User talk:GFauxPas/Archive1

Alternate Proof of Power Rule for POSITIVE (my b) Real Indices
Let $y$ be a real function of $x$ defined as

$y = x^n$ for some constant $n$ such that $n$ and $x$ are not both zero.

Taking the natural logarithm of both sides we have

$\ln y = \ln x^n$

Using Logarithms of Powers this becomes

$\ln y = n \ln x$

Taking the derivative of both sides with respect to $x$ gives us

$\dfrac 1 y \dfrac {\mathrm d y} {\mathrm d x} = n \dfrac 1 x$

Question please to PW veterans, is it necessary to state here that I used derivatives of a constant multiple, chain rule, and derivative of logarithmic function? How much do I need to spell out? It seems that the house style is to say everything out but that seems very wordy. I assume that the correct thing to do is to include them.


 * As it stands, the results you've cited are precisely those which are expected. Good job.
 * Might be worth looking at the eqn template. When that's used, the reasons behind each step can be included in the c column, so the implications flow from one line to the next without the justifications getting in the way.

Multiplying both sides of the equation by $y$ yields

$\dfrac {\mathrm d y} {\mathrm d x} = n \dfrac y x$

Substituting $x^n$ for $y$:

$\dfrac {\mathrm d y} {\mathrm d x} = n \dfrac {x^n} {x}$

From Exponent Combination Laws/Difference of Powers

$\dfrac {\mathrm d y} {\mathrm d x} = n x^{n-1}$

--GFauxPas 21:51, 2 November 2011 (CDT)


 * I am not such a veteran on PW that I know what need, and what needn't be spelled out. Mind though, your proof only works for positive real $x$, as $y$ needs to be in the domain of the natural logarithm. Handling the cases $x=0$ and, in particular, $x$ negative, can be hard. The proof appears correct for positive real $x$, though.
 * Concerning the derivative theorems, I think it might be justified to create a page which gives the standard derivatives of some functions, and the chain rule (and formulae implied like product and quotient rule). But then, maybe this page already exists. --Lord_Farin 04:02, 3 November 2011 (CDT)
 * Good idea. It doesn't and it should. The work on differentiation work was done before we properly developed the technique of transclusion for bagging up related results into one page. As you suggest: Chain Rule, Product Rule, Quotient Rule, perhaps even Leibniz' Rule. Diffs of some standard functions in another page (or series of pages) again, perhaps bagging up trig functions in one page and inverse trig functions on another page. We can categorise and nest as much as we like. I won't do it now as I'm supposed to be at work (just waiting for a long download to finish).--prime mover 04:24, 3 November 2011 (CDT)

Ah silly me, I missed that restriction to positive x. Thank you very much for pointing that out, when I realize what kind of mistakes I make I'm less likely to make them in the future. I'll have to look at the eqn template. In any event, from looking at my syntax, I think I'm getting better at house style. --GFauxPas 05:54, 3 November 2011 (CDT)

Induction
I need more time to work on this and it's getting to unwieldy, so I'm taking it off for the time being while I learn more about induction etc. Thanks for your help Lord_Farin. --GFauxPas 16:42, 3 November 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)

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)

There was a thing here but I moved it to here Definition_talk:Equality --GFauxPas 16:17, 3 November 2011 (CDT)

Equality
Equality is reflexive. That is,


 * Reflexive: $\forall a: a = a$.

This proof depends on Leibniz's Law:

$(1) \quad x = y \dashv \vdash P(x) \implies P(y) \land P(y) \implies P(x)$

We are trying to prove $a = a$. By $(1)$, our assertion is

$(2) \quad P(a) \implies P(a) \land P(a) \implies P(a)$

It is sufficient to show that this statement is interderivable with a tautology.

Statement $(2)$ has the form the form $p \land p$. From the Rule of Idempotence we can write

$(3) \quad P(x) \implies P(x)$

Statement $(3)$ has the form $p \implies p$. From Law of Identity this is a tautology. Thus,

$x = x \dashv \vdash \top$

Prime.mover, you expressed your concern elsewhere that there's not enough of a foundation on PW to have rigorous proofs of statement in pred.calc. Feel free to voice your suggestions here, this is the kind of proof I can make at my level of knowledge and I won't be offended at all if you want me to leave it in my user talk page for the time being. --GFauxPas 17:23, 3 November 2011 (CDT)
 * I'm going to have to do some research, but be aware that a) on this site we don't call $p \land p$ "multiplication" any more, that was a terminology that was introduced by Boole and we have since evolved from that stage. It's Definition:Conjunction now. b) What you call "tautology" is called the Rule of Idempotence on this site, as a taugology is usually referred to nowadays as something completely different. --prime mover 01:37, 4 November 2011 (CDT)

Theorem: Equality is symmetric.

It is sufficient to show that this statement is interderivable with a tautology.

