## Can e^x be thought of as a hyperbola?

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arbiteroftruth
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### Can e^x be thought of as a hyperbola?

The more I look at it, the more an exponential curve seems to be the limit of a hyperbola in much the same way that a parabola is the limit of an ellipse. First off, the intuitive appearance of the curve resembles half of a hyperbola with a horizontal asymptote at y=0 and a vertical asymptote shoved off toward infinity, in much the same way that a parabola is an ellipse with one focus shoved to infinity.

A hyperbola and an ellipse can be described with one equation in which a real value for a given variable yields an ellipse and an imaginary value for the same variable yields a hyperbola. Similarly, eix forms a circle in the complex plane, while ex, as mentioned above, resembles a hyperbola.

Finally, let's look at the area under ex. If we think of the curve as a hyperbola, that would imply that, in a vague hand-wavy sense, there is a -e-x somewhere out beyond the vertical asymptote at infinity, forming the other branch of the hyperbola, and having exactly negative of the area under the more visible branch. Well, the integral of ex from 0 to +inf can be assigned a value in much the same way that a divergent sum can still be assigned a real value. In fact, if you convert the act of integration into a sum of a geometric series(where divergent sums can be easily and coherently defined), and find the limit of the summation as the interval approaches 0(equivalent to shrinking the width of rectangles in an approximation of integration), the integral of ex from 0 to +inf works out to -1. Since the integral from -inf to 0 is 1, this means the total area under ex is precisely 0, consistent with the notion of a hypothetical opposite branch of the hyperbola that cancels out the area.

[EDIT] Also of possible relevance: ln(x), the inverse of ex, is the integral of the right side of 1/x, which is a hyperbola. [/EDIT]

I realize the above analysis isn't exactly rigorous, but that's because my point is simply that this a conjecture/intuition I have about the subject, and I'm open to having all sorts of holes poked in it. What I'm really interested in is this: if ex can be thought of as the limit of a hyperbola as some property tends toward an extreme, how might we define what that property is, and thus define the family of shapes between hyperbolas and exponential curves? That is, if the exponential curve is the result of stretching(or whatever) a hyperbola until one of its asymptotes vanishes into infinity, what is the equation for a hyperbola that has been finitely "stretched" in the same manner?

And if this is something already well-known, somebody just tell me what it's called so I can know where to look to actually find stuff about it. Thanks.

jestingrabbit
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### Re: Can e^x be thought of as a hyperbola?

I often find the most useful way to think about conic sections ie ellipses, parabolas and hyperbolas, as the intersection between a plane and a cone. That's what they are. If your cone is x^2+y^2=z^2, then a mostly horizontal plane, a plane with a normal vector (x, y, z) with x^2 + y^2 < z^2, will produce an ellipse, or a point if it passes through the origin. A mostly vertical plane, with a normal satisfying x^2 + y^2 > z^2, will produce a hyperbola, or a pair of lines if it passes through the origin. The final case, with a normal satisfying x^2 + y^2 = z^2 will produce either a parabola, or a line if it passes through the origin.

The relationship between an ellipse and a parabola is pretty easy to see from here: if you intersect a plane and a cone, and you get an ellipse, and then you tilt the plane in the right way, and the intersection between the plane and the ellipse becomes a parabola, and one of the foci hasn't changed through the entire tilt; if you tilt the plane the other way, and still fix a focus, you get a line. Similarly, if you start with a hyperbola, you can tilt the plane one way to get a parabola, or the other way to get a line.

On the other hand, none of what you've said is remotely as clear. If you want to say that e^x is a degenerate hyperbola, you might want to at least crack open an equation or two, and show that changing a particular parameter gets you e^x.

Also, this

arbiteroftruth wrote:a vertical asymptote shoved off toward infinity

makes baby jesus cry. Its pretty much meaningless, if not absolutely so. Please try to refrain from making these sorts of statements in the future.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

arbiteroftruth
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### Re: Can e^x be thought of as a hyperbola?

jestingrabbit wrote:On the other hand, none of what you've said is remotely as clear. If you want to say that e^x is a degenerate hyperbola, you might want to at least crack open an equation or two, and show that changing a particular parameter gets you e^x.

One may wish to consider the fact that I pretty explicitly stated that the entire point of this thread was to get help in working out precisely that issue.

jestingrabbit wrote:Also, this

arbiteroftruth wrote:a vertical asymptote shoved off toward infinity

makes baby jesus cry. Its pretty much meaningless, if not absolutely so. Please try to refrain from making these sorts of statements in the future.

See above.

But if you wish, I'll try to refrain from asking for help with something unless I already know how to fully explain it and thus don't need the help in the first place.

lightvector
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### Re: Can e^x be thought of as a hyperbola?

