kulisamilhh

2022-05-03

How can be helpful inverse functions determination by integral?

${f}^{-1}(x)=\int \frac{1}{{f}^{\prime}({f}^{-1}(x))}\phantom{\rule{thinmathspace}{0ex}}dx+c.$

${f}^{-1}(x)=\int \frac{1}{{f}^{\prime}({f}^{-1}(x))}\phantom{\rule{thinmathspace}{0ex}}dx+c.$

Eliza Flores

Beginner2022-05-04Added 16 answers

The formula will be useful if ${f}^{\prime}(x)$ can be expressed as a function of $f(x)$. i.e there exists a function $g(y)$ such that ${f}^{\prime}(x)=g(f(x))$. Take the following scenario as an example $f(x)=\mathrm{tan}(x)$, we know

$\frac{d}{dx}\mathrm{tan}(x)=1+\mathrm{tan}(x{)}^{2}\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}g(y)=1+{y}^{2}$

Consider incorporating this into your calculation $f(0)=0$, you get

${f}^{-1}(x)={\mathrm{tan}}^{-1}(x)={\int}_{0}^{x}\frac{dt}{{f}^{\prime}({f}^{-1}(t))}={\int}_{0}^{x}\frac{dt}{g(t)}={\int}_{0}^{x}\frac{dt}{1+{t}^{2}}$

For $|t|1$, we have

$\frac{1}{1+{t}^{2}}=\sum _{n=0}^{\mathrm{\infty}}(-1{)}^{n}{t}^{2n}\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}{\mathrm{tan}}^{-1}(x)=\sum _{n=0}^{\mathrm{\infty}}(-1{)}^{n}{\int}_{0}^{x}{t}^{2n}dx=\sum _{n=0}^{\mathrm{\infty}}(-1{)}^{n}\frac{{x}^{2n+1}}{2n+1}$

Given the relationship, you can create a power series representation of ${\mathrm{tan}}^{-1}(x)$ for $|x|1$.

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