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This Question was formulated somewhat improperly: See here for the correct question: Parametric Equations for A Logarithmic Sine-wave With Alternately Offset Points of Hyperbolic Tangency !!

Note. Please Read Everything Carefully and study the Graph Before Answering! (PLEASE, See second fig. [the close up one] also for more detailed info about the geometry; this is how I think the most self-evident solution will graph but don't feel too constrained by it.) I found the geometry for a sine wave that I've been trying to derive (and have posted about here before). (See my fig.) Everything appears pretty straightforward, however, I seem to lack the skill to derive it. My image should give you most of the information you need. But, here are a few points worth describing explicitly:

  1. The wave-length one side is always Phi to the one on the other side. (I know I'm not using the word 'wave-length' in a standard manner!) That is, if it's 1 on the right, it will be 1.618.... on the left (see fig.)

  2. I'm looking for parametric equations; they should take the form: $x(t)=(FUNCTION)^{-1}*\sin(t), y(t)=(FUNCTION)$ I've given you $y(t)$ and $x(t)$ in the geometry of my image, because the whole thing should fit perfectly on $1/x$, thus the function for $x(t)$ is the one for $y(t)$ raised to the minus 1 or vice versa (Obviously the $\sin(t)$ itself is NOT raised to a negative power or such like!)

  3. If you need points for scale, I think the graph first crosses $y$ at $(0,1)$, in other words, it starts there in the same way that this: https://www.desmos.com/calculator/f53khj12ne Starts at $(0,1)$. I think it next crosses $y$ at $(0, 1.618\dots)$. These are guesses, so don't adhere to them if they don't make sense to you!

  4. Please try to find something where the function for the geometry is NOT inside the sine function ('$\sin(t)$'): I want to play with the curve and trying to get the expression out of the sine function might be a pain for me. I can't wait to see this curve graphed! Thank you all so much; I'm very grateful for your cleverness and effort! enter image description here enter image description here*

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  • $\begingroup$ If you have any questions, please feel free to ask! Once more thank you all so much! $\endgroup$ – Jinny Ecckle Jan 18 '20 at 19:14
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    $\begingroup$ It's great that you're "extremely excited" about this problem (I mean that sincerely), but "bumping" it just to say so is not the appropriate way to bring more attention. Rather, you should make substantive edits to improve the question. Luckily, your question can be improved! :) ... The first image makes clear that you're interested in getting the points of tangency on the left or right to be separated by odd or even powers of $\phi$. Your second image suggests a relation between the left-and-right points (and maybe also the axis-crossings), but the nature of that relation isn't clear. $\endgroup$ – Blue Jan 20 '20 at 0:09
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    $\begingroup$ Also, just to be sure: You want your wave to be tangent to $y=1/x$, correct? ... Also-also: Is one of your points of tangency specifically at $(1,1)$? Does it matter? $\endgroup$ – Blue Jan 20 '20 at 0:10
  • $\begingroup$ Thank you for the response and questions. I'd be honored for you to work on this (though you don't have to)! I'll Try and improve the post. Great Qs! Yes, I want the curve tangent. It's tangent by definition because the function for x(t) is one over the one for y(t) (leaving aside the sin(t) aspect). Odd or even is one way to think about it. I think of it like this: The 'wave-length' grows by PHI from side to side. I hope that makes sense. Thank you once more! $\endgroup$ – Jinny Ecckle Jan 20 '20 at 0:25
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    $\begingroup$ Why is the whole post in italics? $\endgroup$ – Gerry Myerson Jan 22 '20 at 21:08
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I'm going to re-orient things, and shift a phase, for reasons that I hope become clear.

enter image description here

A curve parameterized by $$(x,y)=\left(f(t),\frac{\cos t}{f(t)}\right) \tag{1}$$ meets, and is tangent to, the hyperbola(s) $xy=\pm 1$ when $t$ is an integer multiple of $\pi$. Let $P_k = (x_k,y_k)$ be the point of tangency corresponding to $t = k\pi$.

(Note that the $P_k$ are not the local maxima and minima of the graph, since the tangent lines at those points are not horizontal.)

We want the horizontal offsets between every-other point of tangency to be a power of $\phi$; specifically, we want $$x_{k+1}-x_{k-1} = \phi^k \tag{2}$$

I suspect that OP intends the graph to bounce between the branches of the hyperbolas without crossing the $y$-axis (OP's $x$-axis). Moreover, it seems appropriate —but apparently, it is not; see "Update" below— for the graph to approach the $y$-axis, so that the $x$-coordinate of $P_0$ is the accumulated horizontal offsets in the sum $$x_0 = \phi^{-1}+\phi^{-3}+\phi^{-5} + \cdots = \frac{\phi^{-1}}{1-\phi^{-2}}=\frac{\phi}{\phi^2-1}=\frac{\phi}{(\phi+1)-1} = 1 = \phi^0 \tag{3}$$ (where we have exploited the golden ratio property $\phi^2 - \phi - 1 = 0$). Likewise, $$x_{-1} = \phi^{-2}+\phi^{-4}+\phi^{-6}+\cdots = \frac{\phi^{-2}}{1-\phi^{-2}}=\phi^{-1}\qquad\text{and}\qquad x_1 = 1 + x_{-1} = \phi^1 \tag{4}$$ Interesting. We have three instances where the subscript on $x$ matches the power on $\phi$. Well, if $x_{k-1}=\phi^{k-1}$, relation $(2)$ allows us to write $$x_{k+1} =x_{k-1}+\phi^k = \phi^{k-1}+\phi^{k} = \phi^{k-1}(1+\phi) = \phi^{k-1}\phi^2=\phi^{k+1} \tag{5}$$ so that, by induction, all subscripts on $x$ match the powers on $\phi$. We can extend this notion from integer $k$ to all reals by taking

$$f(k\pi) =x_k= \phi^k \quad\to\quad f(t) = \phi^{t/\pi}\quad\to\quad (x,y) = \left(\phi^{t/\pi},\phi^{-t/\pi}\cos t\right) \tag{$\star$}$$

This certainly seems to give the desired plot:

enter image description here


Update.

