quocdat

the infinite problem

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Lately, I came up with a problem that I cannot solve and seems to be paradoxical.

The problem is simple:

A = x+ x1 + x+ x3 + x4 +...

with x > 1.

The sequence goes on to infinity.

If we multiply A to x, we will get

A.x = x( x+ x1 + x+ x3 + x4 +...)

A.x =  x1 + x+ x3 + x4 +...

A.x = A - x0 = A - 1

A- A.x = 1

So : A = 1/(1 - x)

This clearly cannot be true since if x is 2, A would be -1 instead of infinity. However, I don't know what did I do wrong or does the problem even make sense or not.

I would be thankful to receive help from you guys. 

Thank you.

 

 

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31 minutes ago, quocdat said:

with x > 1.

Your problem lies right here.

32 minutes ago, quocdat said:

So : A = 1/(1 - x)

It is indeed true that the power series expansion of 1/(1-x) is


[math]\frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty  {{x_n}}  = 1 + x + {x^2} + ....[/math]

But this is only true for |x|  < 1

The series for |x|  > 1 is divergent (That is the individual terms get larger and larger as the number of terms increases, so their sum gets larger and larger)

For |x|  < 1 the series is convergent (That is the individual terms get smaller and smaller as the number of terms increases, so their sum gets progressivly closer to some finite limit)

You may only perform the term by term multiplication as in

38 minutes ago, quocdat said:

If we multiply A to x, we will get

A.x = x( x+ x1 + x+ x3 + x4 +...) 

If the series is convergent.

Yours is not.

Congratulations, you have just found out the sort of nonsense that arises when you try this on a divergent series!

You should always, but always, check the restrictions on the series variables before using them.

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It is quite common in enumerative combinatorics to apply this kind of series expansion in situations where you want to count the number of different possible objects of some kind. In your example, the series \(A(x) = \sum_{n=0}^\infty x^n = 1 + 1\cdot x + 1\cdot x^2 + \cdots = \frac{1}{1-x}\) would give the answer to the "problem" ''How many natural numbers are there of size \(n\)?'' You get the answer by looking at the coefficient of \(x^n.\)

In this application it makes no sense to "evaluate" a series at some value of \(x.\) The radius of convergence of the series does not play any role. In fact you may consider a series which counts the number of permutations \(n!\) of the numbers \(\{1,2,\ldots,n\}\):

\[ P(x) = 0! + 1!\cdot x + 2!\cdot x^2 + \cdots = \sum_{n=0}^\infty n!x^n.\]

Which has no convenient shorthand similar to \(A(x) = 1/(1-x)\) for the series \(A(x).\) In fact \(P(x)\) has zero radius of convergence; there is no value \(x \neq 0\) for which \(P(x)\) converges. And yet the series makes complete sense combinatorially. 

Edited by taeto

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33 minutes ago, taeto said:

In this application it makes no sense to "evaluate" a series at some value of x. x.

+1. But I would put it more strongly than that.

I would say it is entirely wrong to do so.

Consider for instance x = 2

Then 1/(1-x) = -1

What does that tell you?

Edited by studiot

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What you have hit upon is a mathematical construction called analytic continuation, where a function is defined for a small domain, i.e. the power series for |z| < 1, is extended to all z, except |z|=1, by using a different expression which coincides with the first expression over its domain (which has to be an open set).

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