Post whacky or neat math stuff

[Harovan]

"If one-and-a-half chickens lay one-and-a-half eggs in one-and-a-half days, how many eggs lay nine chickens in nine days?"

(That's actually pretty easy if you calculate it, but if you ask some random person to answer straight away, you hear either 9 or 81, but very rarely the true solution)

[Dauphin]

Originally posted by Sir Ralph
IIRC that's a good approximation. I made an exact solution (no graphics, no iterations) with 6 digits in the 80's and wrote it down in a notebook, which I hope I can find when I'm at home. If memory serves right, it starts with 1.231...

How did you find it, graphical?

W = width of well
H2 = height at which the 2m stick rests upon the wall
X2 = distance from the wall that 2m stick emerges from water

1) Create formulae for the height at which the two sticks rest on the wall e.g. H2^2 + W^2 = 2^2
2) Create formulaE for the gradient at which the two sticks rest. E.g. G2 = H2/W
3) Create formulae for the value of X. e.g X2 = 1/G2
4) X2 + X3 = W.
5) Solve simultaneous equations

[KrazyHorse]

Originally posted by Sir Ralph
An oldie but goodie...

There's a cylindrical well with some amount of water in it. 2 sticks stand in it, reaching from the bottom at one side to the wall on the other and crossing themselves exactly at the surface of the water. One stick is 3 meters tall (or any other unit), the other 2 meters. The water stands 1 meter high. What's the diameter of the well?

This is equivalent to solving a quartic equation.

[Harovan]

Originally posted by KrazyHorse
This is equivalent to solving a quartic equation.

Yes.

[Ecthy]

54

[BeBMan]

42 is the answer to everything.

[KrazyHorse]

Using double precision, I can get the answer to 1.231185724 or so.

[Harovan]

Sounds right .

I solved it with 6 digits in the 80's, but had only a 8-digit pocket calculator. Precision is secondary, by the way, it always can be increased. But most people I met aren't even able to get the problem into an equation.

[KrazyHorse]

Since it's just a quartic with well-separated roots, it's fairly well-behaved numerically. The quartic I had was in terms of h2, the vertical displacement between the top and bottom of the shorter stick.

The equation is monotonically increasing in the range h2 = [1.093,inf) or so, so most root finding algorithms will do a very good job of pinning it down very quickly. I wrote the dumbest root finding routine ever to solve it (took about 15 seconds to write) and it only evaluated the function 23 times before reaching the limits of double precision. A smarter method (newton-raphson etc) would probably have done it in 6 or 7 function evaluations.

[self biased]

2 + 2 = 5. granted, this only works for specific values of two, but i digress.

[Veritass]

For more calculator fun, you can always multiple 37037037037 by multiples of 3, or 12345679 by multiples of 9.

Hours of fun for the whole family.

[Harovan]

Originally posted by KrazyHorse
Since it's just a quartic with well-separated roots, it's fairly well-behaved numerically. The quartic I had was in terms of h2, the vertical displacement between the top and bottom of the shorter stick.

The equation is monotonically increasing in the range h2 = [1.093,inf) or so, so most root finding algorithms will do a very good job of pinning it down very quickly. I wrote the dumbest root finding routine ever to solve it (took about 15 seconds to write) and it only evaluated the function 23 times before reaching the limits of double precision. A smarter method (newton-raphson etc) would probably have done it in 6 or 7 function evaluations.

As mentioned, I didn't have the luxury of a computer and calculating iterations with a pocket calculator with only 1 memory cell made me shudder. Hence, I had to solve it the hard way, i.e. by reducing it to a cubic and finally solving this. Took me about half an hour, but it was fun.

[Kuciwalker]

Originally posted by KrazyHorse
The equation is monotonically increasing in the range h2 = [1.093,inf) or so, so most root finding algorithms will do a very good job of pinning it down very quickly. I wrote the dumbest root finding routine ever to solve it (took about 15 seconds to write) and it only evaluated the function 23 times before reaching the limits of double precision. A smarter method (newton-raphson etc) would probably have done it in 6 or 7 function evaluations.

Why reinvent the wheel? Most numerical libraries should have some root-finding function built in (or if it doesn't, presumably you've had to write one for your work at some point).

[KrazyHorse]

Originally posted by Kuciwalker


Why reinvent the wheel? Most numerical libraries should have some root-finding function built in (or if it doesn't, presumably you've had to write one for your work at some point).

Because I don't have the numerical recipes library locally on my laptop, and writing the root finder took less time than sshing onto the departmental network, editing a file via terminal emacs and remembering the proper linking syntax.

[KrazyHorse]

      DO 101 I = 0, 100
        GUESS = (TOP+BOT)/2.0
        RES = AFUNKP(GUESS)
        IF(RES.LT.0.0) BOT=GUESS
        IF(RES.GE.0.0) TOP=GUESS
        PRINT *, GUESS, RES, BOT, TOP
 101  CONTINUE