1.13198823...
rpn2calc
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1.13198823...
Divakar Viswanath defines a random Fibonacci sequence:
"...by taking the first two terms to be 1 and by defining the later terms as either the sum or the difference of the previous two terms with probability 1/2. Like the Fibonacci sequence, random Fibonacci sequences increase exponentially, but at a different rate which is equal to 1.13198824... "
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Take the quadratic equation : change a , b , c : gives the golden ratio equation : If we use : change & rearrange : solve for y : |
ax2 + bx + c = 0 a = 1 b = -1 c = -1 x2 - x - 1 = 0 x2 - bx - 1 = 0 x2 - yx - 1 = 0 x2 - 1 = yx x - 1 / x = y |
Solving for x is messy but from 'recipies in c++' book :
x = -.5 * [ y + sgn(y) * sqrt(y^2 + 4)]
where we use the signum function, sgn(y) = [ 1 if y >0 , 0 if Y = 0 , -1 if y < 0 ]
Let us now derive a close approximation of Viswanath's Constant.
If we set y = -1 then solve for x, x = sqrt((y/2)^2 + 1)- y/2
we get x = .5 + sqrt(1.25) or x = 1.618033988749895
now, take the square root, and get 1.272019649514069 set x = 1.272019649514069 and solve for y and get .4858682717566457 add 2 and divide by 10 and get .24858682717566457 and set y = .24858682717566457 solve for x and get 1.131988234781886 this new number is .0000000052181141 different from 1.13198824.
This is a close approximation of Viswanath's Constant.
Another way is take sqrt(sqrt(5)-2) = .4858682717566457 insert a 2 between the decimal point and 4 to get .24858682717566457 and set y = .24858682717566457 solve for x and get 1.131988234781886
or type
((((5^.5-2)^.5+2)/20)^2+1)^.5+(((5^.5-2)^.5)+2)/20
into google
These numbers are fun to play with on calculators with an equation solver.
If you have a solver set y = x - 1/x and see different values of y and x.
The HP 95lx calculator was used in this discovery.
When you use a solver it's easier because you just press keys, and see what happens, like in a game.
This is the sole creation of James Douglass.
This is to see if Google can find this page by
searching for 1.13198823.