No bluetooth devices recognized | problems with drivers (?) I'm quite new to Ubuntu 20.04 (and Linux) and spent some time fixing my problem with bluetooth not being able to find any devices.  Not sure, but I think this is the closest to my problem. dmesg | grep -i 'blue' [    4.542033] Bluetooth: Core ver 2.22 [    4.542054] Bluetooth: HCI device and connection manager initialized [    4.542060] Bluetooth: HCI socket layer initialized [    4.542074] Bluetooth: L2CAP socket layer initialized [    4.542077] Bluetooth: SCO socket layer initialized [    4.711132] Bluetooth: hci0: BCM: chip id 70 [    4.712132] Bluetooth: hci0: BCM: features 0x06 [    4.728141] Bluetooth: hci0: BCM43142A [    4.728145] Bluetooth: hci0: BCM43142A0 (001.001.011) build 0000 [    4.730896] Bluetooth: hci0: BCM: firmware Patch file not found, tried: [    4.730901] Bluetooth: hci0: BCM: 'brcm/BCM43142A0-105b-e065.hcd' [    4.730902] Bluetooth: hci0: BCM: 'brcm/BCM-105b-e065.hcd' [    6.753354] Bluetooth: hci0: command 0x1003 tx timeout [    6.754147] Bluetooth: hci0: unexpected event for opcode 0x1003 [    7.707035] Bluetooth: BNEP (Ethernet Emulation) ver 1.3 [    7.707037] Bluetooth: BNEP filters: protocol multicast [    7.707042] Bluetooth: BNEP socket layer initialized [   21.550716] Bluetooth: RFCOMM TTY layer initialized [   21.550722] Bluetooth: RFCOMM socket layer initialized [   21.550729] Bluetooth: RFCOMM ver 1.11
Bluetooth Broadcom 43142 isn't working Bluetooth is on but can't find or be found by any other devices.  Output of sudo service bluetooth status:  ● bluetooth.service - Bluetooth service    Loaded: loaded (/lib/systemd/system/bluetooth.service; enabled; vendor preset: enabled)    Active: active (running) since ხუთ 2015-06-04 22:33:18 GET; 13min ago  Main PID: 26678 (bluetoothd)    CGroup: /system.slice/bluetooth.service            └─26678 /usr/sbin/bluetoothd -n  ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled   Output of lsusb:  Bus 003 Device 005: ID 0a5c:21d7 Broadcom Corp. BCM43142 Bluetooth 4.0   How can I fix it?
How to scale a tikzpicture to \textwidth Error on \usepackage {geometry} text property The Bash man page describes use of ${!a} to return the contents of the variable whose name is the contents of a (a level of indirection).  I'd like to know how to return all elements in an array using this, i.e.,  a=(one two three) echo ${a[*]}   returns  one two three   I would like for:  b=a echo ${!b[*]}   to return the same.  Unfortunately, it doesn't, but returns 0 instead.  Update  Given the replies, I now realise that my example was too simple, since of course, something like:  b=("${a[@]}")   Will achieve exactly what I said I needed.  So, here's what I was trying to do:  LIST_lys=(lys1 lys2) LIST_diaspar=(diaspar1 diaspar2)  whichone=$1   # 'lys' or 'diaspar'  _LIST=LIST_$whichone LIST=${!_LIST[*]}   Of course, carefully reading the Bash man page shows that this won't work as expected because the last line simply returns the indices of the "array" $_LIST (not an array at all).  In any case, the following should do the job (as pointed out):  LIST=($(eval echo \${$_LIST[*]}))   or ... (the route that I went, eventually):  LIST_lys="lys1 lys2" ... LIST=(${!_LIST})   Assuming, of course, that elements don't contain whitespace. Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$. Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$.  I'm not really sure of how I should proceed to show this, and it's pretty un-intuitive as well.  I've managed to manipulate the RHS into  $$\frac{e^{4\beta}}{\sqrt{\pi}} \int_0^\infty \frac{1}{\sqrt{u}} e^{-(\beta+2u)^2/4u} du$$ which now vaguely resembles a Gaussian. However, substituting $v = \beta + 2u$ in in order to to get something like $e^{-x^2}$ doesn't particularly help. I got $$ \frac{e^{4\beta}}{\sqrt{2\pi}} \int_0^\infty \frac{1}{\sqrt{v - \beta}} e^{-v^2/2(v-\beta)} dv. $$  I'm really unsure of how I should proceed; any help would be appreciated. To my understanding, "Control" can have two meanings in statistics.   Control group: In an experiment, no treatment is given to the member of the control group.  