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"No, they're not equidistant" But you can translate $e^x - 1$ upwards by 1 and they're parallel. All the points, so they are equidistant. Furthermore, I've created (discovered?) a postulate that tests if functions are parallel: If f'(x)=g'(x), and f(x)$≠g(x), then f(x) and g(x) are parallel. If it works for linear functions, why shouldn't it work for exponential functions? Also, assume this is Euclidean Geometry.

Tianyue Ma
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    The answer would depend on your definition for 2 curves to be parallel. You're taking a different version than the standard definition, and so should make it clear that you're talking about "shifted curves" rather than "parallel curves". If so, I'd argue that shifting by 1 in the x direction would also result in "shifted curves". – Calvin Lin Oct 26 '23 at 00:41
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    The graphs of $y = e^x$ and $y = e^x - 1$ are not equidistant; the distance between them is close to $1$ when $x$ is very negative, and approaches $0$ when $x$ is a very large positive number. To go from one graph to the other, you can go diagonally sideways instead of directly upward. – Misha Lavrov Oct 26 '23 at 00:42
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    They are "equidistant" vertically, but not horizontally. In contrast, $e^{x-1}$ is horizontally, but not vertically, "equidistant" from $e^x$. – Geoffrey Trang Oct 26 '23 at 00:50

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There is no objective definition of “parallel functions”. Defining functions to be parallel if they differ by a constant is an interesting option — but is the idea robust? Why dont you try to “push the idea” some? How would you define functions that are at right angles to each other? What is the angle between $x^2$ and $cos(x)$?

When discussing a “geometry of functions” it is common to stay in the context of a function space where the geometry is determined by an inner (dot) product. We say that two vectors in $\mathbb{R}^n$ are parallel if $\tfrac{\bf{u} \cdot \bf{v}}{\|\bf{u}\|\|\bf{v}\|}= 1$, the same can be true for functions if we give a suitable definition of “dot product”, picking $f \cdot g = \int_0^1 f(x)g(x) dx$ is a very common choice but there are infinitely many choices of inner products and they are all useful in different situations.

Lee Fisher
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