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anonymous

  • one year ago

Challenging Hyperbolic Proof:

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  1. anonymous
    • one year ago
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    Prove that \[\frac{ 1+tanhx }{ 1-tanhx }=e ^{2x}\]

  2. anonymous
    • one year ago
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    Medal and fans for first. Already know it.

  3. anonymous
    • one year ago
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    I think it's just tedious algebra. Using the definition of tanh(x) = {e^x - e^(-x)] / (e^x + e^-x), substitute back into the equation and simplify

  4. anonymous
    • one year ago
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    You do not need (e^x-e^-x)/(e^x+e^-x) to solve at all. To prove it, it must be gone.

  5. anonymous
    • one year ago
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    I mean to prove it faster without tedious algebra, it should be gone.

  6. Empty
    • one year ago
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    \[\frac{\cosh x }{\cosh x}\frac{ 1+\tanh x }{ 1-\tanh x }=\frac{\cosh x + \sinh x}{\cosh x - \sinh x} = \frac{e^x}{e^{-x}}\]

  7. anonymous
    • one year ago
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    You missed a between the first and second, but that seems reasonable.

  8. anonymous
    • one year ago
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    a step*

  9. Empty
    • one year ago
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    What step? I skipped several steps but they were simple so no point in wasting time typing :P

  10. Empty
    • one year ago
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    If you want me to elaborate on my reasoning anywhere I can explain myself. :D

  11. anonymous
    • one year ago
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    |dw:1439616015632:dw|

  12. anonymous
    • one year ago
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    Just to clarify. Its all good.

  13. Empty
    • one year ago
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    I think the bigger jump is this one: \[\cosh x - \sinh x = e^{-x}\] since it's kind of subtle to see how this is true. \[\cosh x = \cosh -x\] and \[-\sinh x = \sinh -x\] so we can plug these in to get it.

  14. anonymous
    • one year ago
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    yea and that fact that sinhx+coshx=e^x while coshx-sinhx=e^-x

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