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All Questions 105 views Client authentication on limited hardware I'm developing a authentication and authorization protocol for a Bluetooth device. The device should communicate with an Android app and needs to be able to authenticate the app during the connection ... 643 views Attacks on AES-128 (ECB) based on some knowledge of plaintext I'm working with a third party protocol which employs AES-128 in ECB mode of operation to encrypt a packet composed of 16-byte blocks (it encrypts each block independently). I'm trying to determine if ... 139 views Does OpenTSA support PKCS11 to talk to HSMs? We are planning to implement the time-stamping service (RFC 3161) using OpenTSA wanted to know: Does OpenTSA support PKCS-11 (Cryptoki) to talk to HSMs? BTW: I did not find documentation or much ... 607 views generate hash of all handshake messages for verify data (DTLS) I am trying to create a DTLS client finished packet, where I need to generate verify data for handshake hash is need. And here I want to know how to generate handshake hash. As of now I am doing like ... 90 views Types of cryptography We have cryptography based on hard problems from number theory, like DDH. When we speak about symmetric cryptography with AES, a mode of encryption (CBC, ...), what is the type of problem ? What is ... 406 views Exactly two of the four roots must be greater than N/2 Theorem: Let $y$ be a quadratic residue in $\mathbb{Z}_N$* where $N=pq$. There are exactly four integers $x_1, x_2, x_3, x_4$ where $0 < x_1 < x_2 < \frac{N}{2} < x_3 < x_4 < N$ ... 821 views Pi Message Encryption [duplicate] I created a simple yet efficient message encrypting page that will encrypt messages using Pi. Though not being an expert in this area, I am nowhere near sure how secure this is and how easy it would ... 449 views Encrypting text file line by line with stable encryption I want to store files in a secure manner (i.e. no one should be able to decrypt the contents of the file without the secret key) but I also want a diff tool to be able to show me which lines changed (... 995 views Is there a standard way to use a nonce with HMAC? Many protocols use HMAC on messages that include a nonce, but they don't seem to do it in a consistent way. For example, in OAuth the nonce is in the middle of an URL-encoded key-value string, like ... 359 views How to find out the method used to encrypt [closed] How to find out the method used to encrypt the string 123456 which results in the value: 6g9SkDLIKa8l2VhafFnHvg==? 606 views Choice of multiplication polynomial in Rijndael s-box affine mapping The Rijndael specification details the design choices for the s-box in section 7.2. They describe the choice of affine mapping as follows: We have chosen an affine mapping that has a very simple ... 2k views Does collision resistance imply (or not) second-preimage resistance? I've seen contradictory results. Sometimes hash functions are collision-resistant but not necessarily second-preimage resistant. I've seen this kind of things in papers from Bart Preneel: “Security ... 166 views Condensed ElGamal + AES In normal ElGamal encryption, the encrypted message is a pair (gb, gabM) - that is, the actual "encryption" is simply multiplication by the shared secret ... 391 views How to convert projective to jacobian co-ordinate in ECC? I am doing a small project using elliptic curve in cryptography. My doubt is, can I directly convert a projective to a Jacobian coordinate system without using the affine conversion in elliptic curve ... 105 views 15k views What is Attribute Based Encryption? Can someone explain what attribute based encryption is? I was searching for a book or something that can help me in this regard but so far I have found none. Google also returns practically nothing ... 2k views RSA example-calculation: Public Key = Private Key (e = d) I am a bit confused. I just calculated manually the single steps of RSA for an implementation with small numbers and suddenly $d$ was equal $e$. Please help me understand what I am doing wrong. Here’... 190 views Is it a good idea to use Lagrange/Newtonian interpolation for encryption? I see that Lagrange interpolation is commonly used for secret sharing, but could it be used for encryption? The goal is to reduce database I/O and compute new values on the fly. Suppose the use case ... 