I need to find the Taylor series for $f(x) = \sin x + \cos x$ centered at $0$ up to degree $3$. It's been a long time since the last time I did anything with the Taylor series.
So I was wondering if I could just add the Taylor series for $\sin x$ to the Taylor series of $\cos x$ to find the Taylor series for $\sin x + \cos x$. Any help is appreciated.
On
Inside the radius of convergence the Taylor series is uniquely defined. Hence any (correct) technique for finding the coefficients will end up with the same result.
Another way would be $f(x) = {1 \over 2i} (e^{ix}-e^{-ix}) + {1 \over 2} (e^{ix}+e^{-ix})$ and then use $e^{ix} = \sum_{n=0}^\infty i^n {x^n \over n!}$.
Then $f_n = {1 \over 2i} (i^n -(-i)^n ) + {1 \over 2} (i^n +(-i)^n)) {x^n \over n!}$ :-)
Note that ${1 \over 2i} (i^n -(-i)^n )$ has values $0, 1, 0, -1,...$, and ${1 \over 2} (i^n +(-i)^n) $ has values $1, 0, -1, 0, ...$.
Yes, you can just add the terms of each respective Taylor series together:
$$\sin x + \cos x \approx \left(x+\frac {x^3}{3!}\right)+\left(1-\frac{x^2}{2!}\right)$$
Since the two functions have an interval of convergence of all real numbers, term-by-term addition can be applied.