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Limit of Area

Date: 03/01/98 at 15:44:59
From: Jason
Subject: Geomety POW-Regarding Mathematical Limitsz

Given a square with sides of length 5. Each side of the square is 
divided into three congruent parts, and a square is constructed 
outwardly on the middle part. The new figure has twenty line segments 
determining its boundary. Each of these is further divided into three 
congruent parts, and again a square is formed outwardly on each middle 
part. The process is continued infinitely. What is the limit 
approached by the area of this figure?

I have tried looking up what limit means, but I don't get what the 
area is according to the limit. I know the number of squares 
multiplies by 5 each time. Can you please help me?

Date: 03/02/98 at 17:30:20
From: Doctor Sam
Subject: Re: Geomety POW-Regarding Mathematical Limits


This is a complex problem. Let's hold off on the idea of "limit" and 
just compute some values.

The first square has side length 5. The next square has side length 5/
3. The next squares will be one-third smaller, or 5/9, and the next 
would be one-third of that, or 5/27.

Do you see the pattern?  The side lengths of the squares are the 

   5, 5/3, 5/(3^2), 5/(3^3), ...

These form what are called a geometric sequence. Each number is a 
multiple (in this case, 1/3) of the number that preceeds it.

If you know the lengths of the sides, you can find the area of each 
size square

   5^2, (5/3)^2, (5/(3^2))^2, (5/(3^3))^2, ...,

which can be simplified to 5^2, 5^2/3^2,  5^2/3^4, 5^2/3^6 ...   

Okay so far?

Now, you said that you know the number of squares increases by a 
factor of 5 each time. So you start with 1, then have an additional 4, 

              another 20 = 4*5, then
             another 100 = 4*5^2, etc.

You can find the total area as you go along, adding up the areas at 
each stage of the construction:

   Total Area =        1 square with side length 5
                    + 4 squares with side length 5/3
                  + 4*5 squares with side length (5/3^2)
                + 4*5*5 squares with side length (5/3^3)
              + ... etc.

Now let's write them all down:

             1*5^2 + 4*5^2/3^2 + 4*5*5^2/3^4 + 4*(5^2)*(5^2)/3^6 + ...

Simplify each fraction (but keep them with exponents to show the 

  Total Area = 5^2 + 4*5^2/3^2 +  4*5^3/3^4  + 4*5^4/3^6 + ...

The idea of a limit is that if you keep adding up more and more terms 
in this pattern, the numbers you get will all get closer and closer to 
some number (called the limit of this series of numbers).

The limit can be computed because this pattern is also a geometric 
series, and there is a formula for the sum of such a pattern. The sum 

                       first term
           Sum  =   ----------------
                    1 - common ratio

To see that this is a geometric series, and to discover its common 
ratio, notice that after the first term every term is 5/9 = 5/3^2 
times the preceeding term.

So the common ratio is 5/9, but you need to skip the first area of 25 
because it isn't multiplied by 4. So take the first term as 4*5.

     Total Area = 5^2 + 4*5^2/3^2 + 4*5^3/3^4 + 4*5^4/3^6 + ...
                = 25  +  20(5/9)  + 20(5/9)^2 + 20(5/9)^3 + ...

so the first term is 20(5/9) and the common ratio is 5/9. The total 
area ... the limit as you keep constructing more and more squares ... 

        20(5/9)          20(5/9)            20(5)
  25  + -------  =  25 + -------   =  25 + ------- = 25 + 25 = 50
        1 - 5/9           (4/9)               4
I hope that helps.

-Doctor Sam, The Math Forum   
Associated Topics:
High School Fractals
High School Geometry
High School Sequences, Series
High School Triangles and Other Polygons

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