1. Limited time only! Sign up for a free 30min personal 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!

Question on limits

  1. Mar 25, 2012 #1
    I know you can determine what number the limit of a series approaches, such as 2. Is there a way to do that in reverse? Is there a method where I can I come up with an infinite series that approaches a limit of, say, 7.5? or 11.75? And is the possible different series that will approach, say 5.25, infinite or finite?

    tex
     
  2. jcsd
  3. Mar 25, 2012 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Have you tried to come up with infinite series with a pre-chosen sum? It's a rather trivial exercise... unless you're have a mental block that's preventing you from doing easy things.
     
  4. Mar 25, 2012 #3

    chiro

    User Avatar
    Science Advisor

    Hey thetexan.

    This is pretty much the same as finding the inverse of a function value. You need to aware of a few things:

    There must be a one-to-one relationship between the value you have and the value of x ot t or whatever in the series expansion. The limit must also exist in some neighbourhood and you want to have continuity in an appropriate neighbourhood for when you have to use a numerical routine which will have errors associated with it when you are dealing with a general expression (limited numbers of expressions can be solved analytically).

    If you have these (and maybe other conditions), then if you can assume continuity and convergence in the right neighbourhood then you can use an implicit function algorithm that finds the roots to the equation f(x) = 0 (I'm assuming one dimension is your series, if it's two the same ideas apply but it's more complex and you need more conditions met).

    Then when you find the solutions, you need to take into account the nature of the series itself. If the series is a standard analytic series, then the continuity and convergence should be known before hand and if you get these two properties across the whole domain then you don't have to worry.

    The point where you have to worry is where you don't get this continuity or convergence set of conditions because you can get limits for things that are not analytic like say the function |x| = absolute value of x.

    If you can write your series as some kind of power series and you can gaurantee convergence in the appropriate region, then just use the implicit function solver: it's also known as a root finding algorithm and many popular math packages have this as a library or as a simple function call.
     
  5. Mar 25, 2012 #4
    Well, no. Thus the question. If it is a trivial exercise could you please show me how?
     
  6. Mar 25, 2012 #5

    chiro

    User Avatar
    Science Advisor

    I don't think this is true. Consider the relationship sin(x) = a for a general a. It's not going to be easy to find x for all possible a in the principal branch. This is just one candidate of many in the context of the OP's question.
     
  7. Mar 25, 2012 #6

    chiro

    User Avatar
    Science Advisor

    Are you talking about a general power series? So basically an 'infinite-polynomial' type power series that is gauranteed to converge and has some known inverse region?
     
  8. Mar 25, 2012 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What does anything you've said have to do with the question? I think you're having a mental block too.


    Here are three easy examples:

    [tex]7.5 + 0 + 0 + 0 + 0 + \cdots = 7.5[/tex]
    [tex]4 + 3.5 + 0 + 0 + \cdots = 7.5[/tex]
    [tex]7 + 1/4 + 1/8 + 1/16 + 1/32 + \cdots = 7.5[/tex]​
     
  9. Mar 25, 2012 #8

    chiro

    User Avatar
    Science Advisor

    Yes but that's not a general solution. I think the author wants to analyze a general form of an expression. With the exception of the last one, most of those are more representative of a sequence rather than a series don't you think?

    Also I am referring to an arbitrary sequence that is not based on a formula as opposed to a sequence that is if you need any further clarification.
     
  10. Mar 25, 2012 #9
    It's fairly easy to construct a geometric series that converges to any desired value (actually, you can construct an infinite number of them).
     
  11. Mar 25, 2012 #10

    chiro

    User Avatar
    Science Advisor

    Before I comment further I would like to know from the OP exactly what kinds of series he is referring to, because if he only wants to consider a limit class of geometric series then you and Hurkyl have given a sufficient answer, but if it is more general then the answer will of course change.
     
  12. Mar 25, 2012 #11
    First, I am only a neophyte math hobbiest with not a very good understanding of even this subject.

    When I say series I think about things like the harmonic series, and other series that contain algebraic parts (meaning, to me, terms such as ((x^s+(1-x^3)) ( I just made that up)). Sure 7.5 + 0 + 0 + 0 + 0 is a series but that's not what I mean. I may not even know enough to ask an intelligent question. But when I read about primes and other math work I see these complex series where they describe that they limit to a number such as 2 or 1.5 or such. It occured to me to ask, 'I wonder if you can do that in reverse?'.

