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A general way to find the inverse functions...

chisigma

Well-known member
Feb 13, 2012
1,704
In www.mathhelpforum.com and interesting question has been proposed by the user misiazeska the 05 20 2013...


How to find the inverse of this function?...


$$y= 5\ x^{3} - x^{5}$$


... and the unanimous answer has been '... it doesn't exist any closed formula to find x as function of y...'. In my opinion the proposed problem is a good opportunity to use the following solving procedure to find the inverse of an analytic function. Giving a function...


$$ w = f(z) = a_{1}\ z + a_{2}\ z^{2} + a_{3}\ z^{3} + ... ,\ a_{1} \ne 0\ (1)$$

... which is analytic inside a disk with $|z|< r$, the Taylor expansion of the inverse function is...


$$ z = f^{-1}(w) = b_{1}\ w + b_{2}\ w^{2} + b_{3}\ w^{3} + ...\ (2)$$

... where...


$$b_{n}= \frac{1}{n!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} \{\frac{z}{f(z)}\}^{n}\ (3)$$
Now we consider the original function...


$$y= 5\ x^{3} - x^{5} \implies \sqrt[3]{y}= \sqrt[3]{5}\ x\ \sqrt[3]{1 - \frac{x^{2}}{5}}\ (4) $$

... that, setting $w= \sqrt[3]{y}$ and $x=z$ becomes...


$$w = f(z) = \sqrt[3]{5}\ z\ \sqrt[3]{1 - \frac{z^{2}}{5}}\ (5) $$

... which satisfies the requirements of (1) so that the (2) and (3) can be applied. The coefficients $b_{n}$ are computed as follows...

$$b_{1}= \lim_{z \rightarrow 0} \frac{z}{f(z)}= \lim_{z \rightarrow 0} \frac{1}{\sqrt[3]{5}\ \sqrt[3]{1 - \frac{z^{2}}{5}}} = \frac{1}{\sqrt[3]{5}}\ (6) $$

$$b_{2}= \frac{1}{2}\ \lim_{z \rightarrow 0} \frac{d}{d z} (\frac{z}{f(z)})^{2} = \frac{1}{2}\ \frac{1}{\sqrt[3]{25}}\ \lim_{z \rightarrow 0} \frac{d}{d z} \frac{1}{(1 - \frac{z^{2}}{5})^{\frac{2}{3}}} = \frac{1}{2}\ \frac{1}{\sqrt[3]{25}}\ \lim_{z \rightarrow 0}\ \frac{4\ z}{15\ (1-\frac{z^{2}}{5})^{\frac{5}{3}}} = 0\ (7)$$

$$b_{3} = \frac{1}{6}\ \lim_{z \rightarrow 0} \frac{d^{2}}{d z^{2}} (\frac{z}{f(z)})^{3} = \frac{1}{6}\ \frac{1}{5}\ \frac{d^{2}}{d z^{2}}\ \frac{1}{1-\frac{z^{2}}{5}} = - \frac{1}{30}\ \lim_{z \rightarrow 0}\ \frac{10\ (3 z^{2}+5)}{(z^{2}-5)^{3}} = \frac{1}{75}\ (8)$$

... and now we can write the first three terms of the inverse function...

$$ x= \sqrt[3]{\frac{y}{5}} + \frac{y}{75} + ...\ (9)$$

Of course it is possible to proceed to compute more terms but the main scope is to illustrate a way to find a general way to solve this type of problems. For completeness sake in next post I will try to demonstrate the (3)...

Kind regards

$\chi$ $\sigma$

Questions and comments should be posted here:

http://mathhelpboards.com/commentar...-general-way-find-inverse-functions-4929.html
 
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chisigma

Well-known member
Feb 13, 2012
1,704
Giving a function...


$$ w = f(z) = a_{1}\ z + a_{2}\ z^{2} + a_{3}\ z^{3} + ... ,\ a_{1} \ne 0\ (1)$$

... which is analytic inside a disk with $|z|< r$, the Taylor expansion of the inverse function is...


$$ z = f^{-1}(w) = b_{1}\ w + b_{2}\ w^{2} + b_{3}\ w^{3} + ...\ (2)$$

... where...


$$b_{n}= \frac{1}{n!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} \{\frac{z}{f(z)}\}^{n}\ (3)$$
The reported formula was 'discovered' in the year 1770 by the Italian mathematician and astronomer Giuseppe Luigi Lagrancia [more known in scientific licterature as Joseph Louis Lagrange...]. Some year after the not well known German mathematician Hans Heinrich Burmann found the more general statement of the problem that follows: given a function $w= f(z)$ analytic in a disk of radious r centered in $z=z_{0}$ with $f^{ '} (z_{0}) \ne 0$, so that is...

$$ w = f(z) = w_{0} + f^{\ '} (z_{0})\ (z-z_{0}) + \frac{f^{\ ''} (z_{0})}{2!}\ (z-z_{0})^{2} + ...\ (1)$$

... then the Taylor series of its inverse function $z= g(w)$ is ...

