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closed as off-topic by Arthur Fischer Sep 18 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.

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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
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(+1) For thinking outside the (sand)box. –  cardinal Jul 18 '12 at 19:40
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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
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To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. –  leo Dec 17 '12 at 18:03
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This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. –  Arthur Fischer Sep 18 at 10:37

17 Answers 17

Attempt at evaluating $\displaystyle \int\dfrac{\cos x}{1+e^x}\mathrm{d}x$ using integration by-parts.

$$ \require{cancel} \begin{align} \text{Now, }\int \frac1{1 + e^x}\ \mathrm dx &= \int \frac{\left(1 + e^x - e^x\right)\ \mathrm dx}{1 + e^x}\\ &= \int\mathrm dx - \int \frac{\cancel{e^x}}{1 + e^x}\frac{\mathrm d\left(1 + e^x\right)}{\cancel{e^x}}\\ &= x - \ln(1 + e^x) \end{align} $$ Hence, applying by-parts, we get $$ \int \frac{\cos x}{1 + e^x}\ \mathrm dx = \cos x\Big(x - \ln(1+e^x)\Big) - \int(-\sin x)\Big(x -\ln(1+e^x)\Big)\ \mathrm dx\\ = x\cos x - \cos x \ln(1 + e^x) + \int x\sin x\ \mathrm dx - \int \sin x \ln(1 + e^x)\ \mathrm dx\\ = \cancel{x\cos x} - \cos x \ln(1 + e^x) \cancel{-x\cos x} + \int \cos x\ \mathrm dx - \int \sin x\ \ln(1+e^x)\ \mathrm dx\\ = \sin x - \cos x \ln(1 + e^x) -\sin x\int\ln(1+e^x)\ \mathrm dx + \int\cos x \left({\small \int \ln(1 + e^x)\ \mathrm dx}\right)\mathrm dx\quad $$

Now, let $$y_1 = \int \ln(1 + e^x)\cdot 1 \ \mathrm dx = x\ln(1 + e^x) - \int \frac{x\cdot e^x\ \mathrm dx}{1 + e^x} = \dots $$

Hence, $$\int \frac{\cos x}{1 + e^x} = (1-y)\sin x - \cos x \ln(1+e^x) + \int y\cos x \ \mathrm dx $$

The rest is trivial and left as an exercise to the reader.

Who am I kidding, it's impossible to use by-parts on this.

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@Committingtoaname , sorry, forgot this was here. Is not going so well, can't find a way of clearly proving an important argument. –  Edvin Orlov Oct 11 at 3:46

$\frac{2t}{1-t^2}=\frac B{A-C}$

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Answer for this question : Evaluating $\displaystyle\int_0^\infty\frac{\sin^4x}{x^4}\mathrm dx$

Applying multiple times integration by parts, we get \begin{align} \int_0^\infty\frac{\sin^4x}{x^4}\mathrm dx&=-\frac{1}{3}\left.\frac{\sin^4x}{x^3}\right|_0^\infty+\frac{4}{3}\int_0^\infty\frac{\sin^3x\cos x}{x^3}\mathrm dx\\[7pt] &=0-\frac{2}{3}\left.\frac{\sin^3x\cos x}{x^2}\right|_0^\infty+\frac{2}{3}\int_0^\infty\frac{3\sin^2x\cos^2 x-\sin^4x}{x^2}\mathrm dx\\[7pt] &=0+\frac{2}{3}\int_0^\infty\left(\frac{\sin^22x}{x^2}-\frac{\sin^2x}{x^2}\right)\mathrm dx\\[7pt] &=\frac{2}{3}\int_0^\infty\frac{\sin^2x}{x^2}\mathrm dx\\[7pt] &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi}{3}}} \end{align} This is the simple and elementary way you can get. Not too messy and confusing, though.

