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# How To Start Qemu On Windows

exe has stopped working. Did you have any trouble with the Raspberry Pi emulation? If any of the above didn't work as described, and you were unable to create a Raspberry Pi emulation, leave us a comment. Then the system can boot straight into Windows or start FreeBSD where Windows can be started in a virtual machine. Using QEMU to quickly test an ISO or bootable USB drive When you want to quickly boot from a drive or image in a virtual machine you do not need to setup up a VirtualBox machine. The GNU MCU Eclipse QEMU is a fork of the public open-source QEMU project, customised for more support of Cortex-M cores, and a better integration with the GNU MCU QEMU Debugging plug-in. qemu -boot c -cdrom /dev/cdrom -hda hd. Just install a modern version of QEMU from the repos and start it with the argument -spice and any of the arguments discussed later on the page. Qemu Puppy Linux on Windows puppy. Optional - If using the AC97 sound device. Debugging gPXE with QEMU The QEMU monitor ( CTRL+ALT+2 ) supports debug commands to inspect registers and memory (try help or tab complete). After successful FreeBSD installation, we need the way to connect your Windows to your Qemu. In our qemu-screamer directory, we will use qemu-img to create the disk image. In the VM (referred to as "guest machine") : In the boot options choose to boot from the CD or See installing plan 9 on qemu. Restore VHD Image to Disk/Partition with Linux dd Convert a VHD image from a native Windows backup to raw format using qemu-img, and write it directly to a disk or partition with the Linux dd command I've recently been evaluating native Windows Server Backup as an option for bare-metal backup and recovery for our remaining physical servers at. Normally, QEMU (/usr/libexec/qemu-kvm) uses SDL to display the VGA output. img' with the name of your disc image. 50) under Windows 10 (64-bit) (basically following these instructions) with these commands:. sudo apt-get install qemu-kvm libvirt-bin bridge-utils virt-manager. The open-source routers we will use in a GNS3 simulated network must run on either a Qemu or a VirtualBox hypervisor. After the process finish you should test the Windows VM to make sure it is operating normally. 1) on Win7 Host, qemu vs VirtualBox I hope friends you've read a bit about GNS3 and know how it works. AS supported machine list. Unmount the VHDX using Disk Manager. For example, freezing and thawing filesystems, entering suspend. First boot is going off now. img -fda m078. So, we can install it with front-end tool Virt Manager using command: sudo apt-get install qemu-kvm qemu virt-manager virt-viewer libvirt-bin. Get unlimited access to the best stories on Medium — and support writers while you. Simulate Your Image Using QEMU: Once this particular image is built, you can start QEMU and run the image: $runqemu qemux86 If you want to learn more about running QEMU, see the " Using the Quick EMUlator (QEMU) " chapter in the Yocto Project Development Tasks Manual. will appreciate if somone can give me some hints or tips on how to get the whole thing set up. Setting Up QEMU on the PCs What you need to do to set up qemu on a PC so that you and your project partner can start working. This can also be a URL to an IMG or QCOW2 file, in which case QEMU will boot directly from it. From the Windows guest (win7x86, for example) install the qemu-ga-x64. 90) on Windows (XP) and a couple of things cropped up: * with --nographic, it complains it can't open monitor device 'stdio' and doesn't start. Thoughts The packet capture in GNS3 is such a great learning tool that allows you to see at a very low level the packets coming across the simulated wire as they would in a real production network. It makes sense. By using the QEMU emulator, you can run FreeDOS inside a window on a host operating system. Being able to run Debian in a virtual machine is a great way to play around with it without having to worry about affecting your native system. Just install a modern version of QEMU from the repos and start it with the argument -spice and any of the arguments discussed later on the page. which will boot the virtual machine on your newly created windows image: windowsxp. Unmount the VHDX using Disk Manager. But the new menu offers a few surprises with its dual-personality of part Start menu and part Start screen. Install KVM (QEMU) on CentOS 7 – Virt Manager – VM Console Manage Virtual Machine. QEMU needs a lot of other stuff to run, I would not recommend it to someone who start with virtualization for the first time. qemu -nographic binary. sh script, when our Windows friend would simply need to double-click the. This method uses the portable tool xpm, the xPack Package Manager, and can be used on Windows, macOS and GNU/Linux. I always say that Unix is the generic plumbing of information technology, and Windows is actually among the last remaining major platforms that remains proprietary hasn't fallen into line. Installing QEMU in Windows. While writing this comment I Just mounted a NFS share and started a 1. The Idiot installs the QEMU agent on a Windows 10 VM on Proxmox So yeah, this one's on me. From novice to tech pro — start learning today. For the builds hosted on this page, I employ Arch Linux and I use the docker image build method described here. Get started today Stand Out as the employee with proven skills. iso -boot d -net nic,model=rtl8139 -net tap,ifname=TAP_OpenVPN -soundhw sb16 -localtime" to start plan9, this command can be inserted into a batch file. So my Problem was, that Windows 7 get's stucked, with the bootup screen saying Windows is starting and the Windows 7 Logo, but i recognized the desktop sound, when Windows arrives @Loginscreen, so i don't know if that's just an driver issue within the graphicsdriver or another one. virsh start hostname Step 4 – Extend the partition in Window Windows has a really good partition management utility built into it. Here is a quick-start guide to setting up ARM QEMU emulation under libvirt management: Installing and starting the virtualization software. To resize a VMDK using qemu-img # First convert the vmdk to raw format # NB.$ qemu -m 1G -drive cache=writeback,file=hd0. If the winxp. This Enables the user to have both systems running at the same time eliminating the need to restart the PC and set your BIOS options to boot Linux from USB. img # same as above qemu -nographic -curses binary. KVM, QEMU start or stop virtual machine from command line (CLI) Written by Guillermo Garron Date: 2013-02-11 16:59:13 00:00. qemu:///session has a serious drawback: since the libvirtd instance does not have sufficient privileges, the only out of the box network option is qemu's usermode networking, which has nonobvious limitations, so its usage is discouraged. start QEMU Guest Agent service on Local Computer. However, the image can not boot successfully. I thought I'd take a stab at it, and didn't have much luck. For what it's worth, I find that Qemu 0. This same manager can be used to monitor the status of the service. img file(s) into the window, and click "Start QEMU Test". I've tried Windows 3. This URL can be either an HTTP URL or a file URL (or path to a file). 20 and later. GitHub Gist: instantly share code, notes, and snippets. As soon as you run the above command, the guest Linux OS will start booting and can be tested in the live environment but for its installation, we need to attach virtual hard disk. How to run Ubuntu in qemu on Windows XP media centre? We need you to answer this question! If you know the answer to this question, please register to join our limited beta program and start the. Start with the Installing MS-DOS on Qemu tutorial, then come back here. The QEMU setup on the source and destination hosts have to be identical. exe in this thread; How TO install Windows 98 with Qemu Hardy/ Gusty x86/ AMD64 Here's one from 2015, yes it's a raspberry pi thing but it runs linux so bascially same thing as the second post mentions using qemu. The "QEMU Guest Agent" service cannot be started on Windows 10 RS2 preview build. Open it again via cmd using the “start. img' with the name of your disc image. You do that by: Make sure the window has the focus (label should say QEMU - Press Ctrl-Alt to exit grab ). The qemu-guest-agent is a helper daemon, which is installed in the guest. The QEMU addon in IPFire is specially adapted for this system. org QEMU-Buch / QEMU-Book - a quite versatile book on qemu, provided in German and English (partly as a google translation), its describing e. UPDATE: as of qemu 4. In fact, most cloud providers use KVM as their Hypervisor of choice. QEMU is a hosted virtual machine monitor: It emulates CPUs through dynamic binary translation and provides a set of device models, enabling it to run a variety of unmodified guest operating systems. 10 0 Virtualization Thursday, October 08, 2015 In this tutorial we are going to see how to install the KVM virtualization tool on Ubuntu or Linux Mint. If you plan to send characters back via netconsole or you want to stop and start qemu a lot of times, you should have qemu use the same source port each time by using something like -serial udp::[email protected]:4556to qemu. qemu-windows-10. Start with the Installing MS-DOS on Qemu tutorial, then come back here. However, the image can not boot successfully. If the number is smaller, close some applications or start your virtual machine with less memory (number_of_pages x 2): $qemu-system-x86_64 -enable-kvm -m 1024 -mem-path /dev/hugepages -hda [] Note the -mem-path parameter. 1 (find it at Macintoshrepository. qemu-system-x86_64 -m 1G,slots=3,maxmem=4G Do not use decorations for SDL windows and start them using the whole available screen space. QEMU is available in the default repositories of Ubuntu. Download an ISO image of the Linux distribution you want to run. It works on Linux, Windows, FreeBSD and Mac OS X. enter command "change cdrom isofilename. This way you can easily test a new operating system or try a Live CD on your system without any troubles and dangers. The next thing is that you have to install Windows XP on blank. Fall 2011 CS380L Setting up QEMU. Regardless of the OS, we specify which file is our kernel, which is our initramfs, what to use as the primary hard drive, and finally, we pass some additional boot options to. From the Windows guest (win7x86, for example) install the qemu-ga-x64. qemu -nographic binary. virsh start hostname. If the winxp. It allows you to create and start hardware-accelerated virtual machines under Linux using the QEMU tools. Do so, mapping it to drive Z, and click OK. If you still want to use it on windows and then you can use routers or virtual machines as hosts, instead of qemu. There are 2 ways that QEMU can be compiled for Windows from source: Cross-compiling using MinGW-W64 on a Linux distribution or native MinGW-W64 through MSYS2. Modern builds of QEMU (since 1. The documentation says Ctrl-Alt-[1, 2, 3] switch between the display modes, but it doesn't work for me. Don’t believe the hype! It is entirely possible to mount a VirtualBox VDI image, just like a loopback filesystem… all you need are the right tools and know-how. iso -net nic,model=rtl8139 -net user Now at his point do the regular install using hd0 as your harddrive. How to install ReactOS into QEMU on Windows. Creating and installing windows 10 image using QEMU March 9, 2016 February 7, 2018 Today we are going to create an empty image using Qemu and then we are going to install Windows 10 on it. Start Minix over QEMU; Just double click on the "Minix over QEMU" shortcut on your Desktop or Start Menu. tgz) installed inside of a Qemu virtual machine with FreeBSD running inside, all under Win 7 64bit OS environments. 1 with OVFM boot loader: qemu-system-x86_64 -nographic -vga none \. 5 Requirements * Qemu program. 10 0 Virtualization Thursday, October 08, 2015 In this tutorial we are going to see how to install the KVM virtualization tool on Ubuntu or Linux Mint. QEMU Code Overview Architecture & internals tour Build and start contributing Written in C, runs on POSIX and Windows hosts Code at qemu-project. For the builds hosted on this page, I employ Arch Linux and I use the docker image build method described here. In order to start kqemu, you must do: net start kqemu Usage. This run-time must be 'installer-free' (err. lots of guests and hosts. Qemu is a generic and open source virtual machine emulator. The main thing you need is VT-d support on your processor. The next thing is that you have to install Windows XP on blank. 1,95,98,98SE and could install all of those without problems, I've not tried ME. Compatibility with this software may vary, but will generally run fine under Microsoft Windows 10, Windows 8, Windows 8. to run a virtual machine). It is strange to use an operating system with no web browser, but it does have Paint and Writer as well as the venerable DOS prompt available whenever you need it. However the PCI support from the first version of Windows 95 is. img' with the name of your disc image. cp hostname. When QEMU starts you can use ctrl+alt+f to toggle fullscreen and ctrl+alt to exit mouse grab. img -cdrom debian-7. Downloading, building and booting Linux is explained elsewhere. 0 can fix qemu-system-x86_64. Even networking should work out of box and thus you can use it to run Internet Explorer (for web site testing etc). Qemu-system-ppc can run PPC Mac OS 9. Only one program can benefits of the hardware acceleration. The documentation says Ctrl-Alt-[1, 2, 3] switch between the display modes, but it doesn't work for me. Get unlimited access to the best stories on Medium — and support writers while you. How to use QEMU QEMU is a very flexible virtualization technology however it is quite slow and it is recommended that you understand and evaluate alternative solutions before picking this one. Refer to Getting started with virtualization. Install KVM (QEMU) on CentOS 7 – Virt Manager. Additionally you are not allowed the partition the Hard Disk. If you plan to send characters back via netconsole or you want to stop and start QEMU a lot of times, you should have QEMU use the same source port each time by using something like "-serial udp::[email protected]" to QEMU. If the installation went fine, you can now start your windows system by typing: qemu -hda windowsxp. I create a project like official tutorial GNU MCU Eclipse. All you need is Qemu. Your issues matter to us. install qemu. Just to make it clearer, though, I added (Arch + QEMU) to the video title. However, when trying to run it on a MSP430 QEMU instance I got:$. 90 is the best overall version of Qemu. Such a waste if we snap a preconfigured servers with more than 4GB RAM. Might need to know your distro choice. 