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Find The Range of Function.

  1. Jun 6, 2013 #1
    A function f :[0,1]→ℝ is twice differentiable with f(0)=f(1)=0 such that

    f''(x) - 2f'(x) + f(x) ≥ e^x, x ε [0,1].

    Then for 0<x<1 which of the following is true
    0<f(x)<∞, -1/2<f(x)<1/2, -1/4<f(x)<1, -∞<f(x)<0.

    I tried to solve it in the following way:
    I tried to take the e^x to the left side and found that the equation becomes,
    [(e^-x)f'(x) - (e^-x)f(x)]' ≥ 1, thus on integration,

    (e^-x)f'(x) - (e^-x)f(x) -f'(0) ≥ x which is same as

    [(e^-x)f(x)]' ≥ x on integration

    (e^-x)f(x) ≥ (x^2)/2 + f'(0)x.
    I don't know how to proceed from here.
     
  2. jcsd
  3. Jun 6, 2013 #2

    tiny-tim

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    hi klen! :smile:

    (try using the X2 button just above the Reply box :wink:)
    or you could say let g(x) = e-xf(x) …

    then g''(x) ≥ 1 :wink:

    (btw, how did you get on with your series question?)
     
  4. Jun 7, 2013 #3
    Hi Tiny Tim,

    Thanks for the help.

    in the series question, we can break the sum as groups of four, like this:
    [tex] S_n = \sum_{k=1}^{4} {(-1)^{k(k+1)/2}}{k^2} + \sum_{k=5}^{8} {(-1)^{k(k+1)/2}}{k^2} + ...[/tex]
    and so on,
    in each of these sums the first two are negative terms and last two are positive,
    so the sum evaluates to,
    [tex] S_n = \sum_{k=0}^{n} {(32k-12)}[/tex]
    this gives the sum to be
    [tex] S_n= {16n^2+4n}[/tex]
    which is possible only for first and fourth values.
     
  5. Jun 7, 2013 #4

    tiny-tim

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    excellent! :smile:

    (good LaTeX too :wink:)
     
  6. Jun 7, 2013 #5
    How do you get to this step? :confused:
     
  7. Jun 7, 2013 #6
    In these sums for the first terms we have k=1,5,9...
    so k is of form k=4i-3, i=1,2,3....
    similarly for the next successive terms k is of form 4i-2, 4i-1, 4i.
    Since the first two terms are positive and last two are negative we have,
    [tex]S_n= \sum_{i=1}^n {-(4i-3)^2} + \sum_{i=1}^n {-(4i-2)^2} + \sum_{i=1}^n {(4i-1)^2} + \sum_{i=1}^n {(4i)^2}[/tex]
    which after solving comes,
    [tex]S_n= \sum_{i=1}^n {(32i-12)}[/tex]
     
  8. Jun 7, 2013 #7
    That was cool! :cool:

    Thanks! :smile:
     
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