(here the two last computations for the root differ only by the 220'th digit)
(here the two last computations for the root differ only by the 220'th digit)
[update2]
With an optimized procedure, using Newton-iteration for the (supposed, highest) root I get this table
lambda=1.3862943611198906188 q=0.25 warning: float precision about 1200 dec digits needed for size n=128 Table of highest eigenvalue/largest characteristic root for matrix-sizes highest root computed using Newton-iteration size : largest root of char.polynomial (time needed) -------------------------------------------------------- 2 : 1.2500000000000000000 ( 10 ms) 4 : 1.2631849865636825686 ( 16 ms) 8 : 1.2631855203242152061 ( 15 ms) 16 : 1.2631855203242152061 ( 63 ms) 32 : 1.2631855203242152061 ( 187 ms) 64 : 1.2631855203242152061 ( 702 ms) 128 : 1.2631855203242152061 (2168 ms) 256 : 1.2631855203242152061 (7286 ms) [update2]
With an optimized procedure, using Newton-iteration for the (supposed, highest) root I get this table
lambda=1.3862943611198906188 q=0.25 warning: float precision about 1200 dec digits needed for size n=128 Table of highest eigenvalue/largest characteristic root for matrix-sizes highest root computed using Newton-iteration size : largest root of char.polynomial (time needed) -------------------------------------------------------- 2 : 1.2500000000000000000 ( 10 ms) 4 : 1.2631849865636825686 ( 16 ms) 8 : 1.2631855203242152061 ( 15 ms) 16 : 1.2631855203242152061 ( 63 ms) 32 : 1.2631855203242152061 ( 187 ms) 64 : 1.2631855203242152061 ( 702 ms) 128 : 1.2631855203242152061 (2168 ms) 256 : 1.2631855203242152061 (7286 ms) In case you've not already seen this, here is a list of the characteristic polynomials for sizes 2,4,6,8 and the generalization-scheme is rather obvious. Perhaps this is of some help. It involves the so-called "q-binomials" with your parameter as base.
For conciseness of notation I've rewritten your expression with the parameter $e^{-\lambda} $ as $z$
For the binomial lets write $(a:b)$
the $n$'th $q$-power to base $z$ is $ \displaystyle z_n={z^n-1 \over z-1} $, let's write the q-factorial as $ \displaystyle z_{n!} = z_n \cdot z_{n-1} \cdot \ldots \cdot z_1 $ and the q-binomial $ \displaystyle (a:b)_q = {z_{a!} \over z_{b!} z_{(a-b)!}} $
then for the fist few even matrix-sizes we get the characteristic polynomials as $$ \begin{array} {r|llllll} \text{size} & \\ \hline\\ 2&c_2(x)=&1x^2&-z^0(2:1)_q x \\ 4&c_4(x)=&1x^4&-z^0(4:1)_q x^3 &+ z^2(3:2)_q x^2 \\ 6&c_6(x)=&1x^6&-z^0(6:1)_q x^5 &+ z^2(5:2)_q x^4 &- z^6(4:3)_q x^3\\ 8&c_8(x)=&1x^8&-z^0(8:1)_q x^7 &+ z^2(7:2)_q x^6 &- z^6(6:3)_q x^5&+ z^{12}(5:4)_q x^4\\ \end{array}$$ (The exponents at the free powers of $z$ are $2$ times the binomials $0,1,3,6,10,...$ . Also I hope I didn't mess up the powers of $z-1$ which might or might not occur additionally in this notation of the q-factorials/q-binomials; I'll check this later)
If we are lucky then this gives some possibility to estimate the largest polynomial root depending on your parameter $\lambda$
**[update]** This formal description is very useful; we do no more need to compute the matrix - just construct the characteristic polynomial (represented only by a vector of size $n+1$ !) and determine the largest root. This is the table for matrix-sizes $n \times n$ with $n=2,4,6...,32$; the maximal root approximates well with increasing $n$ for the given $\lambda$ (in the example = $\log(4)$).
This is the table for matrix-sizes $n \times n$ with $n=2,4,6...,32$; the maximal root approximates well with increasing $n$ for the given $\lambda$ (in the example = $\log(4)$).
lambda=1.3862943611198906188 q=0.25 Table of highest eigenvalue/largest characteristic root for matrix-sizes size : eigenvalue (time needed) --------------------------------------- 2 : 1.2500000000000000000 (0 ms) 4 : 1.2631849865636825686 (0 ms) 6 : 1.2631855203241351201 (0 ms) 8 : 1.2631855203242152061 (0 ms) 10 : 1.2631855203242152061 (16 ms) 12 : 1.2631855203242152061 (46 ms) 14 : 1.2631855203242152061 (78 ms) 16 : 1.2631855203242152061 (94 ms) ... 26 : 1.2631855203242152061 (312 ms) 28 : 1.2631855203242152061 (359 ms) 30 : 1.2631855203242152061 (437 ms) 32 : 1.2631855203242152061 (515 ms) In case you've not already seen this, here is a list of the characteristic polynomials for sizes 2,4,6,8 and the generalization-scheme is rather obvious. Perhaps this is of some help. It involves the so-called "q-binomials" with your parameter as base.
