×for every function f:
f(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + ... + anxn + ...
so first ex.
f(x) = ex = a0 + a1x + a2x2 + a3x3 + a4x4 + ...
f(0) = e0 = 1 = a0 + a10 + a202 + a303 + a404 + ... = a0 so a0 = 1
f'(x) = ex=a1 + 2*a2x + 3*a3x2 + 4*a4x3 + ... because (xn)' = nxn-1
f'(0) = e0 = 1 = a1 + 2*a20 + 3*a303 + 4*a403 + ... = a1 so a1 = 1
f"
(x) = ex = 2*a2 + 2*3*a3x + 3*4*a4x2 + ...
f"
(0) = e0 = 1 = 2*a2 + 2*3*a3*0 + 3*4*a4*02 + ... = 2*a2 So a2 = 1/2
f"
'(x) = ex = 2*3*a3 + 2*3*4*a4x + ...
f"
'(0) = e0 = 1 = 2*3*a3 + 2*3*4*a40 + ... = (2*3)a3 So a3 = 1/(2*3)
f"
"
(x) = ex = 2*3*4*a4 + ...
f"
"
(0) = e0 = 1 = 2*3*4*a4 + ... = 2*3*4*a4 so a4 = 1/(2*3*4)
So:
a0 = 1/1, a1 = 1/1, a2 = 1*2, a3 = 1/(1*2*3), a4 = 1/(1*2*3*4)
in short n! = 1*2*3*...*n, 0! = 1 so we get an = 1/n!
ex = 1 + x + (1/2!)x2 + (1/3!)x3 + (1/4!)x4 + ... + (1/n!)xn + ...
Ok, now sin(x)
f(x) = sin(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + ...
f(0) = sin(0) = 0 = a0 + a10 + a202 + a303 + a404 + ... = a0 So a0 = 0
f'(x) = cos(x) = a1 + 2*a2x + 3*a3x2 + 4*a4x3 + ... (xn)' = nxn-1
f'(0) = cos(0) = 1 = a1 + 2*a20 + 3*a303 + 4*a403 + ... = a1 So a1 = 1
f"
(x) = -sin(x) = 2*a2 + 2*3*a3x + 3*4*a4x2 + ...
f"
(0) = -sin(0) = 1 = 2*a2 + 2*3*a3*0 + 3*4*a4*02 + ... = 2*a2 So a2 = 0
f"
'(x) = -cos(x) = 2*3*a3 + 2*3*4*a4x + ...
f"
'(0) = -cos(0) = -1 = 2*3*a3 + 2*3*4*a40 + ... = (2*3)a3 So a3 = -1/(2*3)
f"
"
(x) = sin(x) = 2*3*4*a4 + ...
f"
"
(0) = sin(0) = 0 = 2*3*4*a4 + ... = 2*3*4*a4 So a4 = 0
We can write it in short kinda like above with e, with an = 0 if n = even, andn an = (-1)(n-1)/2/n! if n = odd.
sin(x) = x - (1/3!)x3 + (1/5!)x5 - (1/7!)x7 + ...
You can also do it cos(x), (but I'm not goning to type it
)
cos(x) = 1 + (1/2!)x2 + (1/4!)x4 + (1/6!)x6 + ...
So we've got:
ex = 1 + x + (1/2!)x2 + (1/3!)x3 + (1/4!)x4 + ...
sin(x) = x - (1/3!)x3 + (1/5!)x5 - (1/7!)x7 + ...
cos(x) = 1 - (1/2!)x2 + (1/4!)x4 - (1/6!)x6 + ...
Like you see, they look alike
now we gonna use: i
we know i²=-1
We put the number i in the formula ex So we get:
eix = 1 + ix + (1/2!)(ix)2 + (1/3!)(ix)3 + (1/4!)(ix)4 + ...
To make it easier, we make a tabel of i:
i = i
i2 = -1
i3 = i2 * 1 = -1 * i = -i
i4 = i2 * i2 = (-1) * (-1) = 1
i5 = i4 * i = 1 * i = i
If we fill it in you get:
eix = 1 + ix - (1/2!)x2 - i(1/3!)x3 + (1/4!)x4 + ...
This looks nice, but try it with: cos (x) + i * sin(x)
cos(x) + i * sin(x) = 1 + ix - (1/2!)x2 - i(1/3!)x3 + (1/4!)x4 + ... = eix !!!!!!
And that's what we're looking for.
eix = cos(x) + i * sin(x) !
Aso now it's simple, just put p in it:
ei*p = cos(p) + i * sin(p) = -1 + i * 0
ei*p = -1
So, you want -(e^π
i), means:
-ei*p = 1
.