Using taylor series with Big O notation

Question

I found out that:

$$\tan(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)$$

If I wanted to find a value of $\tan(x)$, how would I use it to find a value for $x$?

Answer 1

I supppose that what you mean is the following : knowing that, for small values of $x$ we can approximate the tangent by the truncated Taylor expansion $$\tan(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)$$ how could we find the value of $x$ knowing the value of $\tan(x)=a$.

This is called series reversion. The idea is to write $$x=\sum_{n=1}^k A_n\, a^n$$ and to replace. Using $k=5$ we then have $$a=\left(\sum_{n=1}^k A_n\, a^n\right)+\frac 13\left(\sum_{n=1}^k A_n\, a^n\right)^3+\frac2 {15} \left(\sum_{n=1}^k A_n\, a^n\right)^5$$ and expand, ignoring the terms of order higher than $5$. This gives $$a=a A_1+a^2 A_2+a^3 \left(\frac{A_1^3}{3}+A_3\right)+a^4 \left(A_2 A_1^2+A_4\right)+a^5 \left(\frac{2 A_1^5}{15}+A_3 A_1^2+A_2^2 A_1+A_5\right)+\cdots$$

Now, by identification, we then have $$A_1=1 \qquad A_2=0 \qquad \frac{A_1^3}{3}+A_3=0\qquad A_2 A_1^2+A_4=0\qquad\frac{2 A_1^5}{15}+A_3 A_1^2+A_2^2 A_1+A_5=0$$ from which $$A_1=1 \qquad A_2=0 \qquad A_3=-\frac 13 \qquad A_4=0 \qquad A_5=\frac 15$$ meaning, by the end $$x=\tan(x)-\frac 13 \tan^3(x)+\frac 15\tan^5(x)+O(\tan^7(x))$$

Try using $\tan(x)=\frac{1}{\sqrt{3}}$; this will give $$x=\frac{41}{45 \sqrt{3}}\approx 0.52603$$ while $\frac \pi 6 \approx 0.523599$