyle \frac{3}{2}, \frac{16}{11}, \frac{137}{96}, \frac{1642}{1157}, \frac{8429}{5950}, \frac{67952}{48001}, \frac{1509939}{1066976}, \frac{38701726}{27353183}, \frac{1124000429}{794502270}, \frac{36478904464}{25787223797} \;...$

$\displaystyle \frac{36478904464}{25787223797} = 1.41461$ which is close to $\displaystyle \sqrt 2 = 1.41421$

As a Mathematica program this is:

(*x is the value to calculate the square root of*)
Clear[x, t, tt, ttt, M1, M2, M3, A, B, b, c, n, k, i, j, 
(*x=2; uncomment this to calculate sqrt of 2 or any other number*)
(*nn is size of matrix*)
nn = 5;
(*Variant of the Pascal triangle*)
t[n_, 1] = 1;
t[n_, k_] := t[n, k] = If[n > 1, t[n - 1, k - 1] + x*t[n - 1, k], 0];
M1 = Table[Table[t[i, j], {j, 1, nn}], {i, 1, nn}];
tt[n_, k_] := 
  tt[n, k] = If[Or[n == k, n == k + 2], 1, If[n == k + 1, x, 0]];
M2 = Table[Table[tt[i, j], {j, 1, nn}], {i, 1, nn}];
ttt[n_, k_] := 
  ttt[n, k] = 
   If[n == k, 1, If[Or[n == k + 1, n == k + 2], (x - 1)*x/2, 1]];
M3 = Table[Table[ttt[i, j], {j, 1, nn}], {i, 1, nn}];
(*Elementwise multiplication and division of the 3 matrices*)
A = M1*M2/M3;
B = Inverse[A];
b = Range[1, x*(nn - 2) + 1, x]
c = Ratios[B[[All, 1]]]^-1;
d = squarerootofx = x + b*c
  • $\begingroup$ @t-laarhoven Your assumption about the notation is correct. I am used to the notation $T(n,k)$ that is often seen for matrices in Neil Sloane's OEIS, therefore I used $n$ as index. The Mathematica program should be correct however. Regarding an expression for the entries in matrix $B$, no I have not thought about that. And yes, I agree that there are easier ways to calculate $\sqrt x$. $\endgroup$ – Mats Granvik Aug 24 '11 at 21:25
  • $\begingroup$ I am not quite sure I understand what the rule is for generating $M_1$. It looks like each row may be able to be generated recursively from the previous one(s); can you write down the relation? $\endgroup$ – Willie Wong Aug 25 '11 at 1:55
  • $\begingroup$ @Willie Wong: Yes, $M_1$ is generated recursively. The recurrence is: $M_1 (n,1)=1\;$,$n>1:\;M_1 (n,k)=M_1(n-1,k-1)+x \cdot M_1(n-1,k)$ $\endgroup$ – Mats Granvik Aug 25 '11 at 8:47

This reminds me a lot of Do these series converge to the von Mangoldt function? :-). I'm sure you have an explanation for how you managed to come up with this hypothesis in the first place, as you did in that case...

The sequence $d$ does indeed converge to $\sqrt x$. The solution of the recurrence

$$M_1 (n,k)=\begin{cases}1&k=1\\M_1(n-1,k-1)+x \cdot M_1(n-1,k)&k>1\\\end{cases}$$


$$ M_1(n,k)=\sum_{j=0}^{n-k}\binom{k+j-2}jx^j\;, $$

where the $k=1$ case works out if the binomial coefficient is generalized to $\binom{j-1}j$. But most of the matrix isn't used because of the zeros in $M_2$; so we will only need the cases $0\le n-k\le2$:

$$ \begin{eqnarray} M_1(n+2,n+2) &=& 1\;, \\ M_1(n+2,n+1) &=& 1+nx\;, \\ M_1(n+2,n+0) &=& 1+(n-1)x+\frac{n(n-1)}2x^2\;. \end{eqnarray} $$

Here and in the following $n\ge1$. Then the corresponding entries of $A$ are

$$ \begin{eqnarray} A(n+2,n+2) &=& 1\;, \\ A(n+2,n+1) &=& \frac2{x-1}(1+nx)\;, \\ A(n+2,n+0) &=& \frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)\;. \end{eqnarray} $$

The first column of the inverse $B$ is determined by the condition


Since $A(n+2,n+2)=1$, we can directly solve for $B(n+2,1)$ to get a recurrence for $B$:

$$ \begin{eqnarray} B(n+2,1) &=& -A(n+2,n+1)B(n+1,1)-A(n+2,n)B(n,1) \\ &=& -\frac2{x-1}(1+nx)B(n+1,1) \\ && -\frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)B(n,1) \;. \end{eqnarray} $$

Dividing through by $B(n+1,1)$ yields

$$ \begin{eqnarray} \frac1{c(n+1)} &=& -A(n+2,n+1)-A(n+2,n)c(n) \\ &=& -\frac2{x-1}(1+nx) -\frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)c(n) \;, \end{eqnarray} $$

which is a one-term recurrence for $c$. With $d(n)=x+(1+(n-1)x)c(n)$, and thus


this yields a one-term recurrence for $d$:

$$ \frac{1+nx}{d(n+1)-x} = -\frac2{x-1}(1+nx) -\frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)\frac{d(n)-x}{1+(n-1)x} \;, $$

$$ \begin{eqnarray} \frac1{d(n+1)-x} &=& -\frac2{x-1}\left(1+\frac{1+(n-1)x+\frac12n(n-1)x^2}{(1+nx)(1+(n-1)x)}\frac{d(n)-x}x\right) \\ &=& -\frac2{x-1}\left(1+\gamma(n)\frac{d(n)-x}x\right) \;, \end{eqnarray} $$

where $\gamma(n)$ contains the dependence on $n$.

