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Hard to define function and Conv/Dive of Integral/Series

  1. Jun 7, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \mbox {1. If f(x) is a positive and continuous function in [1,\infty)\ and\ \lim_{x\rightarrow \infty} f(x)=0 } [/tex]

    [tex] \mbox {Prove or contradict:}\\
    \mbox { If } \sum_{n=1}^\infty f(n) \ \mbox {is converget then} \ \int_{1}^\infty f(x)dx \ \mbox {is also convergent.} [/tex]

    [tex] \mbox {2. If f(x) is a positive and continuous function in [1,\infty)}[/tex]

    [tex] \mbox {Prove or contradict:}\\
    \mbox { If } \int_{1}^\infty f(x)dx \ \mbox {is converget then} \ \sum_{n=1}^\infty f(n) \ \mbox {is also convergent.} [/tex]


    3. The attempt at a solution

    [Edit] Counterexample for (1):

    [tex] \sum_{n=1}^\infty \frac{\sin^2{(n\Pi)}}{n} \ \mbox{is convergent but } \int_{1}^\infty \frac{\sin^2{(x\Pi)}}{x}dx\ \mbox{ is divergent} [/tex]


    [Edit] Counterexample for (2): (Still only idea)

    I should somehow formally define function which is 0 most of the time but every time x is near integer n the function "draws" a triangle with area of [tex] (1/2)^n [/tex].
     
    Last edited: Jun 7, 2010
  2. jcsd
  3. Jun 7, 2010 #2

    LCKurtz

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    In the thread where you posted this question previously:

    https://www.physicsforums.com/showthread.php?p=2751033#post2751033

    I gave you this suggestion:

    How about something easier like this -- suppose an= 0 for all n so the series obviously converges. Now can you build an f(x) such that f(n) = 0 for integers n but whose area under the curve goes to infinity? Once you do that then start thinking about the other way.

    Why don't you begin by addressing that.
     
  4. Jun 7, 2010 #3
    This is what I tried to do, but I have trouble making it formal.
     
  5. Jun 7, 2010 #4

    LCKurtz

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    Can't you think of a continuous function that is zero at the integers?
     
  6. Jun 7, 2010 #5
    Contradiction for 2:

    [tex]\int_{1}^\infty \sin^2{(x\Pi)}dx \ \mbox{is divergent but }
    \sum_{1}^\infty \sin^2{(n\Pi)} \ \mbox{is convergent} [/tex]

    Don't know how I missed it.
    Now I'll think about 1
     
  7. Jun 7, 2010 #6

    LCKurtz

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    But that isn't quite a counterexample to your problem yet. [Why?]
     
  8. Jun 7, 2010 #7
    I corrected my mistake, thanks.
     
  9. Jun 7, 2010 #8

    LCKurtz

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    I wasn't referring to the converget vs divergent typo.
     
  10. Jun 7, 2010 #9
    Contradiction for 1:

    [tex]\sum_{n=1}^\infty \frac {\sin^2{(n\Pi)}}{n} \ \mbox{is convergent but }
    \int_{1}^\infty \frac {\sin^2{(x\Pi)}}{x}dx \ \mbox{is divergent} [/tex]

    Seems right this time.
     
  11. Jun 7, 2010 #10
    Now I should look for a positive function without a limit at [tex] \infty [/tex], right?
     
  12. Jun 7, 2010 #11

    LCKurtz

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    Don't make it so difficult. Apparently you noticed that the problem requires that f(x) can't be negative, and squaring it fixes that. But why put the new denominator in there and make a big problem out of showing the integral diverges? You want to be able to calculate it easily.

    Also, you still don't have a counterexample if the problem really requires f(x) to be positive. Will nonnegative do?
     
  13. Jun 7, 2010 #12

    LCKurtz

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    You need to state more precisely what you need to do. You have to find an example of a series that does what and an integral that does what for a counterexample to 1?

    [Edit] I mean a counterexample to the other one, which is 2.
     
    Last edited: Jun 7, 2010
  14. Jun 7, 2010 #13
    I made the denominator because of this requirement in (1): [tex] \lim_{x\rightarrow \infty} f(x)=0 [/tex]
    Nonnegative will be enough, but I'll try to think about positive function.
    But Ill try to give counterexample to (2).
     
  15. Jun 7, 2010 #14

    LCKurtz

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    Woops, I didn't see the [tex] \lim_{x\rightarrow \infty} f(x)=0 [/tex] thing. Has that always been there? Anyway, what you now have seems to work for the nonnegative case, assuming you are actually going to show the integral diverges (not difficult).
     
  16. Jun 7, 2010 #15
    [tex] \mbox {Now I need to prove or contradict: If f(x) is a nonnegative and continuous function in [1,\infty)}[/tex]

    [tex] \mbox {Prove or contradict:}\\
    \mbox { If } \int_{1}^\infty f(x)dx \ \mbox {is converget then} \ \sum_{n=1}^\infty f(n) \ \mbox {is also convergent.} [/tex]

    I think I have one...
    [Edit] No, I don't have one. Now I'm trying to think about something that will fail the n'th term test.
     
    Last edited: Jun 7, 2010
  17. Jun 7, 2010 #16

    LCKurtz

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    Think about an=1. That will give a divergent series by failing the nth term test. Now build an f(x) ... This is the real meat of this problem.
     
  18. Jun 7, 2010 #17
    I'm trying to think about a function which is 0 almost always but every time x is near integer the function builds very very thin triangle (triangle height is 1)...
     
  19. Jun 7, 2010 #18

    LCKurtz

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    Good, you're onto it. Make it so the areas of the triangles converge when you add them all up. I think you can take it from here.
     
  20. Jun 7, 2010 #19
    I actually have bad sense for making function from my mind formal, can you give me a tip?

    trying something like this [tex]\ \sin^2{(e^x)} [/tex]
     
  21. Jun 7, 2010 #20

    LCKurtz

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    Don't make it so complicated. Use the triangles you mentioned above. Fix it so the triangle at integer n has area an where ∑an is a convergent series of your choice.
     
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