1 with OVFM boot loader: qemu-system-x86_64 -nographic -vga none \. It is the recommended operating system for Raspberry Pi. Yes even the hokey Windows 95a. FreeDOS is mostly compatible with MS-DOS, so you will be able to use FreeDOS and QEMU to run some old MS-DOS programs. Increasing the Size of an Existing Disk Image. I tried for last couple of hours to start the Raspbian in QEMU on Windows. On windows, if you run 1 qemu host, it works. I no longer have the cycles to work on this project. My apologies, that was the wrong illustration. It is the recommended operating system for Raspberry Pi. Posted in Android Tutorial Tagged Android Error. I have some rpm for aix 4. How do I check my KVM version on Linux operating system using the command line? qemu-kvm is an open source and free virtualizer. Yes even the hokey Windows 95a. QtEmu is a graphical user interface for QEMU written in Qt4. On this post, however, I w. Jump to Make sure network service is set to start automatically: SDL-devel libX11-devel gtk2-devel bridge-utils PyXML qemu. Get started today Stand Out as the employee with proven skills. I've downloaded the latest 64-bit version of QEMU for Windows and it installed fine. This is often needed since Windows is quite hungry for memory space. Once Windows is installed you should start it using the following flags. iso Qemu is being run with KVM and non efi bios Specific Problems. cp hostname. Follow the standard audio guide on the Arch Wiki instead. In the first line, the Ethernet device is disabled. After making that change, startup Windows XP again, and when Windows loads, it will reinstall the Viostor driver for that hard drive. When QEMU is compiled with KQEMU support, the following option is added to QEMU: `-no-kqemu' Disable the usage of the QEMU Accelerator module (KQEMU). So my Problem was, that Windows 7 get's stucked, with the bootup screen saying Windows is starting and the Windows 7 Logo, but i recognized the desktop sound, when Windows arrives @Loginscreen, so i don't know if that's just an driver issue within the graphicsdriver or another one. Start with the Installing MS-DOS on Qemu tutorial, then come back here. QEMU Project; This program is provided as is without any warranty. It is the recommended operating system for Raspberry Pi. Next you will want to start up QEMU and begin the installation process. It can also be used to provide virtual hosting of several virtual PCs on a single server. However, the image can not boot successfully. Sometimes you might want to try something "virtually" before doing it on a real Raspberry Pi. Running multiple QEMU VMs of the same sample ¶ If you find yourself wanting to run multiple instances of the same Zephyr sample application, which do not need to talk to each other, use the QEMU_INSTANCE argument. All you need is Qemu. What Qemu GUI works best for you? Pretty simple to use a command line but I use qemu manager in windows. Here's the command for that: qemu -cdrom cdimage. Raspbian and QEMU for Windows. 8GB database and tried some queries. This run-time must be 'installer-free' (err. I want to start my Debian install from within Windows through Qemu and the Hax hypervisor. QEMU and Linux on a USB Device. BUT when I run the GNS3 WorkBench image on a ESXi server, It’s not possible to connect any GNS3 router to the local lan…. In Windows: -Make sure you have some folder shared -Bridge your tap device with your default network interface (see above)-Start Mac OS 9 through Qemu with tap networking -Use DHCP on the Mac side. Click here to visit our frequently asked questions about HTML5 video. The command to start the emulated machine then become:. Now you should have Windows up and running. Qemu is a generic and open source virtual machine emulator. QEMU disk image utility for Windows. 6th step: Start GNS3, Edit -> Preferences, and select the Qemu sub-window. cp hostname. Testing code on QEMU An Anonymous poster asked that I write up a brief tutorial on how to get your ARM code running in QEMU to see how it runs without having to have hardware. I have 2 GPU's in my system a GTX 1080 and a rx 550. I have some rpm for aix 4. Guide to running Windows 7 in QEMU I was trying to get this working earlier and found it a bit confusing, so here's a little guide that might come in handy for people who are forced to use Windoze for work or school. ovl installation is ruined, you can delete it, and start Qemu using the winxp. Using QEMU to quickly test an ISO or bootable USB drive When you want to quickly boot from a drive or image in a virtual machine you do not need to setup up a VirtualBox machine. 1,95,98,98SE and could install all of those without problems, I've not tried ME. 1) Download the Windows version of QEmu, and the latest version of the Ubuntu LTS (Long term stable) iso installer image. Current plan is to modify the Qemu GA MSI installer, to disable the attempt to start the service following installation, if the VirtIO-Serial is not installed. In this post, I’m going to introduce a useful technique to people who are using, or are interested in using, QEMU on Windows. Qemu Launcher utilizes the full system emulation mode of QEMU that allows you to run unmodified operating system on virtual hardware. © 2017 Picoaccel. Am I mistaken, or are you running Windows XP on there somewhere? Ummm, perhaps that's a contributor to your system instability? Anyway, QEMU is hardly an entry-level virtualization software. How To Run Debian In A QEMU KVM Virtual Machine Using Ubuntu or Debian A virtual machine is a piece of software that emulates a real operating system. You can start with examples from the Web, or use mine at the end of this document. qemu-nbd is an NBD server which can handle any disk format supported by qemu (eg. Open it again via cmd using the "start. When searching the web pages to find a way to be able to. The next thing I did was to connect a VGA cable to the VGA port of the graphics card I passed through to the virtual machine. Hardware Requirements. I am running Qemu Manager off of a USB stick, and have no administrative privileges. Logging In You and your partner will be assigned one of the PCs in the CS Lab on which you will store your virtual qemu machine and kernel source in a subdirectory in /local that only you and your partner can access. Hi Stuart, I have problem with Windows nodes in unetlab. Recently we wanted to start a Windows virtual machine from a physical hard disk using a Fedora w/ GNOME 3 host machine to change the domain password of a user. exe The most I have done was to enter bash by init=/bin/bash and boot the Raspbian nearly to the shell, but didn't managed to boot it as far as it would on original RPi. When using the VNC display, you must use the -k parameter to set the keyboard layout if you are not using en-us. It is the recommended operating system for Raspberry Pi. exe - Install grub4dos and install to the MBR and the PBR (do not overwrite the grldr file on the E2B drive). Start Qemu Host by right clicking and pressing Start button Qemu window will open ; Assign IP address for eth0 interface using command: sudo su ifconfig eth0 10. QEMU needs a lot of other stuff to run, I would not recommend it to someone who start with virtualization for the first time. Additionally, it seems to be end-of-life and 64bit support is experimental. img files, then just give it the path to the files. QEMU (Quick EMUlator) is a generic, open source hardware emulator and virtualization suite. Then the system can boot straight into Windows or start FreeBSD where Windows can be started in a virtual machine. This same manager can be used to monitor the status of the service. 50) under Windows 10 (64-bit) (basically following these instructions) with these commands:. I thought I'd take a stab at it, and didn't have much luck. In order to start kqemu, you must do: net start kqemu Usage. Each VM will also TCP forward its SSH port for easy interaction. For me it seems, that the it must be a problem between. You can resize the existing partitions or add new ones. It is used for converting, creating and consistency checking of various virtual disk formats. QEMU is a hosted virtual machine monitor: It emulates CPUs through dynamic binary translation and provides a set of device models, enabling it to run a variety of unmodified guest operating systems. You have to tell Windows to use the netkvm driver from the driver image. So, I assumed that you already have the Qemu or KVM on your Ubuntu system. Reboot the virtual machine, this time without the. This is often needed since Windows is quite hungry for memory space. First boot is going off now. Some notes and a starter script on getting a Windows 10 guest running on a Linux (Debian) host using QEMU. iso in this case). Your issues matter to us. When I entered the file path to the start. Only compiling QEMU for Windows remains, and it's time to start learning about the GNU C compiler toolchain on Windows. It should take less then 16 minutes, including installation. Personally, I prefer to use Arch Linux over Debian. Hi, I try the tutorial using the F4 project and running it in QEMu and everything is working as expected. 3 and Virtualization Station 3. When we would like to run the two Windows machines on different host computers, we can easily do so by using the serial console based on TCP. Running CoreOS Container Linux on QEMU These instructions will bring up a single Container Linux instance under QEMU, the small Swiss Army knife of virtual machine and CPU emulators. Installing QEMU in Windows. If you have a DHCP server on your network, even if your host has a static address, the guest O/S should be able to receive an address from that server, otherwise, you will manually configure the guest's IP address just as you would on any Linux, BSD or Windows system. In Proxmox VE, the qemu-guest-agent is used for mainly two things: To properly shutdown the guest, instead of relying on ACPI commands or windows policies. I have Debian Linux and Windows 10 installed alongside on my first hard drive. Debian 6 0 10 Fixing link 2 years ago If you feel really unconfortable with the command line on Windows then you can run the setupqemuk70 exe installer It can be downloaded QEMU can be installed from. So this roject also provide Qemu with different command names but with the more features. If you successfully converted an existing Windows image as described in the above section, skip to the next section. To resize other image formats, convert it to raw with qemu-img convert first. service Set up a bridge. RHEL6 Xen4 Tutorial. Start QEMU Execute QPU804. The result is very pleasing and AVD runs significantly faster and smoother than without hypervisor. Use the command line below to start QEMU, replacing “path\to” with the real path to the RTOSDemo. dll with newer Intel-CPUs. The daemon libvirt-bin daemon will start automatically at boot time and load the appropriate kvm modules, kvm-amd or kvm-intel, which are shipped with the Linux kernel Debian package. First boot is going off now. 5 Requirements * Qemu program. 5-create new lab and add windows node, edit the node and set the qemu version to 2. Once QEMU is running in Windows, usage is the same as in Linux. In Windows: -Make sure you have some folder shared -Bridge your tap device with your default network interface (see above)-Start Mac OS 9 through Qemu with tap networking -Use DHCP on the Mac side. Hi Stuart, I have problem with Windows nodes in unetlab. img -m 64 -boot a pauseharddisk. I have Qemu Manager 6. You can start with examples from the Web, or use mine at the end of this document. I installed qemu for Windows to my computer and I tried to launch android on qemu. Running multiple QEMU VMs of the same sample ¶ If you find yourself wanting to run multiple instances of the same Zephyr sample application, which do not need to talk to each other, use the QEMU_INSTANCE argument. QEMU still emulates the hardware but also leverages KVM to execute the guest operating system. All you need is Qemu. On Windows only, we need to specify the absolute path to the QEMU VM executable, and the -L flag is needed to specify the path to our QEMU Bios files (incuded with QEMU). 5 Requirements * Qemu program. QEMU/KVM Guide #2 - HowTo Qemu-Monitor : VM 관리/제어 실습 및 활용TIP QEMU/KVM Guide #3 - HowTo Linux. QEMU has a number of really nice ways to set up networking for its guests. img -cdrom debian-7. How to start it? Never Miss a Hacking or Security Guide Get new Null Byte guides every week. Another script for KVM - Init scripts for kvm, using it with time scheduled start order (German) simple shell script to manage your virtual machine with bridged networking Hosting your VMs in a DMZ - a management and configuration script to assist with setting up a VM in a semi-secured demilitarized zone. Often QEMU does the job well enough. I managed to compile QEMU (3. A pair of newly disclosed security flaws could allow malicious virtual machine guests to break out of their hypervisor's walled gardens and execute malicious code on the host box. QEMU for Windows is experimental software and might contain even serious bugs, so use the binaries at your own risk. The basic idea is: Download qemu for windows and unzip it anywhere. 0, Patching is no longer necessary. \$ systemctl enable libvirtd. If you don't want to start QEMU in graphic mode, you can use the -nographic option, and ask the kernel to use ttyAMA0 as the console. Downloading, building and booting Linux is explained elsewhere. Related Links. In my last post about QEMU I showed how easy it was to emulate the Raspberry Pi in Windows. 2 netmask 255. Be aware that only windows server supports multiple cpu sockets. If you used a different version of QEMU and/or Raspbian, you should also mention the exact versions. 0-3 is working! I will try an additional thing: I will install qemu on my private computer to eliminate possible settings of our company windows computer. Run the following command to add your user account to the libvirtd group: sudo adduser name libvirtd. Recently we wanted to start a Windows virtual machine from a physical hard disk using a Fedora w/ GNOME 3 host machine to change the domain password of a user. Even networking should work out of box and thus you can use it to run Internet Explorer (for web site testing etc). 0, and I can get Windows 98 up and running 100%. I've tried Windows 3. qemulator a solution for easy setup and management of qemuQemulator is a full featured emulation suite for the qemu virtual engine, with on demand job control written in python GTK/Glade2. In this article, I want to share how to resize the Raspbian image on Windows. On the download page click the link with the name: qemu-. How to use QEMU QEMU is a very flexible virtualization technology however it is quite slow and it is recommended that you understand and evaluate alternative solutions before picking this one. Hi everyone, Thanks for your replies, they have been of great help. 5 (you lose -localtime and a few other things). Press Alt + Ctrl + 2 to enter the QEMU Monitor (may need shift as well on some computers: Alt + Ctrl + Shift + 2 ). For 32 bit Windows 7 a sound driver for the Intel 82801AA AC97 exists. Since we are not using a explicit boot loader with QEMU right now, you can use its -dtb argument to pass the device tree binary file to the kernel. Helpful links https://www. The open-source routers we will use in a GNS3 simulated network must run on either a Qemu or a VirtualBox hypervisor. Excuse the 'orrid ascii art :-) VLANs and NICs A VLAN is a network switch running in the context of a qemu process. Basically, you can make the most of your hardware to accelerate QEMU virtual machines on Windows: starting with its 2. 04 - Virt Manager - Summary and Networking Step 6 : KVM will start to create a virtual machine depends on our input, once VM is created. This is often needed since Windows is quite hungry for memory space. When searching the web pages to find a way to be able to. Using qemu on Windows (Basics) Have you ever felt an urge to use Linux in your office but you are not allowed to install it without permission. In difference to VirtualBox or VMware Player you can emulate several CPUs, which is necessary for old 16bit applications with EGPFault in Win87EM.


