# Homework Help: Infinite summations: area of polygon

1. Sep 17, 2011

### BrownianMan

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.

2. Sep 17, 2011

### LCKurtz

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. Sep 18, 2011

### BrownianMan

Yes. How do I find the area though?

4. Sep 18, 2011

### LCKurtz

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?

5. Sep 18, 2011

### BrownianMan

The vertices are:

$$(-1,0), \ (-1+2^{-n},1-(-1+2^{-n})^{2}), \ (-1+2 \cdot 2^{-n},1-(-1+2\cdot2^{-n})^{2}), \ (-1+3\cdot2^{-n},1-(-1+3\cdot2^{-n})^{2}), ...,(1,0)$$

I have to show that the area of the polygon is 1 + 4^(-1) + 4^(-2) + ... + 4^(-n).

6. Sep 18, 2011

### LCKurtz

What is your response to the red part?

7. Sep 19, 2011

### BrownianMan

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. Sep 19, 2011

### LCKurtz

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. Sep 19, 2011

### BrownianMan

I triple checked to make sure I typed it correctly. Must be an error in the book then...

10. Sep 19, 2011

### BrownianMan

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. Sep 19, 2011

### LCKurtz

I don't see where your sequence has anything to do with a parabola.

12. Sep 23, 2011

### BrownianMan

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. Sep 23, 2011

### LCKurtz

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:

$$1-(-1+3\cdot2^{-n})^{2}$$

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. Sep 23, 2011

### BrownianMan

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.

Last edited by a moderator: May 5, 2017
15. Sep 23, 2011

### SammyS

Staff Emeritus
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.

Last edited: Sep 23, 2011
16. Sep 23, 2011

### LCKurtz

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.

Last edited by a moderator: May 5, 2017
17. Sep 23, 2011

### LCKurtz

Those aren't the x coordinates he gave. His aren't equally spaced and the points do indeed lie in a straight line.

18. Sep 23, 2011

### SammyS

Staff Emeritus
That's strange, because if you do consider them to be (x, y) coordinates, it all works out as it should.

19. Sep 23, 2011

### SammyS

Staff Emeritus
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 .

The interval [-1, 1] is cut in half n+1 times.

20. Sep 23, 2011

### BrownianMan

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. Sep 23, 2011

### SammyS

Staff Emeritus
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. Sep 24, 2011

### LCKurtz

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.

23. Sep 24, 2011

### BrownianMan

I don't get why there are two second triangles.

24. Sep 24, 2011

### BrownianMan

Nevermind, I got it!

25. Sep 24, 2011

### SammyS

Staff Emeritus
Let's look at the verticies as given by BrownianMan in post # 5 0f this thread.
For n = 0 , let's plug in values starting with point (-1, ) until we get to point (1, 0), remembering that 20 = 1. $$(-1,0), \ (-1+2^{0},1-(-1+2^{0})^{2}), \ (-1+2 \cdot 2^{0},1-(-1+2\cdot2^{0})^{2}), \ (-1+3\cdot2^{0},1-(-1+3\cdot2^{0})^{2}), ...,(1,0)$$ Using only the first three of these verticies gets us to the right hand vertex, (1, 0). $$(-1,0), \ (-1+(1),1-(-1+(1))^{2}), \ (-1+2 \cdot (1),1-(-1+2\cdot(1))^{2})$$ $$=(-1,0), \ (0,1), (1,\ 0)$$
These verticies give us an isosceles triangle of area 1.

For n=1, the verticies are: $(-1,0),\ (-\frac{1}{2},\ \frac{3}{4}),\ (0,1),\ (\frac{1}{2},\ \frac{3}{4}), (1,\ 0)\,.$ 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