User talk:GFauxPas/Archive1

Thing part 2
Part two:

$a^x ≡ \underbrace{a\centerdot a\centerdot a\ldots}_{x \ of \ these \ things} \vdash a^x ≡ e^{x \centerdot lna}, x\in\Z^+$

Thing
Not sure if this is theorem worthy or not, but I'm playing with it here.

Theorem: The definition $a^x ≡ e^{x \centerdot lna}$ Is consistent with the definition $a^x ≡ \underbrace{a\centerdot a\centerdot a\ldots}_{x \ of \ these \ things}$ For $x\in\Z^+$

Part one: $a^x ≡ e^{x \centerdot lna}\vdash a^x ≡ \underbrace{a\centerdot a\centerdot a\ldots}_{x \ of \ these \ things}, x\in\Z^+$

Base step: $a^1 = e^{1 \centerdot lna} = e^{lna} = \underbrace{a}_{1 \ thing}$

Fixed Inductive Hypothesis:

$a^x = \underbrace{a\centerdot a\centerdot a\ldots}_{x \ of \ these \ things}$

Inductive step:

By definition,

$a^x = e^{x \centerdot lna}$

$a^{x+1} = e^{(x+1) \centerdot lna}$

From

$e^{x+n} = e^x \centerdot e^n$

We have

$a^{x+1} = e^{(x+1) \centerdot lna}$

$= e^{x \centerdot lna + lna}$

$= e^{x \centerdot lna} \centerdot e^{1 \centerdot lna}$

$= \underbrace{a\centerdot a\centerdot a\ldots}_{x \ of \ these \ things} \centerdot \underbrace{a}_{one \ of these \ things}$

$= \underbrace{a\centerdot a\centerdot a\ldots}_{x \ + \ 1 \ of \ these \ things}$

Which is what we wanted to prove.

for part 1.

I don't know if I'll finish this or if it will go into the wiki but in any event I wanted to practice mathematical induction. --GFauxPas 15:16, 23 October 2011 (CDT)


 * Let me comment on this. The induction hypothesis is in fact $a^x = \underbrace{a\centerdot a\centerdot a\ldots}_{x \ of \ these \ things}$
 * Subsequently one proves that this implies the same for $x+1$. Sometimes this is too weak (or too hard), and one uses Second Principle of Mathematical Induction.
 * The theorem itself, in some modified form, may appear in proving that Definition:Power (Algebra) is consistent (as that is only mentioned ATM). --Lord_Farin 15:29, 23 October 2011 (CDT)

Aah thank you Lord Farin it's exactly because of things like this that I felt I needed to practice induction :) I don't understand what you meant by the theorem appearing, did you mean this quote?

"This agrees with the definition as given in Powers of Group Elements, which is appropriate as, under multiplication, the real numbers (less zero) form a group."

I'm sorry if my questions are unsophisticated.

--GFauxPas 15:37, 23 October 2011 (CDT)


 * That's where my comment originated from, yes. But I meant more something like 'included', as that statement is (IMO) too informal for PW.
 * I have no problems with your questions; people who are willing to learn I will rarely find annoying plainly because they haven't had the chance to know more (otherwise being a TA would be hard ;) ). --Lord_Farin 15:43, 23 October 2011 (CDT)

Oh sure I definite agree that "x of these things" is very informal but I find it's helpful for me to write down the general framework of the proof, using casual language, and then going back and making it sound more mathy. Often I find out that a proof requires more advanced things than I know. With that in mind, I want to rush to the end of the proof even if I have to be sloppy about it, and then I can make sure it's in my domain. Thanks for your patience and understanding --GFauxPas 15:54, 23 October 2011 (CDT)