For conciseness of notation I've rewritten your expression with the parameter $e^{-\lambda} $ as $z$
For the binomial lets write $(a:b)$
the $n$'th $q$-power to base $z$ is $ \displaystyle z_n={z^n-1 \over z-1} $, let's write the q-factorial as $ \displaystyle z_{n!} = z_n \cdot z_{n-1} \cdot \ldots \cdot z_1 $ and the q-binomial $ \displaystyle (a:b)_q = {z_{a!} \over z_{b!} z_{(a-b)!}} $
then for the fist few even matrix-sizes we get the characteristic polynomials as $$ \begin{array} {r|llllll} \text{size} & \\ \hline\\ 2&c_2(x)=&1x^2&-z^0(2:1)_q x \\ 4&c_4(x)=&1x^4&-z^0(4:1)_q x^3 &+ z^2(3:2)_q x^2 \\ 6&c_6(x)=&1x^6&-z^0(6:1)_q x^5 &+ z^2(5:2)_q x^4 &- z^6(4:3)_q x^3\\ 8&c_8(x)=&1x^8&-z^0(8:1)_q x^7 &+ z^2(7:2)_q x^6 &- z^6(6:3)_q x^5&+ z^{12}(5:4)_q x^4\\ \end{array}$$ (The exponents at the free powers of $z$ are $2$ times the binomials $0,1,3,6,10,...$ . Also I hope I didn't mess up the powers of $z-1$ which might or might not occur additionally in this notation of the q-factorials/q-binomials; I'll check this later)
If we are lucky then this gives some possibility to estimate the largest polynomial root depending on your parameter $\lambda$
**[update]** This formal description is very useful; we do no more need to compute the matrix - just construct the characteristic polynomial and determine the largest root. This is the table for matrix-sizes $n \times n$ with $n=2,4,6...,32$; the maximal root approximates well with increasing $n$ for the given $\lambda$ (in the example = $\log(4)$).
lambda=1.3862943611198906188 Table of highest eigenvalue/largest characteristic root for matrix-sizes size : eigenvalue (time needed) --------------------------------------- 2 : 1.2500000000000000000 (0 ms) 4 : 1.2631849865636825686 (0 ms) 6 : 1.2631855203241351201 (0 ms) 8 : 1.2631855203242152061 (0 ms) 10 : 1.2631855203242152061 (16 ms) 12 : 1.2631855203242152061 (46 ms) 14 : 1.2631855203242152061 (78 ms) 16 : 1.2631855203242152061 (94 ms) ... 26 : 1.2631855203242152061 (312 ms) 28 : 1.2631855203242152061 (359 ms) 30 : 1.2631855203242152061 (437 ms) 32 : 1.2631855203242152061 (515 ms) In case you've not already seen this, here is a list of the characteristic polynomials for sizes 2,4,6,8 and the generalization-scheme is rather obvious. Perhaps this is of some help. It involves the so-called "q-binomials" with your parameter as base.
For conciseness of notation I've rewritten your expression with the parameter $e^{-\lambda} $ as $z$
For the binomial lets write $(a:b)$
the $n$'th $q$-power to base $z$ is $ \displaystyle z_n={z^n-1 \over z-1} $, let's write the q-factorial as $ \displaystyle z_{n!} = z_n \cdot z_{n-1} \cdot \ldots \cdot z_1 $ and the q-binomial $ \displaystyle (a:b)_q = {z_{a!} \over z_{b!} z_{(a-b)!}} $
then for the fist few even matrix-sizes we get the characteristic polynomials as $$ \begin{array} {r|llllll} \text{size} & \\ \hline\\ 2&c_2(x)=&1x^2&-z^0(2:1)_q x \\ 4&c_4(x)=&1x^4&-z^0(4:1)_q x^3 &+ z^2(3:2)_q x^2 \\ 6&c_6(x)=&1x^6&-z^0(6:1)_q x^5 &+ z^2(5:2)_q x^4 &- z^6(4:3)_q x^3\\ 8&c_8(x)=&1x^8&-z^0(8:1)_q x^7 &+ z^2(7:2)_q x^6 &- z^6(6:3)_q x^5&+ z^{12}(5:4)_q x^4\\ \end{array}$$ (The exponents at the free powers of $z$ are $2$ times the binomials $0,1,3,6,10,...$ . Also I hope I didn't mess up the powers of $z-1$ which might or might not occur additionally in this notation of the q-factorials/q-binomials; I'll check this later)
If we are lucky then this gives some possibility to estimate the largest polynomial root depending on your parameter $\lambda$
**[update]** This formal description is very useful; we do no more need to compute the matrix - just construct the characteristic polynomial (represented only by a vector of size $n+1$ !) and determine the largest root.
This is the table for matrix-sizes $n \times n$ with $n=2,4,6...,32$; the maximal root approximates well with increasing $n$ for the given $\lambda$ (in the example = $\log(4)$).
lambda=1.3862943611198906188 q=0.25 Table of highest eigenvalue/largest characteristic root for matrix-sizes size : eigenvalue (time needed) --------------------------------------- 2 : 1.2500000000000000000 (0 ms) 4 : 1.2631849865636825686 (0 ms) 6 : 1.2631855203241351201 (0 ms) 8 : 1.2631855203242152061 (0 ms) 10 : 1.2631855203242152061 (16 ms) 12 : 1.2631855203242152061 (46 ms) 14 : 1.2631855203242152061 (78 ms) 16 : 1.2631855203242152061 (94 ms) ... 26 : 1.2631855203242152061 (312 ms) 28 : 1.2631855203242152061 (359 ms) 30 : 1.2631855203242152061 (437 ms) 32 : 1.2631855203242152061 (515 ms) Loading
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