Talk:Sophomore's dream

Cool!

Title added. —Nils von Barth (nbarth) (talk) 06:32, 12 February 2011 (UTC)

Cool!

(But this article is not linked-to enough. I'll see if I can do something about that.) Michael Hardy 23:36, 6 February 2007 (UTC)

Why "sophomore's dream"?

Title added. —Nils von Barth (nbarth) (talk) 06:32, 12 February 2011 (UTC)

But why "sophomore's dream"? 24.137.126.62 09:08, 28 February 2007 (UTC)

Agreed... a reason for the name would be good to add. Privong 04:20, 16 March 2007 (UTC)

I'm wondering the same thing. There's probably an interesting story behind it. Roger 18:11, 14 May 2007 (UTC)

I came to the talkpage to see if this was being discussed actually. The etymology would add to the article. Mehmet Karatay 13:00, 20 June 2007 (UTC)

I agree that some etymology would flesh this out a bit. —Preceding unsigned comment added by 143.44.71.171 (talk) 01:28, 1 October 2007 (UTC)

I added something, but I couldn't find an adequate reference so at the moment it's unsourced. skeptical scientist (talk) 05:18, 12 May 2008 (UTC)

I don’t know the origin of the name; I’ve listed the earliest reference which is given (2004, as noted on Wolfram, for instance), and this seems to be the origin of the name – any more formal references would be appreciated!

—Nils von Barth (nbarth) (talk) 06:32, 12 February 2011 (UTC)

Expression

Wouldn't the second be simpler as follows?

$\sum_{n=1}^\infty -(-1)^nn^{-n} = \sum_{n=1}^\infty -(-1)^{-n}n^{-n} = - \sum_{n=1}^\infty (-n)^{-n}$

I feel like the rightmost expression is much better. --Cheeser1 16:17, 9 July 2007 (UTC)

True, but I think the current expression is in the standard form for an alternating series. Roger 16:25, 9 July 2007 (UTC)

It's not though - it starts at a different index. --Cheeser1 01:39, 1 October 2007 (UTC)

Images

I don't think the graphs of the functions are very accurate - it appears that as x approaches zero, y tends to a value either slightly above or slightly below 1, when the the actual limit should be exactly 1. Maybe someone with good quality graphing software could confirm my thoughts and upload better images? Pscholl (talk) 12:57, 11 May 2008 (UTC)

Agree with Pscholl. It is sufficiently misleading that it would be better to remove these images. AdamWGibson (talk) 05:22, 4 June 2008 (UTC)

I made a more accurate graph showing both functions. If anyone has any suggestions it would not be difficult to fix... Blues shuffle (talk) —Preceding undated comment was added at 01:06, 12 February 2009 (UTC).

Note

The note added to the mention of Freshman's Dream seems hardly relevant. The term Freshman's Dream, in my experience, refers to that equation with real numbers $x$ and $y$, and the arithmetic operations taking on their standard denotations for real numbers, in which case it is, plain and simple, not generally true. I recommend removing the note, but if anyone feels it's necessary to discuss the exceptions that exist in modern abstract algebra I'd like to hear the rationale. 107.5.98.72 (talk) 21:59, 18 June 2011 (UTC)

I think the note is useful, if only to prevent people from changing the text of the article proper to handle that technicality. If the note weren't there, I would have been tempted to change "the incorrect equation" to something like "the (in general) incorrect equation" or "the (over the reals) incorrect equation", both of which are clunkier and less informative. Perhaps a better rewording is possible, though. 24.220.188.43 (talk) 04:19, 3 September 2011 (UTC)

Computing ∫₀¹ xⁿ (logx)ⁿdx .

The way it is currently written is neat and clear, but I think a quicker way (by linking to Euler integral or to Gamma function) wolud be preferable. Precisely, by the change variable $\scriptstyle x=\exp\, {-\frac{u}{n+1}}$, with $\scriptstyle 0 < u < \infty$, the integral

$\int_0^1 x^n(\log\, x)^n dx$

writes

$(-1)^n (n+1)^{-(n+1)} \int_0^\infty u^n e^{-u} du$

and here we may just put a link to the Euler integral's identity to conclude

$\int_0^\infty u^n e^{-u} du=n!$

Note that the current computation of $\scriptstyle \int_0^1 x^n(\log\, x)^n dx$ by iterated integration by parts, is exactly what one does for $\int_0^\infty u^n e^{-u} du$, so there is no loss of content if we directly refer to the more popular identity for the Euler integral. --pma 07:48, 9 July 2011 (UTC)

done --pma 17:09, 23 November 2011 (UTC)