New proofs
Added http://www.proofwiki.org/wiki/Definite_integral_of_an_even_function and http://www.proofwiki.org/wiki/Definite_integral_of_an_odd_function, please proofread! Also, it's not a typo that I only sourced the proof for the even function and gave no source for the odd function, the book gave the second proof as an assignment after giving the first proof, so I don't think I should cite it? Correct me if I'm wrong. --GFauxPas 21:14, 17 October 2011 (CDT)
 * Yes, and also cite the chapter and section etc. that they come from.
 * Please take time, by the way, to study the format in which other pages have been written. It would be instructional for you to note the changes that have been made so far to what you have entered, and to take note for future entries. --prime mover 00:18, 18 October 2011 (CDT)

Okay thank you for the advice, I'm very new at this but I hope I can learn it --GFauxPas 06:04, 18 October 2011 (CDT)
 * So are many others. You'll pick it up. You'll be expected to. You're a mathematician. --prime mover 11:28, 18 October 2011 (CDT)

It seems a bit presumptuous to call myself a mathematician, I'm just a freshman math student D: but thanks for the compliment, I think? Question please: I have my own proof that came about in physics, an object travelling in a parabolic path will have the same magnitude of velocity at any two points with the same elevation, but will differ in sign. Mathematically, given a second order polynomial and a horizontal line intersecting it at two points, the derivatives at the two points of intersection will be equal in magnitude but differ in sign. Is this theorem-worthy? What is the name of this theorem if I want to type up a proof? What do I search for to see if it's already in the wiki? --GFauxPas 11:37, 18 October 2011 (CDT)
 * See those links on the left? See where it says "Proof Index"? I invite you to press it and explore. --prime mover 12:32, 18 October 2011 (CDT)

Welcome
Welcome to ProofWiki! Since you're new, you may want to check out the general help page. It's the best first stop to see how things are done (next to reading proofs, of course!). Please feel free to contribute to whichever area of mathematics interests you, either by adding new proofs, or fixing up existing ones. If you have any questions please feel free to contact one of the administrators, or post your question on the questions page.

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 * --Your friendly ProofWiki WelcomeBot.

sin x over x
Your edits in Sandbox look promising! Graphic is superb. But we might want to rename the graphic once it has been finished to match the name of the proof it goes with. --prime mover 14:42, 23 September 2011 (CDT)


 * No problems with replying to my comment here by putting a response on my own talk page - but it is better to keep the conversation in one place by replying on the same page it starts. I make sure that all pages I edit are on my watchlist, so I am notified whenever a response is made on the same page I made the comment being replied to.


 * There's a mention of this issue on the main talk page, and I can see why it's been raised - if the conversation goes on a long time and spreads over several talk pages it gets difficult to follow it. --prime mover 16:51, 23 September 2011 (CDT)

How do I rename an uploaded file, do I have to reupload it with a new name?--GFauxPas 22:54, 24 September 2011 (CDT)


 * I believe you don't have the authorisation to rename pages. I've done the renaming of the file in question - see what's in the sandbox. --prime mover 03:19, 25 September 2011 (CDT)

Primemover, I can't find the accepted template for how to cite sources. How do I do it? Is there a standard way?--GFauxPas 17:19, 25 September 2011 (CDT)


 * There isn't a standard way of doing it. There's BookReference which references an entry in the Books section (but for that you need to have set up a page for the book in question), or there's just the technique of describing the entity and someone will go through and tidy it. For a link on the web just link to it. There are citation links to Planetmath and a couple of others. --prime mover 18:07, 25 September 2011 (CDT)

Alternate proof for FToC
Let $f$ have a primitive $F$ continuous on the closed interval $\left[{a..b}\right]$.

