HOME RUN
Mark, "
the proof includes the application of Vieta's formulas"
:wink:
He is right. Firstly, note that, since x = 0 is not a root,
a and b are roots of x^4 - x^3 + x + 1 = 0
1/a and 1/b are roots of (1/x)^4 - (1/x)^3 + (1/x) + 1 = 0.
Or, equivalently, multiplying by x^4, 1/a and 1/b are roots of x^4 + x^3 - x + 1 = 0. (1)
Now let a' = 1/a, b' = 1/b, c' = 1/c, and d' = 1/d be the roots of equation (1). Using Vieta's formulas, we can write down
a' + b' + c' + d' = -1,
a'b' + a'c' + a'd' + b'c' + b'd' + c'd' = 0,
a'b'c' + a'b'd' + a'c'd' + b'c'd' = 1,
a'b'c'd' = 1.
Now we express the above relations in terms of s = a' + b', t = c' + d', p = a'b', and q = c'd'. We seek the equation satisfied by s = a' + b' = 1/a + 1/b. Thus:
s + t = -1 t => -1 - s,
st + p + q = 0, (2)
sq + tp = 1, (3)
pq = 1 q => 1/p.
Now substitute for t and q in (2) and (3):
-s(1 + s) + p + 1/p = 0, (4)
s/p - p(1 + s) = 1. (5)
Multiplying (4) by s, subtracting from (5), and simplifying, we obtain
p = (s^3 + s^2 - 1)/(2s + 1). (6)
Now multiply (4) by p and substitute for p from (6):
-s(1 + s)(s^3 + s^2 - 1)/(2s + 1) + (s^3 + s^2 - 1)^2/(2s + 1)^2 + 1 = 0.
Multiplying by -(2s + 1)^2, and simplifying, we obtain
S^6 + 3s^5 + 3s^4 + s^3 - 5s^2 - 5s - 2 = 0.
Therefore, if the equation x^4 - x^3 + x + 1 = 0 has roots a, b, c, and d, 1/a + 1/b is a root of x^6 + 3x^5 + 3x^4 + x^3 - 5x^2 - 5x - 2 = 0.
If that is not clear, please direct all correspondence to Stormy at the pool hall.
Mark:
RED AND GREEN
Using Excel (instead of actually doing math), the expected return converges to e-1 = 1.718281828. As this is less than two, the game should not be played.
It's probably a series that sums to e, but is missing the first term (1) or something like that.
Is there anything Excel can't do?
Let X be the number of red discs in the bag when play stops.
P(X = 1) = 1/2
P(X = 2) = 1/2 2/3
P(X = 3) = 1/2 1/3 3/4
...
P(X = k) = 1/2 1/3 ... 1/k k/(k+1) = k/(k+1)!
Therefore E(X) = 1/2! 1 + 2/3! 2 + 3/4! 4 + ...
= 1^2/2! + 2^2/3! + 3^2/4! + ...
To evaluate this series we need to consider the general term, k^2/(k+1)!, but before this we will evaluate a different series.
k/(k+1)! = (k+1 -1)/(k+1)! = 1/k! -1/(k+1)!
Hence the original series becomes a telescoping series:
Therefore 1/2! + 2/3! + ... = (1/1! -1/2!) + (1/2! -1/3!) + (1/3! -1/4!) + ... = 1
Now we write, k^2/(k+1)! = (k(k+1) k)/(k+1)! = 1/(k -1)! -k/(k+1)!.
Therefore E(X) = 1^2/2! + 2^2/3! + 3^2/4! + ...
= (1 + 1 + 1/2! + 1/3! + ...) - (1/2! + 2/3! + ...)
= e -1
Hence the return on each game will be 2 -e -1 = 3 -e =28 cents in the ?'bankers' favor.
Stormy, (taking a break from pool) asks, "What made Try disappear from Trivia?"
Dontcha know this is the Dingo roundup season? I don't got no time to play games.
No - wait! On second thoughts, I will be right over, let the games begin
I love Scrabble.
Two missiles are 41,620 kilometers apart aimed directly at each other.
One is traveling at 38,000 kilometers per hour. The other is going 22,000 kph.
They are primed to explode in the center of; let us choose a place where no one will notice, say; Paris, the area of which is 15 kilometers in diameter.
How far apart are they ONE minute before the collision
Which is not a fundamental unit
(1) kilogram (2) meter (3) second (4) watt
