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#### 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...

... 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

*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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