Expressing $y(x)$ explicitly

Question

I want to find the explicit solution for $y$ from the equation $y=-x^{\sqrt[]{2}}+(x^2+1-y)^\frac{1}{{\sqrt[]{2}}}$.

Asking Wolframalpha gives no answer, so should i even bother trying? Thank you.

Answer 1

As Simply Beautiful Art commented, the is absolutely no hope to get any explicit solutions.

The only thing you could do is to solve , for a given value of $x$ equation $$f(y)=y+x^{\sqrt[]{2}}-(x^2+1-y)^\frac{1}{{\sqrt[]{2}}}$$ Because of the term $x^{\sqrt[]{2}}$, you are restricted to $x \geq 0$.

Using Newton method starting with $y_0=0$ and using $$f'(y)=1+\frac 1{\sqrt 2}\left(x^2-y+1\right)^{\frac{1}{\sqrt{2}}-1}$$ you could generate tables like $$\left( \begin{array}{cc} x & y \\ 0 & 0.559793 \\ 1 & 0.396430 \\ 2 & 0.315123 \\ 3 & 0.268608 \\ 4 & 0.237668 \\ 5 & 0.215192 \\ 6 & 0.197911 \\ 7 & 0.184089 \\ 8 & 0.172706 \\ 9 & 0.163121 \\ 10 & 0.154906 \\ 11 & 0.147763 \\ 12 & 0.141478 \\ 13 & 0.135892 \\ 14 & 0.130886 \\ 15 & 0.126366 \\ 16 & 0.122258 \\ 17 & 0.118504 \\ 18 & 0.115055 \\ 19 & 0.111873 \\ 20 & 0.108926 \\ 21 & 0.106186 \\ 22 & 0.103629 \\ 23 & 0.101237 \\ 24 & 0.098993 \\ 25 & 0.096882 \end{array} \right)$$ and try to curve fit the data.

A quick and dirty regression work for $0\leq x \leq 100$ shows that $$y=\frac{0.559974+0.0793612\,x+0.000334669\, x^2}{1+0.598575\, x+0.0198399 \,x^2}$$ is quite decent $(R^2=0.999996)$. $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.559974 & 0.000230 & \{0.559518,0.560430\} \\ b & 0.079361 & 0.000971 & \{0.077434,0.081288\} \\ c & 0.000335 & 0.000012 & \{0.000312,0.000358\} \\ d & 0.598575 & 0.002734 & \{0.593148,0.604003\} \\ e & 0.019840 & 0.000416 & \{0.019015,0.020665\} \\ \end{array}$$

For this range of $x$, the error on $y$ is less than $0.0004$.

Update

In a comment, you asked about a possible explicit Taylor expansion.

What I did was to develop $f(y)$ around $y=0$ and obtained $$f(y)=\left(x^{\sqrt{2}}-\left(x^2+1\right)^{\frac{1}{\sqrt{2}}}\right)+\left(\frac{\left(x^2+1\right)^{\frac{1}{\sqrt{2}}-1}}{\sqrt{2}}+1\right) y+O\left(y^2\right)$$ from which $$y=\frac{\sqrt{2} \left(x^2+1\right) \left(\left(x^2+1\right)^{\frac{1}{\sqrt{2}}}-x^{\sqrt{2}}\right)}{\left(x^2+1 \right)^{\frac{1}{\sqrt{2}}}+\sqrt{2} (x^2+1)}$$ is very decent for $x>1$.

May I confess my surprise ?

You had a quite good idea, indeed.

New update

It is still possible to improve the approximation using the same approach but with the simplest Pade approximant. The next table compare the two approximations for the same values as in the previous table $$\left( \begin{array}{ccc} x & \text{exact}& \text{Taylor} & \text{Padé} \\ 0 & 0.559793 & 0.585786 & 0.565685 \\ 1 & 0.396430 & 0.401048 & 0.396784 \\ 2 & 0.315123 & 0.316039 & 0.315146 \\ 3 & 0.268608 & 0.268891 & 0.268611 \\ 4 & 0.237668 & 0.237783 & 0.237669 \\ 5 & 0.215192 & 0.215248 & 0.215192 \\ 6 & 0.197911 & 0.197942 & 0.197912 \\ 7 & 0.184089 & 0.184107 & 0.184089 \\ 8 & 0.172706 & 0.172718 & 0.172706 \\ 9 & 0.163121 & 0.163129 & 0.163121 \\ 10 & 0.154906 & 0.154911 & 0.154906 \\ 11 & 0.147763 & 0.147767 & 0.147763 \\ 12 & 0.141478 & 0.141481 & 0.141478 \\ 13 & 0.135892 & 0.135895 & 0.135892 \\ 14 & 0.130886 & 0.130888 & 0.130886 \\ 15 & 0.126366 & 0.126367 & 0.126366 \\ 16 & 0.122258 & 0.122259 & 0.122258 \\ 17 & 0.118504 & 0.118504 & 0.118504 \\ 18 & 0.115055 & 0.115056 & 0.115055 \\ 19 & 0.111873 & 0.111874 & 0.111873 \\ 20 & 0.108926 & 0.108926 & 0.108926 \\ 21 & 0.106186 & 0.106186 & 0.106186 \\ 22 & 0.103629 & 0.103630 & 0.103629 \\ 23 & 0.101237 & 0.101238 & 0.101237 \\ 24 & 0.098993 & 0.098993 & 0.098993 \\ 25 & 0.096882 & 0.096882 & 0.096882 \end{array} \right)$$

The formulae to be used in the present case are $$y_{\,\text{Taylor}}=-\frac{F(x,0)}{F^{(0,1)}(x,0)}$$ $$y_{\,\text{Padé}}=\frac{2 F(x,0) F^{(0,1)}(x,0)}{F(x,0) F^{(0,2)}(x,0)-2 F^{(0,1)}(x,0)^2}$$ with $$F(x,0)=x^{\sqrt{2}}-\left(x^2+1\right)^{\frac{1}{\sqrt{2}}}$$ $$F^{(0,1)}(x,0)=\frac{\left(x^2+1\right)^{\frac{1}{\sqrt{2}}-1}}{\sqrt{2}}+1$$ $$F^{(0,2)}(x,0)=-\frac{\left(\frac{1}{\sqrt{2}}-1\right) \left(x^2+1\right)^{\frac{1}{\sqrt{2}}-2}}{\sqrt{2}}$$