Now to get our bearings let's first see without full rigour what happens for large $n$. Then $\gamma(n)$ goes to $1/2$, and we get

$$ \begin{eqnarray} \frac1{d(n+1)-x} &=& -\frac2{x-1}\left(1+\frac12\frac{d(n)-x}x\right) \\ &=& \frac{d(n)+x}{x(1-x)} \;, \\ d(n+1) &=& x+\frac{x(1-x)}{d(n)+x} \\ &=& x\frac{d(n)+1}{d(n)+x}\;. \end{eqnarray} $$

Now we can look at what happens to the difference from $\sqrt x$:

$$ \begin{eqnarray} d(n+1)-\sqrt x &=& x\frac{d(n)+1}{d(n)+x}-\sqrt x \\ &=& \frac{x-\sqrt x}{d(n)+x}(d(n)-\sqrt x)\;. \end{eqnarray} $$

So if we could show that $d(n)$ is always positive, then we might be able to show that the sequence of the differences is dominated by a convergent geometric sequence. To make the approximation with $\gamma(n)=1/2$ rigorous, we will also need to bound $d(n)$ from above.

So let's go back to the full recurrence for $d$:

$$ d(n+1) = x-\frac{x-1}2\left(1+\gamma(n)\frac{d(n)-x}x\right)^{-1}\;. $$

Let's see how $\gamma(n)$ differs from its limit:

$$ \begin{eqnarray} \gamma(n)-\frac12 &=& \frac{1+(n-1)x+\frac12n(n-1)x^2}{(1+nx)(1+(n-1)x)}-\frac12 \\ &=& \frac12\frac{1-x}{(1+nx)(1+(n-1)x)} \end{eqnarray} $$

Thus we have $0<\gamma(n)\lt1/2$ for $x\gt1$ and $1/2\lt\gamma(n)\lt1$ for $x\lt1$. Also note that for $d(n)>0$ we have


Now first assume $x\gt1$. Then for $d(n)\gt0$,

$$ \begin{eqnarray} d(n+1) &\gt& x-\frac{x-1}2\left(1+\gamma(n)\frac{0-x}x\right)^{-1} \\ &=& x-\frac{x-1}2\left(1-\gamma(n)\right)^{-1} \\ &\gt& x-\frac{x-1}2\left(\frac12\right)^{-1} \\ &=& x-(x-1) \\ &=& 1\;. \end{eqnarray} $$

On the other hand, for $d(n)\lt x$,

$$ \begin{eqnarray} d(n+1) &\lt& x-\frac{x-1}2\left(1+\gamma(n)\frac{x-x}x\right)^{-1} \\ &=& x-\frac{x-1}2 \\ &=& \frac{x+1}2\;. \end{eqnarray} $$

Since $1\lt d(1)=(x+1)/2\lt x$, it follows that $1\lt d(n)\le(x+1)/2\lt x$ for all $n$.

Now assume $x<1$. Then $d(n+1)\gt x$. On the other hand, for $d(n)\gt x$,

$$ \begin{eqnarray} d(n+1) &\lt& x-\frac{x-1}2\left(1+\frac12\frac{d-x}x\right)^{-1} \\ &\lt& x-\frac{x-1}2\left(1+\frac12\frac{-x}x\right)^{-1} \\ &=& x-\frac{x-1}2\left(\frac12\right)^{-1} \\ &=& x-(x-1) \\ &=& 1\;. \end{eqnarray} $$

Since $x\lt d(1)=(x+1)/2\lt1$, it follows that $x\lt d(n)\lt1$ for all $n$.

Now we have all the ingredients for the proof of convergence. Let's put $\gamma(n)=\frac12(1+\epsilon(n))$, with $\epsilon(n)\to0$ as $n\to\infty$, so

$$ \begin{eqnarray} d(n+1) &=& x-\frac{x-1}2\left(1+\frac12(1+\epsilon(n))\frac{d(n)-x}x\right)^{-1} \\ &=& x-\frac{x-1}2\frac{x}{x+\frac12(1+\epsilon(n))(d(n)-x)} \\ &=& \frac{x(d(n)+x+\epsilon(n)(d(n)-x))-(x-1)x}{d(n)+x+\epsilon(n)(d(n)-x)} \\ &=& \frac{x(d(n)+1)+\epsilon(n)x(d(n)-x)}{d(n)+x+\epsilon(n)(d(n)-x)}\;, \end{eqnarray} $$

and consider the difference from $\sqrt x$ again:

$$ \begin{eqnarray} d(n+1)-\sqrt x &=& \frac{x(d(n)+1)+\epsilon(n)x(d(n)-x)}{d(n)+x+\epsilon(n)(d(n)-x)}-\sqrt x \\ &=& \frac{(x-\sqrt x)(d(n)-\sqrt x+\epsilon(n)x(d(n)-x))}{d(n)+x+\epsilon(n)(d(n)-x)} \\ &=& \frac{(x-\sqrt x)(1+\epsilon(n)x(d(n)+\sqrt x))}{d(n)+x+\epsilon(n)(d(n)-x)}(d(n)-\sqrt x) \;. \end{eqnarray} $$

Since $d(n)$ is bounded and positive (and $x$ is constant and positive), we have

$$ \limsup_{n\to\infty}\left|\frac{(x-\sqrt x)(1+\epsilon(n)x(d(n)+\sqrt x))}{d(n)+x+\epsilon(n)(d(n)-x)}\right| =\limsup_{n\to\infty}\left|\frac{x-\sqrt x}{d(n)+x}\right| \le\left|\frac{x-\sqrt x}x\right|\lt1\;, $$

and so the sequence of the differences is dominated by a convergent geometric sequence. That completes the proof.


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