Hi I stumbled along the derivation for the taylor expansion for $$f(x+h)$$
How is the Taylor expansion for $f(x + h)$ derived?
And I was wondering whether it works the same as $$f(x+h)g(x+h)$$
For example, if we had $$(x+h)\cos(x+h)$$
Would the taylor expansion be something like this
$$(x+h)\cos(x+h) = x\cos(x) + h(\cos(x)-x\sin(x)) + \frac{h^2}{2}(-2\sin(x)-x\cos(x)) +O(h^3)$$
Put $k(x)=f(x)g(x)$. Then $k(x+h)=f(x+h)g(x+h)$ and you just apply the general procedure of expanding a function into its Taylor series: $$k(x+h)=k(x)+hk'(x)+\frac{h^2}{2}k''(x)+O(h^3)$$ Writing out the derivatives of $k$ in terms of $f,g$, we have $$f(x+h)g(x+h)=f(x)g(x)+h(f'(x)g(x)+f(x)g'(x))+\frac{h^2}{2}(f''(x)g(x)+2g'(x)f'(x)+f(x)g''(x))+O(h^3)$$ On the other hand, you could also multiply the Taylor expansions directly: $$(f(x)+hf'(x)+\frac{h^2}{2}f''(x)+O(h^3))(g(x)+hg'(x)+\frac{h^2}{2}g''(x)+O(h^3))$$ and collecting the powers of $h$ you get exactly the same expansion as above, where $O(h^3)$ will absorb all the powers of $h$ higher than $2$.
The first method is of course easier. The second one shows that the first is equivalent to multiplying the separate series.
