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

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Answer 1 · 2018-02-27T15:14:02

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 · 2018-02-27T19:33:07

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

#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)#