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Infinite summations: area of polygon

  • #1
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Prove that the area of the polygon with vertices at (-1,0), (-1+2^(-n), 1-(-1+2^(-n))^2), (-1+2(2)^(-n), 1-(-1+2(2)^(-n))^2),..., (1,0) is 1 + 4^(-1) + 4^(-2) + ... + 4^(-n).

I tried using the formula for the area of a polygon but could not get the answer. Now sure how to prove this.
 

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  • #2
LCKurtz
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Are the coordinates of the kth point

(-1+k2-n, 1-(-1+k(2)-n)2)

for k = 0 to n, plus the point (1,0)?
 
  • #3
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Yes. How do I find the area though?
 
  • #4
LCKurtz
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Are the coordinates of the kth point

(-1+k2-n, 1-(-1+k(2)-n)2)

for k = 0 to n, plus the point (1,0)?
Yes. How do I find the area though?
Are you sure neither you or I have mistyped anything? Your original post is very hard to parse. This sequence of points lie in a straight line. Is that right? Did you know that? If that is correct, isn't calculating the area straightforward? :confused:
 
  • #5
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The vertices are:

[tex](-1,0), \ (-1+2^{-n},1-(-1+2^{-n})^{2}), \ (-1+2 \cdot 2^{-n},1-(-1+2\cdot2^{-n})^{2}), \ (-1+3\cdot2^{-n},1-(-1+3\cdot2^{-n})^{2}), ...,(1,0)[/tex]

I have to show that the area of the polygon is 1 + 4^(-1) + 4^(-2) + ... + 4^(-n).
 
  • #6
LCKurtz
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Are you sure neither you or I have mistyped anything? Your original post is very hard to parse. This sequence of points lie in a straight line. Is that right? Did you know that? If that is correct, isn't calculating the area straightforward? :confused:
The vertices are:

[tex](-1,0), \ (-1+2^{-n},1-(-1+2^{-n})^{2}), \ (-1+2 \cdot 2^{-n},1-(-1+2\cdot2^{-n})^{2}), \ (-1+3\cdot2^{-n},1-(-1+3\cdot2^{-n})^{2}), ...,(1,0)[/tex]

I have to show that the area of the polygon is 1 + 4^(-1) + 4^(-2) + ... + 4^(-n).
What is your response to the red part?
 
  • #7
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Wouldn't the area be 0 if all of the points lie in a straight line? Or are you referring to all of the points other than (1,0)? If so, that would make it a triangle?
 
  • #8
LCKurtz
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Wouldn't the area be 0 if all of the points lie in a straight line? Or are you referring to all of the points other than (1,0)? If so, that would make it a triangle?
I'm referring to everything except (1,0). And the area of that triangle is nowhere near the sum it is supposed to be. Something is very wrong with your problem.
 
  • #9
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I triple checked to make sure I typed it correctly. Must be an error in the book then...
 
  • #10
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I assume the answer "1 + 4^(-1) + 4^(-2) + ... + 4^(-n)" has to do with Archimedes' argument that 4/3 = 1 + 4^(-1) + 4^(-2) + ... + 4^(-n) is the approximation of the area of a segment bounded by a parabolic region. I'm just not sure how to get it.
 
  • #11
LCKurtz
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I don't see where your sequence has anything to do with a parabola.
 
  • #12
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What do you get for the area of the triangle, out of curiosity? Apparently there is no error in this question, so it should be able to be solved. I still haven't been able to though.
 
  • #13
LCKurtz
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The vertices are:

[tex](-1,0), \ (-1+2^{-n},1-(-1+2^{-n})^{2}), \ (-1+2 \cdot 2^{-n},1-(-1+2\cdot2^{-n})^{2}), \ (-1+3\cdot2^{-n},1-(-1+3\cdot2^{-n})^{2}), ...,(1,0)[/tex]

I have to show that the area of the polygon is 1 + 4^(-1) + 4^(-2) + ... + 4^(-n).
What do you get for the area of the triangle, out of curiosity? Apparently there is no error in this question, so it should be able to be solved. I still haven't been able to though.
The base of the triangle is [-1,1] and it's height if you stop after the nth point (which is the maximum so far) is:

[tex]1-(-1+3\cdot2^{-n})^{2}[/tex]

So 1/2 base x height just gives this value, which goes to 0 as n → ∞. Like I said before, something is wrong with this problem.
 
