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)

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).]

...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.

posted
Seems like a two-body orbit problem, which I don't have the trigonometric math to solve, sort of like the shuttle increasingly overcompensating while attempting docking with the ISS. A diverging progessive integer sequence might answer such a challenging problem that resembles the proverbial word problem two steps backward for every step forward, with escalating delta-V at each interval of correction. However, a valid trigonometric formula for that kind of sequence requires a complicated equation with inverse and absolute sign operators and an incrementing factor that are beyond my math skills. Something like this invalid algebra equation; xn1 + [(|yn1|+1)(-1)] = xn2..., by no means a valid trigonometric equation, put simpler from Kentuck windage math;

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.

posted
Thanks everyone. I think I have what I need. The Achilles number doesn't quite work as the sequence doesn't represent the actual gaps between each step.

I think I'll stick with astrostewart's formula--not elegant, but it works for what I need.

posted
I'm still not exactly clear what you were after here... or how a formula like extrinsic/Astro came up with will work into a narrative. Oh well, best of luck!
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posted
A two-body orbit problem assumes that a moving object vectors relative to a stationary object. In skadder's problem, a person "instantaneously teleports" repeatedly in an oscillating proportionate divergent sequence relative to a starting place.

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.

posted
Ah-hah! Temporal displacement. I've been wrangling with a similar premise for a story idea. Capturing a tachyon consequent in a hypercritical temperature superfluid propels a vessel across the metaverse in an instant of objective time, but instantaneously in subjective time. However, there's a temporal displacement quotient relative to tardyon universe distance traveled. The destination must be of sufficient gravitational pull to pull the vessel down from the metaverse, and must be optically selected in order to accurately target on it. Too far away, though, and the occupants will come out the other end as corpsicles. Metal fatigue hazards from heat-cold oscillations too.
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posted
So what I'm not clear about is if you go out one (+1) and then you come back one (-1), don't you find yourself in the original place (0)? And so on?
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posted
If you need the explanation for a sequence of integers that looks more "elegant," you could try just picking out the numbers as you please and entering them into the On Line Encyclopedia of Integer Sequences.

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.

posted
If you're looking for a formula, according to the link I just posted that sequence (-10, 10, -30...) is an early subsequence of the sequence created by:

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: FORMULA

Euler 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).]

posted
Dag, I feel thick! Algebra, Trig, and Calc I, II and III in college, you'd think I'd be able to follow along. Seven years not using it... sigh.
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