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Definition A: Given $b\in\Bbb R$ we define a map $m\mapsto b^m$ from $\Bbb Z^{\ge 0}$ to $\Bbb R$ recursively by $b^0:=1,b^{m+1}:=b^m\cdot b.$

Observe that this map is increasing for $b>1,$ decreasing for $0<b<1,$ and constant for $b=1.$

Claim 1: For $b>1,$ the map from Definition A is unbounded above.

Proof: Given $m\in\Bbb Z^{>0},$ let $a_m:=b^m-b^{m-1},$ so that each $a_m>0.$ It is readily shown by induction on $m$ that for all $m\in\Bbb Z^{>0},$ we have $a_{m+1}=a_m\cdot b$ and $$b^m=1+\sum_{k=1}^m a_k,$$ so in particular, $$b^m\ge 1+\sum_{k=1}^m a_1=1+ma_1>ma_1.$$ Take any $y\in\Bbb R.$ By the Archimedean Property of Reals, since $a_1>0,$ then there is some $m\in\Bbb Z^{>0}$ such that $ma_1>y,$ and so $b^m>y.$ Since this holds for all $y\in\Bbb R,$ then the map is unbounded above, as desired. $\Box$

Claim 2: Given $m\in\Bbb Z^{\ge0},$ the map $\Bbb R^{>0}\to\Bbb R^{>0}$ given by $b\mapsto b^m$ is non-decreasing, and is increasing for $m\ne 0$.

Proof: It is readily seen that the map $b\mapsto b^0=1$ is nondecreasing, and that the map $b\mapsto b^1=b$ is increasing. Suppose for some $m>0$ that the map $b\mapsto b^m$ is increasing, and take $b,c\in\Bbb R$ with $0<b<c.$ We know that $b^m<c^m$ by inductive hypothesis. Since $0<b$ and $b^m<c^m,$ then $b^{m+1}:=b^m\cdot b<c^m\cdot b.$ Since $b<c$ and $0<c^m,$ then $c^m\cdot b<c^m\cdot c=:c^{m+1},$ and so $b^{m+1}<c^{m+1},$ as desired. $\Box$


Definition B: Given $c\in\Bbb R^{\ge1}$ and $n\in\Bbb Z^{>0},$ we define $c^{\frac1n}:=\sup\{b\in\Bbb R:b^n\le c\}.$ Note in particular that $1^n=1\le c,$ so $\{b\in\Bbb R:b^n\le c\}$ is non-empty, and by Claim 1 is bounded above, so that $c^{\frac1n}$ is well-defined for $n\in\Bbb Z^{\ge 2}$. Furthermore, observing that $b^1=b$ for any $b\in\Bbb R,$ we have $$c^{\frac11}:=\sup\{b\in\Bbb R:b^1\le c\}=\sup\{b\in\Bbb R:b\le c\}=c=c^1,$$ so that Definitions A and B agree, and $c^{\frac1n}$ is well-defined for all $n\in\Bbb Z^{>0}.$

Claim 3: Given $c\in\Bbb R^{\ge 1}$ and $n\in\Bbb Z^{>0},$ $x=c^{\frac1n}$ is a positive real solution to $x^n=c.$ Moreover, it is the unique such solution by Claim 2.

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@cameronbuie How did this go? –  Committing to a challenge Oct 22 at 0:43
    
@Committingtoachallenge: I've not yet finished. Would you like me to tag you once I've done so? –  Cameron Buie Oct 22 at 2:14
    
@CameronBuie Yes :), what is it for? –  Committing to a challenge Oct 22 at 2:16

This answer is free for anyone to use.