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# Partial Fractions Decomposition and Inverse Z Transform of H(z) = (z^2) / (z^2 - 1)

1. Jul 19, 2009

### mdmman

1. The problem statement, all variables and given/known data
What is the inverse Z transform of
$$H(z)=\frac{z^2}{z^{2}-1}$$

2. Relevant equations
Z transform table can be found at http://en.wikipedia.org/wiki/Z-transform" [Broken]

3. The attempt at a solution

$$H(z)=\frac{z^2}{z^{2}-1} = 1+\frac{1}{z^{2}-1}$$

$$\frac{1}{z^{2}-1}=\frac{A}{z+1}+\frac{B}{z-1}$$

$$1=A(z-1)+B(z+1)=z(A+B)+(-A+B)$$

$$0=(A+B), 1=(-A+B)$$

$$A=-0.5, B=0.5$$

$$H(z)=\frac{z^2}{z^{2}-1} = 1+\frac{1}{z^{2}-1} = 1+\frac{-0.5}{z+1}+\frac{0.5}{z-1}$$

$$h(k)=Z^{-1}\{H(z)\} = Z^{-1}\{1\} + Z^{-1}\{\frac{-0.5}{z+1}\} + Z^{-1}\{\frac{0.5}{z-1}\} = \delta(n) + ? + ?$$

I don't know how to figure out what the inverse of $\frac{-0.5}{z+1}$ and $\frac{0.5}{z-1}$. Any help?

Last edited by a moderator: May 4, 2017

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# A quickie for the cryptic crossword enthusiasts: Here is the clue. / Cut cut! / What crazy, bloodthirsty creature remains? (5 letters)

Here is the clue.

Cut cut!

What crazy, bloodthirsty creature remains? (5 letters)

EDIT I added the word 'crazy'. Apologies that I missed this out before.

Please use typical cryptic crossword strategies to solve this but some lateral thinking may be required.

I have made this open-ended so, in the unlikely event that there are multiple valid answers then the one that gets the most up-votes in the next couple of days will be accepted.

If you are not sure that your answer is correct, then it isn't! If you get the intended solution then you will be 100% certain of the reason.

The answer is a single word. Please give the word and explain the cryptic techniques you used. I used two very straightforward ones.

Hints

1. Don't ignore the lateral thinking tag.

2. Make sure you use all the information provided.

3. No technical knowledge is needed (apart from a familiarity with cryptic crosswords) and this is a word that I'm sure everyone will know.

• I hope the answer doesn't involve inserting an extra letter into "cut" to get something where blood is sometimes involved... – Rand al'Thor Aug 24 '15 at 9:26
• @randal'thor Awww, you take the fun out of everything... – dmg Aug 24 '15 at 9:27
• @randal'thor - If that is your answer, submit it and see if you win! It doesn't really fulfil the conditions if I have understood you correctly. – chasly from UK Aug 24 '15 at 9:28
• @chaslyfromUK Nah, I'd probably get flagged as offensive (maybe rightly so). No number in brackets at the end to tell us how long the clue is? :-( – Rand al'Thor Aug 24 '15 at 9:29
• if it weren't for the one word requirement, my guess would've been "the IS" (Islamic State) , you can get there by removing two ("Cut cut!" ) of the words in the first sentence ("Here is the clue."). They're crazy and bloodthirsty, but I don't know about creature, ;) – dramzy Aug 24 '15 at 16:37

## 3 Answers

Here is the answer:

If you cut the head from the word 'cut' (leaving 'ut') then cut these two letters from 'the clue', you are left with 'hecle'. The word 'crazy' implies an anagram, so rearranging these letters gives 'leech' which is a five-letter bloodsucking creature.

• That was my answer. As promised I'll wait for a while to see what other answers there may be and accept whatever gets the most votes. – chasly from UK Aug 24 '15 at 21:13
• @chaslyfromUK - But shouldn't the clue be "Cut cut cut!"? Since you "cut cut" to get "ut", and then "cut ut" to get "he cle"? – Alconja Aug 24 '15 at 23:09
• @Alconja - You are of course right. I only realised that that would be better after I had already posted the puzzle and people had started answering. In future I shall have to wait 24 hours before setting a puzzle. It's so easy to miss something if you do it in a rush. – chasly from UK Aug 24 '15 at 23:14
• Or if you want to make it obvious- cut(cut(cut),....) – Rohcana Aug 25 '15 at 9:44

The answer is

SLICE

Cut cut!
What bloodthirsty creature remains?

When you slice something, you are cutting into it. Also, once you chop off the first letter of SLICE, you are left with LICE, bloodsucking insects.

And of course it is 5 letters long.

• If this is the answer, I don't like the question: it sounded as though the answer was meant to be a bloodthirsty creature. It is clever though. – Rand al'Thor Aug 24 '15 at 9:57
• Except that 'lice' is plural (of louse). I specified 'creature' which is singular. Also 'slice' is not a creature. 'louse' has five letters but it is far too distant from the clue to be valid (IMO) – chasly from UK Aug 24 '15 at 10:02

The answer is

TROLL.

Explanation:

the "cut" command on a computer is implemented by Control+X. In cryptic-crossword lingo, "cut" can suggest either an abbreviation or a shortening of a word by removing some letters. "Cut"="Control+X" can be thought of as an abbreviation, and we can remove some letters from "Control+X" to get trol. Since we want a 5-letter creature, the answer looks like TROLL - we all know those can be bloodthirsty!

Wow ... have I managed to answer a cryptic-crossword question before Bailey M? :-D

• – Rand al'Thor Aug 24 '15 at 9:45
• @randal'thor Not to mention that the larger part of the internet trolls just "Control+X" once and "Control+V" a million times the same thing. – dmg Aug 24 '15 at 9:50
• As per the conditions, I'm not entitled to exclude this answer. As a hint to others, here is the main problem with this answer as I see it. In cryptic-crossword terms, to get from 'trol' to 'troll', would need a clue that used the phrase 'looks like' or 'sounds like' or that in some way specified adding a letter was needed. My question has none of those. – chasly from UK Aug 24 '15 at 9:57
• Another troll question! – d'alar'cop Aug 27 '15 at 5:12


text:
# Generating pgfplots legend from gnuplot data

I’m having trouble getting pgfplots to typeset very small numbers, e.g. 6.82e-08. I don’t know if the problem lies in it being a very small number, or if it’s because of the interaction with reading variable data generated from gnuplot.

The example below will generate two plots. The first is generated from “external” data, and has gnuplot generate the exponential regression. It saves the fitted variables in a file that pgfplots should read and pretty-print in the plot legend. It works as intended for variables b and c but not for a.

At first I thought it was related to the number not being rendered in scientific notation, but forcing it through \pgfmathprintnumber[sci] still outputs wrongly, which leads me to believe the problem lies with the gnuplot interaction.

I want the first graph to have the same legend as the second, but still have it be generated as I’m using this in a macro for different sets of data. What should I do?

