I was wondering about if, say you have: $$\arctan(\sin(x))$$ and you know the Taylor's series representation of both since $\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$ and $\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ then that you could do, instead of the usual expressing the arctangent function as a series of sin, and that sin to be later expressed in powers of x, if I could do something like this: $$\arctan(\sin(x))=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\right)^{2n+1}$$ and maybe do some algebraic manipuations with the sums to get it easier, but I don't think that looks good or if it is already a thing, just a curiosity that popped out.
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1Does this answer your question? composition of power series – mr_e_man Nov 30 '22 at 01:52
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You can certainly do this, but it is often difficult to "simplify" the complicated expression. It is helpful that $\sin(0)=0.$ It means the resulting power series $\sum a_ix^i$ has $a_i$ as the sum of a finite set of values, and thus you don't have to worry about convergence of infinite sums. – Thomas Andrews Nov 30 '22 at 04:32
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1Indeed, it is difficult to even get things we know are true. If we know the power series for $f(x)=\log(1+x)$ and $g(x)=e^x,$ it is tricky to show $g(f(x))=1+x$ just using the power series. It can be done, but it is work. – Thomas Andrews Nov 30 '22 at 04:36
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1By the way, you shouldn't use $n$ in both sum in the final expression. it risks confusion. – Thomas Andrews Nov 30 '22 at 04:43
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See here. – Gary Nov 30 '22 at 07:59