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Hyperbolic cosine looks like a parabola

  1. Oct 18, 2012 #1
    Hello,

    I wanted to know why the graph of the hyperbolic cosine function (1/2(e^x)+1/2(e^-x)) looks like a parabola. Is there any reason for this? I suppose the individual exponential functions both go to infinity in a symmetric way.... but I wanted a better reason :).

    Thanks,
    Mathguy
     
  2. jcsd
  3. Oct 18, 2012 #2

    phyzguy

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    They are only qualitatively similar. Cosh(x) looks like a parabola in the sense that it is concave upward and increases faster than linear, but in fact it is very different from a parabola. The height of a parabola (like y=x^2) increases as x^2 as you move away from 0. y=cosh(x) increases exponentially as x gets large. At x=100, the value x^2 is 10^4, while the value of cosh(x) is about 10^43, which is hugely larger.
     
  4. Oct 18, 2012 #3

    AlephZero

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    The Taylor series for ##\cosh x## is ##1 + x^2/2! + x^4/4! + x^6/6! + \cdots##.

    So the graph of ##y = \cosh x## does look similar to the parabola ##y = 1 + x^2/2## when ##|x|## is small, but not when ##|x|## is big.
     
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