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October 16

Epsilon delta

bon giourno all :) I have a t-shirt that I had a personalized logo put on, and the nerd that I am decided to put (ε,δ) on it. Unfortunately for me whoever processed my personalized design got it backwards and put (δ,ε) instead (I wonder how many years of undergrad real analysis you need to be a minimum-wage secretary at a t-shirt factory :S) and now it does not refer to limits anymore. What can I say from calculus (or math in general) it refers to, for which you pick delta first then epsilon? I don't want to go bland and say it's "just a point". 186.216.224.10 (talk) 02:35, 16 October 2011 (UTC)[reply]

On Wikipedia both Delta-epsilon and Epsilon-delta redirect to Limit of a function. Although while (ε, δ)-definition of limit also redirects to Limit of a function, (δ, ε)-definition of limit is currently a redlink. -- 110.49.27.51 (talk) 04:08, 16 October 2011 (UTC)[reply]

Analyticity test

If a function has the Taylor series around zero $\sum _{k=0}^{\infty }{\frac {f^{(k)}(0)}{k!}}x^{k}$ and $\lim _{k\rightarrow \infty }{\frac {f^{(k)}(\xi )}{k!}}x^{k}=0$ for all real x, does that prove that the function is analytic for all real x, or do you separately have to calculate the radius of convergence?

Specifically, it can be shown that $\lim _{k\rightarrow \infty }\sin ^{(k)}(\xi ){\frac {x^{k}}{k!}}=0$ for all real x, is this sufficient to prove that the series converges to sin(x) for all real x? Widener (talk) 07:39, 16 October 2011 (UTC)[reply]

Consider the function f(x) = e^(-1/x^2), except for x=0, and f(0)=0. Figure out its Taylor series around 0. See if that answers your question. --Trovatore (talk) 07:54, 16 October 2011 (UTC)[reply]

Or to pull back a level, observe that Trovatore is answering your question with an example. Conclude what the answer must be.--121.74.125.249 (talk) 08:00, 16 October 2011 (UTC)[reply]

Well, that's not reliable. Trovatore might have thought he had an example, but been wrong. Or he could have something more devious in mind. You never know. --Trovatore (talk) 08:02, 16 October 2011 (UTC)[reply]

That function you described doesn't have the property

\lim _{k\rightarrow \infty }f^{(k)}(\xi ){\frac {x^{k}}{k!}}=0

for all real x. I don't see how that answers the question. Widener (talk) 08:12, 16 October 2011 (UTC)[reply]

I assume you were trying to provide a counterexample. Widener (talk) 08:14, 16 October 2011 (UTC)[reply]

In case it wasn't clear,

{\frac {f^{(k)}(\xi )}{k!}}x^{k}

is the remainder term for the Taylor polynomial approximation of degree k-1. According to Taylor's theorem, there exists

\xi \in [0,x]

such that

f(x)=\left(\sum _{k=0}^{n-1}f^{(k)}(0){\frac {x^{k}}{k!}}\right)+{\frac {f^{(n)}(\xi )}{n!}}x^{n}

Widener (talk) 10:16, 16 October 2011 (UTC)[reply]

Are we to understand that in your limit, ξ depends in this way on k? 96.46.205.140 (talk) 10:19, 16 October 2011 (UTC)[reply]

In that case, you're stating that everywhere, the difference between the Taylor series and the function goes to zero. So yes, the Taylor series converges to the function everywhere--121.74.125.249 (talk) 11:16, 16 October 2011 (UTC).[reply]

What about the example that Trovatore gave? Widener (talk) 11:21, 16 October 2011 (UTC)[reply]

Let's go back to your various questions. Your originally formulation of the question contained a constant

\xi \,

; and I think that Trovatore may have tried to address the question you actually put. With your new formulation, you implicitly define

f^{(n)}(\xi )\,

to be an expression not involving

\xi \,

. If this is used in a substitution, it turns out that your intended question, correctly formulated, is the following:

If we know that $\lim _{n\rightarrow \infty }{\frac {x^{n}}{n!}}\cdot {\frac {n!}{x^{n}}}\cdot \left(f(x)-\sum _{k=0}^{n-1}f^{(k)}(0){\frac {x^{k}}{k!}}\right)=0$ , will it then follow that $\lim _{n\rightarrow \infty }\left(f(x)-\sum _{k=0}^{n-1}f^{(k)}(0){\frac {x^{k}}{k!}}\right)=0?$

I (and the IP supra) are inclined to answer "Yes, obviously!" to the question, when it is put in this form. Do you agree? JoergenB (talk) 14:01, 16 October 2011 (UTC)[reply]

May I use wikipedia to publish the continuum hypothesis?

