Difference between revisions of "Pythagorean Tree"
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<math>a^2 = (s * cos\theta)^2</math> | <math>a^2 = (s * cos\theta)^2</math> | ||
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Which means: | Which means: | ||
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The area of a Pythagorean Tree is actually quite simple to find. Simply put, it is the area of the original square (Iteration "O") multiplied by the number of iterations plus one. Or ''A=s<sup>2</sup>*n+1'', where ''s'' of course is the length of one of the sides, and n is the number of iterations. | The area of a Pythagorean Tree is actually quite simple to find. Simply put, it is the area of the original square (Iteration "O") multiplied by the number of iterations plus one. Or ''A=s<sup>2</sup>*n+1'', where ''s'' of course is the length of one of the sides, and n is the number of iterations. | ||
− | [[ | + | [[Image:Iteration0.png]] |
As you can see, for 0 iterations, A.K.A the original square, the area is ''s<sup>2</sup>''. Or, when using the formula, ''s<sup>2</sup>*0+1=s<sup>2</sup>'' | As you can see, for 0 iterations, A.K.A the original square, the area is ''s<sup>2</sup>''. Or, when using the formula, ''s<sup>2</sup>*0+1=s<sup>2</sup>'' | ||
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Now, on the first iteration, a trend that will become apparent for the every iteration onward appears. | Now, on the first iteration, a trend that will become apparent for the every iteration onward appears. | ||
− | [[ | + | [[Image:Iteration1pytho.png]] |
NOTE:The triangles in a Pythagorean tree are NOT actually there, in the sense that there is no "interior". The sides are all there, but where the interior would be, there is nothing. As such, the triangles are not counted when looking at the area of a Pythagorean Tree. | NOTE:The triangles in a Pythagorean tree are NOT actually there, in the sense that there is no "interior". The sides are all there, but where the interior would be, there is nothing. As such, the triangles are not counted when looking at the area of a Pythagorean Tree. | ||
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The specific trend is that, with every coming iteration, the areas of the squares will be much lower than that of the previous, but there will be twice as many, so the total sum of the areas of the newly iterated squares will become equivalent to the area of the original square. The total sum of the areas of every square in an iteration is equal to the area of the original square. In this case, for example, there was one square of ''s<sup>2</sup>'', and the new iteration had two squares. Since the triangle used to make the tree is a 45-45-90, the two smaller squares are congruent, but even if they weren't, the end result would be the same. The sum of their areas would be equal to the area of the original square. To confirm the formula, ''s<sup>2</sup>+(.5s<sup>2</sup>+.5s<sup>2</sup>)=2s<sup>2</sup>'' This being iteration 1, our formula would be ''A=s<sup>2</sup>*1+1=s<sup>2</sup>*2=2s<sup>2</sup>''. The formula would continue to increase at a similar rate indefinitely. It is a fractal, and can therefore go on to any value. As such, the formula can calculate it for any amount of iterations. | The specific trend is that, with every coming iteration, the areas of the squares will be much lower than that of the previous, but there will be twice as many, so the total sum of the areas of the newly iterated squares will become equivalent to the area of the original square. The total sum of the areas of every square in an iteration is equal to the area of the original square. In this case, for example, there was one square of ''s<sup>2</sup>'', and the new iteration had two squares. Since the triangle used to make the tree is a 45-45-90, the two smaller squares are congruent, but even if they weren't, the end result would be the same. The sum of their areas would be equal to the area of the original square. To confirm the formula, ''s<sup>2</sup>+(.5s<sup>2</sup>+.5s<sup>2</sup>)=2s<sup>2</sup>'' This being iteration 1, our formula would be ''A=s<sup>2</sup>*1+1=s<sup>2</sup>*2=2s<sup>2</sup>''. The formula would continue to increase at a similar rate indefinitely. It is a fractal, and can therefore go on to any value. As such, the formula can calculate it for any amount of iterations. | ||
+ | A 45-45-90 tree will be used to explain the way the perimeter works, because such a tree shows the most basic way it changes. In a 45-45-90 tree, each and every square in an iteration is congruent. Not so much true for other types of trees, and therefore a bit more difficult to find the perimeter. Referring back to.. | ||
+ | |||
+ | [[Image:Iteration1pytho.png]] | ||
+ | |||
+ | |||
+ | This image, we can see that the area of the smaller squares is ''s<sup>2</sup>/2''. Of course, the area of a square is one of the side lengths squared, so if we take ''s<sup>2</sup>/2'', and find the square root, we get one of the side lengths, which is ''s/√2'. Now, if we were to iterate further... | ||
+ | |||
+ | We'd find that the area of the newly iterated squares is ''s<sup>2</sup>/4''. | ||
+ | If you take the square root of that, you'd find that it comes out to a clean ''s/2.'' Also known as ''√2*√2''. Iterate further, the denominator in the area becomes 8. Square root of 8? 2√2. It goes like this indefinitely. For each iteration, you just multiply the denominator of the side length by √2. The numerator is always s. Of course, this only applies to 45-45-90. Although that's only for one side, the formula remains unchanged. All that happens is that you multiply the fraction by 4, since all four sides of a square are congruent. The perimeter of the squares changes wildly when different right triangles are used. | ||
+ | |||
+ | ---- | ||
+ | History of the Tree. | ||
+ | |||
+ | Albert E. Bosman was the one who created the Pythagorean Tree. He did so in 1942. He was a Dutch Mathematician, who was born in 1891 and died in 1961. | ||
+ | |||
+ | [[Image:Images-1.jpg|Albert Einstein, NOT Albert Bosman]] | ||
Revision as of 10:51, 19 June 2013
Pythagorean Tree, in 2 Dimensions |
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Pythagorean Tree, in 2 Dimensions
- A Pythagorean Tree is a fractal that is created out of squares. The space between the squares in each iteration creates a right triangle. The top line of the square becomes the hypotenuse of the triangle above it.
