Let k = 4n, where n = positive integer.
Let a = x + √(x² - k²) and b = x - √(x² - k²). Prove that as positive integer n -> ∞,
(1/2) (1/k^k) (a^k + b^k) -> Cos(x)
Check out plot of both functions for n = 10, a small number:
http://i254.photobucket.com/albums/hh120/Scythian1...
Update:Eugene and Pauley Morph, both of your proofs reference the complex form of Cosine. But the neat thing about this expression (1/2) (1/k^k) (a^k + b^k) is that it yields a real polynomial with rational coefficients. And it can generate very accurate values of Cos(x) very fast even for "large" x > 0 values.
Update 3:Indica, you fixed Pauley Morph's problem. He should have used θ = arcsin(x/k).
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lim(n -> ∞) (1/2)(1/k^k) (a^k + b^k)
= (1/2) lim(n -> ∞) [(a/k)^k + (b/k)^k]
= (1/2) lim(n -> ∞) ([x/(4n) + √((x/4n)^2 - 1)]^(4n) + [x/(4n) - √((x/4n)^2 - 1)]^(4n))
= (1/2) lim(y -> 0) ([y + √(y^2 - 1)]^(x/y)) + [y - √(y^2 - 1)]^(x/y))), letting y = x/(4n)
= (1/2) lim(y -> 0) (exp{x • ln(y + √(y^2 - 1))/y} + exp{x • ln(y - √(y^2 - 1))/y})
= (1/2) lim(y -> 0) (exp{x • ln(y + √(y^2 - 1)/y} + exp{-x • ln(y + √(y^2 - 1))/y})
= (1/2) (exp{x • lim(y -> 0) ln(y + √(y^2 - 1)/y} + exp{-x • lim(y -> 0) ln(y + √(y^2 - 1)/y}
= (1/2) (exp{x • lim(y -> 0) 1/√(y^2 - 1)} + exp{-x • lim(y -> 0) 1/√(y^2 - 1)}), by L'hospital's rule
= (1/2) (exp{x • 1/i} + exp{-x • 1/i})
= (1/2) (exp{-ix} + exp{ix})
= cos(x).
@Scythian thanks for spotting the mistake.
Note that ab = k^2 where k = 4n is an even integer.
So (1/2) (a^k + b^k)/k^k =
(1/2) (a^k + b^k) / (ab)^(k/2) =
(1/2) ((a/b)^(k/2) + (b/a)^(k/2))
Since n, and therefore k, is going to approach infinity, we may
as well write
a = x + iâ(k² - x²) and b = x - iâ(k² - x²) with |a| = |b| = k
Then we can assume that
a = k exp(i θ) and b = k exp(-i θ)
where θ = arccos(x/k) for large enough values of k.
Then (1/2) (a^k + b^k)/k^k =
(1/2) ((a/b)^(k/2) + (b/a)^(k/2)) =
(1/2) (exp(i k θ) + exp(- i k θ)) =
cos(k θ)
I was doing great up to here and then things fell apart
For large enough k let x/k=sin(α) so αâ0 as kââ
Writing x={ sin(α)/α }{ kα } and taking limits as kââ gives x = 1*lim(kα) = lim(kα)
From definitions of a & b, a/k=exp(i(Ï/2âα)) and b/k=exp(âi(Ï/2âα))
ⴠ½( (a/k)áµ + (b/k)áµ ) = cos(k(Ï/2âα)) = cos(kα) since k is a multiple of 4
lim cos(kα) = cos( lim(kα) ) = cos(x)
If k=4n+1, 4n+2, 4n+3 the limit is sin(x), âcos(x),âsin(x)
That's a pretty powerful function you have there.