Hatrack River Writers Workshop: Any maths experts?

This is topic Any maths experts? in forum Open Discussions About Writing at Hatrack River Writers Workshop.

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Posted by skadder (Member # 6757) on :

I need a number series (that has a logic to it) that will work around a problem I have to solve in a story.

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.

Posted by Wolfe_boy (Member # 5456) on :

Just to be clear.... is this what you're sort of talking about?

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?

Posted by Kitti (Member # 7277) on :

I would have phrased all my numbers in relationship to the origin point (0) which would make the sequence:

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).]

Posted by skadder (Member # 6757) on :

Yes, it's all from the origin point, so that the sequence may look like (if you just looked at the amount of each step):

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...

Posted by Wolfe_boy (Member # 5456) on :

Well, a little work with my friend Wikipedia bought me to this page...

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.

Posted by extrinsic (Member # 8019) on :

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 by AstroStewart (Member # 2597) on :

In case you were looking for an actual formula for this series, I've managed to summarize extrinsic's solution into an actual formula.

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.

Posted by skadder (Member # 6757) on :

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 by Wolfe_boy (Member # 5456) on :

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!

Posted by extrinsic (Member # 8019) on :

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).]

Posted by skadder (Member # 6757) on :

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 by extrinsic (Member # 8019) on :

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.

Posted by Corky (Member # 2714) on :

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?

Posted by extrinsic (Member # 8019) on :

Come back to minus one yielding a minus interval of two from plus one.

Posted by micmcd (Member # 7977) on :

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 by skadder (Member # 6757) on :

10, 20, 40, 80, 160, 320, 640

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?

Posted by extrinsic (Member # 8019) on :

Not a proper noun name that I know of. Authorial license doesn't preclude inventing one.

Posted by micmcd (Member # 7977) on :

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 by MrsBrown (Member # 5195) on :

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.