Basic Description
The Pythagorean Tree begins with a square that has a right triangle branching off of it. The hypotenuse of the triangle must always be the one that is directly connected to the square. When the right triangle is created, the legs of said triangle then become one of the sides of two brand new squares. Important to note is that the length of the legs is not changed during this creation, so the squares are smaller than the big one. The sum of the areas of the two smaller squares is equal to the area of the big square. The interesting thing about the tree is that the right triangle can have any valid value of the non right angles. When the angles of the triangle are changed, one is made bigger, and the other is made smaller. Length of sides corresponds to measure of angles, so the sides change too. Since the leg is bigger, the square created using that leg is also bigger, creating the illusion of a tilt.
A More Mathematical Explanation
- Note: understanding of this explanation requires: *Basic Algebra and Geometry, Trigonometry
In this image, CFD is a right triangle. The original square has an area of s^{2}, meaning its side length is s, because the area of a square is its side length squared (s^{2}). This can be put into the Pythagorean theorem: . This means the sum of the areas of the two branched-off squares will always be equal to the original square. This is because the Pythagorean theorem states that the sum of the side lengths squared (in other words, the sum of the areas) equal the hypotenuse, s, squared. These areas are dependent on the side lengths of the right triangle in the middle, which in turn are dependent the angles.
With the hypotenuse s and angle θ_{1}, the length of side b can be found with . This means that . The other side length, a, can be found similarly: , or .
To find the areas of the 2 branched-off squares, square the side lengths:
Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): b^Proxy-Connection: keep-alive Cache-Control: max-age=0 oxy-CProxy-CProxy-Connection: keep-alive Cache-Control: max-age=0 nection: keep-alive Cache-Control: max-age=0 oxy-Connection: keep-alive Cache-Control: max-age=0 oxy-Connection: keep-alive Cache-Control: max-age=0 nection: keep-alive Cache-Control: max-age=0 = (s Proxy-CProxy-CProxy-Connection: keep-alive Cache-Control: max-age=0 nection: keep-alive Cache-Control: max-age=0 nectiProxy-Connection: keep-alive Cache-Control: max-age=0 : keep-alive Cache-Control: max-age=0 oxy-Connection: keep-alive Cache-Control: max-age=0 sin\thetProxy-Connection: keep-alive Cache-Control: max-age=0 ^2
Which means:
The area of a Pythagorean Tree is actually quite simple to find. Simply put, it is the area of the original square (Iteration "O") multiplied by the number of iterations plus one. Or A=s^{2}*n+1, where s of course is the length of one of the sides, and n is the number of iterations.
As you can see, for 0 iterations, A.K.A the original square, the area is s^{2}. Or, when using the formula, s^{2}*0+1=s^{2}
Now, on the first iteration, a trend that will become apparent for the every iteration onward appears.
NOTE:The triangles in a Pythagorean tree are NOT actually there, in the sense that there is no "interior". The sides are all there, but where the interior would be, there is nothing. As such, the triangles are not counted when looking at the area of a Pythagorean Tree.
The specific trend is that, with every coming iteration, the areas of the squares will be much lower than that of the previous, but there will be twice as many, so the total sum of the areas of the newly iterated squares will become equivalent to the area of the original square. The total sum of the areas of every square in an iteration is equal to the area of the original square. In this case, for example, there was one square of s^{2}, and the new iteration had two squares. Since the triangle used to make the tree is a 45-45-90, the two smaller squares are congruent, but even if they weren't, the end result would be the same. The sum of their areas would be equal to the area of the original square. To confirm the formula, s^{2}+(.5s^{2}+.5s^{2})=2s^{2} This being iteration 1, our formula would be A=s^{2}*1+1=s^{2}*2=2s^{2}. The formula would continue to increase at a similar rate indefinitely. It is a fractal, and can therefore go on to any value. As such, the formula can calculate it for any amount of iterations.
A 45-45-90 tree will be used to explain the way the perimeter works, because such a tree shows the most basic way it changes. In a 45-45-90 tree, each and every square in an iteration is congruent. Not so much true for other types of trees, and therefore a bit more difficult to find the perimeter. Referring back to..
This image, we can see that the area of the smaller squares is s^{2}/2. Of course, the area of a square is one of the side lengths squared, so if we take s^{2}/2, and find the square root, we get one of the side lengths, which is s/√2'. Now, if we were to iterate further...
We'd find that the area of the newly iterated squares is s^{2}/4. If you take the square root of that, you'd find that it comes out to a clean s/2. Also known as √2*√2. Iterate further, the denominator in the area becomes 8. Square root of 8? 2√2. It goes like this indefinitely. For each iteration, you just multiply the denominator of the side length by √2. The numerator is always s. Of course, this only applies to 45-45-90. Although that's only for one side, the formula remains unchanged. All that happens is that you multiply the fraction by 4, since all four sides of a square are congruent. The perimeter of the squares changes wildly when different right triangles are used.
History of the Tree.
Albert E. Bosman was the one who created the Pythagorean Tree. He did so in 1942. He was a Dutch Mathematician, who was born in 1891 and died in 1961.
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