We are trying to prove $x = y \iff y = x$. Taking Leibniz's law as the definition of equality, our assertion is

$P(x) \implies P(y) \land P(y) \implies P(x) \dashv \vdash P(y) \implies P(x) \land P(x) \implies P(y)$

The right hand side has the form $p \land q$. From Rule of Commutation this is equivalent to $q \land p$ and we can say

$P(x) \implies P(y) \land P(y) \implies P(x) \dashv \vdash P(x) \implies P(y) \land P(y) \implies P(x)$

This has the form $p \dashv \vdash p$. From Law of Identity this is a tautology. Thus,

$x = y \iff y = x \dashv \vdash \top$

Theorem: Equality is transitive.

Assume

$(1) \quad a = b$

$(2) \quad b = c$

Our assertion is then

$(3) \quad a = c$.

Using Leibniz's law as the definition of equality,

$(1) \quad P(a) \implies P(b) \land P(b) \implies P(a)$

$(2) \quad P(b) \implies P(c) \land P(c) \implies P(b)$

We wish to show

$(3) \quad P(a) \implies P(c) \land P(c) \implies P(a)$

From Axiom:Rule of Simplification statement $(1)$ implies

$(4) \quad P(a) \implies P(b)$

From Axiom:Rule of Simplification statement $(2)$ implies

$(5) \quad P(b) \implies P(c)$

From Hypothetical Syllogism on $(4)$ and $(5)$

$P(a) \implies P(c)$

A similar proof will show $P(c) \implies P(a)$ (Am I allowed to pull this card in PW? I don't know if it's formal.)

From Axiom:Rule of Conjunction we can write

$P(a) \implies P(c) \land P(c) \implies P(a)$

Which by Leibniz's law is the definition of $a = c$. This is our desired result. --GFauxPas 22:19, 3 November 2011 (CDT)

Theorem:

Things that are equal to the same are equal to each other. That is,

$b=a \land c=a \vdash b=c$

Assume

$(1) \quad b=a$

$(2) \quad c=a$

From Equality is reflexive $(2)$ is equivalent to

$(3) \quad a=c$

From Equality is transitive on $(1)$ and $(3)$

$b=c$

This theorem is (correction) mentioned in a dialogue by Lewis Carroll "What the Tortoise Said to Achilles". A good read if you like meta-logic, you can find it for free online. --GFauxPas 22:59, 3 November 2011 (CDT)
 * From what I remember (haven't read it for a few years) this sketch of Carroll's was not directly to do with this, it was an analysis of the necessity to understand the assumptions behind the modus ponens, and the infinite regress that you can get into, nothing to do directly with this particular result.
 * You might also want to investigate Euclid's take on this (see his Common Notions at the start of Book 1).--prime mover 01:37, 4 November 2011 (CDT)

Roger that (both points), Prime.mover. Thank you for correcting me. --GFauxPas 06:26, 4 November 2011 (CDT)

Differentiating both sides
I don't plan on making a PW page for this but my Calc book takes it for granted so I'll prove it to myself.

Theorem:

Let $p$ and $q$ be differentiable functions of $x$ such that $p=q$. Then

$\frac{\mathrm{d}{p}}{\mathrm{d}{x}} = \frac{\mathrm{d}{q}}{\mathrm{d}{x}}$

Proof:

By Leibniz's law, $p=q$ can be written

$(1) \quad{P(p)} \iff P(q)$

Let $P_D$ be the property of having the derivative $\frac{\mathrm{d}{p}}{\mathrm{d}{x}}$ Then by the given that $\frac{\mathrm{d}{p}}{\mathrm{d}{x}}$ exists,

${P_D}(p)$.

From MPP on $(1)$

${P_D}(q)$

A similar proof can be shown to hold in the other direction. Then from the rule of implication,

${P_D}(p) \iff {P_D}(p)$

Which by Leibniz's law and the given definition of ${P_D}$

$\frac{\mathrm{d}{p}}{\mathrm{d}{x}} = \frac{\mathrm{d}{q}}{\mathrm{d}{x}}$


 * While this is a correct proof, I wonder what the added value is. In my opinion, the only thing that is not entirely rigorous (however intuitive) in mathematics is the definition of the equality sign. From there, we can rigorously define everything we want, and then work on our favourite subject. Furthermore, I wouldn't like a page for every possible statement $P$ we'd like or care to write down, showing that $p=q \wedge P(p) \implies P(q)$. In my opinion, such would be an utter waste of time. But then, opinions exist to be disagreed with. --Lord_Farin 10:41, 4 November 2011 (CDT)

As I said, I'm doing this only for my own benefit, I stated that I don't plan on creating a PW page for these sorts of things. I just wanted to show myself why I'm allowed to, e.g., multiply both sides of an equation by 2, or subtract an equation from another equation. Perhaps taking it for granted is a better view than making a proof for it if it's something so basic, I dunno. I don't have a teacher to ask about these kinds of things so I don't know what's worth thinking about and what's not. And as to the definition of of $=$, prime.mover and I discuss it here http://www.proofwiki.org/wiki/Definition_talk:Equality. --GFauxPas 11:19, 4 November 2011 (CDT)

A while back I wrote a proof that all integers are even or odd. The reason I wrote it was just to practice mathematical induction. The theorem itself seems rather unimportant, and so I don't see a reason to make a page for it, but at someone's request I can make it. --GFauxPas 15:56, 4 November 2011 (CDT)