Author Topic: putting the geometric back into geometetric series (Read 833 times) Tweet Share

kamil9876 · « **on:** September 12, 2009, 11:29:21 pm »

Let OA=1 and OP is perpendicular to the big hypotenuse. In fact all the little hypotenuses are perpendicular to the big hypotenuse. Let $\theta=<POA$ .

One can now show with simple trigonometry and geometry that a consequence of this is that $AD=sin^2\theta$ . Because the geometrical nature is recursive(ie the pattern of the triangles keeps repeating infinitely) one can show that the base of the next triangle is $sin^4\theta$ and the next one is $sin^6\theta$ etc....

Therefore the length of the base of the big triangle is equal to the sum of the little bases:

$S=1+sin^2\theta+sin^4\theta.... $

With geometry one may now work out that: $OB=\frac{1}{sin\theta cos\theta}$ and that the big hypotenuse is: $\frac{OB}{cos\theta}=\frac{1}{sin\theta cos^2\theta}$ .

Using Pythagoras' theorem on the big triangle we get:
$ S^2=\frac{1}{sin^2\theta cos^4\theta}-\frac{1}{sin^2\theta cos^2\theta}$

$=\frac{1-cos^2\theta}{sin^2\theta cos^4\theta} $

$S=\frac{1}{cos^2\theta}$

therefore:
$ 1+sin^2\theta+sin^4\theta....=\frac{1}{cos^2\theta}=\frac{1}{1-sin^2\theta}$

letting $x=sin^2\theta$ just so that it's clear.
$ 1+x+x^2...=\frac{1}{1-x}$

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Author Topic: putting the geometric back into geometetric series (Read 833 times) Tweet Share

kamil9876

putting the geometric back into geometetric series

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