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Is there a non-parametric closed form for a function looking like a sine rotated 45 degrees?

I have encountered also a similar question but it asks for a function resembling the rotated sine, but not necessarily exact.

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Anixx
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  • The second answer in your linked question explains exactly what you want. – flawr Dec 23 '14 at 08:48
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    @flawr no, I do not want parametrically defined curve, I want a *closed form*. – Anixx Dec 23 '14 at 08:49
  • This parametric expression is in closed form. What you mean is a nonparametric expression. – flawr Dec 23 '14 at 09:00
  • A parametrical definition is a closed form. - But I (and I think most others) know what you mean, I won't rule out that it exists, but it ain't going to be pretty. – Henrik supports the community Dec 23 '14 at 09:01
  • If you are looking for an expression of the form $f(x)=..$ then you have to ask yourself if the "rotated" sine is actually a function: would each $x$ have only one corresponding value $f(x)$? – megas Dec 23 '14 at 09:03
  • @m.a. yes, it is a function, everywhere defined, having a unique value. – Anixx Dec 23 '14 at 09:04
  • You will only encounter some points where it is not differentiable at the original zeros of the sine (depending on what is the dependent variable this will be at the even / odd multiples of $\pi$). – flawr Dec 23 '14 at 09:07
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    Essentially, if you want to express $t+\sin t$ as a function of $t-\sin t$, you need a closed-form expression for the inverse function of $t-\sin t=\frac{t^3}{6}-\frac{t^4}{24}+\frac{t^5}{120}-\ldots$. – Jack D'Aurizio Dec 23 '14 at 09:26
  • @Jack D'Aurizio yes. – Anixx Dec 23 '14 at 09:27
  • Then, have a look at http://en.wikipedia.org/wiki/Lagrange_inversion_theorem. – Jack D'Aurizio Dec 23 '14 at 09:28
  • @Henrik parametric expression is a closed form but for curve rather than for the function. Similarly, closed form for inverse function is not a closed form for straight function. – Anixx Dec 24 '14 at 10:15
  • @Anixx Doesn't look like a function to me... – Jack M Dec 24 '14 at 12:13
  • @Jack M this graph is wrong. It is rotated 60 degrees – Anixx Dec 24 '14 at 13:21
  • @Anixx You're correct, my software was in radians rather than degrees, so that's $45$ radians or about $60$ degrees. The $45$ degree one does indeed look plausibly function-like. – Jack M Dec 24 '14 at 14:27

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Let $(X_0,Y_0)$ be a point on the rotated graph. Then, converting to polar coordinates, the point is $(\sqrt{X_0^2+Y_0^2},arctan \frac{Y_0}{X_0})$. Rotating 45 deg. gives ($\sqrt{X_0^2+Y_0^2},arctan \frac{Y_0}{X_0}$+45 deg.$)$. That is $(\sqrt{X_0^2+Y_0^2}cos(arctan \frac{Y_0}{X_0}+45$ deg.$),\sqrt{X_0^2+Y_0^2}sin(arctan \frac{Y_0}{X_0}+45$ deg.$))$ in Cartesian coordinates. If this point is $(x,y)$, then $y=sin$ $x$, so the closed form $\sqrt{x^2+y^2}sin(arctan \frac{y}{x}+45$ deg.$)=sin(\sqrt{x^2+y^2}cos(arctan \frac{y}{x}+45$ deg.$)$ works.
For $45$ deg. rotation clockwise instead, change $+45$ deg. to $+45$ deg.

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