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$$\begin{align} I &=\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx\tag{1}\\ &=\frac12\int_0^1 \frac{t}{2(2-t)(1+t) + 3\sqrt{(2-t)(1+t)}}\,\mathrm dt\tag{2}\\ &=\frac13\int_{\sqrt2}^{1/\sqrt2} \frac{u^2-2}{(1+u)^2(u^2+1)}\,\mathrm du\tag{3}\\ &=\frac13\left[\int_{\sqrt2}^{1/\sqrt2} \frac{3 u}{2 \left(u^2+1\right)}\,\mathrm du -\int_{\sqrt2}^{1/\sqrt2} \frac{3}{2 (u+1)}\,\mathrm du -\int_{\sqrt2}^{1/\sqrt2} \frac{1}{2 (u+1)^2}\,\mathrm du\right]\tag{4}\\ &=\frac13\left[\frac{3}{4} \log \left(u^2+1\right)-\frac{3}{2} \log (u+1)+\frac{1}{2 (u+1)}\right]_{\sqrt2}^{1/\sqrt2}\tag{5}\\ &=\frac{1}{3}\Bigg[\frac{3-2 \sqrt{2}}{2}\Bigg]\tag{6}\\ \end{align}$$

$$\large\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx=\frac{1}{6}\left(3-2 \sqrt{2}\right)$$


$\text{Explanations:}$

$(2)$ Substitute $x^2=t\iff2x\,\mathrm dx=\,\mathrm dt$

$(3)$ Using Type $\rm II$ Euler Substitution

$$\small \sqrt{(2-t)(t+1)}=(t+1)u \iff t+1=\frac{3}{u^2+1} \iff t=\frac{2-u^2}{u^2+1} \iff \,\mathrm dt=-\frac{6 u}{\left(u^2+1\right)^2}\,\mathrm du$$

$(4)$ Using Partial Fraction Decomposition

$$\small\frac{2-u^2}{(1+u)^2(u^2+1)}=-\frac{3 u}{2 \left(u^2+1\right)}+\frac{3}{2 (u+1)}+\frac{1}{2 (u+1)^2}$$

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@committingtoaname: no, and I'd actually forgotten this was here. For one, I realized that there were some errors in what is here; moreover, achielle hui's comments in that question both shows the location I was heading and why it's hard to proceed further. So I abandoned this direct route. –  Semiclassical Oct 8 at 2:04

Since $\mathbb{Z} \oplus \mathbb{Z}_p$ is not isomorphic to $\mathbb{Z} \times \mathbb{Z}_{p^2}$ (one has an element of order $p^2$; the other does not), then your reasoning can't be correct. In fact, the pullback $P$ is actually isomorphic to a proper subgroup of the product $\mathbb{Z} \times \mathbb{Z}_{p^2}$. Since the diagram must commute, given $(x,y) \in P \subseteq \mathbb{Z} \times \mathbb{Z}_{p^2}$, we must have $g(x) = f(y)$, i.e., $x \equiv y \pmod{p}$. Can you show that $P = \{(x,y) \in \mathbb{Z} \times \mathbb{Z}_{p^2} : x \equiv y \pmod{p}\}$ and that $P$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_p$?

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We can split the set of bit strings of length $n$ into two groups.

  1. Those has at least $k$ consecutive $1$s.
  2. Those has at most $k-1$ consecutive $1$s.

Let $N_n$ and $M_n$ be the number of bit strings in each group. It is clear the number we want to calculate is

$$N_n = 2^n - M_n$$

Let $X_n$ stand for any bit string in the second group. We have following possibilities:

$$\begin{array}{rcl} X_n &=& 0 X_{n-1} \\ &\text{or}& 10 X_{n-2} \\ &\vdots&\\ &\text{or}& \underbrace{1..1}_{k-1}0 X_{n-k} \end{array}\quad\text{ for } n \ge k $$ This translates to a recurrence relation of $M_n$:

$$M_n = M_{n-1} + \cdots + M_{n-k}\quad\text{ for } n \ge k\tag{*1}$$

Notice for $0 \le n < k$, we have $M_n = 2^n$. We can extend the validity of $(*1)$ to $n \ge 0$ by defining

$$M_{-1} = 1\quad\text{ and }\quad M_{-2} = \cdots M_{-k} = 0$$

If one compare this to the recurrence relation satisfied by

We find

$$M_n = F^{(k)}_{n+2} \quad\implies\quad N_n = 2^n - F^{(k)}_{n+2}$$

where $F^{(k)}$ is the $k$-step Fibonacci sequence.

