Can I Construct an Infinite Series that Approaches a Specific Limit?

[thetexan]

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

 

[Hurkyl]

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.

 

[chiro]

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

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.

 

[thetexan]

Well, no. Thus the question. If it is a trivial exercise could you please show me how?

 

[chiro]

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.

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.

 

[chiro]

Well, no. Thus the question. If it is a trivial exercise could you please show me how?

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?

 

[Hurkyl]

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.

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]​

 

[chiro]

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]​

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.

 

[Number Nine]

It's fairly easy to construct a geometric series that converges to any desired value (actually, you can construct an infinite number of them).

 

[chiro]

It's fairly easy to construct a geometric series that converges to any desired value (actually, you can construct an infinite number of them).

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.

 

[thetexan]

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 occurred 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

 

[Robert1986]

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).

 

[chiro]

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 occurred 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

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.

 

[Robert1986]

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).

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.

 

[Robert1986]

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.

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?

 

[chiro]

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?

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).

 

[Robert1986]

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).

Ahhh, yes, very true.