264 views Encrypt-then-MAC: full random keys or keys derived from master key? I have this scenario where I use Encrypt-then-MAC (AES256-CBC and HMAC-SHA256) with keys generated by a CSPRNG (specifically, SecureRandom in Java). I'd like to know which is better: Use the CSPRNG ... 105 views Exchanging session keys over symmetric-key channel? I know session keys are exchanged (in SSL e.g.) over public-key cryptography channels but I was wondering if there are already known algorithms/methods for exchanging session keys over a symmetric-key ... 477 views How come Shamir Secret Sharing uses Lagrange interpolation? I've read that Newton polynomials have better computational complexity, but Shamir's uses Lagrange polynomials instead. Does anyone know if there are particular reason why Newton polynomials aren't ... 395 views LFSR Output Sampling for Berlekamp-Massey Looking at the use of Linear Feedback Shift Registers in cryptographic algorithms, I have learned that the Berlekamp-Massey algorithm can be used to find the (shortest) LFSR that generates a given ... 149 views Pohlig-Hellman Algorithm: Adding up the solution via CRT I have a question about the Pohlig-Hellman Algorithm for the discrete log problem. I understand the concept, but doing the exact calculations I get confused at one point; to illustrate, let's look at ... 44 views Minimum length of PKI key for signature [duplicate] I am looking for data which shows how time it would take to brute force (or crack by any other method) different sizes of RSA/DSA keys used for a digital signature. I am looking to use as small a ... 880 views Paillier can add and multiply, why is it only partially homomorphic? I've seen that it's widely accepted that before Gentry's breakthrough (which is not practical yet) in 2009 there were no known full homomorphic encryption scheme. I've read here in another answer ... 139 views What is the quantum-resistant signature scheme with the smallest signature + pubkey? After playing cat and mouse with an number of quantum-resistant signature schemes and coming up with nothing with small enough signatures, I'm turning to Crypto.SE. I need a quantum-resistant ... 318 views Elliptic Curve Crypto, is a distributed signing method possible using Shamir's Secret Sharing? Note: A distributed signature scheme exists for RSA: Practical Threshold Signatures, Victor Shoup. Is it possible to adapt such scheme for ECC? A centralized signing machine is vulnerable to ... 133 views Is there a way to compress multiple signatures of the same data? So, I'm working with a system that allows individual users to each have their own private-public key pair. I would like to allow multiple users to use their private key to sign the same piece of ... 90 views A question or few about Mental Poker I have spent some time studying the "Mental Poker" protocol (sometimes called SRA), initially proposed by Shamir, Rivest, and Adleman -- http://people.csail.mit.edu/~rivest/ShamirRivestAdleman-...	{}
Since 18 of December 2019 conferences.iaea.org uses Nucleus credentials. Visit our help pages for information on how to Register and Sign-in using Nucleus. 28th IAEA Fusion Energy Conference (FEC 2020) 10-15 May 2021 Nice, France Europe/Vienna timezone The Conference will be held virtually from 10-15 May 2021 Validation of GAE simulation and theory for NSTX(-U) and DIII-D 11 May 2021, 08:30 4h Nice, France Nice, France Regular Poster Magnetic Fusion Theory and Modelling Speaker Elena Belova (PPPL) Description New theoretical study of Alfvén eigenmodes (GAEs) in the sub-cyclotron frequency range explains the observed GAE frequency scaling with beam parameters in experiments across devices (1-3). Global Alfvén eigenmodes are frequently excited during neutral beam injection (NBI) in the National Spherical Torus Experiment (NSTX/NSTX-U) (4,5), as well as other beam-heated devices such as MAST and DIII-D (6). These modes are driven unstable through the Doppler shifted cyclotron resonance with the NBI ions, and can be excited in ITER due to super-Alfvénic velocities and strong anisotropy of the beam ions. They can also be excited by alpha particles near the outer edge of the ITER plasma due to anisotropies in the alpha particle distribution. Observations link these modes to flattening of electron temperature profiles and anomalously low central temperature at high beam power in NSTX (7), therefore, the ability to control them will have significant implications for