It happens that, for any $m\geq 1$, $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}\tag{1}$$ where $E_{2m}$ is an integer number.     My proof works through the following lines: the LHS is: $$\frac{1}{(2m)!}\int_{0}^{1}\frac{(\log x)^{2m}}{1+x^2}dx = \frac{1}{2\cdot(2m)!}\int_{0}^{+\infty}\frac{(\log x)^{2m}}{1+x^2}dx,$$ so we just need to compute: $$\left.\frac{d^{2m}}{dk^{2m}}\int_{0}^{+\infty}\frac{x^k}{1+x^2}\right|_{k=0},\tag{2}$$ but: $$ \int_{0}^{+\infty}\frac{x^{1/r}}{1+x^2}\,dx = r\int_{0}^{+\infty}\frac{y^r}{1+y^{2r}}\,dy = \frac{\pi/2}{\cos(\pi/(2r))}$$ by the residue theorem, so $E_{2m}$ is just the absolute value of an Euler number, that belongs to $\mathbb{N}$.    Can someone provide a real-analytic proof of $(1)$?
I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$ so maybe I could simplify the above more?
I got my dad a Logitech LS21 system for Christmas. It's a 2.1 system, but it's recognized in Windows 8 as a 2.0 system (it says there is no subwoofer connected).    How would I go about getting the sub recognized?  It plays some sound now, but it's so quiet you can't hear it without putting your ear right next to it. My linear algebra professor proved that      Diagonalizable matrices with complex values are dense in set of $n \times n$ complex matrices.   He defined a metric (I believe) that was somehow related to the usual metric on $\mathbb{R}^{n^2}$.  Then somehow proved that diagonalizable matrices were dense because for any matrix $A$ if $\det(A - \lambda I) = 0$ on an open subset, then $\det(A - \lambda I)$ was the zero polynomial.  I Googled around a bit and found some stuff talking about the Zariski topology, and I am not sure this is what I want. Most of what I have found on this topology is much more general than what he was doing.  Does anyone know any resources that would explain how to go about proving this statement?  Edit: I found out how to prove this the way my professor did.  We define the norm of a matrix by $$ |A| = \max\{|Ax| : |x| = 1 \}.$$ Now we have a distance $d(A, B) &lt; \epsilon$. You can prove that if $(A - B)_{ij} &lt; \epsilon/n$, then $|A - B| &lt; \epsilon$.  Let $p(x)$ be the characteristic polynomial of $A$, an $n \times n$ matrix.  $$p(x) = \prod_1^n (x - x_i) = x^n + (-1)\sigma_1 x^{n - 1} + \cdots + (-1)^n\sigma_n$$ where $\sigma_1, \ldots, \sigma_n$ are the elementary symmetric polynomials of $x_1, \ldots, x_n$ which are the eigenvalues.  The discriminant of the characteristic polynomial is a symmetric polynomial, therefore it can be written in terms of the elementary symmetric polynomials, which in turn can be written in terms of the entries of the matrix.  But since the discriminant is a polynomial, it only has finitely many roots. Therefore for $\epsilon &gt; 0$, by changing the entries of the matrix less than $\epsilon / n$ we can find a new matrix $B$ such that $|B - A| &lt; \epsilon$ and the discriminant is not zero.   The discriminant being not zero means $B$ has distinct eigenvalues, thus has a basis of eigenvectors. Therefore, it is diagonalizable.  Thus the set of diagonalizable matrices is dense in the set of matrices with respect to that metric. I'm trying to configure a ubuntu server network. [Edit] My ifconfig: lo: flags=73&lt;UP,LOOPBACK,RUNNING&gt; mtu 65536 inet 127.0.0.1 netmask 255.0.0.0 inet6 ::1 prefixlen 128 scopeid 0x10&lt;host&gt; loop txqueuelen 1000 (Loopback Local) RX packets 352 bytes 26160 (26.1KB) RX erors 0 dropped 0 overruns 0 frame 0 TX packets 352 bytes 26160 (26.1KB) TX errors 0 dropped 0 overruns 0 carrier 0 collisions 0  I edited the /etc/network/interfaces file and here is how it is: auto lo auto eth0 iface eth0 inet static address 192.168.15.141 netmask 255.255.255.0 gateway 192.168.15.1  I'm trying to restart the server and I'm using: sudo ifdown --exclude=lo -a &amp;&amp; sudo ifup --exclude=lo -a  This command is an answer from this link:  And I'm getting the following error: Cannot find device &quot;eth0&quot; Failed to bring up eth0.  I'm pretty lay in configuring a server, so I have no idea what's going on. Can someone help me please? Writing a git post-receive hook to deal with a specific branch Using the properties of real numbers, verify commutativity on complex numbers. I have just started self-studying Linear Algebra after a 3 year hiatus from mathematics (in an extended neuroscience foray), and I want to ensure that I am sufficiently verifying commutativity of complex numbers by using the properties of real numbers.  My attempts...  (1) Commutativity of Addition:   Verify that $w + z = z + w$ for all $w, z \in \mathbb{C}$.  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{16mm} = (a + c) + (bi + di) \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{19mm} = (c + a) + (di + bi) \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{13mm} = c + (a + di) + bi \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = c + (di + a) + bi \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{17mm} = (c + di) + (a + bi) \hspace{11mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{13mm} = z + w \hspace{34mm}\text{by definition of }\mathbb{C}$$  $$\blacksquare$$  This seems a bit verbose to me - is it acceptable to combine steps? For example, rather than:  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ can I instead write: $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{48mm} = a + (c + bi) + di \hspace{10mm}\text{by associativity and commutativity in }\mathbb{R}$$?  (2) Commutativity of Multiplication:  Verify that $wz = zw$ for all $w, z \in \mathbb{C}$.  $$\hspace{-14mm}wz = (a + bi)(c + di) \hspace{24mm}\text{by definition of }\mathbb{C}$$ $$\hspace{4mm} = a \cdot c + a \cdot di + bi \cdot c + bi \cdot di \hspace{10mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{9mm} = a \cdot c + bi \cdot c + a \cdot di + bi \cdot di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{35mm} = c \cdot a + c \cdot bi + di \cdot a + di \cdot bi \hspace{10mm}\text{by multiplicative commutativity in }\mathbb{R}$$ $$\hspace{1mm} = c(a + bi) + di(a + bi) \hspace{20mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{5mm} = (c + di)(a + bi)\hspace{34mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{4mm} = zw \hspace{55mm}\text{by distribution in }\mathbb{R}$$ $$\blacksquare$$  This also seems a bit wonky to me. Any and all suggestions on clearing up my verification, reasoning, etc. would be greatly appreciated! Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice? Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?  What would it imply if there was such an example?  EDIT: For instance, maybe this will help understand the kind of example I had in mind, it is known that an important feature to determine that the AC is needed in the Banach-Tarski case is the non-transitivity of rotations on Euclidean space, one might find an example of a transitive group for some given space keeping the rest equal and make the AC unnecessary?  I guess then it might be said that this transitivity will be in this particular case equivalent to the AC or some amount of it, but I guess that it would be important to show this if it hadn't been realized before. I'm trying to install Wifite from , but I can't install cowpatty.  See below:  root@vitor-hp:~# apt-get install cowpatty Reading package lists ... ready Building dependency tree Reading state information ... ready E: Could not find package cowpatty root@vitor-hp:~#    Translated from Portuguese  I use Ubuntu 13.10 64-bit. A question about E12 series resistors I'm trying to calculate E12 resistor series myself.  Above is an illustration of the idea where such equation can be written:  x = r3 + r3 * %tol = r4 - r4 * %tol  form that we can derive:  r4 = r3* [(1 + %tol)/(1 - %tol)]  hence in generalized form we can write (%tol = 0.1 for E12 series);   r(i+1) = ri* [(1 + %tol)/(1 - %tol)]  But when I iterate the above formula in a for loop I obtain the following results for the first resistor values:  1.0000    1.2222    1.4938    1.8258    2.2315    2.7274    3.3335    4.0743    4.9797    6.0863    7.4388  Then I take [(1 + %tol)/(1 - %tol)] = 1.2 rounded value instead of 1.222222 and write the formula as:  r(i+1) = r(i)*1.2  But again when I iterate the above formula in a for loop I obtain the following results for the first resistor values:  1.0000    1.2000    1.4400    1.7280    2.0736    2.4883    2.9860    3.5832    4.2998    5.1598    6.1917  But the real values for E12 should be as:  1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8  What am I doing wrong here? A surjective endomorphism (of a Noetherian ring) is injective. The problem is stated as follows:  "Let $R$ be a Noetherian ring and $\theta$ be a ring homomorphism from $R$ to $R$. Show that if $\theta$ is surjective then it is also injective."  Regardless of the right solution, I don't understand why is the following wrong:     We have $\theta: R\to R$. By the isomorphism theorem $R/\ker\theta\cong\operatorname{Im}\theta$. Since $\operatorname{Im}\theta = R$, it follows $\ker\theta =0$, so it's injective.   Harsh criticism will be appreciated. Thanks. Most efficient way to append arrays in C#? I am pulling data out of an old-school ActiveX in the form of arrays of doubles. I don't initially know the final number of samples I will actually retrieve.  What is the most efficient way to concatenate these arrays together in C# as I pull them  out of the system? Mach-O 64-bit format does not support 32-bit absolute addresses. NASM Accessing Array How does one prove that every finite group is isomorphic to a subgroup of an alternating group? What can you say about the range space and null space Let $V$ be a vector space over a field $F$ and $T$ a linear operator on $V$. If $T^2$$=$ $0$, what can you say about the relation of the range of $T$ to the null space of $T$? Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$   My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this? How to pass anonymous types as parameters? Will reformatting my hard drive fix bad sectors on it? Are +1/+1 counters different from +1/+1 effects? What does the comma operator , do? I would like to be able to write a Java class in one package which can access non-public methods of a class in another package without having to make it a subclass of the other class. Is this possible? Calibre stopped working recently. Tried uninstalling and reinstalling. When I open a terminal and call $calibre I get the following error message: Traceback (most recent call last):   File &quot;/usr/bin/calibre&quot;, line 20, in &lt;module&gt;     sys.exit(calibre())   File &quot;/usr/lib/calibre/calibre/gui_launch.py&quot;, line 73, in calibre     main(args)   File &quot;/usr/lib/calibre/calibre/gui2/main.py&quot;, line 543, in main     listener = create_listener()   File &quot;/usr/lib/calibre/calibre/gui2/main.py&quot;, line 514, in create_listener     return Listener(address=gui_socket_address())   File &quot;/usr/lib/calibre/calibre/utils/ipc/server.py&quot;, line 110, in __init__     self._listener._unlink.cancel() AttributeError: 'NoneType' object has no attribute 'cancel'  Do I need to install some dependency that the sudo apt install didn't get? I tried installing via command line and the software installer/updater GUI. I'm on version Ubuntu 20.04.1 LTS. I made sure that my software in general is up to date also.