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Show me the trick

  1. May 1, 2008 #1
    Hi,I have been trying to solve this equation for days.

    x=y+bsinh(cy) I wanted to find the variation of y as a function of x.i wanted to use the Lambert method which is great and inline with my problem but i can't get it right folks.

    i can solve for any type of x=y+bexp(cy) but i am not successful for that one.

    can you help me.
     
  2. jcsd
  3. May 1, 2008 #2

    rock.freak667

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    I am not sure you can make the subject of that formula.
     
  4. May 3, 2008 #3
    hallo

    i am sorry i don't get what you mean.
     
  5. May 3, 2008 #4
    It means rearrange it to become y=something
     
  6. May 3, 2008 #5

    HallsofIvy

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    I suspect he understood that! He was referring to rock.freak667's reply that he didn't think it could be done! rock.freak667, did you notice the reference to "the Lambert method", by which I think he meant Lambert's W function.

    Yoseph, write sinh(cy) as (ecy- e-cy)/2 and you should be able to do that in the same way as y+ becy
     
  7. May 3, 2008 #6

    rock.freak667

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    ah...I didn't even see the words Lambert method...
     
  8. May 6, 2008 #7
    Ok, I did try it using that way.but the result i got and the one i see at the journal have a difference.i have tried it this way:
    x=y+bsinh(cy)=>x/2+x/2=y/2+c/2*exp(cy) + y/2+c/2*exp(-cy)

    then i have equated
    1) x/2=y/2+c/2*exp(cy)
    2) x/2=y/2+c/2*exp(-cy)

    and then i have solved both for y and might i say i linearly combined the two answers and came up with the one below,

    y=1/c*W((bc)*exp(-bx))-1/c*W((bc)exp(bx))+2*x

    in the paper i referred the answer is different it is:

    y=1/c*W((bc/2)*exp(-bx))-1/c*W((bc/2)exp(bx))+x

    i have done it again and again and i happen to tumble over to the same answer.

    what do you say.
     
  9. May 13, 2008 #8
    Has anyone not been able to solve this problem?
     
  10. May 13, 2008 #9

    uart

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    I can't see how that last step is justified. You decomposed the LHS into two (equal) components and the RHS in two (unequal) components then you equated those components. How do you justify doing that?
     
  11. May 13, 2008 #10

    uart

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    BTW. Neither of the solutions you posted (yours or the one quoted from the "paper") worked numerically for me when I took b=c=1 and a random value for x.
     
  12. May 13, 2008 #11
    This is the title of the paper" Exact analytical solution of channel surface potential as an explicit function of gate voltage in undoped-body MOSFETs using the Lambert W function and a threshold voltage definition therefrom"

    Source: Solid-State Electronics, Volume 47, Number 11, November 2003 , pp. 2067-2074(8)

    I have seen the result it does give a solution,the curve looks like an extended "s" and "b &c"
    are not one in the real problem,if that might make a difference.
     
  13. May 13, 2008 #12
    Yes you are correct.there is no justification.

    any idea how?
     
  14. May 16, 2008 #13

    uart

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    Without seeing the derivation my best guess is that despite the title of "Exact analytical solution" that their solution makes some approximation (like some term being small compared with the others etc) to come up with an approximate solution that is fairly accurate for certain ranges of parameters (b,c) and/or varialbe x . Like I said before, you can easily test their quoted "solution" numerically (pick any numbers you like for b,c,x) and it simply doesn't work.
     
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