User talk:Cornince

‡ Jarlaxle 23:39, July 17, 2005 (UTC)

Hi, I reverted your change to this article. I'm not sure if you misunderstood what is written, but it is definitely correct as it was (left then right then middle/bottom). Let me know if you have any questions. Akihabara 11:15, 10 January 2007 (UTC)[reply]

Well, I was thinking of Kanji like chika(ku), where you start with the upper right part, then you work with the enclosure starting on the left. Perhaps my knowledge isn't as great? Cornince 11:32, 10 January 2007 (UTC)[reply]

You're right about that character, but the sentence in question is talking about three-part kanji; two such kanji are given as examples in the sentence. Akihabara 11:34, 10 January 2007 (UTC)[reply]

Hi. I've just moved the text you recently added to our Pajamas Media article to it's talk page for discussion: see Talk:Pajamas Media#Ron Paul dropped from polls. Please comment on my suggested changes and on whether (and, if so, how) we should cover this stuff in the article. Best wishes, CWC 07:14, 22 March 2007 (UTC)[reply]

Howdy. This comment is in response to this edit, where you mention the binomial identities

${\displaystyle \sum _{k=1}^{n}{k}{n \choose k}={n}2^{n-1}\qquad (6)}$

and

${\displaystyle \sum _{j=0}^{k}(-1)^{j}{n \choose j}=(-1)^{k}{{n-1} \choose k}.\qquad (11)}$

requesting proofs or sources of proofs.

The first formula is actually proved in the article — it's just extremely terse. I'll explain what's going on in the proof here, then give a proof of the second formula. I think the current proof of the first formula is the right size for an encyclopedia article. See the References section of the article for possible links to proofs.

First of all, let's look at what the article says:

The formula

${\displaystyle \sum _{k=1}^{n}{k}{n \choose k}={n}2^{n-1}\qquad (6)}$

follows from expansion (2), after differentiating with respect to either x or y and then substituting x = y = 1.

Expansion (2) is the binomial theorem:

${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}.\qquad (2)}$

If we differentiate the left-hand side of (2) with respect to y, we get

${\displaystyle {\frac {\partial }{\partial y}}(x+y)^{n}=n(x+y)^{n-1},}$

while if we differentiate the right-hand side with respect to y, we get

${\displaystyle \sum _{k=0}^{n}{n \choose k}x^{n-k}{\frac {\partial }{\partial y}}[y^{k}]=\sum _{k=1}^{n}{n \choose k}x^{n-k}ky^{k-1}.}$

Notice that the term with no y in it drops out, so the summation ends up going from k = 1 to k = n. Thus by differentiating both sides of expansion (2) with respect to the same variable y, we get the new equality

${\displaystyle n(x+y)^{n-1}=\sum _{k=1}^{n}{n \choose k}x^{n-k}ky^{k-1}\qquad (2')}$

Substituting x = y = 1 throughout gives

${\displaystyle n2^{n-1}=\sum _{k=1}^{n}{n \choose k}k,}$

which is precisely identity (6).

The proof of identity (11) is based on the fact that the alternating sum of binomial coefficients is zero:

${\displaystyle 0=(1+(-1))^{n}=\sum _{j=0}^{n}(-1)^{j}{n \choose j}\qquad (\alpha )}$

combined with Pascal's rule:

${\displaystyle {n \choose k}+{n \choose k+1}={n+1 \choose k+1}\qquad (3)}$

We can rewrite (α) as the identity

${\displaystyle \sum _{j=0}^{k}(-1)^{j}{n \choose j}=\sum _{j=k+1}^{n}(-1)^{j+1}{n \choose j}\qquad (\beta )}$

By applying Pascal's rule we can reduce the right-hand side of (β) to the right-hand side of identity (11).

First we can reindex the right-hand side of (β):

${\displaystyle \sum _{j=k+1}^{n}(-1)^{j+1}{n \choose j}=\sum _{j=0}^{n-k-1}(-1)^{j+k}{n \choose j+k+1}}$

Applying Pascal's rule gives

${\displaystyle \sum _{j=k+1}^{n}(-1)^{j+1}{n \choose j}=\sum _{j=0}^{n-k-1}(-1)^{j+k}\left[{n-1 \choose j+k}+{n-1 \choose j+k+1}\right].}$

The thing to notice here is that we are summing a telescoping sequence. The middle terms appear once with positive sign and once with negative sign. So only the extreme terms are left:

${\displaystyle \sum _{j=0}^{n-k-1}(-1)^{j+k}\left[{n-1 \choose j+k}+{n-1 \choose j+k+1}\right]=(-1)^{k}{n-1 \choose k}+(-1)^{n}{n-1 \choose n}=(-1)^{k}{n-1 \choose k},}$

the last step following because there are zero ways to choose n out of n - 1 objects.

