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What should i do after using Taylor series?

  1. Jul 20, 2013 #1
    1. The problem statement, all variables and given/known data
    The first equation on the uploaded paper converts to the last equation.


    2. Relevant equations
    when i substitute ln (1-u)=-u-(1/2)(u^2) into the first equation, i can get the first term in (3rd equation).
    but the second term of the 3rd equation ?


    3. The attempt at a solution
    I tried ln(1-u)= -u-(1/2)(u^2)-(u^3)/3, then by completing the square. yet cant get. i wonder is that any way that i not yet learn.
    please guide what to do. Thank you.
     

    Attached Files:

  2. jcsd
  3. Jul 21, 2013 #2
    I don't understand your question. Could you rephrase what you are asking?

    Though, I'll point out (as a fun fact) that, analytically, ##x_{\text{max}}=\dot{x}_0\frac{g \operatorname{W}\left(\frac{-e^{\frac{-\gamma \dot{z}_0}{g}-1} (g+\gamma \dot{z}_0)}{g}\right)+g+\gamma \dot{z}_0}{\gamma (g+\gamma \dot{z}_0)}##, where ##\text{W}## is the Lambert W function. It's pretty. :tongue:
     
  4. Jul 21, 2013 #3
    He's asking why he didn't get the second term in ##x_{max}## equation after he had substituted the natural logarithm with its Taylor expansion. He managed only to recreate the first term. Is there anything that he missed?
     
  5. Jul 22, 2013 #4
    Let's see what I can do:

    $$\left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma^2}\right)\frac{\gamma x_{\text{max}}}{\dot{x}_0}+\frac{g}{\gamma^2}\ln\left(1-\frac{\gamma x_{\text{max}}}{\dot{x}_0}\right)=0 \\ \left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma^2}\right)\frac{\gamma x_{\text{max}}}{\dot{x}_0}=\frac{g}{\gamma^2}\left(\frac{\gamma x_{\text{max}}}{\dot{x}_0}+\frac{1}{2}\left(\frac{\gamma x_{\text{max}}}{\dot{x}_0}\right)^2+\cdots\right)\\ \text{For clarity and ease of notation, I now substitute the variable } u \text{ for}\frac{\gamma x_{\text{max}}}{\dot{x}_0}. \\ \left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma ^2}\right)u=\frac{g}{\gamma^2}\left(u+\frac{u^2}{2}+\frac{u^3}{3}\cdots\right)\\ 0=\frac{\dot{z}_0 u}{\gamma}+\frac{g}{\gamma^2}\left(\frac{u^2}{2}+\frac{u^3}{3}+\cdots \right)$$

    Can you take it from there, Outrageous?
     
  6. Jul 23, 2013 #5
    Thanks for replies.
    My question is exactly like what Seydlitz said.
    Then after that, what should I do next?
     
  7. Jul 23, 2013 #6
    Isolate ##u##, substitute back ##u=\frac{\gamma x_{\text{max}}}{\dot{x}_0}##, isolate ##x_{\text{max}}##.
     
  8. Jul 23, 2013 #7
    Sorry. Can I ask how to isolate u? By using?
     
  9. Jul 23, 2013 #8
    There are three solutions (if you negect terms higher than O(u^4) ). One of them is u=0.
     
  10. Jul 23, 2013 #9
    One will be u= 0 , then what should I going to do next in order to get ......last equation?
     

    Attached Files:

  11. Jul 23, 2013 #10

    CompuChip

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    Science Advisor
    Homework Helper

    If u is not 0 you can divide it out and you are left with a quadratic equation.
    Do you know how to solve something of the form ax² + bx + c = 0?
     
  12. Jul 23, 2013 #11
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