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3rd degree exponential polynomial

  1. Mar 30, 2008 #1
    [SOLVED] 3rd degree exponential polynomial

    1. The problem statement, all variables and given/known data

    Derived from the original question:

    "Reduce to find x"

    [tex]2e^{3x} - e^{2x} - 2e^x = 1[/tex]

    3. The attempt at a solution

    Inspection fails, since the answer is transcendental. I have the answer from my CAS, but I can't figure it out myself.
     
  2. jcsd
  3. Mar 30, 2008 #2

    cristo

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    If you make the substitution u=e^x, then the equation reduces to [itex]2u^3-u^2-2u-1=0[/itex]. Can you solve this?
     
  4. Mar 30, 2008 #3
    I haven't worked the thing out , but you might try letting [tex]u=e^x[/tex] and try that.
    What was the original question?
    I'll play with it more here after we eat.
    CC
     
  5. Mar 30, 2008 #4
    No, that's my problem. :)
     
  6. Mar 30, 2008 #5

    cristo

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    Are you sure that's the right equation? It doesn't have "nice" solutions. What is the original question?
     
  7. Mar 30, 2008 #6
    It really doesn't, lol.
     
  8. Mar 31, 2008 #7
    The original question was "Find x, using the definitions of the hyperbolics in terms of exponentials"

    ...from this expression:

    [tex]2 = {\mathop{\rm Cosech}\nolimits} \left( x \right) - 2{\mathop{\rm Coth}\nolimits} \left( x \right)[/tex]

    Here's my working so far, reducing down to the final polynomial:

    [tex]\displaylines{
    {\mathop{\rm Cosech}\nolimits} \left( x \right) = {2 \over {e^x - e^{ - x} }} \cr
    {\mathop{\rm Coth}\nolimits} \left( x \right) = {{{\mathop{\rm Cosh}\nolimits} \left( x \right)} \over {{\mathop{\rm Sinh}\nolimits} \left( x \right)}} = {{{\textstyle{1 \over 2}}\left( {e^x + e^{ - x} } \right)} \over {{\textstyle{1 \over 2}}\left( {e^x - e^{ - x} } \right)}} = {{e^{2x} + 1} \over {e^{2x} - 1}} \cr}[/tex]

    [tex]\displaylines{
    2 = {\mathop{\rm Cosech}\nolimits} \left( x \right) - 2{\mathop{\rm Coth}\nolimits} \left( x \right) \cr
    = {2 \over {e^x - e^{ - x} }} - 2\left( {{{e^{2x} + 1} \over {e^{2x} - 1}}} \right) \cr
    1 = {1 \over {e^x - e^{ - x} }} - {{e^{2x} + 1} \over {e^{2x} - 1}} \cr
    = {{e^{2x} - 1} \over {\left( {e^x - e^{ - x} } \right)\left( {e^{2x} - 1} \right)}} - \left( {{{e^{2x} + 1} \over {e^{2x} - 1}} \times {{e^x - e^{ - x} } \over {e^x - e^{ - x} }}} \right) \cr
    = {{e^{2x} - 1} \over {e^{3x} - 2e^x + e^{ - x} }} - {{e^{3x} - e^{ - x} } \over {e^{3x} - 2e^x + e^{ - x} }} \cr
    = {{e^{2x} - 1 - e^{3x} + e^{ - x} } \over {e^{3x} - 2e^x + e^{ - x} }} \cr
    e^{3x} - 2e^x + e^{ - x} = e^{2x} - 1 - e^{3x} + e^{ - x} \cr
    e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \cr
    2e^{3x} - e^{2x} - 2e^x = 1 \cr}[/tex]
     
  9. Mar 31, 2008 #8

    cristo

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    Your mistake's in your last line!
    [tex]e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \Rightarrow 2e^{3x}- e^{2x} - 2e^x=-1 [/tex], so then the equation you want to solve, letting u=e^x, is [itex]2u^3-u^2-2u+1=0[/itex], which does have nice solutions!
     
  10. Mar 31, 2008 #9
    Indeed, that solved it. Thank you.

    I have one more question to ask:

    I need to show that:

    [tex]{\mathop{\rm Artanh}\nolimits} \left( {{\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right)} \right) = {\mathop{\rm Ln}\nolimits} \left( {1 + \sqrt 2 } \right)[/tex]

    Here's my working so far, I don't know where to go from the bottom line:

    [tex]$\displaylines{
    {\mathop{\rm Artanh}\nolimits} \left( x \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{x + 1} \over {x - 1}}} \right) \cr
    {\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right) = {\textstyle{1 \over {\sqrt 2 }}} \cr
    {\mathop{\rm Artanh}\nolimits} \left( {{\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right)} \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{1 + {\textstyle{1 \over {\sqrt 2 }}}} \over {1 - {\textstyle{1 \over {\sqrt 2 }}}}}} \right) \cr
    = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{{{\sqrt 2 + 1} \over {\sqrt 2 }}} \over {{{\sqrt 2 - 1} \over {\sqrt 2 }}}}} \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{\left( {\sqrt 2 + 1} \right)\sqrt 2 } \over {\left( {\sqrt 2 - 1} \right)\sqrt 2 }}} \right) \cr
    = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{\sqrt 2 + 1} \over {\sqrt 2 - 1}}} \right) \cr
    {{\sqrt 2 + 1} \over {\sqrt 2 - 1}} = {{\sqrt 2 + 1} \over {\sqrt 2 - 1}} \times {{1 + \sqrt 2 } \over {1 + \sqrt 2 }} \cr
    = {{3 + 2\sqrt 2 } \over 1} = 3 + 2\sqrt 2 \cr
    {\mathop{\rm Artanh}\nolimits} \left( {{\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right)} \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {3 + 2\sqrt 2 } \right) \cr
    = {\mathop{\rm Ln}\nolimits} \left( {\sqrt {3 + 2\sqrt 2 } } \right) \cr
    = \cr} $

    [/tex]
     
    Last edited: Mar 31, 2008
  11. Mar 31, 2008 #10
    WOW, way to leave that out. Next time you ask for help, make sure you post the whole question.
     
  12. Apr 1, 2008 #11
    for your second question, you only need to show that
    [tex]\sqrt {3 + 2\sqrt 2 } =1+\sqrt2
    [/tex]
    and indeed:
    [tex]\sqrt {3 + 2\sqrt 2 }=\sqrt {1 + 2\sqrt 2 + 2 }=\sqrt {1^2 + 2\sqrt2 + (\sqrt2)^2 }=\sqrt {(1+\sqrt2)^2}=1+\sqrt2[/tex]
     
  13. Apr 1, 2008 #12
    Whoa, that's voodoo. How did you figure that out?
     
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