No, e^x cannot be thought of as a hyperbola. Just about the only similarity is that the graph of e^x and a branch of a hyperbola both "look" like convex curves. Just like all of the dozens of other common functions that "look convex", their actual properties are wildly different, and it is not useful at all to think of one as the other.

One way in which a parabola can be considered a limit of ellipses is that for any parabola, there exists a sequence of ellipses that converges uniformly to the parabola within any bounded region. This is not true of hyperbolas and e^x. Every hyperbola whose asymptotes are horizontal and vertical is of the form y = c/(x-x_0) + y_0, for some constants c, x_0, and y_0. That is, nothing more than y = 1/x translated and scaled. It should then be obvious (and it should be easy to show rigorously too) that the graph e^x cannot be a limit of hyperbolas.

arbiteroftruth
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### Re: Can e^x be thought of as a hyperbola?

I should revise the relationship I'm trying to describe. It is indeed clear that e^x cannot be constructed strictly with true hyperbolas. It would be more accurate to say I think that both e^x and hyperbolas can be described as special cases of a more general family of curves.

lightvector, you bring up that all hyperbolas with horizontal/vertical asymptotes are linear transformations of 1/x, or x-1. One way to generalize this to a form that would include exponential(or in this particular case, logarithmic) curves would be to use A+B*x-C, rather than A+B*x-1).

The hyperbolas formed by transformations of 1/x would obviously be the special case where C=1. ln(x) is obtained by considering the family of cases in which A=-B=1/C, and taking the limit as C approaches 0. That is, (1-x-C)/C as C approaches 0.

I'm wondering what the more generalized version of that family would look like, in terms of focii/directrices etc.

EDIT: On a side note, I'd like to mention that the specific family of curves I mentioned, (1-x-c)/c, when c is positive, has a horizontal asymptote at 1/c, and this asymptote gets shoved toward infinity as c approaches 0.

Qaanol
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### Re: Can e^x be thought of as a hyperbola?

Here is a relationship between hyperbolæ and exponentials.
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Timefly
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### Re: Can e^x be thought of as a hyperbola?

arbiteroftruth wrote:One way to generalize this to a form that would include exponential(or in this particular case, logarithmic) curves would be to use A+B*x-C, rather than A+B*x-1).

I don't know what you're trying to do here, but the function f(x) = k(1-x^(1/k)) is not equal to the log function for any k.

jestingrabbit
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### Re: Can e^x be thought of as a hyperbola?

Timefly wrote:I don't know what you're trying to do here, but the function f(x) = k(1-x^(1/k)) is not equal to the log function for any k.

No, but

$\ln(x) = \lim_{h\to 0} \frac{x^h - 1}{h} \qquad \text{ and } \qquad e^x = \lim_{n\to \infty} (1+x/n)^n.$

arbiteroftruth wrote:On a side note, I'd like to mention that the specific family of curves I mentioned, (1-x-c)/c, when c is positive, has a horizontal asymptote at 1/c, and this asymptote gets shoved toward infinity as c approaches 0.

Sure, but its important to note that properties that a sequence has are not necessarily shared by the limit of the sequence. Talking about "asymptotes at infinity" isn't meaningful.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

chenille
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### Re: Can e^x be thought of as a hyperbola?

jestingrabbit wrote:Also, this
arbiteroftruth wrote:a vertical asymptote shoved off toward infinity

makes baby jesus cry. Its pretty much meaningless, if not absolutely so. Please try to refrain from making these sorts of statements in the future.

Not at all. There are perfectly reasonable ways to extend things to infinity, you just have to be careful about what you mean. Here there is a natural idea of zeroes and poles at infinity which comes up all the time in complex analysis, where they are defined in terms of the Riemann sphere. For instance at x = 0 the function f(x)=x has a first-order zero and f(x)=1/x has a first-order pole, while at x = ∞ they are reversed. You can use rational functions to move these things around in a straightforward way.

The behavior of ex at infinity, or e1/x at the origin if you want to look at it there, is something very different from the pole of any x-c. Those give double-sided asymptotes when defined over the reals, whereas this goes to infinity from one side and zero from the other, and has periodic behavior over other complex numbers. So from this perspective, I don't think it's useful to think of it as anything like the pole of a hyperbola moved there. You can definitely look at it as a limit of poles of increasing degree, though, which is basically what's happening in the series ex = 1 + x + x2/2! + x3/3! + ...

Edit: It occurs to me you can do the same thing with the limit given above:
$e^x = 1/e^{-x} = \lim_{n\to \infty} (1-x/n)^{-n}$This sequence of functions starts with a hyperbola and has poles that move to infinity, so I think it answers the original question. Note, though, they also have zeroes of increasing degree at infinity, so the end result is really a very different thing from either. In the other limit, the zeroes get moved and the poles stay in place.