Question

Expand f(z)=Sin z in a taylor series about z=pi/4

Answer 1

Answer:

sin z = $\frac{1}{\sqrt{2} }  + \frac{z-\pi /4}{\sqrt{2} }  -  \frac{(z-\pi /4)^{2} }{2\sqrt{2} }  - \frac{(z-\pi/4)^3 }{6\sqrt{2} }$

please ignore the A^ in the above formula shown.

Step-by-step explanation:

From the taylor series we know,

F(z) = ∑ $\frac{f^n (z0)}{n!}  * (z-z0)^n$

where ∑ is from n= 0 to ∞

we have been given as,

f(z) = sin z  and

z0 = π/4

please see the attachment for the detailed explanation of the answer.

After solving and proceeding in the same manneras shown in the attachment we get:

sin z = $\frac{1}{\sqrt{2} }  + \frac{z-\pi /4}{\sqrt{2} }  -  \frac{(z-\pi /4)^{2} }{2\sqrt{2} }  - \frac{(z-\pi/4)^3 }{6\sqrt{2} }$

Answer 2

how to do this, answer is different in my book