Ex: Placebo vs Drug: You give drugs to one group and not to the other (control),  which is also referred as "controlled experiment".  Control for a variable: Technique of separating out the effect of a particular independent variable. Some of the other names given to this techniques are, " accounting for", "holding constant", "controlling for", some variable.  For example: In a Football viewing study (like or not like),  you may want to take out the effect of gender as we think gender causes bias, that is, male may like it more than female.     So, my question is for point (2). Two questions:  How do you "control"/ "account for" variables, in general. What techniques are used? (In terms of regression, ANOVA framework).   In above example, does choosing male and female randomly constitutes control? That is, is "randomness" one of the techniques for controlling other effects? Windows Power Options Clarification So I was looking through the Windows 7 power options via Control Panel today, and I realized I had a couple of unresolved questions as to the nature of various choices. I was wondering if someone could give me a clear and concise explanation of the difference between the following four selections:   Sleep Mode Hibernate Mode Standby Mode Turn off hard disks after...   For example, regarding option 4 - does this mean that after the elapsed time limit, the computer will go into standby, hibernate, sleep, or just turn off? Below is an image of the options (albeit on a Windows XP system, but I think the same concepts should apply): Use property to show that if $f_n \rightarrow f$ in measure on finite measure set $E$ then $f_n^2 \rightarrow f^2$ in measure on $E$. Consider the following problem:     Let $E$ have finite measure, $\{f_n\} \rightarrow f$ in measure on $E$ and $g$ be a measurable function on $E$ that is finite almost everywhere. Use the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ to prove that $\{f_n^2 \} \rightarrow f^2$ in measure on $E$. Infer from this that if $\{g_n\} \rightarrow g$ in measure on $E$ then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure on $E$.   I'm struggling understand how I can apply the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ in measure to prove that $\{f_n^2 \} \rightarrow f^2$ in measure. I've been able to come up with a proof of the fact that if $\{g_n\} \rightarrow g$ in measure then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure, and this clearly implies the claim that $\{f_n^2 \} \rightarrow f^2$ in measure. Unfortunately my proof doesn't rely on the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ so it doesn't answer the problem since it explicitly asks us to prove it a certain way.  So far my best attempt to utilize the fact is as follows. Let $\eta &gt; 0$ then by triangle inequality and monotonicity of measure  \begin{align*}     \lim_{n\rightarrow \infty}\mu\left\{ \left|f_n^2 - f^2\right| &gt; \eta \right\} &amp;\leq \lim_{n\rightarrow \infty}\mu \left\{\left|f_n - f\right|\cdot \left|f_n\right| + \left|f_n - f\right| \cdot \left|f\right| &gt; \eta \right\} \\     &amp;\leq \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right|\cdot \left|f_n\right| &gt; \eta/2 \right\} + \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| &gt; \eta/2 \right\} \end{align*}  By the fact that $\{f_n \cdot f\} \rightarrow f \cdot f$ in measure \begin{equation*} \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| &gt; \eta/2 \right\} = \mu\left\{\left|f_n \cdot f - f\cdot f\right| &gt; \eta/2 \right\} = 0. \end{equation*} I don't see how to deal with the other limit without first proving the most general statement $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure.   Any help or advice is greatly appreciated. Word for obscuring bad or immoral acts with verbiage I have been in many scenarios where people try to obscure/reduce the intensity of a seemingly bad or immoral act using grammar. E.g. Assuming Thomas killed a dog, you could hear something like:     There's essentially nothing wrong about palliating the miserable life   of a despondent being. If there was any issue here, I'd say it's not   consulting a veterinary doctor first but I've come to understand that some people are just straightforward.   