Weird trees

Reply #21

eddessaknight — Wed, 22 Mar 2023 20:33:30 GMT

PS:

R: All # lmitations on all Sreaks

Even tall trees do not grow to haven.

~NYX stock market floor trading wisdom precaution

Reply #20

eddessaknight — Mon, 20 Mar 2023 00:56:06 GMT

NOTA BENE:

All Trees have been blessed with the power to grow up tall against the forces of gravity

Reply #19

eddessaknight — Wed, 15 Mar 2023 16:37:45 GMT

Doesn't that beautiful butterfly in Hong Kong by flapping it's delicate wings still have it's affect in New York, Sully

Reply #18

sully16 — Tue, 14 Mar 2023 00:23:21 GMT

Interesting, ponder this.

What is the probablity of...

A. Husband rakes leaves when they fall.

B. Son rakes leaves when they fall.

C. Wife/Mom ends up hiring neighbor boy to rake leaves when they fall.

D.Leaves sit on ground until spring.

you have 10 seconds to guess.

Reply #17

eddessaknight — Mon, 06 Mar 2023 01:20:21 GMT

Thanks Joe ala DB101

Your equations could have alignments with Phagorean theories, a worth while checkout

Reply #16

eddessaknight — Wed, 01 Mar 2023 21:26:28 GMT

Side Bar:

In all of the Creator's trees world wide; there are no two leaves that are duplicates

Reply #15

Wavepack — Tue, 07 Feb 2023 23:30:59 GMT

Yes, definitely if it's a choice between operator error or an error in Wolfram Alpha, I'd go with operator error.

That's how I got 30 first for red steady state trees with WolframAlpha.

Reply #14

cottoneyedjoe — Tue, 07 Feb 2023 22:55:47 GMT

WolframAlpha is giving me the correct eigensystem and the correct answer for the original problem done with matrix^23 * initial state vector.

https://www.wolframalpha.com/input?i=eigensystem+of+%5B%5B4%2F5%2C+0%2C+1%2F3%5D%2C+%5B1%2F5%2C+1%2F2%2C+0%5D%2C+%5B0%2C+1%2F2%2C+2%2F3%5D%5D

https://www.wolframalpha.com/input?i=+%5B%5B.8%2C+0%2C+1%2F3%5D%2C+%5B.2%2C+.5%2C+0%5D%2C+%5B0%2C+.5%2C+2%2F3%5D%5D%5E23+*+%5B%5B100%5D%2C+%5B0%5D%2C+%5B0%5D%5D

Reply #13

Wavepack — Tue, 07 Feb 2023 22:35:24 GMT

I used Wolfram Alpha to form and solve the characteristic equation for the poles. I blame WA and not the guy entering the state update matrix.

Reply #12

cottoneyedjoe — Tue, 07 Feb 2023 21:33:48 GMT

Your solution is close, but notice that your R(n) eventually goes to 0 as n goes to infinity. It hits the correct values for the first several values of n, but then goes toward 0 when it should be stabilizing at 30.

Reply #11

cottoneyedjoe — Tue, 07 Feb 2023 21:25:03 GMT

You got it right. As you guessed, it's an eigenvalue and eigenvector problem from a linear algebra book. Several ways you can do it. For example, if you notice that the system has a steady state, you can get the answer with middle school math as you did.

It could also be set up as a system of recurrence relations:

G(n) = (4/5)*G(n-1) + (1/3)*R(n-1)

Y(n) = (1/2)*Y(n-1) + (1/5)*G(n-1)

R(n) = 100 - G(n) - Y(n)

with characteristic polynomial x^3 - (59/30)x^2 + (19/15)x - 3/10 = 0... [ More ]

Reply #10

Wavepack — Tue, 07 Feb 2023 19:02:33 GMT

db101, yeah, once you realize steady state has been reached up to the nearest integer, then your steady state solution is the easiest solution.

Also, my equation above for R(n) needs to be multiplied by 100 inside the round() function, or

R(n) = round(250 (4/5)^n + 200 (1/2)^n - 450 (2/3)^n)

Below is the linear algebra which gets the exact closed form solution. The poles are the zeros of the characteristic equation, ie the solution of

det(zI - A) = 0

where I = 3x3 identi... [ More ]

Reply #9

Wavepack — Tue, 07 Feb 2023 18:18:35 GMT

The exact closed form solution for Red(Year) = R(n) is

R(n) = round[(5/2) (4/5)^n + 2 (1/2)^n - (9/2) (2/3)^n]

n = number of years

Reply #8

db101 — Tue, 07 Feb 2023 16:44:09 GMT

After running lots of simulations I discovered the ratio of green to yellow to red stabilizes to a particular ratio, regardless of the initial state. That ratio is 50% green, 20% yellow, and 30% red. On this property the expected numbers of each (after a long enough time) is 50 green, 20 yellow, and 30 red. There's nothing special about 23, it's just a number that's large enough for the steady state (eigenvector?) to materialize. I also realized you got this from a linear algebra book (hardcover... [ More ]

Reply #7

Wavepack — Tue, 07 Feb 2023 15:55:21 GMT

^ Correction. G(n) above is wrong. Ok, I'll come clean. I used wolfram alpha to compute round(A 27

Reply #6

Wavepack — Tue, 07 Feb 2023 15:25:56 GMT

Extra Credit:

Number of green trees at year n = G(n) = 100 (4/5)^n

This is an LTI (linear time invariant) discrete system. Zeros in entries of the state update matrix and initial state saved the day for solving in the time domain. The state update matrix is

4/5 0 1/3

1/5 1/2 0

0 1/2 2/3

with poles at z={4/5, 1/2, 2/3}, so that the LTI system is stable. The initial state is

100

0

0

Using z-transforms to solve for the Z-transform of the

Reply #5

Wavepack — Tue, 07 Feb 2023 08:34:41 GMT

Reply #4

db101 — Tue, 07 Feb 2023 05:48:07 GMT

Reply #3

cottoneyedjoe — Mon, 06 Feb 2023 22:27:03 GMT

Hardcover.

Reply #2

db101 — Mon, 06 Feb 2023 20:27:09 GMT

I'll take a hint. What kind of book was it

Reply #1

cottoneyedjoe — Mon, 06 Feb 2023 05:29:00 GMT

Meant to add that expected number of trees should be rounded to nearest whole number.

Weird trees

cottoneyedjoe — Mon, 06 Feb 2023 05:08:13 GMT

(Adapted this problem from a college textbook. As a hint, I can tell you what kind of book.)

There's a species of weird trees that can change its leaf color once a year in the fall. The tree keeps that leaf color for a full year until the following fall, when it can change color again. The leaves change color randomly according the following rules.

If the leaves are currently green, they can either stay green or turn yellow. The probability the leaves turn yellow is 1/5. If the leaves ar... [ More ]