Minimal example:

\documentclass{article}
\usepackage{pgfplotstable}

\begin{filecontents}{data.csv}
x,  y
2,  0.000058
3,  0.001888
4,  0.058763
5,  1.78986
\end{filecontents}

\begin{document}

\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}
\addplot [raw gnuplot, smooth] gnuplot {
f(x)=(a)*(b^(c*x));
a=6.8e-8;
b=2.7;
c=3.4;
fit f(x) 'data.csv' using 1:2 via a,b,c;
plot f(x);
set print "data.csv.dat";
print a,b,c;
};
\pgfplotstablegetelem{0}{0}\of\parameters \pgfmathsetmacro{\paramA}{\pgfplotsretval}
\pgfplotstablegetelem{0}{1}\of\parameters \pgfmathsetmacro{\paramB}{\pgfplotsretval}
\pgfplotstablegetelem{0}{2}\of\parameters \pgfmathsetmacro{\paramC}{\pgfplotsretval}
$y=\pgfmathprintnumber[sci]{\paramA} \cdot \pgfmathprintnumber{\paramB}^{\pgfmathprintnumber{\paramC} \cdot x}$
}
\end{axis}
\end{tikzpicture}
\end{figure}

\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}
\addplot [raw gnuplot, smooth] gnuplot {
plot (6.82564001285694e-08)*(2.71554805209898^(3.41986746643527*x));
};
$y=\pgfmathprintnumber{6.82564002501364e-08} \cdot \pgfmathprintnumber{2.71554805209898}^{\pgfmathprintnumber{3.41986746643527} \cdot x}$
}
\end{axis}
\end{tikzpicture}
\end{figure}

\end{document}

Result:

data.csv.dat.

6.82564001285694e-08 2.71565816582345 3.41972869952421

erroneous plot:

desired plot:

• According to the manual of pgfplotstable, the precision is limited to 10^{-6} – user31729 Mar 21 '15 at 14:26
• In addition, this has nothing to do with gnuplot or not. It's a pgf feature – user31729 Mar 21 '15 at 14:39
• Would it be possible to alter this by manually copying and editing the .sty file? Maybe do it on-the-fly if that’s possible? – user74027 Mar 21 '15 at 14:41
• You could run the fit data through (for example) sed and format it so you separate the exponent from the factor and reduce the precision of the numbers. Or look into gnuplot if you could change the output format there... – jens_bo Mar 21 '15 at 14:51
• @jens_bo: The problem isn't the format, it's the small value in fact – user31729 Mar 21 '15 at 15:58

Changing the gnuplot output is probably the easiest way to do it and circumvent the precision limitation of pgfplots.

a/10^floor(log10(a)),floor(log10(a)) gives you the parts of the scientific notation in two separate columns. Those can be read into pgfplots and put into the legend.

\begin{tikzpicture}
\begin{axis}
\addplot [raw gnuplot, smooth] gnuplot {
f(x)=(a)*(b^(c*x));
a=6.8e-8;
b=2.7;
c=3.4;
fit f(x) 'data.csv' using 1:2 via a,b,c;
plot f(x);
set print "data.csv.dat";
print a/10^floor(log10(a)),floor(log10(a)),b,c;};
$y=\pgfmathprintnumber{\paramAa}\cdot 10^{\paramAb} \cdot \pgfmathprintnumber{\paramB}^{\pgfmathprintnumber{\paramC} \cdot x}$


text:
### Configuring X window and desktop environments on FreeBSD 6.2

by Forrest Sheng Bao http://fsbao.net

I will assume that your graphic interface card can work well with FreeBSD and the driver program runs ok now. So I skip the process about installing graphic card driver.

To use a desktop environment, such as GNOME or KDE, you need to let X server run first. X server requires a configuration file to initialize it. The configuration file of X server is named as xorg.conf and is located at /etc/X11/xorg or /usr/X11R6/etc/X11/.

You may copy the xorg.conf file from the Linux system that you run on the same computer, located in

/etc/X11/xorg.conf

FreeBSD also provides a program to generate the xorg.conf automatically. As the super user, simply run:
Xorg -configure

This will generate an X11 configuration skeleton file in your home directory called xorg.conf.new
To test whether you have set it correctly, you can use
Xorg -config /path/to/your/xorg.conf

For example,
Xorg -config xorg.conf.new
Maybe you will need to manually modify some parts of the auto-generated configuration file. On my workstation, I have to edit the Screen section. I need to add the
"Defaultdepth 24" and "Modes" 1289x1024
lines into that section.If you can see the display and you can move your mouse, it generally means the configuration is correct. After that, please copy the xorg.conf.new file to /etc/X11/ or /usr/X11R6/etc/X11. Remember to rename it as xorg.conf.

To leave the test screen, press Ctrl+Alt+F1.(Or any F-n that correspond to your original login console.)

Once the X can work, you can go on to your desktop environment. Please make sure that you have installed your preferred desktop environment. You can install them use sysinstall or install them from port collections.

There are two ways to start a desktop environment. One is to associate it will the X starting script. The other is to configure it in the ttys configuration files. To most users, they should choose the later option which will start the desktop environment automatically when the computer starts.

If your default run-level is not graphics but command-line, which means you like to start X when you need it not automatially, you can choose the first option.

To make GNOME as the environment, run

echo "/usr/local/bin/gnome-session" > ~/.xinitrc

To make KDE as the environment, run
echo "exec startkde" > ~/.xinitrc

When typing startx on the shell prompt, the specified desktop environment will be started as X starts.
(You can also try kdm on the shell prompt to start KDE directly.)
If you prefer the GUI to be started automatically when the computer finish booting, you can choose the second option, configuring the /etc/ttys file.

Go to the line starts with ttyv8. You may find it like this:

ttyv8 "/usr/X11R6/bin/xdm -nodaemon" xterm off secure

Comment this line if it exists. (Or if you prefer the default XDM, leave it along and don't operate as following instructions.

Insert this line, if you prefer KDE:

ttyv8 "/usr/local/bin/kdm -nodaemon" xterm on secure
It sounds like no that easy way to start GNOME when the computer starts. But you can start gnome-related programs by editing /etc/rc.conf and adding the line
gnome_enable="YES"

Then choose GNOME as the session when you are facing the KDE login window. Make sure you have set KDE as ttyv8 at /etc/ttys if you wanna start GNOME in this way.
Reference:

FreeBSD GNOME Project: GNOME 2.18 FAQ, http://www.freebsd.org/gnome/docs/faq2.html

Chapter 5 The X Window System, FreeBSD Handbook, http://www.freebsd.org/doc/en_US.ISO8859-1/books/handbook/x11.html

KDE on FreeBSD FAQs, http://freebsd.kde.org/faq.php

#### 1 comment:

Anonymous said...


text:
# Differential Equations¶

In this chapter, some facilities for solving differential equations are described. Currently only simple equations without auxiliary conditions are supported.

OdeSolve(expr1==expr2)

general ODE solver

Param expr1,expr2

expressions containing a function to solve for

This function currently can solve second order homogeneous linear equations with real constant coefficient. The solution is returned with unique constants generated by {UniqueConstant}. The roots of the auxiliary equation are used as the arguments of exponentials. If the roots are complex conjugate pairs, then the solution returned is in the form of exponentials, sines and cosines. First and second derivatives are entered as y', y''. Higher order derivatives may be entered as y(n), where n is any positive integer.

Example

In> OdeSolve( y'' + y == 0 )
Out> C42*Sin(x)+C43*Cos(x);
In> OdeSolve( 2*y'' + 3*y' + 5*y == 0 )
Out> Exp(((-3)*x)/4)*(C78*Sin(Sqrt(31/16)*x)+C79*Cos(Sqrt(31/16)*x));
In> OdeSolve( y'' - 4*y == 0 )
Out> C132*Exp((-2)*x)+C136*Exp(2*x);
In> OdeSolve( y'' +2*y' + y == 0 )
Out> (C183+C184*x)*Exp(-x);

Solve(), RootsWithMultiples()

OdeTest(eqn, testsol)

test the solution of an ODE

Param eqn

equation to test

Param testsol

test solution

This function automates the verification of the solution of an ODE. It can also be used to quickly see how a particular equation operates on a function.

Example

In> OdeTest(y''+y,Sin(x)+Cos(x))
Out> 0;
In> OdeTest(y''+2*y,Sin(x)+Cos(x))
Out> Sin(x)+Cos(x);

OdeOrder(eqn)

return order of an ODE

Param eqn

equation

This function returns the order of the differential equation, which is order of the highest derivative. If no derivatives appear, zero is returned.

Example

In> OdeOrder(y'' + 2*y' == 0)
Out> 2;
In> OdeOrder(Sin(x)*y(5) + 2*y' == 0)
Out> 5;
In> OdeOrder(2*y + Sin(y) == 0)
Out> 0;

WronskianMatrix(func, var)

create the Wronskian matrix

Param func

an $$n$$-dimensional vector of functions

Param var

a variable to differentiate with respect to

The function WronskianMatrix() calculates the Wronskian matrix of $$n$$ functions. The Wronskian matrix is created by putting each function as the first element of each column, and filling in the rest of each column by the $$(i-1)$$-th derivative, where $$i$$ is the current row. The Wronskian matrix is used to verify that the $$n$$ functions are linearly independent, usually solutions to a differential equation. If the determinant of the Wronskian matrix is zero, then the functions are dependent, otherwise they are independent.

Example

In> WronskianMatrix({Sin(x),Cos(x),x^4},x);
Out> {{Sin(x),Cos(x),x^4},{Cos(x),-Sin(x),4*x^3},
{-Sin(x),-Cos(x),12*x^2}};
In> PrettyForm(%)
/                                                 \
| ( Sin( x ) )      ( Cos( x ) )      /  4 \      |
|                                     \ x  /      |
|                                                 |
| ( Cos( x ) )      ( -( Sin( x ) ) ) /      3 \  |
|                                     \ 4 * x  /  |
|                                                 |
| ( -( Sin( x ) ) ) ( -( Cos( x ) ) ) /       2 \ |
|                                     \ 12 * x  / |
\                                                 /

The last element is a linear combination of the first two, so the determinant is zero:

In> A:=Determinant( WronskianMatrix( {x^4,x^3,2*x^4+3*x^3},x ) )
Out> x^4*3*x^2*(24*x^2+18*x)-x^4*(8*x^3+9*x^2)*6*x
+(2*x^4+3*x^3)*4*x^3*6*x-4*x^6*(24*x^2+18*x)+x^3
*(8*x^3+9*x^2)*12*x^2-(2*x^4+3*x^3)*3*x^2*12*x^2;
In> Simplify(A)
Out> 0;


text:
# Why is $FOLLOW$ not necessary for $LL(1)$ grammars with no $\epsilon$ transitions?

I'm aware of how $$FIRST$$ and $$FOLLOW$$ sets are used to construct a parsing table for $$LL(1)$$ grammars.

However, I've encountered this statement from my notes:

With $$\epsilon$$ productions in the grammar, we may have to look beyond the current non-terminal to what can come after it

In my opinion, this suggests that $$FOLLOW$$ is not necessary for $$LL(1)$$ grammars that have no $$\epsilon$$ transition. Am I wrong? And if I'm not, why is this the case?

Thanks

That's correct for $$LL(1)$$: if there are no $$\epsilon$$ productions, the $$LL$$ parser-generation algorithm will never consult $$FOLLOW$$, because it only does that if it finds $$\epsilon$$ in the $$FIRST$$ set for the first non-terminal in the right-hand side of a production. (So it might not need the $$FOLLOW$$ sets even if there are some $$\epsilon$$ productions, provided that none of those productions occur at the beginning of a right-hand side.)
The observation doesn't generalise well to other values of $$k$$. You'll need $$FOLLOW_k(\alpha)$$ if any production can derive a string whose length is less than $$k$$.


text:
# Gradient of l2 norm squared

Could someone please provide a proof for the following rule:

$$\nabla\|x\|_2^2 = 2x$$

I.E. why is the gradient of the $L_2$ norm square of $x$ equal to $2x$?

