Talk:Lidstone series

Limits
Excerpt from the article:
 * $$h(\theta; f) = \lim \sup \frac{1}{r} \log |f(r e^{i\theta})|\,$$
 * is bounded above by t. Thus, the constant N used in the summation above is given by
 * $$t= \lim \sup h(\theta; f)\,$$
 * is bounded above by t. Thus, the constant N used in the summation above is given by
 * $$t= \lim \sup h(\theta; f)\,$$
 * $$t= \lim \sup h(\theta; f)\,$$

The first limsup must have meant
 * $$h(\theta; f) = \underset{r\to\text{something}}{\lim \sup} \frac{1}{r} \log |f(r e^{i\theta})|\,$$
 * $$h(\theta; f) = \underset{r\to\text{something}}{\lim \sup} \frac{1}{r} \log |f(r e^{i\theta})|\,$$

and the second must have meant
 * $$t= \underset{\theta\to\text{something}}{\lim \sup} h(\theta; f).\,$$
 * $$t= \underset{\theta\to\text{something}}{\lim \sup} h(\theta; f).\,$$

What are the two "somethings"? Michael Hardy (talk) 02:13, 3 April 2009 (UTC)


 * Hi Micheal, The first something is \infty, as should be apparent from the article exponential type. The second should not be lim sup, but a simple sup over the interval theta in [0,2pi).  linas (talk) 19:52, 3 April 2009 (UTC)

Got it. Thank you.

Next question: Who's Lidstone? Michael Hardy (talk) 05:46, 4 April 2009 (UTC)
 * OK, a guess: Could it be George James Lidstone, after whom Lidstone smoothing is named? Michael Hardy (talk) 05:50, 4 April 2009 (UTC)
 * .......and I find the answer here. Michael Hardy (talk) 05:54, 4 April 2009 (UTC)


 * So it would seem. FWIW, I did look; the book I have does not hint at who Lidstone was. linas (talk) 20:46, 21 April 2009 (UTC)