How does one react to the above mumbo jumbo? I think there are similar situations also found in law courts. What can this be referred to as? Grammatification, obscure tactics? Duplicate flag lists "you have already raised this flag" after flagging as "Unclear what you're asking" What is the moment of a joint random variable? Simple question, yet surprisingly difficult to find an answer online.  I know that for a RV $X$, we define the kth moment as $$\int X^k \ d P = \int x^k f(x) \ dx$$ where the equality follows if $p = f \cdot m$, for a density $f$ and Lebesgue measure $m$.  So, what is the kth moment of, say, $(X,Y)$? $\int (X,Y) \ d P$ doesn't really seem like the answer to me.... If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong. Misplaced navigation bar and question title Bounded sequence in a metric space I'm having trouble getting basic APA bibliographies to work. The same thing works with nejm style, but changing the style to apa gives errors:  ! Undefined control sequence. &lt;argument&gt; \mkbibdateapalongextra  {year}{month}{day}\iffieldundef {endyear}{...  Package biblatex Warning: Macro 'name:apa:last-first' undefined. (biblatex)                Using \newbibmacro.  Package biblatex Warning: Macro 'name:apa:first-last' undefined. (biblatex)                Using \newbibmacro.   I'm using texlive 2012, as you can see I'm using biber, yes, I reran biber, and also tried delting the .bbl file and rerunning tools. My source markup is:  \documentclass[12pt]{article}  \usepackage[hidelinks]{hyperref} \usepackage[backend=biber,style=apa]{biblatex}  \addbibresource{BibResource.bib}  \begin{document}  test\parencite{ormrod2004human}  \clearpage  \printbibliography  \end{document} If I live in the US, for example, where current interest rates are practically nil, how can I take advantage of the fact that ? Proving that surjective endomorphisms of Noetherian modules are isomorphisms and a semi-simple and noetherian module is artinian. I am revising for my Rings and Modules exam and am stuck on the following two questions:  $1.$ Let $M$ be a noetherian module and $ \  f : M \rightarrow M \  $  a surjective homomorphism. Show that $f : M \rightarrow M $ is an isomorphism.  $2$. Show that if a semi-simple module is noetherian then it is artinian.  Both these questions seem like they should be fairly straightforward to prove but I cannot seem to solve them. the cone is contractible Solving multiple congruences by CRT = Chinese Remainder Theorem How do I get the solution given by CRT to match another solution, e.g. the least positive? For example say I have X = 1234. I choose mi as 5, 7, 11, 13. This satisfies the simple requirements of . More precisely given in my example k = n = 4, and the product of any k - 1 is smaller then X how come simply computing the remainder of each won't give equations that solve to X = 1234. In the case of the example, x = 4 mod 5 x = 2 mod 7 x = 2 mod 11 x = 12 mod 13  Which resolves to  (won't CRT produce the smallest?) Any hints?  Below is the solution linked to on &quot;Math Celebrity&quot; (cached to avoid link rot). Why won't this work?  I'm trying to make an instance of a class delete itself.  &gt;&gt;&gt; class A():     def kill(self):         del self   &gt;&gt;&gt; a = A() &gt;&gt;&gt; a.kill() &gt;&gt;&gt; a &lt;__main__.A instance at 0x01F23170&gt; So I was taught that:  Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy  If E(photon) = IE, then KE=0 of the electron.   What does this physically/theoretically mean?   My current thoughts/interpretation is that enough energy/force is applied to ionize the electron so it is 'sufficiently far' from the atom, and then I guess it just moves with whatever speed it is moving at with natural kinetic laws, since no more energy/force is being applied....?  Any clarifications would be greatly appreciated, thanks in advance.