Question

Find the Taylor series for f(x)=sin(x) centered at c=π/2.sin(x)=∑ n=0 [infinity]On what interval is the expansion valid? Give your answer using interval notation.

Answer 1

Answer:

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

Step-by-step explanation:

given that f(x) = sin x

we have to find the Taylor series for that

$f(x) = sin x : f( = 1\\f'(x) = cos x :(f'\frac{\pi}{2})=0\\f"(x) = -sinx :f" (\frac{\pi}{2}) =-1\\f^4 (x) = -cosx : f^4 (\frac{\pi}{2}) =0$

and so on.

i.e. 2nd, 4th, 6th terms would be 0

and also 1st, 5th, 9th terms would be positive for f value and 3rd, 9th,... would be negative

Using the above we can write Taylor series as

$f(x) = f(a)+\frac{f'(a)}{1!} (x-\frac{\pi}{2}) +...+f^n(a) /n! (x- \frac{\pi}{2})^n+...$

$sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...$

This is valid for all real values of x.

x ∈$(-\infty, infty)$