證明sinx=x-x³/3!+x⁵/5!-x⁷/7!+...

since

sinx = (e^ix - e^(-ix) )/2i

by e^x = 1 +x/1! + x^2/2! + x^3/3! + x^4/4!

we have

sinx = [( 1 + ix/1! + (ix)^2/2! + (ix)^3/3! +....) - ( 1 + (-ix) + (-ix)^2/2! + (-ix)^3/3! +....... ) ] /2i

=[ 1 +ix - x^2/2! - x^3/3! + x^4/4! +....) - ( 1 -ix - x^2/2! + x^3/3! + x^4/4! +...)]/2i

= [2i (x - x^3/3! + x^5/5! - x^7/7! +.....)]/2i

= x - x^3/3! + x^5/5! -x^7/7! +......



希望幫到你!

2011-10-27 23:47:32 補充：
since

e^ix = cosx + i sinx ----- (1)

e^-ix = cosx - i sinx -----(2)

(1)-(2)

we have

sinx = (e^ix - e^(-ix) )/2i

Let z = cos x + i sin x
dz/dx = -sin x +i cos x
dz/dx = iz
∫ 1/z dz =∫ i dx
ln z = ix
e^(ix) = z = cos x + i sin x
(euler's formula)

actually,if u prove euler's formular without this, u will hv done it

唔用 taylor series 咁用咩？