Euler's formula's proofs are kind of awesome.
One of my favs:
Euler's formula:e&*i = cos& + i sin&
(& playing the role of Theta in this exercise.)
MacLaurin's expansion (similar to Taylor series) states that a function can be expressed as the coefficients of a polynomial as follows:
f(x) = f(0) + f'(0)x + f''(0)x2 + f'''(0)x3 + ...
And so on and so forth. I'll spare you the proof of this, but if you think about it, it's fairly intuitive.
So, to prove f(x) = g(x), you need to prove that all of its derivatives at x = 0 are equal.
Follow so far?
So now we take e&i and take the derivatives at & = 0
f(0) = 1
f'(0) = i*1 = i
f''(0) = ii1 = -1
f'''(0) = iii*1 = -i
fIV(0) = i4 * 1 = 1
fV(0) = i5 * 1 = i
And so on. See the pattern. 1, i, -1, -i, rinse and repeat.
Now we take cos& + i*sin(&), which isn't as fun as taking the derivative of e.
f(0) = cos(0) + isin(0) = 1 + 0i = 1
f'(0) = -sin(0) + icos(0) = -0 + 1i = i
f''(0) = -cos(0) - isin(0) = -1 - 0i = -1
f'''(0) = -sin(0) - icos(0) = -0 - 1i = -i
fIV(0) = cos(0) + isin(0) = 1 + 0i = 1
fV(0) = -sin(0) + icos(0) = -0 + 1i = i
And so on. Notice it follows the same pattern: 1, i,-1,-i.
And because they have the same values for fn (0), they are, in fact the same.
QED(ish)
One of my favs:
Euler's formula:e&*i = cos& + i sin&
(& playing the role of Theta in this exercise.)
MacLaurin's expansion (similar to Taylor series) states that a function can be expressed as the coefficients of a polynomial as follows:
f(x) = f(0) + f'(0)x + f''(0)x2 + f'''(0)x3 + ...
And so on and so forth. I'll spare you the proof of this, but if you think about it, it's fairly intuitive.
So, to prove f(x) = g(x), you need to prove that all of its derivatives at x = 0 are equal.
Follow so far?
So now we take e&i and take the derivatives at & = 0
f(0) = 1
f'(0) = i*1 = i
f''(0) = ii1 = -1
f'''(0) = iii*1 = -i
fIV(0) = i4 * 1 = 1
fV(0) = i5 * 1 = i
And so on. See the pattern. 1, i, -1, -i, rinse and repeat.
Now we take cos& + i*sin(&), which isn't as fun as taking the derivative of e.
f(0) = cos(0) + isin(0) = 1 + 0i = 1
f'(0) = -sin(0) + icos(0) = -0 + 1i = i
f''(0) = -cos(0) - isin(0) = -1 - 0i = -1
f'''(0) = -sin(0) - icos(0) = -0 - 1i = -i
fIV(0) = cos(0) + isin(0) = 1 + 0i = 1
fV(0) = -sin(0) + icos(0) = -0 + 1i = i
And so on. Notice it follows the same pattern: 1, i,-1,-i.
And because they have the same values for fn (0), they are, in fact the same.
QED(ish)