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$1=\sum_{n>0} (nx)^n$ Dont have any solution in $\mathbb C$? Are there other types of equations with no solutions in $\mathbb C$? Can one define an object that satisfyes these equations? Does it behave well?

If x is a matrix, and 1 the identity matrix, is there a solution?

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Since the RHS series diverges for nonzero $x$, it can hardly be said to be a function of $x$ in $\mathbb{C}$. – hardmath Apr 29 '12 at 19:15
x isnt restricted to C, so it doesnt have to diverge. – TROLLKILLER Apr 29 '12 at 19:16
You asked about "any solution in $\mathbb{C}$", so I naturally thought you posed the question on that domain. What domain are you asking about? – hardmath Apr 29 '12 at 19:25
I am asking what domains contain an x which satisfy it. – TROLLKILLER Apr 29 '12 at 19:26
For the question to be meaningful one needs a context in which the "equation" makes sense, presumably at least a topological ring with $\mathbb{N}$ embedded in it (to make sense of the coefficient $n$ equally as the "exponents" of the series). – hardmath Apr 29 '12 at 19:39

This has no solution, because $\sum\limits_{n=1}^\infty (nx)^n$ diverges whenever $x\neq 0$. To see this, suppose $x\neq 0$ so $|x|>0$ and let $N\in \mathbb N$ be greater than $1/|x|$. Then for $n\geq 2N$ we have $(nx)^n>2^n$ which is unbounded, hence the terms do not even go to $0$ so the series must diverge.

Edit: There is no solution for matrices over $\mathbb C$ either. Any non-nilpotent matrix $x$ will cause the terms to grow without bound as in the case of complex numbers, while the image of any nilpotent matrix $x$ is a proper subspace and the images of powers of $x$ are contained in it, so the image of the sum will be a proper subspace as well, yet the image of the identity matrix is not.

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sry u didnt understand the question – TROLLKILLER Apr 29 '12 at 19:23
@Übermensch Care to explain? – Alex Becker Apr 29 '12 at 19:25
I have defined x to be the object which satisfyes the equation. What more can be said about x, other then its not in C and not a matrix? – TROLLKILLER Apr 29 '12 at 19:34

The right hand side converges on the maximal ideal of the ring $\mathbb{Z}[[x]]/(1- \sum n^nx^n)$ and is true of the element $x$ there.

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I don't think "converges" applies unless a topology is given. Of course $1 - \sum n^n x^n$ is an element of the ring of formal power series over the integers $\mathbb{Z}\[\[x\]\]$, so I suppose you could specify a topology induced by ideals $(x^n)$ as a basis around the origin. – hardmath Apr 29 '12 at 22:47
There's a problem with taking the quotient as indicated in the ring of formal power series: $1 - \sum_{n \gt 0}n^n x^n$ is a unit there, so the principal ideal is all of $\mathbb{Z}\[\[x\]\]$. – hardmath Apr 29 '12 at 23:03
Ah, he doesn't allow the term $n = 0$. (To your first complaint - I am taking the $x$-adic topology, yes). – user29743 Apr 30 '12 at 14:54

At best Übermensch wants to know if expanding the search for solutions to matrices might improve the situation. It does not.

Assuming that matrices with entries in $\mathbb{C}$ are intended, the RHS series will diverge except on nilpotent matrices. For any nilpotent $x$, since the RHS terms all contain at least one factor $x$, the series can never converge to a non-singular limit. In particular the identity cannot be the "limit" (which I put in quotes since for nilpotent $x$, only finitely many terms of the series are nonzero).

Added: The comments I made on another Answer posted here show that no solution is possible if convergence is taken as in formal power series. Any mapping of $\mathbb{Z}[[X]]$ to ring R such that the image $x$ of $X$ satisfies the equation $1 = \sum_{n \gt 0} n^n x^n$ would necessarily map the unit $1 - \sum_{n \gt 0} n^n X^n$ to zero, and thus the kernel would be all of $\mathbb{Z}[[X]]$.

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