Pythagorean Tree

Mathematical argument to validating the claim that a Pythagoras corner of initial square size of 1 ×
1 units does condition into a 6 × 4 unit box noning some(prenominal) strengths and/or limitations or your argument.


To find the brass length of a Pythagoras tree you use this formula ? ?2? , substitute for the
2
looping you need. To prove that a Pythagoras Tree of initial square size of 1 × 1 fits into a
6 × 4 box, this pattern is used.
1

As is seen the diagonal from the first iteration is the same length as
the side length of the previous iteration. This pattern then continues
on to infinity. From that the pursuance table is set up for find
the width.
#
1
2
3
4
5
6

Side length
Total Length
2
2
2
2+3
3
2 ??1?2 ?2? ? = 1
4
3.5
4+5
2 ??1?2 ?2? ? = .5
6
3.75
6+7
2 ??1?2 ?2? ? = .25
8
3.875
8+9
2 ??1?2 ?2? ? = .125
10
3.9375
10 + 11 2 ??1? ?2? ? = 0.0625
2

1

Stage
0+1


1 ?
? .
? =

1
=
2

0.5 1
=
=
1
2


0.25 1
=
=
0.5
2


? = 2, = 1?2
? =

2

1
1?2
= 4
? 1 × 1 ? 4

1

For finding length the pattern is changed slightly, as
shown. When finding the length iteration 1 has to
be ignored because it is not part of the straight
then diagonal pattern. As the Pythagoras
Tree is stellate simply double each
2

length 2 get both sides e.g. 2 ?? ?2? ?
2
1

2

becomes 4 ??2 ?2? ? .




1

The neighboring table is then
generated from this
pattern again, but after
the infant sum is
figure 2 units need
to be added on because of stage 1.
#
1
2
3
4
5

1

2
1
2 ?? ?2? ?
2

1

1

Side Length
Total Length
2
2&3
2
4 ??1?2 ?2? ? = 2
4
4&5
3
4 ??1?2 ?2? ? = 1
6
6&7
3.5
4 ??1?2 ?2? ? = .5
8
8&9
3.75
4 ??1?2 ?2? ? = .25
10
10 & 11 4 ??1?2 ?2? ? = 0.125
3.875
Stage


1 ?
? .
1
=
2
? =

0.5 1
=
=
1
2


0.25 1
=
=
0.5
2


? = 2, = 1?2
? =

2

1
1?2
= 4
? + 2 = 4 + 2
= 6
? 1 × 1 ? 6

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