Thanks

Use the definition. If $$f(x)=\|x\|^2_2= \left(\left(\sum_{k=1}^n x_k^2 \right)^{1/2}\right)^{2}=\sum_{k=1}^n x_k^2 ,$$ then $$\frac{\partial}{\partial x_j}f(x) =\frac{\partial}{\partial x_j}\sum_{k=1}^n x_k^2=\sum_{k=1}^n \underbrace{\frac{\partial}{\partial x_j}x_k^2}_{\substack{=0, \ \text{ if } j \neq k,\\=2x_j, \ \text{ else }}}= 2x_j.$$ It follows that $$\nabla f(x) = 2x.$$

• Thanks. One follow up question, does this still hold if x is complex? – user167133 Jul 30 '14 at 20:56
• Isn't the L2 norm defined as: $$f(x)=\|x\|^2_2= \left(\left(\sum_{k=1}^n |x_k|^2 \right)^{1/2}\right)^{2}=\sum_{k=1}^n |x_k|^2$$, there is an absolute value sign surrounding $x_k$, so when you take the derivative what should pop out is $2|x_j|$ not just $2x_j$, correct me if I am wrong – Carlos - the Mongoose - Danger Oct 23 '15 at 17:46
• @Lookbehindyou well |t|^2 = t^2 for any $t\in\Bbb R$. Moreover, note that $\frac{d}{t}|t|^2 = 2|t|\operatorname{sign}(t)=2t$, where $\operatorname{sign}$ is the sign function – Surb Feb 29 '16 at 14:49
• Could you please explain what happened to the sigma in the last step? – Gigili Aug 16 '16 at 12:38
• @Gigili: If $k\neq j$, then $\frac{\partial }{\partial x_j}x_k^2=0$. Then, $$\sum_{k=1}^n\frac{\partial }{\partial x_j}x_k^2=0+...+0+2x_j+0+...+0=2x_j.$$ – Surb Aug 16 '16 at 12:44

Another approach that extends to more general settings is to use the connection between the norm and the inner product, $$\|x\|^2 = (x,x).$$

We have the finite difference, \begin{align} \|x+sh\|^2 - \|x\|^2 &= (x+sh,x+sh) - (x,x) \\ &= (x,x) + 2s(x,h) + s^2(h,h) - (x,x) \\ &= 2s(x,h) + s^2(h,h). \end{align}

The gradient acting in the direction $$h$$ is the limit of this finite difference as the stepsize goes to zero, \begin{align} (\nabla\|x\|^2, h) &:= \lim_{s \rightarrow 0} \frac{1}{s}\left[\|x+sh\|^2 - \|x\|^2\right] \\ &= \lim_{s \rightarrow 0} \frac{1}{s}\left[2s(x,h) + s^2(h,h)\right] \\ &= (2x,h). \end{align} Since this holds for any direction $$h$$, the gradient must be $$\nabla \|x\|^2 = 2x$$.


text:
# Properties

 Label 90.c Number of curves $8$ Conductor $90$ CM no Rank $0$ Graph

# Related objects

Show commands for: SageMath
sage: E = EllipticCurve("90.c1")

sage: E.isogeny_class()

## Elliptic curves in class 90.c

sage: E.isogeny_class().curves

LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
90.c1 90c8 [1, -1, 1, -48002, 4059929] [6] 192
90.c2 90c7 [1, -1, 1, -4082, 14681] [6] 192
90.c3 90c6 [1, -1, 1, -3002, 63929] [2, 6] 96
90.c4 90c4 [1, -1, 1, -2597, -50281] [2] 64
90.c5 90c5 [1, -1, 1, -617, 5231] [2] 64
90.c6 90c2 [1, -1, 1, -167, -709] [2, 2] 32
90.c7 90c3 [1, -1, 1, -122, 1721] [12] 48
90.c8 90c1 [1, -1, 1, 13, -61] [4] 16 $$\Gamma_0(N)$$-optimal

## Rank

sage: E.rank()

The elliptic curves in class 90.c have rank $$0$$.

## Modular form90.2.a.c

sage: E.q_eigenform(10)

$$q + q^{2} + q^{4} + q^{5} - 4q^{7} + q^{8} + q^{10} + 2q^{13} - 4q^{14} + q^{16} - 6q^{17} - 4q^{19} + O(q^{20})$$

## Isogeny matrix

sage: E.isogeny_class().matrix()

The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.

$$\left(\begin{array}{rrrrrrrr} 1 & 4 & 2 & 12 & 3 & 6 & 4 & 12 \\ 4 & 1 & 2 & 3 & 12 & 6 & 4 & 12 \\ 2 & 2 & 1 & 6 & 6 & 3 & 2 & 6 \\ 12 & 3 & 6 & 1 & 4 & 2 & 12 & 4 \\ 3 & 12 & 6 & 4 & 1 & 2 & 12 & 4 \\ 6 & 6 & 3 & 2 & 2 & 1 & 6 & 2 \\ 4 & 4 & 2 & 12 & 12 & 6 & 1 & 3 \\ 12 & 12 & 6 & 4 & 4 & 2 & 3 & 1 \end{array}\right)$$

## Isogeny graph

sage: E.isogeny_graph().plot(edge_labels=True)

The vertices are labelled with LMFDB labels.


text:
DeepAI

Local energy estimates for the fractional Laplacian

The integral fractional Laplacian of order s ∈ (0,1) is a nonlocal operator. It is known that solutions to the Dirichlet problem involving such an operator exhibit an algebraic boundary singularity regardless of the domain regularity. This, in turn, deteriorates the global regularity of solutions and as a result the global convergence rate of the numerical solutions. For finite element discretizations, we derive local error estimates in the H^s-seminorm and show optimal convergence rates in the interior of the domain by only assuming meshes to be shape-regular. These estimates quantify the fact that the reduced approximation error is concentrated near the boundary of the domain. We illustrate our theoretical results with several numerical examples.

• 11 publications
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09/01/2021

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Truncation error estimates of approximate operators in a generalized particle method

To facilitate the numerical analysis of particle methods, we derive trun...
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Parameter dependent finite element analysis for ferronematics solutions

This paper focuses on the analysis of a free energy functional, that mod...
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On the linear convergence of additive Schwarz methods for the p-Laplacian

We consider additive Schwarz methods for boundary value problems involvi...
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nPINNs: nonlocal Physics-Informed Neural Networks for a parametrized nonlocal universal Laplacian operator. Algorithms and Applications

Physics-informed neural networks (PINNs) are effective in solving invers...
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Weighted analytic regularity for the integral fractional Laplacian in polygons

We prove weighted analytic regularity of solutions to the Dirichlet prob...

1. Introduction

In this work we consider finite element discretizations of the problem

 (1.1) {(−Δ)su=fin Ω,u=0in Ωc:=Rd∖Ω,

where is a bounded domain and is the integral fractional Laplacian of order ,

 (1.2) (−Δ)su(x):=Cd,s p.v. ∫Rdu(x)−u(y)|x−y|d+2sdy.

The normalization constant makes the integral in (1.2), calculated in the principal value sense, coincide with the Fourier definition of . It is well understood that, even if the data is smooth (for example, if and ), then the unique solution to (1.1) develops an algebraic singularity near (cf. Example 2.2). This is in stark contrast with the classical Laplacian equation.

Nevertheless, in such a case one expects the solution to be locally smooth in , and thus the discretization error to be smaller in the interior of the domain. Our main result (Theorem 5.3) is a quantitative estimate of the fact that the finite element error is concentrated around .

The fractional Laplacian (1.2) is a nonlocal operator: computing requires the values of at points arbitrarily far away from . Nonlocality is also reflected in the variational formulation of (1.1): the natural space in which the problem is set is the zero-extension fractional Sobolev space , and the norm therein is not subadditive with respect to domain partitions. Furthermore, it is not possible to localize the inner product in , because functions with supports arbitrarily far away from each other may have nonzero -inner product. This is also in stark contrast with the local case (i.e., with the inner product in ), and makes the development of local estimates for such a nonlocal problem a more delicate matter, especially in the case of general shape–regular meshes. This is the main purpose of this paper.

In recent years, there has been significant progress in the numerical analysis and implementation of (1.1) and related fractional-order problems. Finite element discretizations provide naturally the best approximation in the energy norm. A priori convergence rates in the energy norm for approximations using piecewise linear basis functions on either quasi-uniform or graded meshes were derived in [2]; similar results, but regarding convergence in in case , were obtained in [7]. The use of adaptive schemes and a posteriori error estimators has been studied in [3, 20, 23, 32, 36]. A non-conforming discretization, based on a Dunford-Taylor representation was proposed and analyzed in [6]. We refer to [5, 8] for further discussion on these methods. In contrast, the analysis of finite difference schemes typically leads to error estimates in the -norm under regularity assumptions that cannot be guaranteed in general [16, 17, 26].

The rest of the paper is organized as follows. In Section 2, we review the fractional-order spaces and the regularity of solutions to (1.1) in either standard and weighted Sobolev spaces. In Section 3, we describe our finite element discretization, review basic energy based error estimates, and combine such estimates with Aubin-Nitsche techniques to derive novel convergence rates in -norm. In Section 4, we provide a proof of Caccioppoli estimate for the continuous problem. In Section 5, which is the central part of the paper, we combine Caccioppoli estimates and superapproximation techniques, to obtain interior error estimates with respect to -seminorms. At the end of this section we show some applications of our interior error estimates. In particular, we discuss the convergence rates of the finite element error in the interior of the domain with respect to smoothness of the domain and the right hand side in the case of quasi-uniform and graded meshes. The results are summarized in Tables 1 and 2. Finally, several numerical examples at the end of the paper illustrate the theoretical results from Section 5.

2. Variational formulation and regularity

In this section, we briefly discuss important features of fractional-order Sobolev spaces that are instrumental for our analysis. Furthermore, we consider regularity properties of the solution to (1.1) and review some negative results that lead to the use of certain weighted spaces, in which the weight compensates the singular behavior of the gradient of the solution near the boundary of the domain. Having regularity estimates in such weighted spaces at hand, we shall be able to increase the convergence rates by constructing a priori graded meshes.

2.1. Sobolev spaces

Sobolev spaces of order provide the natural setting for the variational formulation of (1.1). More precisely, we consider to be the set of -functions such that

 (2.1) |v|Hs(Rd):=(Cd,s2∫Rd∫Rd|v(x)−v(y)|2|x−y|d+2sdxdy)1/2<∞,

where is taken as in (1.2). Clearly, these are Hilbert spaces; we shall denote by the bilinear form that gives rise to the fractional-order seminorms, namely,

 (2.2) (v,w)s:=Cd,s2∫Rd∫Rd(v(x)−v(y))(w(x)−w(y))|x−y|d+2sdxdy.

For the variational formulation of (1.1), we need the zero-extension spaces

 ˜Hs(Ω):={v∈Hs(Rd):supp(v)⊂¯¯¯¯Ω},

for which the form becomes an inner product. Moreover, if , then integration in (2.2) takes place in . We shall denote the -norm by , and remark that the -norm of is not needed because a Poincaré inequality holds in the zero-extension Sobolev spaces.