  • #14
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Why is that point the maximum?

When n = 0, the area of the triangle is 1.

Isn't this question similar to the one found on this link: http://jwilson.coe.uga.edu/EMT668/EMT668.Folders.F97/Edenfield/Convergent%20Sequences/Series.html [Broken]

around the middle of the page where it says:

So, can we find the area of the set of infinite equilateral triangles? Sure. The area of the second triangle is 1/4 of the area of the original triangle. The area of the third triangle is 1/16 of the original, etc.
 
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  • #15
SammyS
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Are the coordinates of the kth point

(-1 + k  , 1-(-1+k(2)-n)2)

for k = 0 to n, plus the point (1,0)?
OK, so let's call this (xk, yk), i.e. xk = -1 + k 2-n and yk = 1-(-1+k(2)-n)2.

Therefore, yk = 1 - (xk)2 , so that the points (xk, yk) lie on a parabola.

Furthermore, there are 2n+1 points, that is to say, k goes from 0 to 2n+1.
xn+1=-1 + (2n+1)(2)-n = -1 + 2 = 1, which gives the right endpoint.

This gives a max. value for the parabola of 1 + (-1 + (2n)(2)-n)2 = 1 + (-1 + 1) = 1 at x = 0 . (This is the (2n)th point.)​

Yes, the x coordinates of the points are equally spaced, and thus are linear in k, but the points themselves do not fall along a straight line.
 
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  • #16
LCKurtz
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The base of the triangle is [-1,1] and it's height if you stop after the nth point (which is the maximum so far) is:

[tex]1-(-1+3\cdot2^{-n})^{2}[/tex]

So 1/2 base x height just gives this value, which goes to 0 as n → ∞. Like I said before, something is wrong with this problem.
Why is that point the maximum?

When n = 0, the area of the triangle is 1.

Isn't this question similar to the one found on this link: http://jwilson.coe.uga.edu/EMT668/EMT668.Folders.F97/Edenfield/Convergent%20Sequences/Series.html [Broken]

around the middle of the page where it says:

So, can we find the area of the set of infinite equilateral triangles? Sure. The area of the second triangle is 1/4 of the area of the original triangle. The area of the third triangle is 1/16 of the original, etc.
You haven't posted any geometry to go with this problem so I don't see any connection. All I see is a triangle. If you take n = 20 and plot the y values for k = 1 to 20 you get, courtesy of Maple:

0.1907347723e-5, 0.3814693628e-5, 0.5722037713e-5, 0.7629379979e-5, 0.9536720427e-5, 0.1144405906e-4, 0.1335139586e-4, 0.1525873085e-4, 0.1716606403e-4, 0.1907339538e-4, 0.2098072491e-4, 0.2288805263e-4, 0.2479537852e-4, 0.2670270260e-4, 0.2861002486e-4, 0.3051734529e-4, 0.3242466391e-4, 0.3433198071e-4, 0.3623929570e-4, 0.3814660886e-4

They are increasing and the last one is the largest. I have said all along something is wrong and I don't know what it is. Why don't you post a picture of the geometry that goes with this problem.
 
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  • #17
LCKurtz
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Are the coordinates of the kth point

(-1+k2-n, 1-(-1+k(2)-n)2)

for k = 0 to n, plus the point (1,0)?
Yes. How do I find the area though?
OK, so let's call this (xk, yk), i.e. xk = -1 + k 2-n and yk = 1-(-1+k(2)-n)2.

Therefore, yk = 1 - (xk)2 , so that the points (xk, yk) lie on a parabola.

Furthermore, there are 2n+1 points, that is to say, k goes from 0 to 2n+1.
xn+1=-1 + (2n+1)(2)-n = -1 + 2 = 1, which gives the right endpoint.

This gives a max. value for the parabola of 1 + (-1 + (2n)(2)-n)2 = 1 + (-1 + 1) = 1 at x = 0 . (This is the (2n)th point.)​

Yes, the x coordinates of the points are equally spaced, and thus are linear in k, but the points themselves do not fall along a straight line.
Those aren't the x coordinates he gave. His aren't equally spaced and the points do indeed lie in a straight line.
 
