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How can I find a parabolic function that mimics a hyperbolic one? How would I find the parabolic function for the hyperbolic function $y=5\cosh(\frac x5)$?

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That depends on what you mean by "mimic". –  Gerry Myerson Feb 21 '13 at 0:32
    
Probably by the Taylor series? –  Berci Feb 21 '13 at 1:09
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4 Answers 4

For the hyperbolic cosine the Taylor series is $\cosh(x)=1+x^2/2!+x^4/4!+\cdots$. Hence $5\cosh x/5 = 5(1+(x/5)^2/2!+\cdots)$ and a close approximation to your function, for values close to $0$ is $y=5+{x^2 \over 10}$.

A hanging cable or chain appears in the shape of a parabola but that is only an approximation. The ideal shape is hyperbolic or catenary.

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This is probably as close as we can get to OP's request. –  Ross Millikan Feb 21 '13 at 14:08
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It sounds like you want a parabola that traces a path similar to $\cosh$. These are very different functions. Far from the origin, parabolas can only rise as $kx^2$, while $\cosh$ is exponential, which rises much faster. Over a particular range, you can find the best parabola that fits $\cosh(x)$ but it is a deception to claim that all even functions that go to $+\infty$ as $x \to \infty$ are similar in any other way.

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For simplicity, let's take parabola opening up with value $y=1$ at $x=0$.

Take the cosh function as $y = b \cosh(x/b)$ and the parabola as $y = a x^2+1$.

To get them as close as possible, we will try to match the second derivative at $x = 0$, since both have zero slope there.

For the parabola, $y''(0) = 2a$.

For the cosh, $y' = \sinh(x/b)$ and $y'' = (1/b)\cosh(x/b)$, so $y''(0) = 1/b$.

To have the parabola match this, we must have $2a = 1/b$ or $a = 1/(2b)$.

Therefore the parabola that most closely matches this hyperbola is $y = x^2/(2b) + 1$. In your case, $b = 5$ so the parabola is $x^2/10+1$.

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The question seems to be about the hyperbolic cosine, not about any hyperbolas. –  Gerry Myerson Feb 21 '13 at 1:33
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The circular trig functions relate to the complex numbers. The hyperbolic ones to the split-complex numbers. Presumably the best analogy for parabolic functions would be the dual numbers.

These are boring. $\exp(\epsilon x) = 1 + \epsilon x$, so we would define

  • $\mathop{\text{cosp}}(x) = 1$
  • $\mathop{\text{sinp}}(x) = x$
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Wow. You don't seriously think this is what the question is about, do you? –  Gerry Myerson Feb 21 '13 at 2:25
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