Geometric proof The Quadrature of the Parabola
1 geometric proof
1.1 dissection of parabolic segment
1.2 areas of triangles
1.3 sum of series
geometric proof
dissection of parabolic segment
archimedes dissection of parabolic segment arbitrary number of triangles.
the main idea of proof dissection of parabolic segment infinitely many triangles, shown in figure right. each of these triangles inscribed in own parabolic segment in same way blue triangle inscribed in large segment.
areas of triangles
in propositions eighteen through twenty-one, archimedes proves area of each green triangle 1 eighth of area of blue triangle. modern point of view, because green triangle has half width , fourth of height:
by extension, each of yellow triangles has 1 eighth area of green triangle, each of red triangles has 1 eighth area of yellow triangle, , on. using method of exhaustion, follows total area of parabolic segment given by
area
=
t
+
2
(
t
8
)
+
4
(
t
8
2
)
+
8
(
t
8
3
)
+
⋯
.
{\displaystyle {\mbox{area}}\;=\;t\,+\,2\left({\frac {t}{8}}\right)\,+\,4\left({\frac {t}{8^{2}}}\right)\,+\,8\left({\frac {t}{8^{3}}}\right)\,+\,\cdots .}
here t represents area of large blue triangle, second term represents total area of 2 green triangles, third term represents total area of 4 yellow triangles, , forth. simplifies give
area
=
(
1
+
1
4
+
1
16
+
1
64
+
⋯
)
t
.
{\displaystyle {\mbox{area}}\;=\;\left(1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots \right)t.}
sum of series
archimedes proof 1/4 + 1/16 + 1/64 + ... = 1/3
to complete proof, archimedes shows that
1
+
1
4
+
1
16
+
1
64
+
⋯
=
4
3
.
{\displaystyle 1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots \;=\;{\frac {4}{3}}.}
the formula above geometric series—each successive term 1 fourth of previous term. in modern mathematics, formula special case of sum formula geometric series.
archimedes evaluates sum using entirely geometric method, illustrated in adjacent picture. picture shows unit square has been dissected infinity of smaller squares. each successive purple square has 1 fourth area of previous square, total purple area being sum
1
4
+
1
16
+
1
64
+
⋯
.
{\displaystyle {\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots .}
however, purple squares congruent either set of yellow squares, , cover 1/3 of area of unit square. follows series above sums 4/3.
Comments
Post a Comment