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Noliec Group Title

Does anyone know where you could find the deduction of the inverse of tanh(x) starting from the standard hyperbolic trigonometry definitions? (Or does anyone possibly know how to deduce it from them?)

  • one year ago
  • one year ago

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  1. Noliec Group Title
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    @hartnn , @TuringTest , @AravindG Please help ;_;

    • one year ago
  2. Noliec Group Title
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    I basically get it, down to tanh(x)=(e^(2x)-1)/(e^(2x)+1)

    • one year ago
  3. AravindG Group Title
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    It is right isnt it?

    • one year ago
  4. Noliec Group Title
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    Indeed, but how do you go from here to the inverse of tanh(x)?

    • one year ago
  5. hartnn Group Title
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    do u know componendo - dividendo ?

    • one year ago
  6. AravindG Group Title
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    Take tanh x=y. Then apply log on both sides.

    • one year ago
  7. AravindG Group Title
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    Solve for x.

    • one year ago
  8. hartnn Group Title
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    if a/b = c/d then (a+b) /(a-b) = (c+d)/ (c-d) did u know this ?

    • one year ago
  9. Noliec Group Title
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    That's what I did, I didn't get anywhere on the log expression though. Hmm, not sure hartnn.

    • one year ago
  10. hartnn Group Title
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    that rule is called componendo-dividendo apply that for y/1= (e^(2x)-1)/(e^(2x)+1)

    • one year ago
  11. Noliec Group Title
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    I'd have to give a proof in order to utilize it though. D: It's supposed to be solved without it.

    • one year ago
  12. hartnn Group Title
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    that will isolate e^2x and hence you can apply log

    • one year ago
  13. hartnn Group Title
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    solved without what ?

    • one year ago
  14. Noliec Group Title
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    No shortcuts. :D

    • one year ago
  15. Noliec Group Title
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    Algebraically

    • one year ago
  16. AravindG Group Title
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    Great! Good work!

    • one year ago
  17. hartnn Group Title
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    lol ok, don't apply that rule (bdw, that was algebraic property) y= (e^(2x)-1)/(e^(2x)+1) y (e^2x +1) = e^2x -1 y e^2x + y = e^2x -1 y e^2x -e^2x = -1-y need i go further ?

    • one year ago
  18. hartnn Group Title
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    hint for next step factor out e^2x from on left side

    • one year ago
  19. hartnn Group Title
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    but did u get what i did ?

    • one year ago
  20. hartnn Group Title
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    \(\huge \dfrac{1}{2}\ln (\dfrac{1+x}{1-x})\) you should get this

    • one year ago
  21. hartnn Group Title
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    -1-y / y-1 = (1+y)/ (1-y) the 1/2 will come outside ln right ? because 2x= ln (...)

    • one year ago
  22. Noliec Group Title
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    What happens after you apply ln?

    • one year ago
  23. hartnn Group Title
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    e^2x =[(1+y)/(1-y) ] clear till here ??

    • one year ago
  24. Noliec Group Title
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    Man I feel stupid. D:

    • one year ago
  25. Noliec Group Title
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    Absolutely.

    • one year ago
  26. hartnn Group Title
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    any more doubts?

    • one year ago
  27. Noliec Group Title
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    Nope, thanks a lot! :)

    • one year ago
  28. hartnn Group Title
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    welcome ^_^

    • one year ago
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