Piecing together the chain of equalities gives the result

${\displaystyle \sum _{j=0}^{k}(-1)^{j}{n \choose j}=(-1)^{k}{n-1 \choose k},}$

which is the same as identity (11) in the article.

I didn't mean for this comment to be quite so long. If you have any questions, just ask, either here or on my talk page. Michael Slone (talk) 08:09, 25 November 2007 (UTC)[reply]

Thank you so much for your help. Is it possible you can also help me with proving

${\displaystyle n\geq 0,\quad \sum _{k=0}^{n}(-1)^{k}{n \choose k}(n-k)^{n}=n!}$ ?

Using induction, I’ve gotten (opening the binomial coefficient):

${\displaystyle (n+1)n!=(n+1)\sum _{k=0}^{n}(-1)^{k}\left[{\dfrac {n!}{k!(n-k)!}}\right](n-k)^{n}}$

Multiplying in the n+1 to the n! and multiplying the function by k+1/k+1 putting the bottom into the k! and closing the binomial coefficient:

${\displaystyle =\sum _{k=0}^{n}(-1)^{k}{n+1 \choose k+1}(k+1)(n-k)^{n}}$

Reindexing the sum:

${\displaystyle =\sum _{k=1}^{n+1}(-1)^{k-1}{n+1 \choose k}k[n-(k-1)]^{n}}$

n-(k-1) becomes n+1-k and then I multiply the function by -1/-1 and put the top -1 into the (-1)^(k-1) and the bottom -1 into the k(n+1-k)^n and then I add and subtract (n+1)(n+1-k)^n:

${\displaystyle =\sum _{k=1}^{n+1}(-1)^{k}{n+1 \choose k}[-k(n+1-k)^{n}+(n+1)(n+1-k)^{n}-(n+1)(n+1-k)^{n}]}$

Adding the first 2 terms together to make (n+1-k)(n+1-k)^n or (n+1-k)^(n+1) and then separating the rest creating 2 separate sums:

${\displaystyle =\sum _{k=1}^{n+1}(-1)^{k}{n+1 \choose k}(n+1-k)^{n+1}-(n+1)\sum _{k=1}^{n+1}(-1)^{k}{n+1 \choose k}(n+1-k)^{n}}$

Now, I’m back to square one in trying to prove that

${\displaystyle -(n+1)\sum _{k=1}^{n+1}(-1)^{k}{n+1 \choose k}(n+1-k)^{n}=(n+1)^{n+1}}$

Is it possible you can provide some assistance in this regard? Thanks. ^_^ Cornince (talk)

Try to prove the stronger claim

${\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}(x-k)^{n}=n!}$.

If you interpret the left-hand side properly, the induction will be almost automatic. Michael Slone (talk) 03:39, 1 December 2007 (UTC)[reply]

I'm sorry, I'm not even sure where to begin with this. I was able to get to the below, using the binomial theorem: ${\displaystyle \sum _{k=0}^{n}\sum _{j=0}^{n}{n \choose k}{n \choose j}(-1)^{k+j}x^{n-j}k^{j}=n!}$ But beyond that, I'm not quite sure, where to go, other than multiply both sides by (n+1) and try to manipulate the form of the left side to the n+1 of the original. Thank you for helping me so far; please give me the slightest nudge in some direction. ^_^ Cornince 07:26, 3 December 2007 (UTC)[reply]

It is likely to be difficult to prove your claim directly from the binomial theorem. Instead, I suggest looking into the calculus of finite differences. For example, if f(x) is a function, then

${\displaystyle \nabla f(x)=f(x)-f(x-1)}$,

${\displaystyle \nabla ^{2}f(x)=f(x)-2f(x-1)+f(x-2)}$,

${\displaystyle \nabla ^{3}f(x)=f(x)-3f(x-1)+3f(x-2)-f(x-3),}$

…

See the pattern? Verify that this pattern works for all n, then compare it to the left-hand side of the equation you're trying to prove. What should f(x) be? Michael Slone (talk) 08:22, 3 December 2007 (UTC)[reply]

OK, let me answer your quesetion, f(x) would be xⁿ, right? Cornince (talk) 01:00, 12 December 2007 (UTC)[reply]

Yes, that is correct. Michael Slone (talk) 02:30, 12 December 2007 (UTC)[reply]

