Theorem: π/4 = 1 - 1/3 + 1/5 - 1/7 + ...
Proof:
(1) D(tan-1x)= 1/(x2 + 1) [See Theorem, here]
(2) So:
∫ dx/(x2 + 1) = tan-1 x [See Theorem 2, here]
(3) Since tan(0) = 0 and tan(π/4) = 1, we have:
(4) If we assume that (x2 + 1) ≠ 0, then we have:
1/(x2 + 1) = 1 - x2 + x4 - x6 + x8 - ...
since:
(a) Let a = 1 - x2 + x4 - x6 + x8 - ...
(b) Then:
x2*a = x2 - x4 + x6 - x8 + ...
(c) Then:
a + x2*a = 1
(d) a + x2*a = a(1 + x2) = 1
(e) So:
a = 1/(1 + x2)
which means that:
1/(1 + x2) = 1 - x2 + x4 - x6 + x8 - ...
(5) But this means that:
(6) Further, we know that:
1 - 1/3 + 1/5 - 1/7 + ...
(7) Putting step #3 together with step #5 and step #6 gives us:
π/4 = 1 - 1/3 + 1/5 - 1/7 + ...
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations
, World Scientific, 2001
2 comments:
This solution is wrong because we can only use 1/(1+x^2)=1-x^2+x^4...... at |x|<1, because it is geometric progression. If you do like that, I can prove 1=0.
|x| is lesser than one in this case because it is actually lim y->1 integral from x=0 to x=y 1/(1+x^2)
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