The sequence is as follows: O, -1, +1, -2, +2, -3, +3, -4
If you imagined your self on a road (0), then the first number (minus 1) is 50(ish) miles behind you, the next number is 100 miles in front etc.,etc.
I need the minus 4 number to be 500 (ish) miles behind.
The numbers need to have some natural logic...obviously this may not be possible.
I am no mathmatics expert (in any way) so I probably haven't made myself clear, but will explain it further if you need me to.
0 = Here.
-1 = 50 behind
+1 = 100 ahead
-2 = 150 behind
+2 = 200 ahead
-3 = 250 behind
+3 = 300 ahead
-4 = 350 behind
+4 = 400 ahead
Right? Except that -4 needs to be 500 instead of 350? The math works out to be approx. 70 as the steps needed to achieve this (71 point something, actually)
0, 70 behind, 140 ahead, 210 behind, 280 ahead, 350 behind, 420 ahead, 490 behind.
Making the steps 71 gets you to 497.
Is this what you wanted? Or am I completely confusing your question?
0, -70, 70, -140, 140, -210, 210, -280, 280, -350, 350, -420, 420, -490
Wolf-boy solved the problem so that the difference between +3 and -4 (210 and -280) is about 500. Is that what you needed? Or did you want -500 with respect to the origin point? In that case, it's a simple case of division (500/4=125) and you get the sequence:
0, -125, 125, -250, 250, -375, 375, -500
[This message has been edited by Kitti (edited October 03, 2009).]
0, 50, 150, etc.
Except, I wanted an elegant series based on some natural elaboration...
0, 1, 4, 16, 256 (0, +1, -3, +13)
I'm confusing myself now...
http://en.wikipedia.org/wiki/Achilles_number
...and it just so happens that the 7th number in the sequence is 500, precisely. The sequence of the numbers isn't exactly elegant, but it is pretty cool. It's also at least one university-level math course over my head.
xn1 + yn1 = xn2...
0 + (-1) = -1, minus y interval of one, -x interval of one
-1 + (2) = 1, plus y interval of three, +x interval of two
1 + (-3) = -2, minus y interval of five, -x interval of three
-2 + (4) = 2, plus y interval of seven, +x interval of four
2 + (-5) = -3, minus y interval of nine, -x interval of five
-3 + (6) = 3, plus y interval of eleven, +x interval of six
3 + (-7) = -4, minus y interval of thirteen, -x interval of seven
-4 + (8) = 4, plus y interval of fifteen, +x interval of eight
4 + (-9) = -5, minus y interval of seventeen, -x interval of nine
In order to end at minus 500 miles from starting point 0 in nine intervals multiply xn and yn number variables in the matrix times 100, travel 100 miles from origin 0 to first stop at minus 100 miles, travel 200 miles for second stop at plus 100 miles, travel three hundred miles to third stop at minus 200 miles, and so on, ninth stop at minus 500 miles after traveling 900 miles from eighth stop. At least it's proportionately valid. y could represent delta-V in terms of energy expenditure.
The distance (y) at position "i" along this sequence (if you define i=0 to correspond to y=0) is given by:
y(i) = [125*i * (-1)^i] + y(i-1).
This sequence will go:
y(0)=0 (by definition)
y(1)=-125
y(2)= 125
y(3)=-250
y(4)= 250
y(5)=-375
y(6)= 375
y(7)=-500
y(8)= 500
...
The 125 factor in the formula is to give you a distance of +/- 500 at i=7,8, as you requested.
I think I'll stick with astrostewart's formula--not elegant, but it works for what I need.
AstroStewart's solution graphed distance over time describes a zigzag course relative to the starting point, where x equals time and y equals distance, the point of origin is (0, 0). The sequential location points on the graph are (xsub1, -125), (xsub2, 125), (xsub3, -250), (xsub4, 250), (xsub5, -375), (xsub6, 375), (xsub7, -500), (xsub8, 500). The values for x equals time aren't given, but arbitrarily assigning increasing values--(1, 2, 3, 4, 5, 6, 7, 8)--are a needed factor for graphing so that the oscillations are visible on the graph instead of one line segment running through all points between and including -500 and 500.
[This message has been edited by extrinsic (edited October 04, 2009).]
quote:
In skadder's problem, a person "instantaneously teleports" repeatedly in an oscillating proportionate divergent sequence relative to a starting place.
Correct except its time travel and the units are years.
Fair warning: if you're sitting about with a bunch of bored number theorists, "stump the guy closest to his PhD" is an addictive timekiller.
I have changed my requirements a little.
I think--if my calculations are correct--that the final jump would be minus 430 years, which is near enough to the point in history that I want.
-10
+10
-30
+50
-110
+210
-430
Does the above sequence have a name?
5*(-1)^n*(Expansion of (1-x)/(1-x-2*x^2))
By "Expansion" of the polynomial fraction listed, I basically mean the coefficients of the (infinite) polynomial such that (1-x-2*x^2) multiplied by that poly would give you 1-x.
Again, from the web site:
quote:
FORMULAEuler expands(1-x)/(1-x-2*x^2) into an infinite series and finds that the coefficient of the n-th term is (2^n + (-1)^n 2)/3. Section 226 shows that Euler could have easily found the recursion relation: a(n) = a(n-1) + 2a(n-2) with a(0) = 1 and a(1) = 0. - V. Frederick Rickey (fred-rickey(AT)usma.edu), Feb 10 2006. [Typos corrected by Jaume Oliver i Lafont, Jun 01 2009]
[This message has been edited by micmcd (edited October 06, 2009).]