Thank you, now I wonder how in induction using what you showed, I would get xⁿ changed to xⁿ⁺¹? I'm still not quite seeing how to go about this, or even where to start. Cornince (talk) 03:03, 12 December 2007 (UTC)[reply]

Thank you for continuing to look at this. I had been thinking about something similar to the difference operator. I used a you notation (用) function z(x), to which 用(m,0) = z(m), and the recurrence relation would be 用(m,n+1) = 用(m+1,n) - 用(m,n) for n >= 0. (An example is that 用(m,1) = 用(m+1,0) - 用(m,0) = z(m+1) - z(m).) This I think is the same as the forward difference operator. I proved the form of the particular equation you're mentioning (which I find is called the binomial transform) here:

http://en.wikipedia.org/wiki/User:Cornince/no2

(Look at the top equation.) In this I demonstrated that the equation in question has the same recurrence relation as the you notation. A quick look at case n=0 for both would also demonstrate that they both have the same starting place.

I did look at it for this in proving the case that where z(a) = (x - a)^n (or your f(x)) then it comes to n!. But the issue is one of raising the power of the function as you raise the number of items. Like, for example, if I wanted to use induction to get from (2)^2 - 2(1)^2 + (0)^2 to (3)^3 - 3(2)^3 + (1)^3 - (0)^3, I know how to use the difference operator to get the extra item and the binomial coefficients changed to the 3rd row, but by which means would the power of 2 be raised to 3 through induction?

Cornince (talk) 04:21, 5 December 2007 (UTC) (this was cornince typing the response)[reply] 

I saw this addition to (one of) your subpage(s) by an IP user and I wanted to let you know about it in case you didn't want it there. Regards. Thingg^⊕⊗ 16:45, 31 January 2008 (UTC)[reply]

Thanks. That was me from a different computer, where I probably forgot to log in. ^_^ Cornince (talk) 19:47, 31 January 2008 (UTC)[reply]

The article May 2010 Tennessee floods has been proposed for deletion because of the following concern:

This is not a major flood. Fails WP:NOT#NEWS.

While all contributions to Wikipedia are appreciated, content or articles may be deleted for any of several reasons.

You may prevent the proposed deletion by removing the {{dated prod}} notice, but please explain why in your edit summary or on the article's talk page.

Please consider improving the article to address the issues raised. Removing {{dated prod}} will stop the proposed deletion process, but other deletion processes exist. The speedy deletion process can result in deletion without discussion, and articles for deletion allows discussion to reach consensus for deletion. Joe Chill (talk) 21:02, 2 May 2010 (UTC)[reply]

For information on what is a major flood, read here: http://en.wikipedia.org/wiki/Wikipedia:Notability_%28events%29

I have nominated May 2010 Tennessee floods, an article that you created, for deletion. I do not think that this article satisfies Wikipedia's criteria for inclusion, and have explained why at Wikipedia:Articles for deletion/May 2010 Tennessee floods. Your opinions on the matter are welcome at that same discussion page; also, you are welcome to edit the article to address these concerns. Thank you for your time.

Please contact me if you're unsure why you received this message. Joe Chill (talk) 19:54, 3 May 2010 (UTC)[reply]

On 4 May 2010, In the news was updated with a news item that involved the article May 2010 Tennessee floods, which you created. If you know of another interesting news item involving a recently created or updated article, then please suggest it on the candidates page.

--HJ Mitchell | Penny for your thoughts? 15:16, 4 May 2010 (UTC)[reply]

You are receiving this because you have commented on either Autogynephilia, Homosexual transsexual, or Blanchard, Bailey, and Lawrence theory in the past two years; all such commenters have received this notice. It has been proposed to merge these three articles to eliminate WP:Redundancy, WP:UNDUE, WP:POV, and to keep the focus on the specific Blanchardian theory of M2F transsexuality (in contrast to Transsexual sexuality, which would be to focus on the subject in general). Please feel free to comment on the proposal at Talk:Autogynephilia#Merger proposal. -- 70.57.222.103 (talk) 20:08, 8 September 2010 (UTC)[reply]

Hello, Cornince. You have new messages at Talk:Schizophrenia.
Message added 20:27, 1 December 2013 (UTC). You can remove this notice at any time by removing the {{Talkback}} or {{Tb}} template.

Lova Falk talk 20:27, 1 December 2013 (UTC)[reply]

Hi,
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Hello, Cornince. Voting in the 2018 Arbitration Committee elections is now open until 23.59 on Sunday, 3 December. All users who registered an account before Sunday, 28 October 2018, made at least 150 mainspace edits before Thursday, 1 November 2018 and are not currently blocked are eligible to vote. Users with alternate accounts may only vote once.

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