  • #18
SammyS
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Those aren't the x coordinates he gave. His aren't equally spaced and the points do indeed lie in a straight line.
That's strange, because if you do consider them to be (x, y) coordinates, it all works out as it should.
 
  • #19
SammyS
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Are the coordinates of the kth point

(-1+k2-n, 1-(-1+k(2)-n)2)

for k = 0 to n, plus the point (1,0)?
The answer that OP gave to this question was YES. So these are indeed coordinates, with xk = -1+k 2-n , etc.

However, as I see it, k goes from 0 to 2n+1 .

Added in Edit:

The interval [-1, 1] is cut in half n+1 times.
 
  • #20
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The answer that OP gave to this question was YES. So these are indeed coordinates, with xk = -1+k 2-n , etc.

However, as I see it, k goes from 0 to 2n+1 .

Added in Edit:

The interval [-1, 1] is cut in half n+1 times.
Ok, so then would it be sufficient to say that the "original triangle" (when n=0) has an area of 1. Then the base of the second triangle is 1/2, and the area is (1/2)*(1/2)*1 = 1/4, and so on. Thus, the area of the polygon in question is sum from i = 0 to n 1/4^i = 1 + 1/4 + 1/16 + 1/64 + ... + 1/4^(-n) as required.
 
  • #21
SammyS
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The "base" of the second triangle is 1/2 of the base of the "original triangle", but the "original triangle" is 2 as I see it. There are two second triangles, 4 at the next level, 8 at the next, etc.

Look at the altitudes.
 
  • #22
LCKurtz
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I still don't get the picture you guys are talking about. The first point is [-1,0] and [1,0] is added at the end. That gives a base 2 units long regardless of the value of n. The other points are in a straight line approaching [-1,0]. Here's the picture I get for n = 20. The points appear as though they are in a vertical line since the x coordinates are so close to -1, but the line isn't exactly vertical.

triangle.jpg
 
  • #23
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The "base" of the second triangle is 1/2 of the base of the "original triangle", but the "original triangle" is 2 as I see it. There are two second triangles, 4 at the next level, 8 at the next, etc.

Look at the altitudes.
I don't get why there are two second triangles.
 
  • #24
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Nevermind, I got it!
 
  • #25
SammyS
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Let's look at the verticies as given by BrownianMan in post # 5 0f this thread.
The vertices are:
[tex](-1,0), \ (-1+2^{-n},1-(-1+2^{-n})^{2}), \ (-1+2 \cdot 2^{-n},1-(-1+2\cdot2^{-n})^{2}), \ (-1+3\cdot2^{-n},1-(-1+3\cdot2^{-n})^{2}), ...,(1,0)[/tex]
For n = 0 , let's plug in values starting with point (-1, ) until we get to point (1, 0), remembering that 20 = 1. [tex](-1,0), \ (-1+2^{0},1-(-1+2^{0})^{2}), \ (-1+2 \cdot 2^{0},1-(-1+2\cdot2^{0})^{2}), \ (-1+3\cdot2^{0},1-(-1+3\cdot2^{0})^{2}), ...,(1,0)[/tex] Using only the first three of these verticies gets us to the right hand vertex, (1, 0). [tex](-1,0), \ (-1+(1),1-(-1+(1))^{2}), \ (-1+2 \cdot (1),1-(-1+2\cdot(1))^{2})[/tex] [tex]=(-1,0), \ (0,1), (1,\ 0)[/tex]
These verticies give us an isosceles triangle of area 1.


For n=1, the verticies are: [itex](-1,0),\ (-\frac{1}{2},\ \frac{3}{4}),\ (0,1),\ (\frac{1}{2},\ \frac{3}{4}), (1,\ 0)\,.[/itex] These verticies form two triangles. The base of each is coincident with each of the sides of equal measure in the above isosceles triangle. Each of these triangles is equal in area to a triangle with a base of length 1, and altitude of 1/4 . The sum of area of these two triangles is 1/4 .


For n=2, there are 9 verticies, forming 4 triangles, each "stacked" on top of the triangles formed by n = 0, and n = 1. Each of these has area equal to a triangle with base of length 1/2 and altitude of 1/16. (i would expect a student solving this would have to show this fact.) The sum of area of these four triangles is 4 (½) (1/2) (1/16) = 1/42


 

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