$[a..b]$ can be divided into any number of subintervals of the form $[x_{k-1}..x_k]$ where

$ a = x_0 < x_1 ... < x_{k-1} < x_k = b $

By repeatedly adding and subtracting like quantities,

$F(x_k) \underbrace{- F(x_{k-1}) + F(x_{k-1})}_{0}... \underbrace{- F(x_1)+ F(x_1)}_{0} - F(x_0)$

$\implies$

$(A) \quad F(b) - F(a) = \sum_{i=1}^{k} F(x_i) - F(x_{i-1}) $

Because $F = f'$, $F$ is differentiable. Because $F$ is differentiable, $F$ is continuous. By the mean value theorem, in every subinterval $[x_{k-1}..x_k]$ there is some $c_i$ where $F'(c_i) = \frac{F(x_i)-F(x_{i-1})}{\Delta x_i}$

where $\Delta x_i \equiv x_{i} - x_{i-1}$

I know it's messy, I'm just getting the framework down.

If I multiply both sides by $\Delta x_i$ I get

$F'(c_i)\Delta x_i = F(x_i) - F(x_{i-1})$

Substituting $F'(c_i)\Delta x_i$ into $(A)$ we get

$F(b) - F(a) = \sum_{i=1}^{k} F'(c_i)\Delta x_i$

Because $F' = f$

$F(b) - F(a) = \sum_{i=1}^{k} f(c_i)\Delta x_i$

Because $f$ and $F$ are both continuous, we can take the limit $||\Delta||→0$ of both sides, where $||\Delta||$ is the magnitude of the largest subinterval $[x_{k-1}..x_k]$

$\lim_{||\Delta|| \to 0} F(b) - F(a) = \lim_{||\Delta|| \to 0} \sum_{i=1}^{k} f(c_i)\Delta x_i$

The $LHS$ is a constant and is unchanged by taking the limit of it. The $RHS$ is the definition of the integral.

$F(b) - F(a) = \int_{a}^{b}f(x)\mathrm{d}{x}$

This guy $\uparrow$ needs proofreading --GFauxPas 10:51, 27 September 2011 (CDT)


 * Enter it into the appropriate page anyway, and add the "proofread" template (see Template:Proofread for usage). --prime mover 06:50, 28 September 2011 (CDT)

Sorry for it being so messy and unstructured, I hope you guys can make something good of it! I'm not going to be able to work on it for the next few days though so I leave it to you until I come back--GFauxPas 16:03, 28 September 2011 (CDT)


 * It's now on the "incomplete" list, so can be attended to as and when anyone cares to. --prime mover 16:37, 28 September 2011 (CDT)

Convergence and other principles of analysis
Currently I am a teaching assistant for an advanced analysis course, so if you have any questions regarding real (multidimensional or not) analysis, feel free to drop a note on my talk page. When you eventually get there, I might be able to help out on Complex Analysis as well. --Lord_Farin 14:27, 23 October 2011 (CDT)

Awesome, thanks a lot --GFauxPas 14:33, 23 October 2011 (CDT)

Notation
I note from your front page you're setting up some copypasta for yourself. Before you go too far down that route, pls note the following:

1. The raw symbols ≡, · and Δ and so on are never used on ProofWiki. The $\LaTeX$ code is always used: $\equiv, \cdot, \Delta$ (or when appropriate $\triangle$ and its variants).

2. For "defined as" we use $:=$ as this is a specific symbol meaning "is defined as". The $\equiv$ symbol has plenty of other meanings and it is best kept for those.

Hope this is OK. --prime mover 16:29, 24 October 2011 (CDT)

Thank you for your insight prime.mover. Point 1 I was aware of and I just put it there for my own reference and for copy pasting when I don't have latex format available.For point 2, a notation is just a notation so I'm glad you pointed out that $:=$ is the house style. $\equiv$ is just what I've been using in my math notes. PW uses $\iff$ for the biconditional so that means that $\equiv$ is used for congruence in modular arithmetic? Thanks for the correction. --GFauxPas 20:04, 24 October 2011 (CDT)

Theorem
Let $x \in \R$ be a real number such that CORRECTION $x < -1$ or $x > 1$

Let $arcsec x$ be the arcsecant of $x$.

Then:
 * $\dfrac {d \left({arcsec x}\right)}{dx} = \dfrac 1 {|x|\sqrt {x^2 - 1}}$.

