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Homework Help: Tricky Convergence question

  1. Apr 13, 2004 #1
    This was a question on one of our past exams. I don't even know where to start and I'm getting very discouraged.

    Given [tex]\int [\frac{1}{\sqrt{1+2x^2}} - \frac{c}{x+1}] = \ln(\frac{\sqrt{1+2x^2} + \sqrt{2}x)}{(x+1)^c}) + k.[/tex]

    Find a value c such that [tex]\int_{0}^{\infty} [\frac{1}{\sqrt{1+2x^2}} - \frac{c}{x+1}][/tex] converges.

    I tried taking the limit as x-> infinity of the large ln function, but I'm getting nowhere.

    Can someone please give me a hint that'll get me started on this question at least?
  2. jcsd
  3. Apr 13, 2004 #2
    as x tends to infinitum, the logarithm is superiorly bounded by

    [tex]\frac{1}{(\sqrt{1+2x^2} - \sqrt{2}x)(x+1)^c}[/tex]

    and being [tex]x \geq 0[/tex] implies that

    [tex]\frac{1}{(\sqrt{1+2x^2} - \sqrt{2}x)(x+1)^c} \leq \frac{1}{(x+1)^c}[/tex]

    and this converges if [tex]c \geq 0[/tex]
    Last edited: Apr 13, 2004
  4. Apr 14, 2004 #3


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    The term (sqrt(1+2x^(2))-sqrt(2)x)->0 as x->inf; hence you must be more careful in the choice of c.
    We should have, in general for e->0, that sqrt(1+e)-1>=1+1/2e-1/8e^(2) (alternating series with terms decreasing in magnitude)
    Setting e=1/(2x^(2)), you should find that the expr. converges for c>=1
  5. Apr 14, 2004 #4


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    Sorry, estimate all wrong.
    You should begin with the original expression.
    sqrt(1+2x^(2))+sqrt(2)x=sqrt(2)x(2+1/(4x^(2))-1/(16x^(4))+-+-)->2sqrt(2)x when x->inf.
    Hence, convergence requires abs(ln(x^(1-c)))<inf as x->inf.
    Only c=1 achieves this result.
  6. Apr 14, 2004 #5
    true... i dont know what i was thinking... that was my orignial estimate but i messed up resolving the first term of the integral... the shame...
  7. Apr 15, 2004 #6
    Wow. Thanks a lot! Just one stupid question - how did you get the alternating series expansion for [tex]\sqrt{1+2x^2} + \sqrt{2}x[/tex]? Did you just take successive derivatives and use Taylor polynomials?
  8. Apr 17, 2004 #7
    What do you mean?

    using the taylor expansion around zero of

    [tex](1+x)^n \sim 1+nx+n(n-1)\frac{x^2}{2!}+...[/tex]

    its easyly shown that

    [tex]\sqrt{1+2x^2} + \sqrt{2}x=\sqrt{2}x(1+\sqrt{1+\frac{1}{2x^2}})=\sqrt{2}x(2+\frac{1}{4x^2}-\frac{1}{16x^4}+...)[/tex]

    as x tends to infinitum this function behave like x... the function [tex](1+x)^c[/tex] behaves like [tex]x^c[/tex] (same argument) thats why you need [tex]c \geq 1[/tex]
    Last edited: Apr 17, 2004
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