What is the Taylor series for #sin 2x#?

1 Answer
George C.
Sep 13, 2015

#sin 2x = sum_(k=0)^oo (-1)^k/((2k+1)!) (2x)^(2k+1)#

Explanation:

Let us start from here:

#d/(dt) sin t = cos t#

#d/(dt) cos t = -sin t#

#sin 0 = 0#

#cos 0 = 1#

Incidentally, a picture of this would be a point moving anticlockwise around the unit circle at a speed of #1# radian per second, starting from #(1, 0)#. The position of the point at time #t# is #(cos t, sin t)# and its velocity (which is tangential) is #(-sin t, cos t)#.

The general formula for the Taylor series for #f(t)# at #0# is:

#f(t) = sum_(n=0)^oo f^((n))(0)/(n!) t^n#

In the case of #sin t#, we find that only the terms for odd values of #n# are non-zero, and the signs on them are alternating:

#f^((0))(t) = sin t, f^((1))(t) = cos t,#

#f^((2))(t) = -sin t, f^((3))(t) = -cos t,...#

So:

#f^((2k))(0) = (-1)^k sin(0) = 0#

#f^((2k+1))(0) = (-1)^k cos(0) = (-1)^k#

So we can write:

#sin t = sum_(k=0)^oo (-1)^k/((2k+1)!) t^(2k+1)#

Now substitute #t = 2x# to get:

#sin 2x = sum_(k=0)^oo (-1)^k/((2k+1)!) (2x)^(2k+1)#

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