See the problem here:http://www.claymath.org/millennium/Riemann_Hypothesis/

I did some work at home, so pretend it's homework.

Here are my first attempts, http://en.wikipedia.org/wiki/Talk:Continuum_hypothesis#Possible_Proof

It applies to wikipedia, and me, and the CMI. http://www.claymath.org/

This is true because they made it a game with a million dollar prize, and I won the game, and posted the proof on wikipedia.

So I already did it, so you might as well say yes... even though someone took it down, and then asked me about it... until no one wanted to ask anymore questions... and they told me where to ask the question above. So I did, and like I said, you might as well say yes, since it has already happened.

Since it has already happened it is not a future event.

So, can I?

WhatisFGH (talk) 14:35, 16 October 2011 (UTC)[reply]

Yes, but only here on the Reference Desk or on your userpage, not in Wiki articles or their talk pages. Count Iblis (talk) 14:58, 16 October 2011 (UTC)[reply]

Works for me, since that proves that wikipedia is a peer reviewed journal, so it can post self referential material about it self... so herp derp I didn't do anything wrong in the first place... since if you can publish something that is peer reviewed on wikipedia.. that is by definition a peer reviewed journal.

HERP DERPWhatisFGH (talk) 15:01, 16 October 2011 (UTC)[reply]

I'm not sure that most scientists or publishers would regard a Wikipedia user page as a peer-reviewed journal, but it is certainly useful for an initial peer review prior to formal publication. Dbfirs 08:10, October 17, 2011‎ (UTC)

Wikipedia is not a journal. Further, it is not peer-reviewed. To be peer-reviewed, the reviewers must have proven technical ability to perform a proper review. Wikipedia editors are just random people on the Internet who may or may not have any clue what they are talking about. Even if one claims to have a PhD in physics, you do not know that is true. -- kainaw ™ 12:56, 17 October 2011 (UTC)[reply]

Besides all the above: the proof that you posted is not a proof. Working worker ant (talk) 23:19, 17 October 2011 (UTC)[reply]

There's your peer review! Dbfirs 06:19, 18 October 2011 (UTC)[reply]

@WhatisFGH: Wikipedia has a policy against original research in articles. And articles' talk pages are not appropriate places to post such things, since their purpose is to discuss how best to maintain and improve the article; other material is off-topic. If you do post something original on your user page and it stays there, that's not peer-review. You might consider posting to the arXiv. They do some initial screening, I think, but nothing that amounts to peer-review. Michael Hardy (talk) 10:40, 22 October 2011 (UTC)[reply]

1 2 3 4 5 6 7 8 9 0

why do our math system only has 10 characters? and can mathematical theory or physics theory can be represented by other numbers, example roman numerals. sorry for bad english. — Preceding unsigned comment added by 203.112.82.128 (talk) 16:50, 16 October 2011 (UTC)[reply]

Our article decimal explains why. Other number systems would be possible, but Roman numerals are not good because it is hard to do arithmetic using them. Looie496 (talk) 16:55, 16 October 2011 (UTC)[reply]

An HP-16C programmer's calculator with dedicated A-F keys to aid in entry of hexadecimal numerals.

Our article hexadecimal gives an example of a commonly used number system with sixteen digit symbols, typically 0-9 & A-F. Reusing the ten decimal symbols characters and adding the first six letters has become standard, but several alternatives have been suggested and are equally valid. -- 49.229.147.145 (talk) 03:07, 17 October 2011 (UTC)[reply]

Note that Roman numerals are a decimal system, but they do not use positional notation and lack a zero, making arithmetic more difficult, but not impossible. -- 49.229.147.145 (talk) 03:07, 17 October 2011 (UTC)[reply]