For examples,

$$\begin{array}{rcl} (M_n)_{k=2} &=& (\color{red}{1,2,3,5,8,13,21,34,55,89,144},\ldots)\\ &\text{ vs }&\\ \text{Fibonacci numbers } (F_n) & = & (0, 1,\color{red}{1,2,3,5,8,13,21,34,55,89,144},\ldots)\\ \\ (M_n)_{k=3} &=& (\color{red}{1,2,4,7,13,24,44,81,149,274,504},\ldots)\\ &\text{ vs }&\\ \text{Tribonacci numbers } (T_n) & = & ( 0, 1, \color{red}{1, 2, 4, 7, 13, 24, 44, 81, 149,274,504},\ldots)\\ \\ (M_n)_{k=4} &=& (\color{red}{1,2,4,8,15,29,56,108,208,401,773},\ldots)\\ &\text{ vs }\\ \text{Tetranacci numbers } (T_n) & = & ( 0, 1, \color{red}{1, 2, 4, 8, 15, 29, 56, 108, 208, 401, 773},\ldots) \end{array} $$

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As an example we take the detailed exact cover sample from wikipedia

$$U=\{1,2,3,4,5,6,7\}$$

$$U=\{1,2,3,4,5,6,7\} \\ \begin{eqnarray} S_1 &=& \{1, 4, 7\} \\ S_2 &=& \{1, 4\} \\ S_3 &=& \{4, 5, 7\} \\ S_4 &=& \{3, 5, 6\} \\ S_5 &=& \{2, 3, 6, 7\} \\ S_6 &=& \{2, 7\} \end{eqnarray} $$

We have to calculate the numbers $a_j$ assigned to every $S_j$.

So first we calculate $a_1$ that is assigned to $S_1$. We write $N_d$ for a number represented in base $d$. So $$26=26_{10}=11010_2=222_3=1A_{1}$$

We use the base $d=7$, because the base must be (at least) one greater then the number of sets of the cover. To find the number assigned to $S_1$ we write the elements of U in descending order : $$\begin{array}\\ 7&6&5&4&3&2&1 \end{array}$$. Now we write 1 $1$ under each element that is contained in the set $S_1$ and get $$\begin{array}\\ 7&6&5&4&3&2&1 \\ 1& & &1& & &1 \end{array}$$ Under the elements of U that are not in $S_i$ we write a $0$ and get

$$\begin{array}\\ 7&6&5&4&3&2&1 \\ 1&0&0&1&0&0&1 \end{array}$$

So the assigned number is $$1001001_7=117993$$

The remaining numbers are calculated in a similar way and we get

$$\begin{array}{lcrcr} \\ a_1&=&1001001_7 &=& 11 7993\\ a_2&=& 1001_7 &=& 344\\ a_3&=&1011000_7 &=& 12 0393\\ a_4&=& 110100_7 &=& 1 9257\\ a_5&=&1100110_7 &=& 13 4512\\ a_6&=&1000010_7 &=& 11 7656\\ \end{array}$$ and further $$ b=1111111_7=13 7257 $$

To solve the exact cover problem we have to solve the 0-1-knapsack problem

$$117993 \, x_1 +344 \, x_2 +120393 \, x_3 +19257 \, x_4 +134512 \, x_5 +117656 \, x_6 =13 7257$$

The only solution for this knapsack is $$344+19257+117656=137257$$ so $$a_2+a_4+a_6=b$$

or, written in base $7$

$$\begin{array}{rr}\\ & 1001_7 \\ +& 110100_7 \\ +&1000010_7 \\ \hline{} &1111111_7 \end{array}$$

(note that for each place there is exactly one row with a $1$ on this place)

This means that $\{S2,S4,S6\}$ is an exact cover, which complies with our result.

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