    My question is can some math method be used to discover a series that when added together to a limit produces a predetermined sum...let's say 13.564...? or 4.1? or .66? or 8?

    tex
     
  13. Mar 25, 2012 #12
    OK, let's say you want to make a geometric series converge to [itex]x[/itex] where we require [itex]0 < x < [/itex], and where we want a series in the form [itex]\sum a^n [/itex] where [itex]|a| < 1[/itex]. Now, we know (or at least some of us know, I don't know if you have covered this) that
    [tex]
    \sum a^n = \frac{1}{1-a}
    [/tex]

    OK, so let's try this:
    [tex]
    \frac{1}{1-a} = \frac{1}{x}
    [/tex]

    And let's solve for a:
    [tex]
    a = 1 - x
    [/tex]

    Since [itex]|x| < 1[/itex] we have that [itex]|1-x| < 1[/itex], so this is a valid [itex]a[/itex]. Now we have this series:
    [tex]
    \sum a^n = \sum (1-x)^n = \frac{1}{1-(1-x)} = \frac{1}{x}
    [/tex]

    Now, this isn't [itex]x[/itex]. So, let's take [itex]r = x^2[/itex]. Now, we know that
    [tex]
    \sum ra^n = \frac{r}{1-a}
    [/tex]

    So, let's take [itex]r = x^2[/itex] so that we have:

    [tex]
    \sum ra^n = \sum x^2(1-x)^n = \frac{x^2}{1-(1-x)} = x
    [/tex]

    So, unless I screwed up somewhere, this is a series that converges to and [itex]x[/itex] such that [itex]0 < x < 1[/itex]. Now, do you see how you can take this method and adjust it by adding a term and/or multiplying the series by a constant that will cause it to converge to any given value?


    NOTE: I think most standard texts reverse the roles [itex]r[/itex] and [itex]a[/itex] from the way that I have done it. I always forget the formula, though, and end up having to re-derive it, so my apologies if this causes any confusion, but you should still be able to follow what I have written (at least, that transposition of [itex]r[/itex] and [itex]a[/itex] shouldn't keep you from understanding).
     
    Last edited: Mar 25, 2012
  14. Mar 25, 2012 #13

    chiro

    User Avatar
    Science Advisor

    In this case if you are looking at a general power series, then what you can do is pick any series that converges and is continuous around a neighbourhood of the point you have, and then after that it can get quite messy.

    You could use the properties of convergence for the series and use a root finding algorithm (provided the conditions are right) to get an approximation for your x, or t or whatever: it will not be exact of course but if you provide a decimal limit with some resolution and your answer for x or t gives an answer that has an error less than the resolution then you are done (you say that there is an error term though).

    Other ways that I can think of is to use relationships between L^2 and l^2 using any kind of integral transform. The Basel problem can be solved this way by considering the function y = x in the region of [-pi,pi] and then using norm identities to show equivalence. Here is the wiki page:

    http://en.wikipedia.org/wiki/Basel_problem

    The tricky part for the above though is that you need to figure out what the function should be for integrating and this is far from easy.
     
  15. Mar 25, 2012 #14
    I realize I was a bit unclear when I mentioned using this method to create a more general sequence. If [itex]|x| > 1[/itex] just note that [itex]|1/x| < 1[/itex] to make a power series in the more general sense.
     
  16. Mar 25, 2012 #15
    Unless I'm just way out of the neighbourhood (pun intended), I think the way I described will do what he wants, won't it?
     
  17. Mar 25, 2012 #16

    chiro

    User Avatar
    Science Advisor

    It depends on how general he wants to go with the series definition.

    Your series expansion has the same coeffecient for each term of a given power in the series. Since the OP mentioned things like harmonic series, it made sense for me to interpret more general forms of series. In that vein, I mentioned what I mentioned above.

    Your example is basically for constructing a very limited class of series and if the OP wants to know what to do for more complex series like a harmonic series (which he mentioned above) then you can't just resort only to examples like that.

    I don't think the OP realizes how difficult it is to expand not only power series with arbitrary coeffecients, but even the general series like say the Riemann Zeta function (I brought this up because he did mention an interest in number theory).
     
  18. Mar 25, 2012 #17
    Ahhh, yes, very true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question on limits
  1. Limits Question (Replies: 14)

  2. Limit question (Replies: 5)

  3. Limit question (Replies: 7)

Loading...