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Timeline for A derivative of sorts?

Current License: CC BY-SA 2.5

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Nov 18, 2010 at 20:47 vote accept Michael Hardy
May 26, 2010 at 4:56 answer added Michael Hardy timeline score: 0
May 25, 2010 at 21:44 comment added Michael Hardy Applied to the function $$ g_k(\theta_1,\theta_2,\theta_3,\dots) = \sum_{|A|=k}\prod_{i\in A}\sin\theta_i\prod_{i\not\in A}\cos\theta_i $$ this seems to yield (but correct me if I'm wrong) $$ -(\cos\theta_1) g_{k-1}(\theta_2,\theta_3,\dots) - (\sin\theta_1)g_k(\theta_2,\theta_3,\dots) - \frac{\partial}{\partial\theta_1} g_k(\theta_1,\theta_2,\theta_3,\dots). $$
May 25, 2010 at 20:13 comment added Willie Wong Being a bit naive here: isn't the limit you wrote equal to $ \lim \frac{f(\triangle \alpha,\alpha,\beta,\ldots) - f(0,\alpha,\beta,\ldots)}{\triangle \alpha} - \frac{f(\alpha + \triangle\alpha,\beta,\ldots) - f(\alpha,\beta,\ldots)}{\triangle \alpha} $ using that $f$ is right-translation invariant? Which formally makes it just the difference of the two derivatives. Also, what is the topology you are using on "$\mathbb{R}^\infty$"? I.e. what do you mean by continuous? Just separately continuous in each of the variables?
May 25, 2010 at 19:55 comment added George Lowther Why would anyone want to consider such a limit?
May 25, 2010 at 19:54 comment added Noah Stein Could you elaborate on the setting you are interested in? Is there a reason you don't just take the usual partial derivative of $f$ with respect to $\alpha$ and keep the other variables fixed? Do you have an example $f$ in mind? It is probably important to pin down whether you are considering only finitely many nonzero variables or not and then to put a topology on the domain of $f$ for this to make any sense.
May 25, 2010 at 19:43 history asked Michael Hardy CC BY-SA 2.5