Question

Higher derivatives of sine form a 4-cycle. Integer powers of `i` form a 4-cycle. What is the connection (if there is one)?

Answer 1

The correlation is given by Euler's formula:

`e^(ix) = cosx+isinx`

from which we can derive the expression of `sinx` as:

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

Differentiating both sides:

`d/dx sinx = (ie^(ix)+ie^(-ix))/(2i) = i (e^(ix)+e^(-ix))/(2i)`

`d^2/dx^2 sinx = i(ie^(ix)-ie^(-ix))/(2i) = i^2 (e^(ix)-e^(-ix))/(2i)`

`d^3/dx^3 sinx = i^2(ie^(ix)+ie^(-ix))/(2i) = i^3 (e^(ix)+e^(-ix))/(2i)`

`d^4/dx^4 sinx = i^3(ie^(ix)-ie^(-ix))/(2i) = i^4 (e^(ix)-e^(-ix))/(2i) = i^4 sinx`

The the fact that `i^4 = i^0 = 1` implies that:

`d^4/dx^4 sinx = sinx`

and in general:

`i^(4n+k) = i^k => d^(4n+k)/dx^(4n+k) sinx = d^k/dx^k sinx`

Answer 2

Thanks to Andrea's answer I have an explanation that I like a lot.

Explanation:

`sinx = 1/(2i)(e^(ix)-e^(-ix))`

So,

`d/dx(sinx) = 1/(2i) (ie^(ix)-(-i)e^(-ix))`

`= 1/(2i) (e^(ipi/2)e^(ix)-(e^(-ipi/2))e^(-ix))`

`= 1/(2i) (e^(i(x+pi/2))-e^(-i(x+pi/2)))`

`= sin(x+pi/2)`