# Sandbox for drafts of long, complex posts [closed]

This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

When you are happy with your draft here, you may simply copy the code and paste it to the desired location.

## Proper Use of the Sandbox

1. Do not post a new answer! We wish all the answers on this page to be owned by the Community user (so that only a non-sentient bot is informed of edits to these answers). Posting a new answer will make you the owner, meaning that you will be notified whenever another user makes an edit to that answer.

The sandbox has been closed to prevent the creation of new answers. There are more than enough existing answers for users to edit over, and this will greatly reduce the frequency at which we request that the answers be disassociated from specific users.

2. Do not delete answers! Deleting seems like a reasonable option, but there are no "hard deletions" on Stack Exchange, and users with sufficient privileges will still see your supposedly deleted postings. Deleted answers will be undeleted and cleared for the use of others.

3. Do look for an answer which indicates that it is free and then edit it to your heart's content. If none appears available, take over the one that has been left unchanged the longest (which will appear at the bottom of the page if you order answers by "activity").

4. Do not expect your draft to remain untouched for days. There are no guarantees that your draft will be the latest revision if you return days later. While users will try not to step over others' toes, it may happen that an unfinished draft is edited out. Your draft will, however, still exist as a revision of the answer it was made in. If your drafting is expected to take place over a longer period of time, either

• take note of the URL of the answer provided by clicking the share button, or
• save a copy of your draft locally (or even "in the cloud").
5. Do clear your draft when you are finished. This includes removing all $\LaTeX$ from your answers. Replacing all code with a simple statement like

This answer is free for anyone to use

is sufficient. Periodically users may go through and free up answer slots that have not been edited in, say, over one month. But you can aid in the smooth running of this sandbox by clearing away your drafts when you are finished with them.

6. Do not "claim" multiple answers concurrently. Since this post is closed, the answers are a limited resource. If you really must compose several long, complex posts at the same time, you can still use a single answer, separating the different drafts using Markup: horizontal rules (---) and/or headings (# Header 1 #) are natural choices.

7. Do not create new such sandboxes. The point of having a unique such sandbox is that it minimizes the noise on the front page when the sandbox is edited. If there were multiple sandboxes they will frequently occupy numerous front page slots, pushing other topics off the front page, and increasing noise.

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## closed as off-topic by arjafi♦Sep 18 '14 at 10:37

• This question does not appear to be about Mathematics Stack Exchange or the software that powers the Stack Exchange network within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. – Asaf Karagila Jul 18 '12 at 8:35
(+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! – Grace Note Oct 5 '12 at 14:45
To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. – arjafi Sep 18 '14 at 10:37
PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. – Najib Idrissi Dec 2 '15 at 14:07
@GraceNote Could you please also change to Community owned all of the posts in our Formatting Sandbox., including deleted posts too (I just got many pings when someone started using a deleted post). Thanks. – Bill Dubuque Dec 24 '15 at 19:50

1767225    Set up Sylvester's dialytic eliminant, which will give you a quartic equation in $x\;$. $\begin{vmatrix} c_1&b_1x+e_1&a_1x^2+d_1x+f&0\\ 0&c_1&b_1x+e_1&a_1x^2+d_1x+f\\ c_2&b_2x+e_2&a_2x^2+d_2x+f&0\\ 0&c_2&b_2x+e_2&a_2x^2+d_2x+f \end{vmatrix}=0\;$

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Common Mistakes
The following are some of the most common mistakes people make when writing mathematical proofs.
$\color{#08F}{\text{1. Arguing from examples.}}$
. . .
$\color{#08F}{\text{2. Using the same letter to mean two different things.}}$
...

$\color{#08F}{\text{3. Jumping to a conclusion.}}$
To jump to a conclusion means to allege the truth of something without giving an adequate reason. Consider the following “proof” that the sum of any two even integers is even.
Suppose m and n are any even integers. By definition of even, m = 2r and n = 2s for some integers r and s. Then m + n = 2r + 2s. So m + n is even. The problem with this “proof” is that the crucial calculation
2r + 2s = 2(r + s)
is missing. The author of the “proof” has jumped prematurely to a conclusion.

$\color{#08F}{\text{4. Circular reasoning.}}$
To engage in circular reasoning means to assume what is to be proved; it is a variation of jumping to a conclusion. As an example, consider the following “proof” of the fact that the product of any two odd integers is odd:
Suppose m and n are any odd integers. When any odd integers are multiplied, their product is odd. Hence mn is odd."

$\color{#08F}{\text{5. Confusion between what is known and what is still to be shown.}}$ A more subtle way to engage in circular reasoning occurs when the conclusion to be shown is restated using a variable.

$\color{#08F}{\text{6. Use of any rather than some.}}$
There are a few situations in which the words any and some can be used interchangeably. For instance, in starting a proof that the square of any odd integer is odd, one could correctly write “Suppose m is any odd integer” or “Suppose m is some odd integer.” In most situations, however, the words any and some are not interchangeable.
Here is the start of a “proof” that the square of any odd integer is odd, which uses any when the correct word is some:
Suppose m is a particular but arbitrarily chosen odd integer.
By definition of odd, m = 2a + 1 for any integer a.
In the second sentence it is incorrect to say that “m = 2a + 1 for any integer a” because a cannot be just “any” integer; in fact, solving m = 2a + 1 for a shows that the only possible value for a is (m − 1)/2. The correct way to finish the second sentence is, “m = 2a + 1 for some integer a” or “there exists an integer a such that m = 2a + 1.