NSTX-U, ITER, and other fusion devices where Alfvénic and super-Alfvénic fast ions might be present. Numerical simulations using the HYM code have been performed to study the excitation of GAEs in NSTX, NSTX-U and most recently for DIII-D. The HYM code is an initial value 3D nonlinear, global stability code in toroidal geometry, which treats the beam ions using full-orbit, delta-f particle simulations, while the one-fluid resistive MHD model is used to represent the background plasma. Nonlinear HYM simulations show unstable counter-rotating GAEs with toroidal mode numbers, frequencies and saturation amplitudes that match the experimentally observed unstable GAEs in NSTX-U and NSTX (1,2). New simulations performed for typical DIII-D plasma and beam parameters demonstrate that high-frequency modes with $\omega/\omega_{ci}\sim 0.6$, previously identified as compressional Alfvén eigenmodes (CAEs) (6), have shear Alfven polarization and are in fact the GAEs (Fig.[1]). Simulations show unstable counter-propagating GAEs with $\delta B_\\|<0.1 \delta B_\perp$, high toroidal mode numbers $\|n\|>20$, and frequencies close to the observed $\omega/\omega_{ci} \sim 0.6-0.7$ for NBI injection velocity $V_0/V_A=0.9$. The unstable modes have $k_\perp \rho_b\sim 0.5$, and growth rates $\gamma/\omega_{ci} \sim 0.002-0.003$. These simulation results combined with growth rate calculations (Fig.[2]) based on the local dispersion relation (1,3) explain some of the puzzling DIII-D observations. Namely, high-frequency modes (identified as CAEs in Ref.(6)) observed in DIII-D had small values of $k_\perp \rho_b$ ($k_\perp \rho_b \sim$ 0.8, in some cases $<$0.5), i.e. smaller than previously predicted (5,8) for the unstable CAEs ($1 < k_\perp \rho_b <2$) or GAEs ($2 < k_\perp \rho_b <4$); the ratio $\omega/\omega_{ci}$ remained approximately constant when the toroidal field was varied; the most unstable mode frequency scaling with $n_e$ was weaker than the Alfven speed scaling, i.e. weaker than $1/\sqrt{n_e}$; the observed frequency splitting was not consistent with theoretical predictions for CAEs. Simulation results and growth rate calculations (1,3) show that previously derived and widely cited instability conditions for counter-propagating GAE: $2< k_\perp \rho_b <4$ and for CAE: $1< k_\perp \rho_b <2$ (5,8) are valid only for higher frequency modes with $\omega\sim \omega_{ci}$, when the beam injection velocity is very large compared to the resonant velocity $V_0 \gg V_{res}$. The new theory predicts that the GAE linear growth rate is largest for small values of $k_\perp$ ($k_\perp \rho_b<1$) (Fig.[2]). For a beam ion distribution function $f(\lambda) \sim exp[-(\lambda-\lambda_0)^2/\Delta \lambda^2]$, a sufficient condition for the instability is: $1-V_{\\|res}^2/V_0^2 ≤ \lambda_0$ , where $\lambda=\mu B_0/\varepsilon$ is a pitch related parameter. The most unstable counter-GAEs have frequencies in the range (1): $(1+V_0/V_A)^{-1} < \omega/\omega_{ci} ≤ (1+V_0/V_A\sqrt{1-\lambda_0})^{-1} , (1)$ where the most unstable frequency roughly corresponds to the upper bound in Eq.(1). For DIII-D parameters ($V_0/V_A \sim 1$, $\lambda_0 \sim$0.5-0.7), this gives $0.5<\omega/\omega_{ci} ≤0.65$, consistent with observations and the simulation results. The observed scaling with $B_{tor}$ , $\lambda_0$ and weak $n_e$ scaling also agrees with Eq.(1). For example, “left beam sources” in DIII-D, have $\lambda_0 \approx$ 0.5 and excite modes with $\omega/\omega_{ci} \approx$ 0.56, whereas for “right sources” $\lambda_0 \approx$ 0.78 and $\omega/\omega_{ci} \approx$ 0.69 (6). Our theory predicts $\omega/\omega_{ci}\approx$ 0.58 and 0.68 respectively. In addition, it is shown that CAEs have the same instability condition and range of unstable frequencies as the GAEs in the limit $k_\perp \ll k_\\|$, otherwise CAE’s growth rates are much smaller than that of GAEs, consistent with simulation results and the NSTX(-U) observations. Expression (1) correctly predicts the scaling of the most unstable GAE frequencies with NBI parameters for NSTX, NSTX-U and DIII-D. Thus, for $\lambda_0 \sim$0.6 and large normalized injection velocities in NSTX ($V_0/V_A\sim$3-5), the predicted frequency range $\omega/\omega_{ci}\sim$0.2-0.3 agrees with observations. Due to stronger toroidal field in NSTX-U, and smaller relative injection velocities ($V_0/V_A≲$2), higher frequency GAEs were observed $\omega/\omega_{ci}\sim$0.4 (4). For DIII-D conditions with $V_0/V_A\sim$1, a much higher frequencies predicted consistent with the observed $\omega/\omega_{ci}\sim$0.6 (6). References 1. Belova E.V., E. Fredrickson, J. Lestz, N. Crocker, Phys. Plasmas 26, 092507 (2019). 