Getting inverse of polynoms with trigonometric functions

Question

I'm trying to get the inverse of $$f(x) = \cos(x) + 3x$$

I tried it by definition of $\cos(x)$ with no luck:

$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+...$$

Answer 1

there exists no explicit inverse, but an impplicit one is as follow by exchanging "x-y": $$\large y=\cos x+3x\overbrace{\longrightarrow}^{x\leftrightarrow y} x=\cos y+3y$$ Or in a suitable format as: $$\cos (f^{-1}(x))+3f^{-1}(x)=x$$ As: $$f(f^{-1}x)=x$$ So: $$f'(f^{-1}x).(f^{-1})'(x)=1$$ Since $y$ is a monotonically strictly increasing function and little guessing after drawing its rough sketch maybe you'll recognize: $$f^{-1}(1)=0$$ So: $$f'(f^{-1}(1))(f^{-1}(1))'=1$$ Or: $$(f^{-1}(1))'=1/f'(0)=1/3$$

Answer 2

As Aditya answered, there is no analytical inverse fo your function. However, if,as you tried, you can approximate locally your function by a series, you can inverse the series. Suppose you have $$y=\sum_{i=0}^n a_i x^i$$ (you know the $a_i$'s) you could, at least under some conditions, write $$x=\sum_{j=0}^m b_j y^j$$ replace in the first series, develop and cancel as much terms as you can to get the $b_j$'s.

Taking your example $$y=\cos(x) + 3x=1+3 x-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^7\right)$$ you should arrive to $$x=\frac{y-1}{3}+\frac{1}{54} (y-1)^2+\frac{1}{486} (y-1)^3+\frac{(y-1)^4}{8748}-\frac{(y-1)^5}{78732}-\frac{2 (y-1)^6}{295245}+O\left((y-1)^7\right)$$