$\color{#08F}{\text{7. Misuse of the word if.}}$
Another common error is not serious in itself, but it reflects imprecise thinking that sometimes leads to problems later in a proof. This error involves using the word if when the word because is really meant. Consider the following proof fragment:
Suppose p is a prime number. If p is prime, then p cannot be written as a product of two smaller positive integers.
The use of the word if in the second sentence is inappropriate. It suggests that the primeness of p is in doubt. But p is known to be prime by the first sentence. It cannot be written as a product of two smaller positive integers because it is prime. Here is a correct version of the fragment:
Suppose p is a prime number. Because p is prime, p cannot be written as a product of two smaller positive integers. ”

Source: Discrete Mathematics with Its Applications, Susanna Epp. p.156-157

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$y = f(x)$ instead of $(x,\space y)\space \in \space f$

$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+ a_0$
$g(x) = b_nx^n + b_{n-1}x^{n-1} + \cdots + b_1x+ b_0$

$(f+g)(x)=y$ instead of $(x, y) \in (f+g)$

$f(x) + g(x) = y$ instead of $(x, y) \in (f+g)$

$(f+g)(x)=(a_n +b_n)x^n + (a_{n-1} + b_{n-1})x^{n-1} + \cdots + (a_1+b_1)x+ (a_0 +b_0)$ instead of $[x,\space (a_n +b_n)x^n + (a_{n-1} + b_{n-1})x^{n-1} + \cdots + (a_1+b_1)x+ (a_0 +b_0)] \in (f+g)$

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Für Luise:

$$(a+b)^2\\=(a+b)(a+b)\\=(a+b)a+(a+b)b\\=(a^2+ab)+(ab+b^2)\\=a^2+2ab+b^2$$

$$(a+b)^3\\=(a+b)^2(a+b)\\=(a^2+2ab+b^2)(a+b)\\=(a^2+2ab+b^2)a+(a^2+2ab+b^2)b\\=(a^3+2a^2b+ab^2)+(a^2b+2ab^2+b^3)\\=a^3+3a^2b+3ab^2+b^3$$

$$(a-b)(a+b)\\=(a-b)a+(a-b)b\\=(a^2-ba)+(ab-b^2)\\=a^2-ab+ab-b^2\\=a^2-b^2$$

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available-------------------------------

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## $$AP = \lambda BP$$

Notationally, $XY$ will represent the length of the segment $\overline{XY}$ and $XY^2$ means $(XY)^2$. $X-Y-Z$ means that point $Y$ is on the segment $\overline{XZ}$. We say that $Y$ is between $X$ and $Z$. This is true if and only if $XY + YZ = XZ$.

An expression like $A-B-C-D-E$ would then indicate how the points $B$, $C$, and $D$ are positioned on the segment $\overline{AE}$.

We need to show that $\Gamma_{AB}(\lambda) = \{ P \in \mathbb R^2: AP = \lambda BP\}$ is a circle.

$\Gamma_{AB}(1) = \left\{ \dfrac 12(A+B) \right\}$

Since $AP = \lambda BP \iff BP = \dfrac 1\lambda AP$, then we can assume that points $A$ and $B$ have been chosen so that $\lambda \gt 1$.

LEMMA $1$: For $\lambda > 1$, there are two points, $P_L$ and $P_R$, on the line $\overleftrightarrow{AB}$ such that $AP = \lambda BP$. They can be described by

• $A-P_L-B-P_R$.
• $AP_L = \dfrac{1}{\lambda+1}AB$.
• $AP_R = \dfrac{1}{\lambda - 1}AB$.

DEFINITION $2$: Let $\Gamma$ be the circle with diameter $\overline{P_LP_R}$.

We will show that $\Gamma_{AB}(\lambda) = \Gamma$.

We will assume that the following theorem is already known.

THEOREM $3$: $P \not \in \{X, Y\}$ is on the circle with diameter $\overline{XY}$ if and only if $\angle XPY$ is a right angle.

Let's say that a solution, $P$, to $AP = \lambda BP$ that isn't on the line $\overleftrightarrow{AB}$ is a non trivial solution.

then $\Gamma_{AB}(\lambda) = \Gamma$ will follow if we can prove the following theorem.

THEOREM $4$: $P$ is a non trivial solution to $PA = \lambda PB$ if and only if $\angle P_LPP_R$ is a right angle.

PROOF. First we show that $AP = \lambda BP \implies \angle P_LPP_R$ is a right angle.

We know that $AP = \lambda BP$ and $AP_L = \lambda BP_L$. Hence $\dfrac{AP}{AP_L} = \dfrac{BP}{BP_L}$. This implies that ray $\overrightarrow{PP_L}$ bisects $\angle APB$. So let $m\angle APP_L = m\angle BPP_L = \theta$.

Let $D$ be the point on $\overline{AP}$ such that $PA::PD = \lambda::1$. Then

\begin{align} \lambda = \dfrac{PA}{PD} = \dfrac{PA}{PB} &\implies PB = PD \\ &\implies \triangle PCB \sim \triangle PCD \\ &\implies m \angle DBP = \theta' = \dfrac{\pi}{2} - \theta \end{align}

Also \begin{align} \lambda = \dfrac{PA}{PD} = \dfrac{P_RA}{P_RB} &\implies \triangle ADB \sim \triangle APP_R &\text{By SAS for similar triangles.}\\ &\implies \overline{BD} \parallel \overline{P_RP} \\ &\implies m\angle BPP_R = m\angle DBP = \theta'\\ &\implies m\angle P_LPB + m\angle BPP_R = \dfrac{\pi}{2} \\ &\implies \angle P_LPP_R \text{ is a right triangle.} \end{align}

Finally, we show that $\angle P_LPP_R$ is a right angle $\implies AP = \lambda BP$.

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