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A plane curve is printed on a piece of paper with the directions of both axes specified. How can I (roughly) verify if the curve is of the form $y=a e^{bx}+c$ without fitting or doing any quantitative calculation?

For example, for linear curves, I can choose two points on the curve and check if the midpoint is also on the curve. For parabolas, I can examine the geometric relationship between the tangent at a point and the secant connecting the peak and that point. Does the exponential curve have any similar geometric features that I can take advantage of?

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  • $\begingroup$ Interesting question :) $\endgroup$ – TripleA Jan 12 '17 at 0:23
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    $\begingroup$ Actually you don't need the axes' directions. Just look for the (only) asymptote. $\endgroup$ – Pedro Sánchez Terraf Jan 12 '17 at 0:36
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    $\begingroup$ Do a log plot of $f(x)-f(\pm \infty)$ and eyeball if it's linear. $\endgroup$ – dxiv Jan 12 '17 at 3:45
  • $\begingroup$ If you can replot the data with semilog axes, it should be obvious. $\endgroup$ – Brian Borchers Jan 12 '17 at 23:26
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You can't. No, not just "in theory", but also in practice.

I tried this when doing regression before and I gave up on it once I realized how impossible it is:

Curve

(Ignore the left upwards part of the parabola; pretend you don't have that piece of information when you're trying to tell which is which.)


Update

Since I couldn't reproduce the plot above anymore (I only kept the screenshot, and I'm not sure why the formulas don't seem to be reproducing it), I'll include an artificial one that illustrates the same problem:

Plot 2

To reproduce:

Plot[{Exp[x] / 4, 0.32 x^2 + 0.12 x + 0.26}, {x, 0, 2}, PlotRange -> {Automatic, {0, 2}},
    GridLines -> Automatic, AspectRatio -> 1, BaseStyle -> {FontSize -> 14}]
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    $\begingroup$ This looks an awful lot like what my economics professor called the curse of exponentiality (which as he used it referred to the fact that you can't really tell whether a curve (e.g. US debt) is exponential or something like parabolic until it's too late to do anything about it; when I did a google to make sure I had the right term, it looks like it usually is taken to refer to something slightly different), but the main point being: you might be able to determine better if you extend the curve out further $\endgroup$ – Foon Jan 12 '17 at 21:10
  • $\begingroup$ @Foon: It very well may have been (though the only such "curse" I've heard about is that of "dimensionality" rather than "exponentiality"). It's worth pointing out though that the nice thing is nothing in real life can grow exponentially for very long; if you're observing something physical, it's likely to be e.g. logistic instead. What I was doing regression on was the complexity of an algorithm, which can happily grow exponentially. :-) If I remember correctly I couldn't grow it to the right since I didn't have the time to wait for it to finish with very large inputs... $\endgroup$ – user541686 Jan 12 '17 at 21:16
  • $\begingroup$ You only can't tell between quadratic and exponential in that example because the parameter is varied over just a small range, so mostly you see only two Taylor terms of the exponential anyway. If the parameter reached twice as far, in either direction, then it would be much clearer what's going on. $\endgroup$ – leftaroundabout Jan 12 '17 at 21:32
  • $\begingroup$ @leftaroundabout: I think it's safe to assume the OP isn't asking about the cases where it's obvious which is which... because those cases are obvious. $\endgroup$ – user541686 Jan 12 '17 at 21:35
  • $\begingroup$ The question seems to be about fairly precisely drawn curves rather than dirty data or approximations, so I think this does not answer it. On the other hand it does answer a different interesting question! $\endgroup$ – PJTraill Jan 12 '17 at 21:37
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Are the doubling points evenly spaced?

Assume $a$ and $b$ are positive (if not, it's easy to see and readjust - $a$ is positive if the curve flattens to the left, $b$ has the same sign as $a$ if the curve is increasing). Mentally reset the $x$ axis at the height to which the curve tends to become horizontal to the far left (that's $c=\lim_{x\rightarrow -\infty} ae^{bx}+c$) . Note: there's a way to be really rigorous about this and obtain $c$ with ruler and compass (see below) but then it's not really "eyeballing".

With $c$ set to $0$, take a point $x_1$ on the (new) $x$ axis, and a random multiplier $M>1$; say, $M=2$. Eyeball the point $x_2$ at which $y(x_2)$ is roughly $M$ times $y(x_1)$. Then eyeball the point $x_3$ at which $y(x_3)$ is roughly $M$ times $y(x_2)$, and so on, and check if the points $x_1,x_2,x_3,...$ at which $y$ multiplies by $M$ are evenly spaced, as they must be if your curve is an exponential.

Easier to do than to say... as long as you have found enough "multiplication" points on your piece of paper. If your curve is very "flat" you'd have to choose an $M$ very close to $1$, but it still works in theory (or if you do it with ruler and compass with sufficient accuracy - but again, that's not really "eyeballing"). In practice it will be hard to tell such a "flat" exponential from a parabola or even a line: remember that $e^{bx}$ is very close to $1+bx$ if $bx\ll 1$, i.e. if $\frac{1}{b}$ is much larger than the largest $x$ you have on your piece of paper. Then again, even a "sufficiently flat" parabola is hard to tell from a line...

