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Convergence questions

  1. Apr 10, 2005 #1
    [tex] \forall c \ne 0 [/tex], does the volume generated by function [tex] x\left| y \right| = c [/tex] from x=1 to x=∞ converge to a constant? Or from x=0 to x=1? Or from x=0 to x=∞ ?

    But in general (aside from that), [tex] \forall c \ne 0 [/tex], Do [tex] \exists f\left( x \right){\text{,}}\,g\left( y \right) [/tex] such that [tex] \mathop {\lim }\limits_{y \to \pm \infty } g\left( y \right) = \infty [/tex] and the volume generated by [tex] f\left( x \right)g\left( y \right) = c [/tex] from x=a to x=∞ converges to a constant? (where 'a' and 'c' are constants)

    If [tex] \exists f\left( x \right){\text{,}}\,g\left( y \right) [/tex] (and i'm sure they do exist), what is (or must be) the relationship between [tex] f\left( x \right) [/tex] and [tex] g\left( y \right) [/tex] ?
     
    Last edited: Apr 10, 2005
  2. jcsd
  3. Apr 10, 2005 #2
    What?

    No volume is generated by [tex]x|y|=c[/tex]... It looks very 2-D to me... do you mean area? If so, you should be able to calculate that easily.

    [tex]x|y|=c[/tex]
    [tex]|y|=\frac{c}{x}[/tex]
    [tex]y=\pm\frac{c}{x}[/tex]

    The area of that is equal to twice the area of y=c/x for any interval.

    Am I missing something here?
     
  4. Apr 10, 2005 #3
    If z is constant, why do you need x and y? Wouldn't you just write z=c?
     
  5. Apr 10, 2005 #4
    I'm referring to a 3D cartesian coordinate system (x,y,z);
    I mean, volume from x=a to x=∞ and the "z" from b to c ([tex] b \ne c [/tex]) (sorry! :redface: for the typing (not my best at this hour))

    Disregard my first post; I meant the volume of [tex] f\left( x \right)g\left( y \right) = z [/tex] where [tex] \mathop {\lim }\limits_{y \to \pm \infty } g\left( y \right) = \infty [/tex] , where the volume from x=a to x=∞ and from z=b to z=c would converge to a constant.

    What would be the relationship between f(x) and g(y) ?
     
    Last edited: Apr 10, 2005
  6. Apr 10, 2005 #5
    I still have no idea what this function is supposed to look like...

    Is it z=x|y|?

    I'm very confused.
     
  7. Apr 10, 2005 #6
    For z=x|y|, given that volume increases as we traverse the x-axis, but decreases as the |y| becomes ever thinner (when we multiply it by increasing positive values of (x)), would it converge to a constant if integrated from the x-limits in my first post and from z=0 to z=c ?
    (Why the trouble? I forgot to mention the z-limits of integration of course! :frown: )

    But in general, we have [tex] z = f\left( x \right)g\left( x \right) [/tex] where [tex] \mathop {\lim }\limits_{y \to \pm \infty } g\left( y \right) = \infty [/tex] (as in g(y)=|y|). Would such a f(x) and g(x) exist given that the volume generated from x=a to x=∞ and from z=b to z=c converges to a constant?
    (not considering the trivial case of f(x)=0)
     
    Last edited: Apr 10, 2005
  8. Apr 10, 2005 #7
    Are you sure you don't mean the y limits of integration? I'm still not quite sure what you're getting at.

    The function z=x|y| varies in height: at (0,0), z=0; at (1,1), z=1; at (-2,5), z=-10; at (2,-5), z=10; etc.

    To find the volume under z, one would need x and y limits.
     
  9. Apr 10, 2005 #8
    If the question is about the volume of the solid of revolution, the answer is negative except in case of x=1 to x=inf.
    The second question is answerable in terms of order relations only in the regions where the functions don't have zeroes.( It's easy to construct an example where the volume diverges jut because of a zero of f or g).
    If the lower limit 'a' in your notation is in a region void of zeroes,one can compare f & g with powers of x & y (big or little ohs), then use the condition fg=c to compare y^2 with a power of x,say p<0. If p<-1, the integral will converge.
    The general problem of deciding the convergence the volume generated by the revolution of F(x,y)=c would be very difficult.
    Regards,
    Einstone.
     
  10. Apr 10, 2005 #9
    Yes, but you already have your y-limits; for z=x|y| we know the x-limits (from a to ∞) and the z-limits are z=1 to z=c. The y-limits are already defined from -1/a to 1/a, because ONLY those y-values satisfy initial limits condition 1=a|y| (from initial x-limit x=a and z-limit z=1), and the applicable y-values approach zero as (x) approaches infinity. We are given two limits--the y-limits are just defined by the other two. As you see, we can't really start from z=0 (unless we take a limit as the lower z-limit approaches zero, which I intend to do later).

    In that case, what would be the volume from x=a to x=∞ and from z=1 to z=c ?
     
    Last edited: Apr 10, 2005
  11. Apr 10, 2005 #10
    Hmmm... now I think you mean the graph of [tex]y=\pm\frac{z}{x}[/tex]... Bah! defining it in terms of x and y, but giving the limits in terms of x and z was confusing!

    That volume most certainly won't converge from x=a to ∞ and z=b to c.

    Proof:

    [tex]A=\int_b^c{\int_a^{\infty}{\frac{z}{x}dx}dz}[/tex]
    [tex]=\int_b^c{(zln{\infty}-z\ln{a})dz}[/tex]
    [tex]=\int_b^c{\infty zdz}[/tex]

    Which is obviously ∞. I mean it diverges. (Yeah, I was to lazy to write out the limits. So sue me.)
     
  12. Apr 10, 2005 #11
    But then (since [tex]y=\pm\frac{z}{x}[/tex] didn't work), would it be possible to find two such functions y=g(z)/f(x), such that [tex] \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{z \to \infty } g\left( z \right) = \infty [/tex], but the volume from x=a to x=∞ and from z=b to z=c would converge to a constant?
     
  13. Apr 10, 2005 #12
    Sure. Let [tex]f(x)=x^2[/tex] and [tex]g(z)=1[/tex].

    [tex]\int_b^c{\int_a^{\infty}{\frac{1}{x^2}dx}dz}[/tex]

    Converges.
     
  14. Apr 10, 2005 #13
    BUt for that [tex] \mathop {\lim }\limits_{z \to \infty } g\left( z \right) \ne \infty [/tex]
     
  15. Apr 10, 2005 #14
    But in general though, what must be the relationship (conditions that must be satisfied) between f(x) and g(z) that would show convergence as y=f(x)*g(z) or y=f(x)/g(z) from x=a to x=∞ and z=b to z=c ?
    (I suppose/expect these conditions be similar to those for series convergence:blushing:)
     
  16. Apr 10, 2005 #15
    Whoops, I forgot about that bit.

    Then let [tex]f(x)=x^2[/tex] and [tex]g(z)=z[/tex].

    :)
     
  17. Apr 11, 2005 #16
    [tex] y = x^2 z [/tex]
    ~(works for me)!
     
  18. Apr 11, 2005 #17
    No, I meant [tex]y=\frac{z}{x^2}[/tex].
     
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