$$ z= g(w) = z_{0} + \sum_{n=1}^{\infty} \frac{(w-w_{0})^{n}}{n!}\ \lim_{z \rightarrow z_{0}} \frac {d^{n-1}}{d z^{n-1}} \{\frac{z-z_{0}}{f(z)- w_{0}}\}^{n}\ (2)$$
The coefficients of the expansion of g(*) can be found by the following theorem, generally known as 'Burmann's theorem' : let be $\varphi(*)$ a function defined as...


$$\varphi(z) = \frac{z-z_{0}}{f(z)- w_{0}}\ (3)$$
… then the analytic function g(z) can be expanded, in a certain domain of z, as...

$$ g(z) = z_{0} + \sum_{n=1}^{m-1} \frac{\{f(z) - w_{0}\}^{n}}{n!}\ \frac{d^{n-1}}{d \xi^{n-1}}\ \{g^{\ '} (\xi)\ \varphi^{n} (\xi) \}_ {\xi= z_{0}}\ (4) $$

... where...

$$ R_{m}= \frac{1}{2\ \pi\ i}\ \int_{z_{0}}^{z} \int_{\gamma} \{\frac{f(\zeta)- w_{0}}{f(\xi)- w_{0}}\}^{m-1}\ \frac{g^{\ '}(\xi)\ f^{\ '}(\zeta)}{f(\xi)-f(\zeta)}\ d \xi\ d \zeta\ (5)$$

... being $\gamma$ a contour such that for any $\zeta$ inside $\gamma$ the equation $f(\xi)=f(\zeta)$ has only the solution $\xi = \zeta$. In the case $g(z)=z$ is $g^{\ '}(z)=1$ and the (4) becomes the (2). The (4) and (5) that $f^{\ '} (z_{0}) \ne 0$ is a necessary but non suffcient condition for the validity of (2) because the other condition $\lim _{m \rightarrow \infty} R_{m}=0$ must be satisfied, and that is an important detail that usually is not specified in the lecterature...

Kind regards

$\chi$ $\sigma$
 
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chisigma

Well-known member
Feb 13, 2012
1,704
As reported in...

Unsolved statistics questions from other sites, part II - Page 4

... several years ago I performed the computation of the function $\log_{10} \text{erfc}\ (x)$, where...

$$ \text{erfc} (x) = 1- \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{- t^{2}}\ dt\ (1)$$

... and the results are represented here...



Because is $\log_{10} a = \frac{\ln a}{\ln 10}$ the main task is the computation of the function...

$$\lambda (x) = \ln \{\text{erfc} (x)\}\ (2)$$

Now $\lambda(*)$ is analytic and its McLaurin expansion is given by...


$$\lambda(x) = a_{1}\ x + a_{2}\ x^{2} + a_{3}\ x^{3} + ... \ (3)$$


... where...


$$a_{n} = \frac{1}{n!}\ \lim_{x \rightarrow 0} \frac{d^{n}}{d x^{n}} \lambda(x)\ (4)$$
The computation of the $a_{n}$ is not too diffcult but very tedious and fortunately 'Monster Wolfram' is available...


$$a_{1}= - \frac{2}{\sqrt{\pi}},\ a_{2}= - \frac{2}{\pi},\ a_{3} = - \frac{2\ (4 - \pi)}{3\ \pi^{\frac{3}{2}}},\ a_{4}= \frac{4\ (\pi - 3)}{3\ \pi^{2}},\ a_{5} = - \frac{96 - 40\ \pi + 3\ \pi^{2}}{15\ \pi^{\frac{5}{2}}},\ a_{6}= - \frac{4\ (120 - 60\ \pi + 7\ \pi^{2})} {45\ \pi^{3}}$$
$$a_{7}= - \frac{5760 - 3360\ \pi + 532\ \pi^{2} - 15\ \pi^{3}} {315\ \pi^{\frac{7}{2}}},\ a_{8}= - \frac{8\ (420 - 280\ \pi + 56\ \pi^{2} - 3\ \pi^{3})}{105\ \pi^{4}},\ … $$

From the pratical point of view in many cases is important the inverse of the $\lambda(*)$ function and we are in condition to apply what has been found in previous posts writing...


$$ x = \lambda^{-1} (*) = b_{1}\ \lambda + b_{2}\ \lambda^{2} + b_{3}\ \lambda^{3} + ...\ (5)$$

... where...

$$ b_{n} = \frac{1}{n!}\ \lim_{x \rightarrow 0} \frac{d^{n-1}}{d x^{n-1}} (\frac{x}{\lambda(x)})^{n}\ (6)$$

The $b_{n}$ can be computed using (6) and also in thic case 'Monster Wolfram' is useful...


$$ b_{1}= - \frac{\sqrt{\pi}}{2},b_{2} = - \frac{\sqrt{\pi}}{4},b_{3} = - \frac{\sqrt{\pi}}{24}\ (2 + \pi),b_{4} = - \frac{\sqrt{\pi}}{48}\ (1 + 3\ \pi)$$

$$b_{5}= - \frac{\sqrt{\pi}}{960}\ (4 + 50\ \pi + 7\ \pi^{2}),b_{6}= - \frac{\sqrt{\pi}}{5760}\ (4 + 180\ \pi + 105\ \pi^{2}) (7)$$

... and at this point also 'Monster Wolfram' is unable to proceed further (Sadface)...

The effective utulity of (6) and (7) will be analized in a succesive post...

Kind regards

$\chi$ $\sigma$
 
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