If your piece of paper on the other hand is sufficiently large, as a bonus, you can also gauge $a$ (or more accurately $a \cdot$ e), $b$ (or more accurately $\frac{\ln M}{b}$) and $c$. We know that $c$ is the height to which the curve converges as $x\rightarrow -\infty$, while ($a \cdot$ e) is the distance between said height and the $y$ axis intersect. And the horizontal spacing $\Delta x$ between the points at which you check the curve is $\frac{\ln M}{b}$ since $e^{b\Delta x}=M$.

Let's be rigorous!

First of all, how can we rigorously construct $c$ with ruler and compass? To do it, all we have to do is to take a random positive horizontal spacing $\Delta x$, and obtain $\Delta^0_y=y(0)-y(-\Delta x)$, and $\Delta^1_y=y(-\Delta x) - y(-2\Delta)$. Call $\rho<1$ the ratio $\frac{\Delta^1_y}{\Delta^0_y}$. It's immediate that $c$ is located at distance $\sum_{i=0}^\infty \rho^i \Delta^0_y = \Delta^0_y \frac{1}{1-\rho}$ below the intersect of the curve with the $y$ axis, which we can easily if somewhat laboriously obtain with ruler and compass noting that $\Delta^0_y$ is mid-proportional between said distance and $\Delta^0_y-\Delta^1_y$.

Also, as Rahul correctly points out, to be really formal one would have to choose $M$ sufficiently "randomly" (uniformly at random in any arbitrary small non-degenerate interval suffices), so that the probability of encountering a function in the form $e^{bx+f(x)}$ with $f(x)$ periodic with a period that's exactly an integer multiple of $M$ would be $0$. In practice, since you are only eyeballing for rough exponentiality, checking that your curve does not "wiggle" is enough to rule out such cases!

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    $\begingroup$ If it's very flat, you can do the opposite, $x$ should double at $y_1, y_2, y_3$ etc. $\endgroup$ – Travis Jan 12 '17 at 0:42
  • $\begingroup$ How easy is it to estimate where $c$ lies visually? $\endgroup$ – Erick Wong Jan 12 '17 at 2:34
  • $\begingroup$ The doubling property only holds if $c=0$. Otherwise you have to find the points where the distance from $y=c$ doubles, and we don't know $c$ beforehand. $\endgroup$ – Rahul Jan 12 '17 at 2:43
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    $\begingroup$ Yes. And you can estimate c easily by looking to the far left, since it's the limit of y as x goes to $-\infty$. $\endgroup$ – Anonymous Jan 12 '17 at 2:45
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    $\begingroup$ $ae^{bx+f(x)}$ for any $\bigl(\frac{\ln 2}b\bigr)$-periodic function $f$ also has equally spaced doubling points. $\endgroup$ – Rahul Jan 12 '17 at 5:00
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Assuming $c=0$ (it's not the graph of a exponential function otherwise):

Pick a point on the curve. Draw its tangent and extend it until it meets the $x$-axis. Also drop a vertical from the point to the $x$-axis. Now you have a right triangle.

Do this for lots of points. The bases of all the triangles should have the same length.

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  • $\begingroup$ The tangent to $y=ae^{bx}+c$ at $(x_1,y_1)$ meets the $x$-axis at $(x_1-y_1/y'_1, 0).$ The distance from this point to $(x_1,0)$ is $y_1/y'_1=(1/b)+c/y'_1$ which is not constant unless $c=0.$.... The tangent meets the horizontal line $y=c$ at $(x_1-1/b,c)$.... So if you have enough data points to suggest that some line $y=c$ is a horizontal asymptote, this will work (approximately). $\endgroup$ – DanielWainfleet Jan 12 '17 at 2:34
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    $\begingroup$ Oh, I didn't notice the constant term in the question. Personally I wouldn't call it an exponential curve unless $c=0$. $\endgroup$ – Rahul Jan 12 '17 at 2:38
  • $\begingroup$ Neither would I,,,,,,,,,,,,,, $\endgroup$ – DanielWainfleet Jan 12 '17 at 2:47
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    $\begingroup$ It's easy, just "lift" or "lower" the x axis so that the curve converges to it as it it goes to $-\infty$. If you want to be really picky about that, you can check two equal horizontal intervals, check the relative vertical intervals $\Delta y_1$, $\Delta y_2$ and from those construct rigorously $c$ ... but the post title says "eyeballing"! $\endgroup$ – Anonymous Jan 12 '17 at 2:52
  • $\begingroup$ @Anonymous: Not so easy if you don't have the curve extending to $-\infty$. After all, in your own answer you point out the difficulty if the range of $x$ is not large enough compared to $1/b$. $\endgroup$ – Rahul Jan 12 '17 at 3:02

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