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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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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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(-1+k2

for k = 0 to n, plus the point (1,0)?

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Yes. How do I find the area though?

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(-1+k2^{-n}, 1-(-1+k(2)^{-n})^{2})

for k = 0 to n, plus the point (1,0)?

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?Yes. How do I find the area though?

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[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).

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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?

What is your response to the red part?

[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).

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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.

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

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

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- #13

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[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).

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.

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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.*

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:

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SammyS

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OK, so let's call this (xAre the coordinates of the kth point

(-1 + k , 1-(-1+k(2)^{-n})^{2})

for k = 0 to n, plus the point (1,0)?

Therefore, y

Furthermore, there are 2

x_{n+1}=-1 + (2^{n+1})(2)^{-n} = -1 + 2 = 1, which gives the right endpoint.

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

This gives a max. value for the parabola of 1 + (-1 + (2

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

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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.

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: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.

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

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(-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?

Those aren't the x coordinates he gave. His aren't equally spaced and the points do indeed lie in a straight line.OK, so let's call this (x_{k}, y_{k}), i.e. x_{k}= -1 + k 2^{-n}and y_{k}= 1-(-1+k(2)^{-n})^{2}.

Therefore, y_{k}= 1 - (x_{k})_{2}, so that the points (x_{k}, y_{k}) lie on a parabola.

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

This gives a max. value for the parabola of 1 + (-1 + (2^{n})(2)^{-n})^{2}= 1 + (-1 + 1) = 1 at x = 0 . (This is the (2^{n})^{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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That's strange, because if you do consider them to be (x, y) coordinates, it all works out as it should.Those aren't the x coordinates he gave. His aren't equally spaced and the points do indeed lie in a straight line.

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The answer that OP gave to this question was YES. So these are indeed coordinates, with x

(-1+k2^{-n}, 1-(-1+k(2)^{-n})^{2})

for k = 0 to n, plus the point (1,0)?

However, as I see it, k goes from 0 to 2

Added in

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

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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.The answer that OP gave to this question was YES. So these are indeed coordinates, with x_{k}= -1+k 2^{-n}, etc.

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

Added inEdit:

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

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Look at the altitudes.

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

Look at the altitudes.

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Nevermind, I got it!

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For n = 0 , let's plug in values starting with point (-1, ) until we get to point (1, 0), remembering that 2The 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]

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) = 1/4

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