wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Average value in an infinite series
        
(Message started by: mattian on Oct 22^nd, 2004, 9:11am)

Title: Average value in an infinite series
Post by mattian on Oct 22^nd, 2004, 9:11am

In an infinite series of bounded numbers, the average value is finite X. If a single unbounded value of infinite magnitude is added to the series, what is the new average?

Title: Re: Average value in an infinite series
Post by towr on Oct 22^nd, 2004, 11:43am

::[hide]X + inf/inf, which is undefined..[/hide]::

Title: Re: Average value in an infinite series
Post by Obob on Oct 22^nd, 2004, 1:51pm

If we are taking the average of a series {a_n}, the only natural way to define the average is as the limit of A_n as n goes to infinity, where A_n is the average of the first n terms. In particular, we note that X has to be defined in this way; otherwise we would have to consider dividing an infinite series by infinity, or some other nonsense.

When we add the term of "infinite magnitude," notice that it makes no sense to make it the "final" term in the series, because there is no such thing. Hence there must be some finite n for which a_n is infinity. But then if k>=n we have A_k=infinity (as we are only ever dividing by a finite number), and therefore that the limit of A_k as k goes to infinity is infinity. Hence the average can only be interpreted as being infinite.

Title: Re: Average value in an infinite series
Post by mattian on Oct 22^nd, 2004, 2:01pm

Obob,

The converse of your argument implies that we take the limit of the value of a term as it tends to infinity and work with it as a very large but finite number which, when distributed over an infinite domain, yields an average value of X - a finite result.

Or take the limit of the series domain as it tends to infinity while taking the limit of the very large but finite term as it tends to infinity and if the two limits converge at the same rate, it could be argued that the resultant average is something like X + 1?

Correct my thinking here:

Let N be the number of terms in the series, and let V = N be the value of the "infinite" term.

then A = ((N * X) + V) / (N) - where A is the average.

If V = N then A = (NX + N)/N

.: A = X + 1

Yes/no?

Title: Re: Average value in an infinite series
Post by mattian on Oct 22^nd, 2004, 2:19pm

I suppose - as towr put it - inf/inf is undefined and I have incorrectly arrived at a value of 1 for it.

However, Fourrier Analysis frequently makes use of the concept that
inf * 0 = 1

In that an infinite impulse of energy exerted in an infinitely small amount of time, amounts to a single impulse of unit magnitude. Granted, this is a conventional construct more than a physical property of the world, because in the context of infinite impulses, 1 and 2 (and 42 for that matter) are essentially equivalent except with respect to each other.

The dirac delta is derived by starting out with an integral amount of energy with a magnitude of 1 before the limit is taken as time tends to 0 - so the solution of 1 is a predetermined and contrived result.

But could we not contrive the infinite series problem in the same way by regulating the rate of convergence on the limits that we take?

By the way: Feel free to rate my comments from 1 to 10 on the nonsence scale.

Title: Re: Average value in an infinite series
Post by Obob on Oct 22^nd, 2004, 3:09pm

If you take a sequence and add the "value" infinity to it, you will get infinity for the resulting average, as I showed earlier. However, I suppose if you decide to interpret infinity as a limit, then you could arrive at other answers. But the only way to arrive at other answers in this fashion is to define something like adding N+1 to the Nth average A_N and then taking a limit. There is certainly a large degree of arbitrariness in this choice. For example, let a be a real number. If we decide to add (a-X)(N+1) to the Nth average A_N, then the resulting limit will be equal to a. Moreover, if we attach some other rate of growth, we could just as easily end up with an infinite value.

This illustrates the difference between realizing infinity as a formal mathematical symbol and as the end result of a limiting process. Usually when a question involves both a limit and infinity, a great deal of confusion can arrise when the infinity is treated as a limit instead of as a symbol; this is especially the case when infinity is treated as a limit which is dependent on the variables of the other limit.

On another note, the convention infinity*0=0 is probably more prevalent than the convention infinity*0=1. This is becase the latter convention is really just a statement about the delta function, and everything there can be made much more rigorous and can be turned into a more sensible equation. You can't just treat the equation infinity*0=1 without a great deal of care; otherwise we have the following nonsense: 2=2*infinity*0=infinity*(2*0)=1. The reason for the convention infinity*0=0 is nested in the theory of the Lebesgue integral, and has to do with the fact that the integral of any function over any set of measure zero should be zero, regardless of the finiteness of the function.

Title: Re: Average value in an infinite series
Post by Icarus on Oct 22^nd, 2004, 7:43pm

Neither [infty][cdot]0 = 1 or [infty][cdot]0 = 0 is in any way a "convention". In both situations mentioned, delta functions and integrals of extended functions over sets of measure zero, the appropriate equation is merely one way of describing what that particular application amounts to. But in each case, the result is far from being a convention, because it is true only of that particular, very proscribed, situation.

In particular, you should not think of the idea of delta functions as meaning [infty][cdot]0 = 1, because, for example, the integral of 2[delta] = 2, which amounts to saying [infty]*0 = 2.

Indeterminate expressions: 0/0, 0⁰, [infty]*0, [infty]/[infty], 1^[subinfty], etc, are called "indeterminant" exactly because they can be given different values in different situations. The only convention we have for any of them is that 0⁰ = 1, and this convention only exists because it was easier to specifically exclude 0⁰ from the rule that lim (x^y) = (lim x)^{lim y}, than it was to state an exception for 0 every time a power series is written down.

Title: Re: Average value in an infinite series
Post by william wu on Oct 23^rd, 2004, 10:08am

Often in courses on signal processing, the delta is presented as a strange function which is [infty] at 0, and 0 everywhere else, represented graphically by a spike. However, I think this is more of an instructional tool, and without further clarification it doesn't make much sense. For instance you could try defining the delta function as the limit of a box with width [epsilon] and height 1/[epsilon]. But this limit fails as a pointwise statement.

Consequently it can be helpful to think of delta functions as not functions, but rather tempered distributions -- which is what they really are. You can think of them solely on how they act when paired with other functions in integrals; namely, the sifting property: [int] [delta](x) g(x) dx = g(0). This approach reduces the [delta] function to merely the functional that evaluates a function at zero.

Title: Re: Average value in an infinite series
Post by Icarus on Oct 23^rd, 2004, 7:05pm

That is the most common mathematical way of interpreting the delta function. An alternate approach, more natural in some ways, is to view the delta function (and all related functions) as taking on values in an extension of the real numbers (with far more actual infinities than just [pm][infty]). Then the definition of integration is extended to these functions in such a way that the desired behavior occurs. (I have heard of this approach, but have never seen it developed myself.)