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rloginunix
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Converging Geometric Series
« on: Sep 10th, 2016, 7:34am » |
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Converging Geometric Series
Show that the magnitude of the sum of any converging geometric series:
a + ar + ar2 + ar3 + ... + arn + ...
where:
-1 < r < 1
as:
n  
can be found with Euclidean straight edge and compass by constructing only two line segments whose lengths represent the magnitudes of the first two terms of the above series.
Geometrically, a(=1) and some mysterious equivalent of r are given. Exactly what this equivalent of r is - is for you to figure out. Once you figure it out - consider that it is given (in a geometric form).
- graphic illustrations on wiki are not it
- do not forget the case when -1 < r < 0
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dudiobugtron
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Re: Converging Geometric Series
« Reply #1 on: Sep 11th, 2016, 2:03am » |
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If we let a=1 as you suggest, it will be easier! So let's do that.
Next:
The first and second terms are 1 and r. So r will be represented by the length of the second line you draw.
I assume you start with a line connecting two points A and B. The length of AB is a, or 1. If you don't start with anything, you can construct it yourself, choosing any two non-coincident points as A and B. Extend the line so it continues a decent distance past A and past B as well, since you'll need it later.
Next, set your compass to the length r (you can choose any length 0 < |r| < 1 - it will work whatever you choose. If you are already given the length that corresponds to r, then choose that.) Mark a circle around point A, and mark the points where it crosses the line through A and B as "D'" and "E". D is in between A and B, and E is the other point.
If r is positive, then r is represented by AD. If it's negative, it's represented by AE. (If r is zero, then the sum of the series is just a, so you don't really need to construct any more lines.)
The formula the sum of a converging series is a/(1-r). So we need to find a way to represent that geometrically. We can do this using the ratios of similar triangles.
For positive r:
1-r is represented by the distance BD.
Construct a line through B that is perpendicular to AB. Set the compass to length BD (ie: 1-r), and mark off point F on that line you just constructed, at the set distance (1-r) from point B.
Connect up A to F and you have a right-angled triangle ABF. The ratio of AB to BF is 1/(1-r).
Next, set the compass to distance AB (ie, 1), and mark a new point K on the perpendicular line through B and F that is a distance of 1 from B. (It should be on the same side of line AB as F is.) Then, construct a line through K that is perpendicular to line AF. By connecting it up to the line that passes through AB, and marking the point of intersection as L, you now have a new right angled triangle LBK, which is similar to ABF. Call the length of LB 'x'. The ratio of LB to BK is x/1, or just x.
By similar triangles, x = 1/(1-r), so we have our representation of the sum of the series: LB.
For negative r:
This is basically the same as above; the only real change is that BE (=1-r) is larger than a (=1), so the second triangle you construct will be smaller than the first one. You still use the same trick though, comparing 1/(1-r) to y/1 by using similar triangles.
Conveniently, you already have point A marked off, so it's even easier to construct the answer in this case. I can give the details if needed but it's very similar to the above.
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rloginunix
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Re: Converging Geometric Series
« Reply #2 on: Sep 11th, 2016, 8:10am » |
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Very good.
However, the algorithm that you have outlined works when two linear magnitudes are given: a and r. In this case, only one linear magnitude is given: a line segment of length a, say it is 1. The second parameter that is given is a (constant) common ratio r.
It may be argued that if r is geometrically convenient, 1/2 or 1/3 or 1/4 or 1/n, then the linear equivalent of the second term of the converging geometric series may be constructed with a straight edge and a compass (that algorithm is well known), in which case the above algorithm is applicable (consider it covering the case when r is representable is a rational number p/q in its lowest form).
But we are somewhat ambitious in the above theorem: any r (such that -1 < r < 1).
What if, and this is a hint - not a random number pulled out of the thin air, r = 0.707106781...?
How do we translate "a times 0.707106781..." into geometry?
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dudiobugtron
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Re: Converging Geometric Series
« Reply #3 on: Sep 11th, 2016, 12:49pm » |
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To construct 1/root(2), you can first construct root(2) by making a right-angled triangle of base and height 1. Then you can use the division algorithm I outlined above to get 1/root(2). Or, you could just bisect the line of length root(2).
I don't think this helps me see yet how to solve the issue in general though. I think what you are getting at is: "what if you are given a specific r that you *can't construct*?"
But I don't see how that is solvable - if you can't construct it, then if you could multiply a by it, you could construct it from that result. So therefore you can't multiply a by it.
Edit: wait, I think I see what you mean. Are you talking about being given an angle between 0 and 180, and then having r represented by cosine of that angle? This still has the same problem though: If you can't construct the angle, you can't construct the side length, and vice versa.
So, perhaps you are given a line (the x axis), and are given AB already constructed at a given angle from that line (A is the point of intersection with the x axis, and B is the other endpoint). In that case, you can just construct a perpendicular line from the x axis that passes through B, and label the point where it passes the x axis as C. Then the r you are looking for will be AC.
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| « Last Edit: Sep 11th, 2016, 3:41pm by dudiobugtron » |
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rloginunix
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Posts: 1029
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Re: Converging Geometric Series
« Reply #4 on: Sep 12th, 2016, 8:52am » |
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on Sep 11th, 2016, 12:49pm, dudiobugtron wrote:| Edit: wait, I think I see what you mean. Are you talking about being given an angle between 0 and 180, and then having r represented by cosine of that angle? This still has the same problem though: If you can't construct the angle, you can't construct the side length, and vice versa |
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That's right: r varies continuously, within a bounded range we interpret it as a value of cos(alpha) where alpha is an angle between zero and one eighty, an arbitrary r then corresponds to a cosine of exactly one angle = arccos(r) (sin(x) does not work).
Do not confuse constructing with given. Remember, that the above angle, an equivalent of r (as we now established), is given (in a geometric form - say, literally drawn on paper. It is possible then to make a copy of a given angle, 4 circles - but there is no need to be so picky. Assume that an arbitrary angle between zero and one eighty is given and we are free to simply draw it anywhere we need to (so now we have one linear and one angular measure as input).
(imho: this is one of these situations when using GeoGebra or Cinderella is quite beneficial - it is possible to vary the angle in question and observe the result)
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