How do you find the taylor series for #sinx# in powers of #x-pi/4#?

1 Answer

According to the definition of Taylor series it should be

#\sum_{n\geq 0}\frac{f^{(n)}(\pi/4)}{n!}*(x-\frac{\pi}{4})^n#

Now you have to compute the first derivatives at #x=pi/4# for
#f(x)=sinx#

You will see a pattern to arise which is periodic.

Finally (if my calculations are right) you should find tht the Taylor pattern is

#f^{(n)}(\pi/4)=(-1)^{\lfloor n/2\rfloor}\cdot\frac{\sqrt{2}}{2}#

The "strange" bracket sign you see in the formula above is the integer part or floor function

Here #n# is an integer of course.

Footnote

See the definition for Taylor series and integer part function

Taylor Series

Integer Part function