Fractional-order Sobolev spaces can be equivalently defined through interpolation of integer-order spaces; remarkably, if one suitably normalizes the standard

-functional, then the norm equivalence constants can be taken to be independent of [29, Lemma 3.15 and Theorem B.9]. Although the constant in (2.1) is fundamental in terms of continuity of Sobolev seminorms as , we shall omit it whenever is fixed. For simplicity of notation, throughout this paper we shall adopt the convention .

Let denote the dual space to , and be their duality pairing. Because of (2.2) it follows that if then and

 (2.3) (v,w)s=⟨(−Δ)sv,w⟩,∀w∈˜Hs(Ω).

This integration by parts formula motivates the following weak formulation of (1.1): given , find such that

 (2.4) (u,v)s=⟨f,v⟩∀v∈˜Hs(Ω).

Because this formulation can be cast in the setting of the Lax-Milgram Theorem, existence and uniqueness of weak solutions, and stability of the solution map , are straightforward.

2.2. Sobolev regularity

The Lax-Milgram Theorem guarantees the well-posedness of (2.4) in if . A subsequent question is what additional regularity does inherit for smoother . For the sake of finite element analysis, here we shall focus on Sobolev regularity estimates.

By now it is well understood that for smooth domains and data , solutions to (1.1) develop an algebraic singular layer of the form (cf. for example [25])

 (2.5) u(x)≈dist(x,∂Ω)s,

that limits the smoothness of solutions. Indeed, if is locally smooth in but behaves as (2.5), then one cannot guarantee that belongs to ; actually, in general (see Example 2.2).

We now quote a recent result [9], that characterizes regularity of solutions in terms of Besov norms. Its proof follows a technique introduced by Savaré [34], that consists in combining the classical Nirenberg difference quotient method with suitably localized translations and exploiting certain convexity properties.

Theorem 2.1 (regularity on Lipschitz domains).

Let be a bounded Lipschitz domain, and for some . Then, the solution to (1.1) belongs to the Besov space , where , with

 (2.6) ∥u∥Bs+γ2,∞(Ω)≤C(Ω,d,s,γ)∥f∥Hr(Ω).

Consequently, by an elementary embedding, we deduce

 (2.7) ∥u∥Hs+γ−ε(Ω)≤C(Ω,d,s,γ)ε∥f∥Hr(Ω)∀ε>0.

There are two conclusions to be drawn from the previous result. In first place, assuming the domain to be Lipschitz is optimal, in the sense that if was a domain then no further regularity could be inferred. Thus, reentrant corners play no role on the global regularity of solutions: the boundary behavior (2.5) dominates any point singularities that could originate from them; we refer to [22] for further discussion on this point. In second place, in general the smoothness of the right hand side cannot make solutions any smoother than . The expansion (2.5) holds in spite of the smoothness of near . We illustrate these two points with a well-known example [21].

Example 2.2 (limited regularity).

Let and . Then, the solution to (1.1) is

 (2.8) u(x)=Γ(d2)22sΓ(d+2s2)Γ(1+s)(1−|x|2)s+,

where . Therefore, .

We also point out a limitation in the technique of proof in Theorem 2.1 from [9] that is related to the example above. Namely, in case and for some , solutions are expected to be smoother than just ; however, one cannot derive such higher regularity estimates from Theorem 2.1. For smooth domains (i.e., ), the following estimate holds [35]:

 (2.9) f∈Hr(Ω), −s≤r<1/2−s⇒u∈˜H2s+r(Ω).

2.3. Regularity in weighted Sobolev spaces

By developing a fractional analog of the Krylov boundary Harnack method, Ros-Oton and Serra [33] obtained a fine characterization of boundary behavior of solutions to (1.1) and derived Hölder regularity estimates. In order to exploit these estimates and apply them in a finite element analysis, reference [2] introduced certain weighted Sobolev spaces, where the weight is a power of the distance to . Let

 δ(x):=dist(x,∂Ω),δ(x,y):=min{δ(x),δ(y)}.

Then, for and , we consider the norm

 (2.10) ∥v∥2Hkγ(Ω)=∫Ω⎛⎝|v(x)|2+∑|β|≤k|∂βv(x)|2⎞⎠δ(x)2γdx

and define and as the closures of and , respectively, with respect to the norm (2.10).

Next, for , with and , and , we consider

 ∥v∥2Htγ(Ω):=∥v∥2Hkγ(Ω)+|v|2Htγ(Ω),|v|2Htγ(Ω):=∫Ω∫Ω|∇kv(x)−∇kv(y)|2|x−y|d+2sδ(x,y)2γdydx

and the associated space

In analogy with the notation for their unweighted counterparts, we define zero-extension weighted Sobolev spaces by

 (2.11) ˜Htγ(Ω):={v∈Htγ(Rd): v=0 a.e. in Ωc},∥v∥2˜Htγ(Ω):=∥v∥2˜Hkγ(Ω)+|v|2Htγ(Rd).

We have the following regularity estimate in the scale (2.11) [2, Proposition 3.12], [5, Formula (3.6)].

Theorem 2.3 (weighted Sobolev estimate).

Let be a bounded, Lipschitz domain satisfying the exterior ball condition, , for some , , and be the solution of (2.4). Then, it holds that and

 ∥u∥˜Htγ(Ω)≤C(Ω,d,s)√(β+2s−t)(1+2(γ+s−t))∥f∥Cβ(¯¯¯Ω).
Remark 2.4 (optimal parameters).

In finite element applications of Theorem 2.3, discussed in Section 3, we will design graded meshes with a grading dictated by . The optimal choice of parameters and depends on both the smoothness of the right hand side and the dimension of the space. We illustrate this now: let , , , and be sufficiently small, and choose and , to obtain the optimal regularity estimate

 ∥u∥˜Htγ(Ω)≤C(Ω,d,s)ε∥f∥Cβ(¯¯¯Ω).

In contrast, if , we set to be any positive number and take as above to arrive at

 ∥u∥˜Htγ(Ω)≤C(Ω,d,s,β)√ε∥f∥Cβ(¯¯¯Ω).
Remark 2.5 (exterior ball condition).

Taking into account the results from [22], the exterior ball condition could be relaxed. Nevertheless, because the analysis of effects of reentrant corners is beyond our purposes, we leave such an assumption on .

3. Finite Element Discretization

We next consider finite element discretizations of (2.4) by using piecewise linear continuous functions. Let ; for , we let denote a triangulation of , i.e., is a partition of into simplices of diameter . We assume the family to be shape-regular, namely,

 σ:=suph>0maxT∈ThhTρT<∞,

where and is the diameter of the largest ball contained in . As usual, the subindex denotes the element size, ; moreover, we take elements to be closed sets.

We shall also need a smooth mesh function , which is locally comparable with the element size. Note that shape-regularity yields (cf. [31,  Lemma 5.1]), and thus

 (3.1) |h(x)−h(y)|≤C(σ)|x−y|,∀x,y∈Ω.

Let be the set of interior vertices of , be its cardinality , and the standard piecewise linear Lagrangian basis, with associated to the node . With this notation, the set of discrete functions is

 (3.2) Vh:={v∈C0(Ω):v=N∑i=1viφi}.

It is clear that for all and therefore we have a conforming discretization.

3.1. Interpolation and inverse estimates

Fractional-order seminorms are not subadditive with respect to domain decompositions; therefore, some caution must be exercised when localizing them. With the goal of deriving interpolation estimates, we define the star (or patch) of a set by

 SA:=⋃{T∈Th:T∩A≠∅}.

Given , the star of is the first ring of and the star of is the second ring of . The star of the node is .

We have the following localization estimate [18, 19].

 (3.3) |v|2Hs(Ω)≤∑T∈Th[∫T∫ST|v(x)−v(y)|2|x−y|d+2sdydx+C(d,σ)sh2sT∥v∥2L2(T)]∀v∈Hs(Ω).

This inequality shows that to estimate fractional seminorms over , it suffices to compute integrals over the set of patches plus local zero-order contributions. In addition, if these contributions have vanishing means over elements –as is often the case whenever is an interpolation error– a Poincaré inequality allows one to estimate them in terms of local -seminorms. Thus, one can prove the following local quasi-interpolation estimates (see, for example, [2, 10, 12]).

Proposition 3.1 (local interpolation estimates).

Let , , , and be a suitable quasi-interpolation operator. If , then

 (3.4) ∫T∫ST|(v−Πhv)(x)−(v−Πhv)(y)|2|x−y|d+2sdydx≤Ch2(t−s)T|v|2Ht(SST),

where . Moreover, considering the weighted Sobolev scale (2.11), it holds that for all ,

 (3.5) ∫T∫ST|(v−Πhv)(x)−(v−Πhv)(y)|2|x−y|d+2sdydx≤Ch2(t−s−γ)T|v|2Htγ(SST).

For the purpose of this paper, we shall make use of a variant of (3.3). Even though the fractional-order norms can be localized, it is clear that the -inner product of two arbitrary functions cannot: it suffices to consider two positive functions with supports sufficiently far from each other. The following observation is due to Faermann [19, Lemma 3.1]. Since we use it extensively, we reproduce it here for completeness.

Lemma 3.2 (symmetry).

For any and bounded, there holds

 ∑T∈Th∫T∫ScTv(y)w(x)ρ(|x−y|)dydx=∑T∈Th∫T∫ScTv(x)w(y)ρ(|x−y|)dydx.
Proof.

Let

be the characteristic function of the set

. Using Fubini’s Theorem, we equivalently write

 ∑T∈Th∫T∫ScTv(y)w(x)ρ(|x−y|)dydx=∫Ωv(y)ψ(y)dy=∑T′∈Th∫T′v(y)ψ(y)dy,

where

 ψ(y):=∑T∈ThχScT(y)∫Tw(x)ρ(|x−y|)dx.

Let’s fix . Since a.e. provided and otherwise, we deduce

 ψ(y)=∫ScT′w(x)ρ(|x−y|)dx.

This yields the desired identity. ∎

Proposition 3.3 (equivalent fractional inner product).

Let . Then, it holds that

 (v,w)Hs(Ω)=∑T∈Th[∫T∫ST(v(x)−v(y))(w(x)−w(y))|x−y|d+2sdydx+2∫T∫ScTv(x)(w(x)−w(y))|x−y|d+2sdydx].
Proof.

It suffices to write

 (v,w)Hs(Ω)=∑T∈Th[∫T∫ST(v(x)−v(y))(w(x)−w(y))|x−y|d+2sdydx+∫T∫ScT(v(x)−v(y))(w(x)−w(y))|x−y|d+2sdydx]

and notice that

 ∑T∈Th∫T∫ScTv(x)w(x)|x−y|d+2sdydx=∑T∈Th∫T∫ScTv(y)w(y)|x−y|d+2sdydx,

and

 ∑T∈Th∫T∫ScTv(x)w(y)|x−y|d+2sdydx=∑T∈Th∫T∫ScTv(y)w(x)|x−y|d+2sdydx

in view of Lemma 3.2 (symmetry) with , where and we recall that is the diameter of the largest ball contained in . This completes the proof. ∎

Remark 3.4 (fractional inner product on subdomains).

Proposition 3.3 is also valid for any subdomain , i.e.

 (v,w)Hs(D)=∑T∈Th [∫T∩D∫ST∩D(v(x)−v(y))(w(x)−w(y))|x−y|d+2sdydx +2∫T∩D∫ScT∩Dv(x)(w(x)−w(y))|x−y|d+2sdydx].

