proof:(θ) e^iπ +1 = 0 from Maclaurin series; f(x) = f(0) + x f1(0) + x^2/2! f2(0) + x^3/3! f3(0) + x^4/4! f4(0) .... for e^x f(0) = e^0 = 1 f1(0) = e^x = e^0 = 1 f2(0) = e^x= e^0 =1 f3(0) = e^0 = 1 f4(0) = e^0 = 1 so, e^x = 1 + x + x^2/2! + x^3/3! + x^4/4!+ .... e^ix = 1 + ix + - x^2/2! - ix^3/3! + x^4/4!+... (a) when; f(x) = Cosx f(0) = 1 f1(x)= -sinx; f1(0) = 0 f2(x) = -cosx; f2(0) = -1 f3(x) = sinx; f3(0) = 0 f4(x) = cosx; f4(0) = 1 so, cosx = 1- x^2/2! + x^4/4! - x^6/6! + x^8/8! - ..... (i) similarly; sinx = x- x^3/3! + x^5/5! - x^7/7! + ..... isinx = ix - ix^3/3! + ix^5/5! -ix^7/7! +.... (ii) Adding (i) and (ii); cosx + isinx = 1+ ix - x^2/2! - ix^3/3! + x^4/4! + x^5/5! - ..... from (a); e^ix = cosx + isinx e^iπ = cosπ + isinπ e^iπ = -1 +0 e^iπ +1 = 0 (proved) -> Logarithmic and trigonometric functions are closely related. -> Observe the beauty of this equation; e= exponential number π = Circumference/Diameter, i= iota, square root of -1, complex number. 1= base of natural(c0unting number).
Posted on: Wed, 31 Dec 2014 14:08:00 +0000
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