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Noliec

  • one year ago

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?)

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  1. Noliec
    • one year ago
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    @hartnn , @TuringTest , @AravindG Please help ;_;

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

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

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

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

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

  7. AravindG
    • one year ago
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    Solve for x.

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

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

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

  11. Noliec
    • one year ago
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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.

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

  13. hartnn
    • one year ago
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    solved without what ?

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

  15. Noliec
    • one year ago
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    Algebraically

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

  17. hartnn
    • one year ago
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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 ?

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

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

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

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

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

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

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

  25. Noliec
    • one year ago
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    Absolutely.

  26. hartnn
    • one year ago
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    any more doubts?

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

  28. hartnn
    • one year ago
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    welcome ^_^

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