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Critical points

  1. Dec 6, 2009 #1
    i have the following problem:

    find the critical points of:

    P = [tex](x_{1} - 1)^{2} + (x_{n})^{2} + \sum(x_{k+1} - x_{k})[/tex]

    the bounds of the sum are from i = 1 to n-1.

    so i differentiate P with respect to x and i set it equal to zero, and i eventually get the expression:

    [tex]\sum(x_{k+1} - x_{k}) = 1 - x_{1} - x_{n}[/tex]

    what do i do from here / how do i differentiate the summation notation?

    thanks!
     
  2. jcsd
  3. Dec 6, 2009 #2

    Dick

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    How can you differentiate with respect to x when there is no x in the problem? I think you want to differentiate with respect to x1, x2, x3.... xn and set them all equal to zero to get a critical point.
     
  4. Dec 7, 2009 #3
    alright, so how would i do that within the summation notation?
     
  5. Dec 7, 2009 #4

    Dick

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    You can write the summation in a form that's a lot simpler. Take the case n=4. Then the sum is (x4-x3)+(x3-x2)+(x2-x1). Do you see what I'm saying?
     
  6. Dec 7, 2009 #5
    i'll write P as:

    [tex]P = x^{2}_{1} - 2x_{1} + 1 + x^{2}_{n} + (x_{n} - x_{n-1}) + ... + (x_{2} - x_{1})[/tex]

    and now differentiating P wrt x1, ..., xn:

    [tex]P^{'}_{x_{1}} = 2x_{1} -3 = 0, x_{1}= 3/2[/tex]

    [tex]P^{'}_{x_{2}} = -1 + 1= 0, x_{2}= 0[/tex]

    [tex]P^{'}_{x_{n}} = 2x_{n}+ 1 = 0, x_{n}= -1/2[/tex]

    so would the critical points be (3/2, 0, ..., 0, -1/2)? or am i doing something wrong...
     
  7. Dec 7, 2009 #6

    Dick

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    Only a bit wrong. Your P'_x2 doesn't have any x2 in it. It's identically zero. No matter what x2 is. Doesn't that mean x2 can be anything? Same for x3...xn-1?
     
  8. Dec 7, 2009 #7
    yep, that makes sense. why isn't x2, ..., xn-1 = 0 then? are you saying it should just be 1?
     
  9. Dec 7, 2009 #8

    Dick

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    No, I'm saying that if x2...xn-1 don't even occur in your equations, then there is nothing to solve for and they could be ANYTHING.
     
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