Proof
Let $y = arcsec x$ where $x < -1$ or $x > 1$.

Then $x = sec y$ where $y \in [0..\pi] \land y \ne \pi/2$.

Then $\dfrac {dx} {dy} = \sec y \ tan y$ from Derivative of Secant Function.

From Derivative of an Inverse Function it follows that $\dfrac {dy} {dx} = \frac{1}{sec y \ tan y}$.

Squaring both sides we have

$ (\dfrac {dy} {dx})^2 = \frac{1}{sec^2 y \ tan^2 y} $

From corollary to Sum of Squares of Sine and Cosine $1 + tan^2 y = sec^2 y \implies tan^2y = sec^2 y - 1$.

Using this identity we can write

$ (\dfrac {dy} {dx})^2 = \frac{1}{(sec^2 y) \ sec^2 y - 1} $

$x$ was defined as $sec \ y$ so

$ (\dfrac {dy} {dx})^2 = \frac{1}{(x^2) \ x^2 - 1} $

Taking the square root of each side of this equation yields

$ |\dfrac {dy} {dx}| = \frac{1}{|x|\sqrt {x^2 - 1}}$

I'm stuck here. I'm trying to find a way to get rid of the absolute value sign on the left hand side. I'm going to think about it more, hopefully it's not a dead end... --GFauxPas 13:31, 28 October 2011 (CDT)
 * In such cases, it is probably a good idea to do a case distinction on the sign of $\frac{dy}{dx}$ (i.e., handle cases $<0$, $=0$ and $>0$ separately). Also, when $-1 0$, this simplifies to

$ \dfrac {dy} {dx} = \frac{1}{|x|\sqrt {x^2 - 1}}$

Because $\frac{1}{|x|\sqrt {x^2 - 1}}$ is never non-positive, we have exhausted all the cases. ???

--GFauxPas 10:49, 1 November 2011 (CDT)


 * Well, not exactly. The idea is to read off the sign of $\dfrac{dy}{dx}$ from the sign of $\dfrac{dx}{dy}$ (they are the same), depending on $x$. Most generally, one investigates when the absolute value does, and when it does not, coincide with the identity, and separates those cases. In the present case, when the domain is corrected to something where $\operatorname{arcsec}$ in fact takes real values, one has to be incredibly careful to make the statement precise. I suggest you read some more, eg. at MathWorld. --Lord_Farin 14:00, 1 November 2011 (CDT)

Thanks for your help Lord_Farin, my proof-writing skills or lack thereof are self-taught so I'm not very good.

How about this?

Since $\dfrac {dy} {dx} = \frac{1}{sec y \ tan y}$ the sign of $\dfrac {dy} {dx}$ is the same as the sign of $secy tan y$.

Writing $sec y tan y$ as $\frac{siny}{cos^2y}$ it is evident that the sign of $\dfrac {dy} {dx}$ is the same as the sign of $siny$.

From Sine and Cosine are Periodic on Reals $siny$ is never negative on its domain ($y \in [0..\pi] \land y \ne \pi/2$). Thus our absolute value is unnecessary and we have

$ \dfrac {dy} {dx} = \frac{1}{|x|\sqrt {x^2 - 1}}$

Which is what we desired to prove?

??? Thanks for your patience. --GFauxPas 11:02, 2 November 2011 (CDT)


 * I wrote it out today, and arrived at a similar formula. Actually, I even used about the same reasoning. Autodidactism is not a problem, it might just conflict with some standard notions in the literature. I recommend you read a lot of proofs here on PW (where the proof style is somewhat more rigorous than in most literature), and also read a lot of books on subjects you deem interesting. --Lord_Farin 12:29, 2 November 2011 (CDT)

Thanks for the help Lord_Farin, I'll do that. For now, Derivative of arcsecant function. --GFauxPas 13:28, 2 November 2011 (CDT)