0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Looking at that, I can assure you that you will not be solving for $y$ algebraically any time soon. $\endgroup$ – Simply Beautiful Art Mar 15 '17 at 23:09
  • $\begingroup$ After your splendid idea : plotting the two functions, you do not see any difference ! $\endgroup$ – Claude Leibovici Mar 16 '17 at 13:34
  • $\begingroup$ Indeed! Approximations huzzah! $\endgroup$ – Ola Mar 16 '17 at 14:34
  • $\begingroup$ When you send a comment, start with $@user_name$. I just saw your last one by accident. Do you like the work after your splendid idea ? It is just amazing. $\endgroup$ – Claude Leibovici Mar 16 '17 at 14:57
1
$\begingroup$

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 $$F(x,0)=x^{\sqrt{2}}-\left(x^2+1\right)^{\frac{1}{\sqrt{2}}}$$ $$F^{(0,1)}(x,0)=\frac{\left(x^2+1\right)^{\frac{1}{\sqrt{2}}-1}}{\sqrt{2}}+1$$ $$F^{(0,2)}(x,0)=-\frac{\left(\frac{1}{\sqrt{2}}-1\right) \left(x^2+1\right)^{\frac{1}{\sqrt{2}}-2}}{\sqrt{2}}$$

$\endgroup$
  • $\begingroup$ Thank you. Some questions: Could it still be possible to find an explicit taylor expansion? Could it be possible to solve for $x$? $\endgroup$ – Ola Mar 16 '17 at 10:00
  • $\begingroup$ @OlaB. I think it would be a good idea .... to forget it ! However, let me think. If anything comes to my mind, I'll post. Cheers. $\endgroup$ – Claude Leibovici Mar 16 '17 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.