Next, we write some inverse estimates that we shall use in what follows. By using standard scaling arguments, one can immediately derive the estimate

 (3.6) ∥vh∥Ht(T)≤Cinvhs−tT∥vh∥Hs(T),∀vh∈Vh,0≤s≤t≤1.

Let be a fixed smooth function. We shall also need the following variant of (3.6) with , whose proof follows immediately because the space is finite dimensional:

 (3.7) |ηvh|H1(ST)≤Chs−1T|ηvh|Hs(ST)∀vh∈Vh, T∈Th, 0≤s≤1.

3.2. Energy-norm error estimates

The discrete counterpart of (2.4) reads: find such that

 (3.8) (uh,vh)s=⟨f,vh⟩∀vh∈Vh.

Subtracting (3.8) from (2.4) we get Galerkin orthogonality

 (3.9) (u−uh,vh)s=0∀vh∈Vh.

The best approximation property

 (3.10) ∥u−uh∥˜Hs(Ω)=minvh∈Vh∥u−vh∥˜Hs(Ω)

follows immediately. Consequently, in view of the regularity estimates of discussed in Section 2, the only ingredient missing to derive convergence rates in the energy norm is some global interpolation estimate. Even though the bilinear form involves integration over , it is possible to prove that the corresponding energy norm is bounded in terms of fractional-order norms on by resorting to fractional Hardy inequalities (see [2]).

Therefore, for quasi-uniform meshes, if one can simply combine (3.3) and (3.4) with a fractional Hardy inequality [24, Theorem 1.4.4.4] to replace by [2, 10] and obtain

 (3.11) ∥v−Πhv∥˜Hs(Ω)≤C(Ω,d,s,σ,t)ht−s|v|Ht(Ω)∀v∈Ht(Ω).

In case , one cannot apply a fractional Hardy inequality. Instead, one may exploit the precise blow-up of the Hardy constant of as to deduce [2, §3.4], [10, Theorem 4.1]

 (3.12) ∥v−Πhv∥˜Hs(Ω)≤C(Ω,d,s,σ,t)εht−s−ε|v|Ht(Ω)∀v∈Ht(Ω), ε∈(0,t−s).

Alternatively, one could derive either (3.11) or (3.12) by simply interpolating standard global and estimates. However, if we aim to exploit Theorem 2.3 (weighted Sobolev estimate), then we require a suitable mesh refinement near the boundary of . For that purpose, following [24, Section 8.4] we now let the parameter represent the local mesh size in the interior of , and assume that, besides being shape-regular, the family is such that there is a number such that for every ,

 (3.13) hT≤C(σ){hμ,if T∩∂Ω≠∅,hdist(T,∂Ω)(μ−1)/μ,if T∩∂Ω=∅.

This construction yields a total number of degrees of freedom (see

[4, 10])

 (3.14) N=dimVh≈⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩h−d,if μdd−1.

Thus, if the interior mesh size and the dimension of satisfy the optimal relation (up to logarithmic factors if ). As anticipated in Remark 2.4 (optimal parameters), the weight in Theorem 2.3 (weighted Sobolev estimate) needs to be related to the parameter , which satisfies (3.13). To do so, we combine (3.3) with either (3.5) or (3.4), depending on whether intersects or not, to find the relation for . If , it suffices to use a fractional Hardy inequality to replace by [2, 10] and obtain

 (3.15) ∥v−Πhv∥˜Hs(Ω)≤⎧⎨⎩Cht−s|v|Htγ(Ω) if s≠1/2,Cεht−s−ε|v|Htγ(Ω) if% s=1/2,

for all with a constant that depends on and . On the other hand, if , we choose , where is sufficiently small, and exploit the explicit blow-up of the Hardy constant of as , as we did earlier with (3.12), to derive the second estimate in (3.15). We point out that (3.15) does not follow by interpolation of global estimates.

We gather the energy error estimates for quasi-uniform and graded meshes in a single theorem.

Theorem 3.5 (global energy-norm convergence rates).

Let be a bounded Lipschitz domain, and denote the solution to (2.4) and denote by the solution of the discrete problem (3.8), computed over a mesh consisting of elements with maximum diameter . If , then we have

 (3.16) ∥u−uh∥˜Hs(Ω)≤C(Ω,d,s,σ)hα|logh|1+κ∥f∥L2(Ω),

where and if and zero otherwise. Additionally, if satisfies an exterior ball condition, let be such that

 (3.17) β≥{2−2sif d=1,d2(d−1)−sif d≥2,and μ={2−sif d=1,dd−1if d≥2.

Then, if , and the family satisfies (3.13) with as above, we have

 (3.18)

In terms of the number of degrees of freedom , the estimate above reads

 (3.19)
Proof.

If , we combine (3.10), (3.11) with Theorem 2.1 (regularity in Lipschitz domains) with to obtain

 (3.20) ∥u−uh∥˜Hs(Ω)≤Chα−ε|u|Hs+α−ε(Ω)≤Chα−εε∥f∥L2(Ω).

In case , instead of (3.11) we use (3.12) with the same as in (2.7) to get

 (3.21) ∥u−uh∥˜Hs(Ω)≤Cεhα−2ε|u|Hs+α−ε(Ω)≤Chα−2εε2∥f∥L2(Ω).

Moreover, coupling (3.10), (3.15) and Theorem 2.3 (weighted Sobolev estimate) with and if and and if yields for

 (3.22) ∥u−uh∥˜Hs(Ω)≤Cht−s|u|Htγ(Ω)≤⎧⎪⎨⎪⎩Ch2−s−ε∥f∥Cβ(¯¯¯Ω)if d=1,Cεhd2(d−1)−εd∥f∥Cβ(¯¯¯Ω)if d≥2,

and analogous estimates hold if . Upon taking , we end up with (3.16) and (3.18), as asserted. Inequality (3.19) follows by the choice of and (3.14). ∎

Remark 3.6 (exponents of logarithms).

In case , which can only happen if , the exponents of logarithms in Theorem 3.5 can actually be reduced by a factor of (see discussion in Remark 2.4).

Remark 3.7 (optimality).

The convergence rates derived in Theorem 3.5 are theoretically optimal for shape-regular elements. Nevertheless, because we deal with continuous piecewise linear basis functions, one would expect convergence rate with respect to . It is remarkable that such a rate can only be achieved if upon grading meshes according to (3.13). For dimensions , anisotropic meshes are required in order to obtain optimal convergence rates. This limitation stems from the algebraic singular layer (2.5) and becomes more apparent as increases, but comparison of (3.16) and (3.18) shows that in all cases graded meshes improve the convergence rates with respect to .

We also point out that setting the grading parameter to be would lead to a higher rate in (3.18) in terms of the interior mesh size . However, the resulting rate in (3.19) would be the same as for (up to logarithmic factors) but the finite element matrix would turn out to be worse conditioned.

3.3. L2-norm error estimates

Upon invoking new regularity estimates for data , we now perform a standard Aubin-Nitsche duality argument to derive novel convergence rates in . We distinguish between quasi-uniform and graded meshes.

Proposition 3.8 (convergence rates in L2(Ω) for quasi-uniform meshes).

Let be a bounded Lipschitz domain. If , then for all we have

 (3.23) ∥u−uh∥L2(Ω)≤Ch2α|logh|2(1+κ)∥f∥L2(Ω),

where and if and zero otherwise.

Proof.

Let be the error, and let be the solution to (2.4) with instead of the right hand side . Then, the Galerkin orthogonality (3.9) and the Cauchy-Schwarz inequality yield

 ∥e∥2L2(Ω)=(ϕ,e)s=(ϕ−Πhϕ,e)s≤∥ϕ−Πhϕ∥˜Hs(Ω)∥e∥˜Hs(Ω),

where is a quasi-interpolation operator satisfying (3.11) if or (3.12) if . Combining these estimates with (2.7), we deduce

 (3.24) ∥ϕ−Πhϕ∥˜Hs(Ω)≲hα−(1+κ)εεκ∥ϕ∥Hs+α−ε(Ω)≲hα−(1+κ)εε1+κ∥e∥L2(Ω)∀ε>0.

This, in conjunction with the energy error estimates (3.20) and (3.21), implies

 ∥e∥L2(Ω)≲h2(α−(1+κ)ε)ε2(1+κ)∥f∥L2(Ω).

Finally, taking gives rise to (3.23). ∎

In Proposition 3.8, the assumption


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# Answer to Question #130095 in Mechanics | Relativity for Ariyo Emmanuel

Question #130095
The water in a tank is pressurized by air,
and the pressure is measured by a multifluid manometer as shown in the Fig. The
tank is located on a mountain at an
altitude of 1400 m where the atmospheric
pressure is 85.6 kPa. Determine the air
pressure in the tank if ℎ1 = 0.1 m, ℎ2 =
0.2 m, and ℎ3 = 0.35 m. Take the
densities of water, oil, and mercury to be
1000 kg/m3
, 850 kg/m3
, and 13,600
kg/m3
, respectively?
1
2020-08-26T11:24:33-0400

Solution

given data:-

atmospheric pressure( Patm)=85.6kPa

Patm =85600Pa

density of water"(\\rho_w)=1000kg\/m^3"

density of oil"(\\rho_o)=850kg\/m^3"

density of mercury"(\\rho_m)=13600kg\/m^3"

height of water column(h1)= 0.1m

height of oil column from water level(h2)=0.2m

height of mercury column (h3)=0.35m

figure can be drawn like this using data of question

pressure at point B from left side of figure can be written

"P_B =P_{air}+\\rho_ogh_1+\\rho_wgh_2" ........(eq.1)

pressure at point B from right side of figure can be written

"P_B =P_{atm}+\\rho_mgh_3" ................(eq.2)

from equation 1 and 2

"P_{air}+\\rho_ogh_1+\\rho_wgh_2=P_{atm}+\\rho_mgh_3"

"P_{air}=P_{atm}+\\rho_mgh_3-\\rho_ogh_1-\\rho_wgh_2" ...(eq.3)

"\\rho_mgh_3=13600\\times9.8\\times0.35=46648Pa"

"\\rho_ogh_1=850\\times9.8\\times0.1=833Pa"

"\\rho_wgh_2=1000\\times9.8\\times0.2=1960Pa"

by putting the value in eq.3

"P_{air}=85600+46648-833-1960"

=129455Pa

"\\colorbox{aqua}{$P_{air}=129.45KPa$}"

therefore pressure of air in the tank is 129.45KPa.

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text:
### Potential Energy, Coservation Laws

A particle move in a potential energy
$U(x)=3x^2-x^3$.
1. What is the force? draw the graph of  U(x).
2. What is the direction of the force at every point?
3. What are the equilibrium points? are they stable?
4. For given energies, what are the possible trajectories?


text:
# very simple logic question

#### noetsi

##### Fortran must die

I have been asked for the average time customers spent in a given service.

Some customers are in the service multiple times. They way I have calculated the response to the question, is the total time in services divided by the number of times the service is taken. This is really the average time the service takes.

Should I instead sum up the time each customer has gone through services (this will add up the multiple service times for those customers who have multiple services and connect it to a specific customer) and divide the sum of this by the number of customers?

Been working to much recently....this is probably too simple a question to even think about.

#### hlsmith

##### Omega Contributor
Are subsequent times indepenent? Or does the length of time change based on the number of times? Do people with multiple times differ from those with fewer?