In comments below and in a revised question, OP updated the requirements so that (in my re-oriented context) the curve must pass through $(1,0)$; for greater generality, we'll make this $(\beta,0)$. Moreover, the revised question asks that the offsets between tangent points be scaled powers of $\phi$. These changes are not difficult to accommodate. Let's return to the above analysis at $(2)$, adjusting it to include $\alpha$:

$$x_{k+1}-x_{k-1} = \alpha\phi^k \tag{2'}$$

Observing that $$\phi^{k+1}-\phi^{k-1} = \phi^k \left( \phi - \frac{1}{\phi}\right) = \phi^k (\phi-(\phi-1)) = \phi^k \tag{3'}$$ it's reasonable to suspect that $f$ has the form $$f(t) = \alpha\phi^{t/\pi}+c \tag{4'}$$ for some constant $c$ that gets lost in the difference in $(2')$.

Previously, getting the curve to approach the $y$-axis amounted to having $c=0$ (with $\alpha=1$). Now, to pass through $(\beta,0)$, all we need to do is force $f(t)$ to be $\beta$ when $\cos(t)$ is $0$; specifically, OP wants the curve to meet $(\beta,0)$ between my $P_1$ and $P_{-1}$, so we take $t=-\pi/2$. Solving gives $$\beta = f\left(-\frac{\pi}{2}\right) = \alpha\phi^{-\pi/2/\pi}+c \qquad\to\qquad c = \beta-\frac{\alpha}{\sqrt{\phi}} \tag{5'}$$ whence

$$f(t) = \alpha\phi^{t/\pi} - \frac{\alpha}{\sqrt{\phi}} + \beta \tag{$\star$'}$$

For $\alpha=\beta=1$, the plot is as follows:

enter image description here

The substitution $t\to t-\pi/2$ shifts the phase of things so that $(\beta,0)$ occurs at $t=0$. Moreover, it trades $\cos t$ for $\sin t$ in the parameterization, so that, calling the shifted function $f_0$, we have

$$f_0(t) = \alpha\phi^{(t-\pi/2)/\pi} + \beta - \frac{\alpha}{\sqrt{\phi}} = \frac{\alpha}{\sqrt{\phi}}\left(\phi^{t/\pi}-1\right) + \beta \quad\to\quad (x,y) = \left(f_0(t),\frac{\sin t}{f_0(t)}\right)$$

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    $\begingroup$ @JinnyEcckle: If all your new question does is scale the $x$-values by $\alpha$, won't this work? $$\left(\alpha\cdot\phi^{t/\pi}, \left(\alpha\cdot\phi^{t/\pi}\right)^{-1}\cos t\right)$$ $\endgroup$ – Blue Jan 20 '20 at 21:53
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    $\begingroup$ @JinnyEcckle: As for the sine-vs-cosine thing: you can simply phase-shift back, replacing $t$ with $t-\pi/2$. ... I chose to use $\cos t$ because it made the tangency points occur at integer multiples of $\pi$ (that is, even multiples of $\pi/2$) rather than odd multiples of $\pi/2$. This made the relations easier to formula without the messy $(2k+1)$ factors in my comment to your question. $\endgroup$ – Blue Jan 20 '20 at 22:01
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    $\begingroup$ @JinnyEcckle: Scaling and de-phasing works for me! For instance, for $\alpha=2$, this gives a compressed version of the graph: $$\left(2p^{(t -\pi / 2) / \pi}, \left(2p^{(t - \pi/ 2) / \pi}\right)^{-1} \sin t \right)$$ $\endgroup$ – Blue Jan 20 '20 at 22:35
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    $\begingroup$ @JinnyEcckle: (Whoops. I wrote "$p$" where I meant "$\phi$".) BTW, note that $$\phi^{(t-\pi/2)/\pi}=\phi^{t/\pi-1/2}=\frac{\phi^{t/\pi}}{\sqrt{\phi}}$$ By interesting coincidence, $\sqrt{\phi}=1.272\ldots$, which seems to align with your "mental calculation" of scale factor $1.28$ (so this probably isn't a coincidence). $\endgroup$ – Blue Jan 20 '20 at 22:41
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    $\begingroup$ @JinnyEcckle: I'm glad to help. :) ... That said, I urge to you always give as much information about a problem from the get-go ("study the graph" is insufficient) to avoid misunderstandings and lost time. I'm concerned that, when you write "My original graph (sin(t)*PHI^t/PI, PHI^-t/PI) ...", you already had this solution in mind, making my (re-)derivation unecessary; and you could/should have mentioned "start at (0,1)" when I asked for specifics. ... Remember: we're mathematicians, not mindreaders. $\endgroup$ – Blue Jan 20 '20 at 23:06
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I found something that seems to be heading in the right direction. However, does not really fit your curve nd doesn't uise PHI, but it might inspire you to think up the answer. i hope this helps:

https://www.desmos.com/calculator/7bhjuv91c1

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  • $\begingroup$ Thanks for putting some effort. Maybe it will help! :) $\endgroup$ – Jinny Ecckle Jan 18 '20 at 20:24
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    $\begingroup$ Sorry I couldn't post this in the comments. it wouldn't let me. Thanks. $\endgroup$ – EEEEsl Jan 18 '20 at 20:34

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