Author |
Topic: Average value in an infinite series (Read 7522 times) |
|
mattian
Senior Riddler
 
Gender: 
Posts: 404
|
 |
Average value in an infinite series
« on: Oct 22nd, 2004, 9:11am » |
 Quote  Modify
|
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?
|
|
 IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Average value in an infinite series
« Reply #1 on: Oct 22nd, 2004, 11:43am » |
Quote Modify
|
::X + inf/inf, which is undefined..::
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Obob
Guest
|
|
Re: Average value in an infinite series
« Reply #2 on: Oct 22nd, 2004, 1:51pm » |
Quote Modify
 Remove
|
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.
|
|
IP Logged |
|
|
|
mattian
Senior Riddler
Gender:
Posts: 404
|
|
Re: Average value in an infinite series
« Reply #3 on: Oct 22nd, 2004, 2:01pm » |
Quote Modify
|
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?
|
| « Last Edit: Oct 22nd, 2004, 2:08pm by mattian » |
IP Logged |
|
|
|
mattian
Senior Riddler
Gender:
Posts: 404
|
|
Re: Average value in an infinite series
« Reply #4 on: Oct 22nd, 2004, 2:19pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
Obob
Guest
|
|
Re: Average value in an infinite series
« Reply #5 on: Oct 22nd, 2004, 3:09pm » |
Quote Modify
Remove
|
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.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Average value in an infinite series
« Reply #6 on: Oct 22nd, 2004, 7:43pm » |
Quote Modify
|
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 (xy) = (lim x)lim y, than it was to state an exception for 0 every time a power series is written down.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
william wu
wu::riddles Administrator
Gender:
Posts: 1291
|
|
Re: Average value in an infinite series
« Reply #7 on: Oct 23rd, 2004, 10:08am » |
Quote Modify
|
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.
|
|
IP Logged |
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Average value in an infinite series
« Reply #8 on: Oct 23rd, 2004, 7:05pm » |
Quote Modify
|
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.)
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print