1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Tricky Convergence question
  1. Tricky question (Replies: 4)

  2. Tricky Question (Replies: 4)