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In his Paper The Euler Characteristic of acyclic Categories Kazunori Noguchi gives the following definition:

Let $f(t)$ be a formal power series over $\mathbb{Z}$. If there exists a rational function $g(t)/h(t)$, such that $f(t)=g(t)/h(t)$, then define $$f_{|_{t=-1}} = g(-1)/h(-1)$$ if $h(-1)\not= 0$

Now - given my pretty limited knowledge about formal power series - I'd assume that $f(t)=g(t)/h(t)$ yields $f_{|_{t=-1}} = g(-1)/h(-1) = f(-1)$ which makes me wonder why he's using the quotient construction in the first place, instead of just evaluating $f(t)$ at $t=-1$ (provided f(t) converges at $t=-1$).

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The possibility of non-convergence of the formal power series for $f$ is exactly the issue. –  paul garrett Oct 31 '12 at 12:21
    
@paul Yeah, I thought about that, but the construction seems unnecessary complicated to me if it's only purpose is to state that f(t) converges at $t=-1$. –  roman Oct 31 '12 at 12:27
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Ah, but as in GEdgar's answer, the series for $f$ need not converge to be evaluable. This is a kind of "algebraic" analytic continuation. –  paul garrett Oct 31 '12 at 14:46
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up vote 2 down vote accepted

Example. $f(x) = \sum_{n=0}^\infty x^n$, $g(x)=1$, $h(x) = 1-x$. Then $f(x) = g(x)/h(x)$, but $f(x)$ diverges at $-1$. Nevertheless, we want to plug in $-1$ in the rational function and get $$ f_{|_{t=-1}} = g(-1)/h(-1)=1/(1+1)=1/2. $$ But despite this, we do not write this calculation as $$ f(-1) = 1 - 1 + 1 - 1 + 1 - 1 + \dots = \frac{1}{2} $$

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But $\sum_{n=0}^\infty = \frac{1}{1-x}$ only holds for $\left|x\right|<1$. So $f(x)\not=g(x)/h(x)$ for $\left|x\right|\ge 1$ and thus the prerequisites for the definition aren't satisfied in the first place. –  roman Oct 31 '12 at 14:59
    
@roman: Dear roman, You are missing the point. The defintion means that $f(x) = g(x)/h(x)$ *as formal expressions in $x$*; the whole point is to apply it in situations where the power series doesn't converge at $x = -1$. Regards, –  Matt E Oct 31 '12 at 15:49
    
What matt said. $f(x) = g(x)/h(x)$ as formal power series. Convergence not required. We could similarly set up a case where the radius of convergence of $f$ is strictly less than $1$, so that $-1$ is not even on the boundary of the disk of convergence. –  GEdgar Oct 31 '12 at 18:08
    
Ah ok, I see. Thank you. –  roman Nov 1 '12 at 8:54
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