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 $x = \frac{\pi}{4}$ for
$f \left(x\right) = \sin x$

You will see a pattern to arise which is periodic.

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

${f}^{\left(n\right)} \left(\setminus \frac{\pi}{4}\right) = {\left(- 1\right)}^{\setminus \lfloor \frac{n}{2} \setminus \rfloor} \setminus \cdot \setminus \frac{\setminus \sqrt{2}}{2}$

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