Today's proof for Euler's Formula is based on the Taylor's Series. Euler's Formula is the equation:
eix = cosx + isinx
In a previous blog, I spoke about Euler's Identity which is derived from Euler's Formula. Richard Cotes was the first person to provide a proof but the great popularizer of this result was Leonhard Euler. Euler's Identity is used in the construction of cyclotomic integers which are used in Kummer's proof of Fermat's Last Theorem for regular primes.
Lemma 1: Maclaurin's Series
f(x) = f(0) + (x/1!)f'(0) + (x2/2!)f''(0) + (x3/3!)f'''(0) + ... + (xn/n!)fn(0)
(1) Taylor's series gives us:
f(x) = f(a) + f'(a)(x-a) + [f''(a)/2!](x-a)2 + .... + [f(n)(a)/n!](x-a)n + ....
[See here for its proof]
(2) Now, if a=0, then we have:
f(x) = f(0) + (x/1!)f'(0) + (x2/2!)f''(0) + (x3/3!)f'''(0) + ... + (xn/n!)fn(0).
QED
Lemma 2: ex = 1 + (x/1) + (x2/2!) + (x3/3!) + ....
(1) Let f(x) = ex
(2) From the properties of ex [See here for details]:
f(x) = ex → f(0) = e0 = 1
f'(x) = ex → f'(0) =e0 = 1
fn(x) = ex → fn(0) = e0 = 1
(3) We know that ex is continuous since it has a derivative at each point. [See here for details of why this is true]
(4) By Lemma 1 above, we have:
ex = 1 + (x/1!)(1) + (x2/2!)(1) + ... (xn/n!)(1)
QED
Lemma 3: sinx = x - (x3/3!) + (x5/5!) - (x7/7!) + ...
(1) Let f(x) = sin x
(2) From the properties of sin, we know:
f(0) = sin(0) = 0 [See here for details if needed]
f'(x) = cos x → f'(0) = 1 [See here for proof if needed]
f''(x) = -sin(x) → f'(0) = 0 [See here for proof if needed]
f'''(x) = -cos x → f'(0) = -1
(3) From this, we see that:
fn(0) = 0 if n is even.
fn(0) = 1 if (n-1)/2 is even
fn(0) = -1 if (n-1)/2 is odd
(4) Putting this all together gives us:
sin x = (x/1)(1) + (x2/2!)(0) + (x3/3!)(-1) + ...
QED
Lemma 4: cos x = 1 - (x2)/2! + (x4/4!) - (x6/6!) + ...
(1) Let f(x) = cos x
(2) From the properties of cos, we know:
f(0) = cos(0) = 1 [Details if needed are found here]
f'(x) = -sin x → f'(0) = 0 [Details if needed are found here]
f''(x) = -cos(x) → f'(0) = -1
f'''(x) = sin x → f'(0) = 0
(3) From this, we see that:
fn(0) = 0 if n is odd.
fn(0) = 1 if (n/2) is even
fn(0) = -1 if (n/2) is odd
(4) Putting this all together gives us:
cos x = 1 + (x2/2!)(-1) + (x3/3!)(0) + (x4/4!)(1) + ...
QED
Theorem: Euler's Formula
eix = cos x + isin x
(1) From Lemma 2, we have:
eix = 1 + ix + (ix)2/2! + (ix)3/3! + ...
(2) Since i2 = -1 and i4 = 1, this gives us: (for details on i, see here)
eix = (1 - x2/2! + x4/4! + ...) + i(x - x3/3! + x5/5! + ...)
(3) From Lemma 4 above, we see that:
cos(x) = (1 - x2/2! + x4/4! + ...)
(4) From Lemma 3 above, we see that:
isin(x) = i(x - x3/3! + x5/5! + ...)
(5) Combining step #2 with step #3 and step #4 gives us:
eix = cos x + i sin x.
QED
Corollary: De Moivre's Formula
(cos x + isin x)n = cos(nx) + isin(nx)
Proof:
(1) (eix)n = einx
(2) (cos x + i sin x)n = cos(nx) + isin(nx) [Applying Euler's formula above]
QED
Friday, February 24, 2006
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5 comments:
This formula can be interpreted as saying that the function eix traces out the unit circle in the complex number plane as x ranges through the real numbers. Here, x is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counter clockwise and in radians The formula is valid only if sin and cos take their arguments in radians rather than in degrees.
source: Wikipedia
For x=pi, the equation is
e^ipi = -1 or e^ipi + 1 = 0 or
e^ipi = 0 - 1 ; ln [e^ipi] = ln [0/1]; hence, ipi = ln [0] = 0 ;
e^0 = 1 ; 1 + 1 = 0 (mod 2)= 2
See 'Proof' by David Auburn(p.73-4)
"Let X equal the quantity of all quantities of X. Let X equal the cold ... months [11, 12, 1, 2] ... and four of heat [5, 6, 7, 8] leaving four months of in-/determine temperature [2,3,9,10].
[The months are a unit circle, and the Euler equation is its design. The indetermined area is undefined as is the definition of infinite.]
"Let X equal the month of full bookstores [infinite or undefined as month 10]. The number of books approaches infinity as the number of months of cold approaches four."
The 'proof' is also a statement of:
cos x = e^ix + e^[-ix] (over 2)and
sin x = e^ix - e^[-ix] (over 2i)
solving for both cos x and sin x
cos pi = e^ipi + 1/1 = 2 =
0 (mod 2) = -1
sin pi = 1 - 1 = 0
e^ipi = cos pi + isin pi = -1 + 0
sin pi - cos pi = 2 = 0 (mod 2)
The 'proof' is a restatement of Fermat's Last Theorem for modular formulation of infinite number of primes, in the transformation of finite to infinite, using a modular clock function:
a unit circle in the complex number plane. "Let X equal the number ...."
12 04 06
What a great blog you have here!!! I did a similar derivation of the Inverse tangent function using these same principles. But the explicit representation for Tan^-1 came out algebraically. See here. Meanwhile I will blogroll you:)
When compared to other silly blogs, yours seems to me like "God among insects". Keep the good work bro.
This theorem resulted in a breakthrough in mathematics. Before, imaginary exponents were undefined. This theorem introduced imaginary exponents, as well as logarithms of numbers that are not positive and real.
Though this proves the desired result, it doesn't come out of intuition like the previous simpler proof..
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