[VA] SRC#003 New Year 2019 Special

01282019, 07:34 PM
Post: #21




RE: [VA] SRC#003 New Year 2019 Special
(01282019 12:46 PM)Albert Chan Wrote: I PM Thomas last week for how his estimated iterations work. 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. Wikis are great, Contribute :) 

01282019, 11:43 PM
Post: #22




RE: [VA] SRC#003 New Year 2019 Special
(01282019 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 numbercrunching 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 relearning stuff I already supposedly know. Bob Prosperi 

01282019, 11:59 PM
Post: #23




RE: [VA] SRC#003 New Year 2019 Special
(01282019 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 

01292019, 02:04 AM
Post: #24




RE: [VA] SRC#003 New Year 2019 Special
(01282019 11:59 PM)Thomas Klemm Wrote:(01282019 11:43 PM)rprosperi Wrote: I will also check out some other videos in the same 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. Bob Prosperi 

02082019, 06:46 PM
Post: #25




RE: [VA] SRC#003 New Year 2019 Special
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 

02082019, 09:36 PM
Post: #26




RE: [VA] SRC#003 New Year 2019 Special
(02082019 06:46 PM)Albert Chan Wrote: Tried doing √3 with this matrix power method, noticed a pattern: 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. 

02092019, 01:55 AM
(This post was last modified: 02092019 02:41 AM by Albert Chan.)
Post: #27




RE: [VA] SRC#003 New Year 2019 Special
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 

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