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+--- Thread: [VA] SRC#003- New Year 2019 Special (/thread-12206.html)
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RE: [VA] SRC#003- New Year 2019 Special - pier4r - 01-28-2019 07:34 PM
(01-28-2019 12:46 PM)Albert Chan Wrote: I PM Thomas last week for how his estimated iterations work.
Sadly, I still don't understand the geometric intuition ...
I know that is difficult to gauge whether a question is of general interest, but such explanations may help many (also those future readers!) and not only you, so if I were you I would ask in the public thread rather than in PM.
RE: [VA] SRC#003- New Year 2019 Special - rprosperi - 01-28-2019 11:43 PM
(01-28-2019 06:17 PM)Thomas Klemm Wrote: Maybe this video can give some geometric intuition:
Awesome Video Thomas, thanks for recommending it and the link! I certainly never learned Eigenvectors and Eigenvalues from this perspective, and it provides MUCH clarity for the mechanics underlying the number-crunching techniques I was taught. Having a better "feel" of what's going on, as this video provides, gives one much better insight into how to resolve issues when the 'normal equations/tools' don't work.
I will also check out some other videos in the same series. I'm always eager to learn stuff that's well presented, even if it is re-learning stuff I already supposedly know.
RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-28-2019 11:59 PM
(01-28-2019 11:43 PM)rprosperi Wrote: I will also check out some other videos in the same series.
I bet you will like: Visualizing quaternions
An explorable video series
RE: [VA] SRC#003- New Year 2019 Special - rprosperi - 01-29-2019 02:04 AM
(01-28-2019 11:59 PM)Thomas Klemm Wrote:(01-28-2019 11:43 PM)rprosperi Wrote: I will also check out some other videos in the same series.
I bet you will like: Visualizing quaternions
An explorable video series
Thanks, I'll check these out as well.
After the original above, I stumbled onto this video, with the first explanation of Euler's Identity I can honestly say I understood. So, while it now has a tiny bit less magic, it's still beautiful, and I can say I understand it.
RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-08-2019 06:46 PM
Tried doing √3 with this matrix power method, noticed a pattern:
let M = {{1,3}, {1, 1}}
M^2 = {{3*1+1, 3*(1+1)}, {1+1, 3*1+1}} = {{4,6}, {2,4}}
M^3 = {{3*2+4, 3*(2+4)}, {2+4, 3*2+4}} = {{10,18}, {6,10}}
Right diagonal ratio stayed at 3.0, and left diagonal same numbers.
-> only need to do bottom row. Top row can be deduced if needed.
-> each matrix multiply required only 2 add, and 1 multiply
row 2 of M^4 = {6+10, 3*6+10} = {16, 28}
row 2 of M^5 = {16+28, 3*16+28} = {44, 76}
row 2 of M^6 = {44+76, 3*44+76} = {120, 208} ...
Doing the average of ratios, for M^6: √3 ~ ½(208/120 + 360/208) ~ ½(1.733333 + 1.730769) = 1.732051
RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-08-2019 09:36 PM
(02-08-2019 06:46 PM)Albert Chan Wrote: Tried doing √3 with this matrix power method, noticed a pattern:
let M = {{1,3}, {1, 1}}
M^2 = {{3*1+1, 3*(1+1)}, {1+1, 3*1+1}} = {{4,6}, {2,4}}
M^3 = {{3*2+4, 3*(2+4)}, {2+4, 3*2+4}} = {{10,18}, {6,10}} ...
To prove that the ratio converge to √3, noticed above actually does Farey Fraction:
M^1: √3 between 1/1 and 3/1, so (1+3)/(1+1) = 4/2 is better estimate.
M^2: √3 between 4/2 and 3/(4/2) = 6/4, so (4+6)/(2+4) = 10/6 is better estimate
M^3: ...
Newton's method, does the same thing, but converge faster: x = ½(x + 3/x)
1: ½(1/1 + 3*1/1) = 2/1 = 2
2: ½(2/1 + 3*1/2) = 7/4 = 1.75
3: ½(7/4 + 3*4/7) = 97/56 ~ 1.732143
4: ½(97/56 + 3*56/97) = 18817/10864 ~ 1.73205081
The fractions are so good that all above (and at least 6 more !) are √3 continued fraction convergents.
RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-09-2019 01:55 AM
The idea of only doing only *last* row work for 3x3 matrix too.
Let M = {{k, n, n}, {1, k, n}, {1, 1, k}}
-> M^p = {{c, n*a, n*b},
{b, c, n*a},
{a, b, c}}, for some a, b, c
-> For M^(p+1), last row = {k*a+b+c, n*a+k*b+c, n*a+n*b+k*c}
Example: this is result of M^200 last row ratios:
Code:
lua> k, n = math.pi, 2019
lua> a, b, c = 1, 1, k
lua> for i=2,200 do
: a, b, c = (k*a+b+c)/n, a+(k*b+c)/n, a+b+k*c/n
: end
lua> =a, b, c
2.916424658351884e-212 3.686063969271537e-211 4.658809733574611e-210
lua> =b/a, c/b
12.638982319380704 12.638982319385288