[VA] SRC#003- New Year 2019 Special
01-28-2019, 12:46 PM (This post was last modified: 01-28-2019 12:57 PM by Albert Chan.)
Post: #18
 Albert Chan Senior Member Posts: 1,894 Joined: Jul 2018
RE: [VA] SRC#003- New Year 2019 Special
(01-28-2019 11:56 AM)pier4r Wrote:  "The time spent on any item of the agenda will be in inverse proportion to the knowledge and understanding effort involved."

+1
I PM Thomas last week for how his estimated iterations work.
Sadly, I still don't understand the geometric intuition ...

Thomas Klemm Wrote:
Albert Chan Wrote:How is above formula derived ?

If the matrix $$M$$ is diagonalized the powers can be calculated easily as:

$$\begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0\\ 0 & 0 & \lambda_3 \end{bmatrix}^n= \begin{bmatrix} \lambda_1^n & 0 & 0\\ 0 & \lambda_2^n & 0\\ 0 & 0 & \lambda_3^n \end{bmatrix}$$

Thus assuming the initial vector $$v$$ in a general position the ratio of the 2nd to the 1st coordinate after iterating $$n$$ times will be roughly:

$$\frac{\lambda_2^n}{\lambda_1^n}$$

We want that to be so small that it doesn't affect the display:

$$1+10^{-10}=1$$

For the given example I have a geometric intuition.
A vector is stretched in direction of the 1st eigenvector by factor 175.524449043.
The other two eigenvectors define a plane where the projection upon is rotated and stretched by 152.076171921.
Now you can wonder how often you have to iterate such that stretching in the dominant direction is so much bigger that the rest becomes irrelevant.

I assumed you are familiar with the concepts of eigenvalue and eigenvectors from linear algebra.
Otherwise it's not possible to explain the formula.

HTH
Thomas
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 Messages In This Thread [VA] SRC#003- New Year 2019 Special - Valentin Albillo - 01-15-2019, 10:42 PM RE: [VA] SRC#003- New Year 2019 Special - Paul Dale - 01-16-2019, 03:51 AM RE: [VA] SRC#003- New Year 2019 Special - Carsen - 01-16-2019, 03:57 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Okken - 01-16-2019, 04:24 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-16-2019, 09:33 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-16-2019, 09:50 AM RE: [VA] SRC#003- New Year 2019 Special - DavidM - 01-17-2019, 06:04 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-18-2019, 05:04 AM RE: [VA] SRC#003- New Year 2019 Special - DavidM - 01-18-2019, 07:15 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-18-2019, 05:44 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-18-2019, 08:28 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-20-2019, 11:33 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-20-2019, 06:13 PM RE: [VA] SRC#003- New Year 2019 Special - Valentin Albillo - 01-20-2019, 07:40 PM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-09-2019, 01:55 AM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 01-20-2019, 09:28 PM RE: [VA] SRC#003- New Year 2019 Special - Valentin Albillo - 01-27-2019, 09:38 PM RE: [VA] SRC#003- New Year 2019 Special - pier4r - 01-28-2019, 11:56 AM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 01-28-2019 12:46 PM RE: [VA] SRC#003- New Year 2019 Special - pier4r - 01-28-2019, 07:34 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-28-2019, 05:40 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-28-2019, 06:17 PM RE: [VA] SRC#003- New Year 2019 Special - rprosperi - 01-28-2019, 11:43 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-28-2019, 11:59 PM RE: [VA] SRC#003- New Year 2019 Special - rprosperi - 01-29-2019, 02:04 AM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-08-2019, 06:46 PM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-08-2019, 09:36 PM

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