If I go to H&R block two years in a row for my taxes, it is likely the second time is faster since I am familiar with the process and they already have data loaded into the system

#### Dragan

##### Super Moderator

I have been asked for the average time customers spent in a given service.

Some customers are in the service multiple times. They way I have calculated the response to the question, is the total time in services divided by the number of times the service is taken. This is really the average time the service takes.

Should I instead sum up the time each customer has gone through services (this will add up the multiple service times for those customers who have multiple services and connect it to a specific customer) and divide the sum of this by the number of customers?

Been working to much recently....this is probably too simple a question to even think about.

I suspect that you should compute a weighted average; which is based on what you intimated in Part 3 of your first post.

#### hlsmith

##### Omega Contributor
Is time value normally distributed?

May think about something like this if time variabe may be dependent on person:

Code:
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#### noetsi

##### Fortran must die
I am not sure whether the times are independent or not in honesty. They go to different vendors and different training so effectively one training probably has no impact on the time of the next training.

I wanted to thank hlsmith and dragan for their comments. As is often the case the decision was administrative rather than logic based. They asked us for the simplest answer, we gave them average time in service which was quickest to compute. But I learned from this thread which will help in the future.


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# Keppler's third law and the speed of cellestial bodies.

1. Aug 9, 2011

### Nikitin

Hi. Keppler's third law implicates that the velocity of an object orbiting around our sun is directly dependant on the object's average distance to the sun... Afterall, if the velocity of be more irregular, then no constant would exist.

I realize this probably has something to do with the average centripetal force provided by the gravity of the sun...

I was thinking that this is because the longer away an object is the bigger its orbit and thus the smaller the average centripetal force and thus one could assume that it is possible that the speed must be smaller as well? mass*(v^2)/r=centripetal force. But then again the radius r is also bigger if the orbit is bigger and that allows for a larger speed..

bah.

And:

The next sub-chapter of my physics book it is about Newton's work with gravity and using that jazz to prove keppler's laws.. So maybe I will find out eventually tho still I'd appreciate it if you people could make some simple explanations.

Last edited: Aug 9, 2011
2. Aug 9, 2011

### Drakkith

Staff Emeritus
The larger the radius, the lower the velocity must be in order to sustain an orbit. In an elliptical orbit the orbiting body moves faster as the distance between it and the larger body decreases, with a max velocity at its closest approach, perigee. After that it slows down until it reaches apogee, or farthest approach.

3. Aug 9, 2011

### Janus

Staff Emeritus
In a simple case, consider a circular orbit where the centripetal force needed to maintain the circular path is equal to the force of gravity or:

$$\frac{GMm}{r^2}= \frac{mv^2}{r}$$

This reduces to

$$\frac{GM}{r} = v^2$$

Given that the period(P) is the time it would take for the object to travel a distance of $2\pi r$

A little algebra will get you

$$P^2 = 4 \pi^2 \frac{r^3}{GM}$$

Which demonstrates Kepler's third law.

4. Aug 10, 2011

### Nikitin

thanks=) everyone

janus: I'll be learning about that formula you posted (force of gravity) in the next sub-chapter chapter

5. Aug 10, 2011

### Nikitin

After reading the next chapter, I understand everything, Janus. A very elegant explanation imo^^


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## Chemistry (4th Edition)

(a) 1. HCl is a strong acid, therefore: $[H_3O^+] = [HCl] = 0.0010M$ $pH = -log[H_3O^+]$ $pH = -log( 1 \times 10^{- 3})$ $pH = 3$ (b) 1.KOH is a strong base, therefore: $[OH^-] = [KOH] = 0.76M$ $pOH = -log[OH^-]$ $pOH = -log( 0.76)$ $pOH = 0.119$ $pH + pOH = 14$ $pH + 0.119 = 14$ $pH = 13.88$


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# Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$

Is there a good approximation (or useful bounds) for the median $$\nu_\alpha$$ of a $$\Gamma(\alpha,1)$$ distribution with $$0<\alpha<1$$?

I have only been able to find things like Berg & Pedersen (2006, Methods and Applications of Analysis), who generalize earlier work of Choi (1994, Proceedings of the American Mathematical Society) and give a good asymptotic expression for $$\nu_\alpha$$ as $$\alpha\to\infty$$, which works pretty well for $$\alpha\geq 1$$ - but nothing whatsoever for $$0<\alpha<1$$.

(I am interested in this because would like to update my earlier Q&A about minimizing MAPEs and other forecasting errors for gamma distributed future outcomes. So far, it only works if $$\alpha\geq 2$$, and the case $$1<\alpha<2$$, which leads to getting $$\nu_\alpha$$ for $$0<\alpha<1$$ as above, is missing.)

• Since most stats packages implement an inverse-cdf (quantile function) for the gamma, even when $\alpha<1$, presumably that could be used. Or were you after a closed-form expression? – Glen_b -Reinstate Monica Nov 27 '19 at 11:28
• @Glen_b-ReinstateMonica: yes, and that works well numerically, and yes, I was hoping for a closed form similar to the one in section 6 in Berg & Pedersen (2006). – Stephan Kolassa Nov 27 '19 at 11:34
• This is probably extremely naive, but couldn'' you calculate the medians numerically with qgamma for $\alpha$ between $0$ and $1$ and then fit some sort of polynomial model or something else to these? – COOLSerdash Nov 27 '19 at 19:20
• @COOLSerdash: that would be a possibility, yes. The convexity of the median as a function of $\alpha$ (Berg & Pedersen, 2008) would likely help. What I am looking for is error guarantees, so perhaps someone has already done something along these lines using interpolation. – Stephan Kolassa Nov 27 '19 at 20:06
• @whuber: That sounds interesting. Do you have pointers to any of those results? Do the simple approximations come with bounds? The Berg & Pedersen expansion seems to work well even non-asymptotically, it's the missing bounds that bothers me. – Stephan Kolassa Feb 25 at 7:57

Let $$\mu_\alpha$$ be the median of a $$\Gamma(\alpha)$$ distribution. This means the area under the density $$f_{\Gamma(\alpha)}(x) = \frac{x^{\alpha-1}}{\Gamma(\alpha)} \,e^{-x}$$ between $$x=0$$ and $$x=\mu_\alpha$$ equals $$1/2.$$ A graph of $$f_{\Gamma(\alpha)}$$ is sketched here in black (for $$\alpha=0.3$$), understanding the graph extends infinitely upwards as $$x$$ approaches $$0$$ and flattens down to $$0$$ as $$x$$ grows large:

The median $$\mu_\alpha$$ separates the left half of the area (darkened) from the right half. The dotted red curve is an upper bound for the Gamma density, enabling a lower bound for the median to be found by using the area under the red curve instead of the gray area.

$$1 - x \le e^{-x} \le 1$$ for $$x \ge 0$$ implies

$$\frac{x^{\alpha-1}}{\Gamma(\alpha)} \,(1-x) \le f_{\Gamma(\alpha)}(x) \le \frac{x^{\alpha-1}}{\Gamma(\alpha)}$$

which in turn gives $$\int_0^{\mu_\alpha}\frac{x^{\alpha-1}}{\Gamma(\alpha)} \,(1-x)\,\mathrm{d}x \le \int_0^{\mu_\alpha}f_{\Gamma(\alpha)}(x)\,\mathrm{d}x =\frac{1}{2} \le \int_0^{\mu_\alpha}\frac{x^{\alpha-1}}{\Gamma(\alpha)}\,\mathrm{d}x.$$

Evaluating the integrals produces

$$\frac{\mu_\alpha^\alpha}{\alpha\Gamma(\alpha)}\left(1 - \frac{\alpha}{\alpha+1}\mu_\alpha\right) \le \frac{1}{2} \le \frac{\mu_\alpha^\alpha}{\alpha\Gamma(\alpha)}.$$

This can be simplified a little by solving for $$\mu_\alpha$$ and recalling $$z\Gamma(z) = \Gamma(z+1)$$ for any $$z:$$

$$\mu_\alpha \ge \left(\frac{\Gamma(\alpha+1)}{2}\right)^{1/\alpha} \ge \mu_\alpha\left(1 - \frac{\alpha}{\alpha+1}\,\mu_\alpha\right)^{1/\alpha}.$$

One way to exploit the second inequality is to find a bound for $$\mu_\alpha/\alpha$$ when $$0 \lt \alpha \le 1$$. This expression is increasing: we have

$$\frac{\mu_\alpha}{\alpha} = e^{-\varphi(\alpha)}$$

for $$\varphi$$ as defined in equation (3) in Berg & Pedersen (2006), and Proposition 3.6 in the same paper shows that $$\varphi$$ is decreasing.

Therefore the value of $$\frac{\mu_\alpha}{\alpha}$$ at $$\alpha=1$$, equal to $$\log(2)$$, provides an upper bound

$$\mu_\alpha \le \alpha \log(2).$$

Crude as this is, it enables us to eliminate $$\mu_\alpha$$ from the denominator of the right hand side by replacing it with its upper bound, yielding the bounds

$$\left(\frac{\Gamma(\alpha+1)}{2}\right)^{1/\alpha} \le \mu_\alpha \le \left(\frac{\alpha + 1}{\alpha + 1 - \log(2) \alpha^2}\right)^{1/\alpha}\,\left(\frac{\Gamma(\alpha+1)}{2}\right)^{1/\alpha}.$$

Dividing these bounds by $$\mu_\alpha$$ gives the relative error plotted here:

The lower bound clearly is an accurate approximation for $$\alpha \lt 0.2,$$ yielding at least two significant decimal digits, while the upper bound gives at least one significant digit throughout.

Having obtained definite bounds (useful for analysis), we may adjust them to produce even more accurate estimates. For instance, $$0.9075$$ times the upper bound approximates $$\mu_\alpha$$ to within one percent relative accuracy when $$0.16\le\alpha\le 1$$ and $$1.0035$$ times the lower bound achieves $$0.4\%$$ relative accuracy for $$\alpha \lt 0.16.$$


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# How do you differentiate g(x) = (6x+9)sin(6x) using the product rule?

Apr 25, 2018

$g ' \left(x\right) = 18 \left(2 x + 3\right) \left(\cos 6 x\right) + 6 \left(\sin 6 x\right)$
$g \left(x\right) = \left(6 x + 9\right) \left(\sin 6 x\right)$
$g ' \left(x\right) = \left(6 x + 9\right) \times 6 \left(\cos 6 x\right) + \left(\sin 6 x\right) \times 6$
$g ' \left(x\right) = 18 \left(2 x + 3\right) \left(\cos 6 x\right) + 6 \left(\sin 6 x\right)$


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Volumes of revolution about 2 curves

I'm fairly okay with volumes of revolutions in general but what if I have to rotate it about 2 curves?

The question here asks:

"The region bounded by the given curves is rotated about a specific axis. Set up, but do not evaluate, an integral which gives the volume of the resulting solid by any method."

$y=4x-x^2$,$y=8x-2x^2$ about $x=-2$ and then about $y=-5$

If this asked me how to rotate about the line $x=-2$, that would be fine. How about if I am asked to rotate the curve about $x=-2$ and then $y=-5$? How would I go about doing that?

1. rotate the given region about the axis $x=-2$;
2. rotate the given region about the axis $y=-5$.
With these problems, I always find drawing a graph of both functions to be helpful. Rotating about a $y=$ line, you use the washer method with the given functions. Rotating about a $x=$ line, you will need to change the equations to functions of $y$ rather than $x$.