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How can I express the following function as $y$ as a function of $a$ and $x$? $$2\arcsin \left (\frac{\frac{x}{2}}{\frac{x^2}{8y}+\frac{y}{2}}\right )\cdot\frac{x^2}{8y}+\frac{y}{2}=ax$$ Reason for asking is: friends and I were laying a laminate floor and discussed how bad it would be if we left no room around the edge and the floor expanded a few %, my thesis was and is: pretty bad. But I'd love to be able to express how bad.

I approximated the problem by disregarding 2 of the four walls and looking at a cross section of bulging floor. I then decided that it looks somewhat similar to a section of a circle (although in hind-sight a parabola might be more accurate). I know the distance between the walls $x$ and the expansion factor $a$ which will give me the length of the floor after it has bulged: $ax$. I'm interested in the height of the bulge in the middle of the room $y$.

Now I need to rewrite the above function that I obtained by performing more steps than I care to write out (unless people here want to know) in order to get an expression for $y$ that depends only on $a$ and $x$. But I'm stuck on the arcsin, I can't figure out how to get all terms with $x$ and $a$ to one side and $y$ to the other.

Any help is greatly appreciated!

Blue
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XiB
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  • You mean $y(a,x)=2{\left(\dfrac{x^2}{8y}+\dfrac{y}{2}\right)}\arcsin \left (\dfrac{\frac{x}{2}}{\dfrac{x^2}{8y}+\dfrac{y}{2}}\right )=ax$? – Narasimham Jul 04 '22 at 20:22
  • I'm not 100% sure but it looks to me like $y(a,x)$ is dependent upon itself. I was hoping to find an expression for y that does not depend upon $y$, and instead only on $x$ and $a$. – XiB Jul 05 '22 at 22:21

3 Answers3

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In the same spirit as @Narasimham, for small values of $t$, the simplest Padé approximant is $$\sin ^{-1}(t)=\frac{6 t}{6-t^2}$$ whose error is $\frac{17 }{360}t^5$.

Just to give an idea $$I(a)=\int_{-a}^{+a} \Bigg[\sin ^{-1}(t)-\frac{6 t}{6-t^2} \Bigg]^2\,dt$$ some results $$\left( \begin{array}{cc} a & I(a) \\ 1.0 & 6.083 \times 10^{-3} \\ 0.9 & 6.818 \times 10^{-4} \\ 0.8 & 1.135 \times 10^{-4} \\ 0.7 & 1.854 \times 10^{-5} \\ 0.6 & 2.634 \times 10^{-6} \\ 0.5 & 2.919 \times 10^{-7} \end{array} \right)$$

Using $$t=\frac{4 x y}{x^2+4 y^2}$$ $$2{\left(\frac{x^2}{8y}+\frac{y}{2}\right)}\arcsin \left (\frac{\frac{x}{2}}{\frac{x^2}{8y}+\frac{y}{2}}\right )\sim \frac{3 x \left(x^2+4\right) y^2 \left(x^2+4 y^2\right)}{3 x^4+16 x^2 y^2+48 y^4}$$ leads to a quadratic equation in $z=y^2$ $$12 \left(x^2+4(1-a)\right)z^2+x^2 \left(3 \left(x^2+4\right)-16 a\right)z-3 a x^4=0$$

Edit

It is possible to improve the approximation minimizin with respect to $k$

$$\Phi(k)=\int_{-1}^{+1} \Bigg[\sin ^{-1}(t)-\frac{ t}{1-k~t^2} \Bigg]^2\,dt$$ which is analytic (the longish expression will not be reported here). Using $k=\sin^2(t)$, the optimum value corresponds to the solution of $$\sin ^5(t) (2 \pi \cos (t)-5)+\sin ^3(t) ((5-2 \pi ) \cos (t)+8)+3 (\cos (t)+1) \cos ^5(t) \tanh ^{-1}(\sin (t))-\sin (t) (\cos (t)+1) \left(4 \pi \cos ^5(t) \log \left(\frac{2}{\sec (t)+1}\right)+3\right)=0$$ which, numerically, is $$k_{\text{opt}}=0.258134535583254629273188966092605183439\cdots$$ This number is not identified by inverse symbolic calculators.

The norm decreased by a factor of $5.35$ and the maximum error by a factor of $3$.

A good approximation $$t_{\text{opt}}=\frac{19 \pi ^2+411 \pi+402}{741 \pi ^2-1119 \pi-269 }$$ gives the first $20$ significant figures for $k_{\text{opt}}$.

Claude Leibovici
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  • This looks very impressive! I'm going to need a while to partially understand what you are doing and why. I have never heard of a pade approximant, I don't really understand why you picked

    $$t=\frac{4xy}{x^2+4y^2}$$

    as a value for t

    And the things you added in your edit are a bit beyond my ability to understand without diving into my books for help haha.

    But thank you tremendously for taking the time to work through my problem, in time I might be able to use your solution :).

     – XiB Jul 05 '22 at 22:32
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If $$\dfrac{x}{2}<<{\left(\frac{x^2}{8y}+\frac{y}{2}\right)} $$

then the slope of sine curve tends to a constant $a=1$ near x=0,

$$y(a,x)=2{\left(\frac{x^2}{8y}+\frac{y}{2}\right)}\arcsin \left (\frac{\frac{x}{2}}{\frac{x^2}{8y}+\frac{y}{2}}\right )\approx x$$

Arcsin series expansion

We can expand $2u \sin^{} \dfrac{x/2}{u} $when $ u={\left(\dfrac{x^2}{8y}+\dfrac{y}{2}\right)} $

using power series to any desired accuracy and numerically evaluate implicit polynomials.

Narasimham
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  • In my toy example that I use to check if I'm going wrong somewhere, I use x=8, y=2 and a is somewhere between 1.1 and 1.3. So

    $$ \frac{x}{2} </< (\frac{x^2}{8y} + \frac{y}{2}) $$

    for very small values of $y$ this is true, but my hypothesis is that with an expansion of the flooring of only a few % $y$ will grow to extreme proportions very quickly.

     – XiB Jul 05 '22 at 22:34
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I prefer to add a separate answer.

Using $t=\frac{4 x y}{x^2+4 y^2}$, the problem is to solve for $t$ the equation $$\frac{\sin ^{-1}(t)}{t}=a$$

Expanding the lhs as a series around $t=0$ and using series reversion

$$t=b+\frac 3{40}\sum_{n=1}^\infty (-1)^n\alpha_n \,b^{2n+1} \qquad \text{with} \qquad b= \sqrt{6(a-1)}$$

The first coefficients $\alpha_n$ make the sequence $$\left\{3,\frac{323}{560},\frac{21197}{201600},\frac{1030999}{55193600},\frac{1407584257}{430510080000},\frac{409960705481}{723256934400000},\cdots\right\}$$

For a quite severe test, trying for $a=\frac 43$, that is to say $b=\sqrt 2$, we have for the above truncated series $$t=\frac{102045347517601}{75339264000000 \sqrt{2}}=0.957760$$ while the solution is $0.956774$.

When $t$ is known, then, for a given $x$, $y$ is known.

Claude Leibovici
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