2. Belova, E.V., Gorelenkov, N.N., Crocker, N.A., et al., Phys. Plasmas 24, 042505 (2017). 3. Lestz, J., N.N. Gorelenkov, E.V. Belova, et al., Phys. Plasmas 27, 022513 (2020). 4. E. D. Fredrickson et al., Phys. Rev. Lett. 118, 265001 (2017). 5. Gorelenkov, N.N., Fredrickson, E, Belova, et al., Nucl.Fusion 43, 228 (2003). 6. Heidbrink, W., E. Fredrickson, N. Gorelenkov, et al., Nucl. Fusion 46, 324 (2006). 7. D. Stutman, et al., Phys. Rev. Lett. 102, 115002 (2009). 8. Kolesnichenko, Ya.I., White R.B., Yakovenko, Yu.V., Phys. Plasmas 13, 122503 (2006). Affiliation Princeton Plasma Physics Laboratory United States Co-authors Jeff Lestz (Princeton University) Neal Crocker (University of California Los Angeles) Eric Fredrickson (PPPL) Presentation Materials There are no materials yet.	{}
## Relaxed Locally Correctable Codes with Improved Parameters by Vahid Reza Asadi and Igor Shinkar I am naturally very interested in this paper, although it does fall short of separating LDCs from RLDCs. While is constructs a $q$-query RLDC that encodes a message of length $k$ using a codeword of block length $O(k^{1+O(1/{q}}))$, the LDC lower bound asserts block length $\Omega(k^{1+(2/(q-2)}))$, for any even $q\geq4$. Adding to the account in the abstract, I would mention that a recent lower bound by Gur and Lachish that asserts that $q$-query RLDC that encodes a message of length $k$ must use codewords of block length $n = \Omega(k^{1+1/\Omega(q^2)})$. Correction: All foregoing bounds refer to non-adaptive algorithms. This makes the upper bounds stronger, but the lower bounds weaker (although $q$ adaptive queries can be emulated by $2^q$ non-adaptove queries). This gap is closed in a recent work. #### The original abstract Locally decodable codes (LDCs) are error-correcting codes $C : \Sigma^k \to \Sigma^n$ that admit a local decoding algorithm that recovers each individual bit of the message by querying only a few bits from a noisy codeword. An important question in this line of research is to understand the optimal trade-off between the query complexity of LDCs and their block length. Despite importance of these objects, the best known constructions of constant query LDCs have super-polynomial length, and there is a significant gap between the best constructions and the known lower bounds in terms of the block length. For many applications it suffices to consider the weaker notion of relaxed LDCs (RLDCs), which allows the local decoding algorithm to abort if by querying a few bits it detects that the input is not a codeword. This relaxation turned out to allow decoding algorithms with constant query complexity for codes with almost linear length. Specifically, [Ben+06] constructed an $O(q)$-query RLDC that encodes a message of length $k$ using a codeword of block length $n = O(k^{1+1/\sqrt{q}})$. In this work we improve the parameters of [Ben+06] by constructing an $O(q)$-query RLDC that encodes a message of length $k$ using a codeword of block length $O(k^{1+1/{q}})$. This construction matches (up to a multiplicative constant factor) the lower bounds of [KT00; Woo07] for constant query LDCs, thus making progress toward understanding the gap between LDCs and RLDCs in the constant query regime. In fact, our construction extends to the stronger notion of relaxed locally correctable codes (RLCCs), introduced in [GRR18], where given a noisy codeword the correcting algorithm either recovers each individual bit of the codeword by only reading a small part of the input, or aborts if the input is detected to be corrupt. Available from ECCC TR20-142. Back to list of Oded's choices.	{}
• 2 Vote(s) - 4 Average • 1 • 2 • 3 • 4 • 5 Searching for an asymptotic to exp[0.5] tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 08/01/2014, 11:36 PM Some further thought : $a_n = \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1}) + ln(n-1))$ There is good news and bad news. Basicly both are that this is in a way very similar to the Original estimate for a_n combined with our estimate for a_(n+1). $a_n = (n-1) \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1})$ $a_n / (n-1) = \exp(\exp^{0.5}(h_{n+1}) - (n+1) h_{n+1})$ Just as our estimate for a_(n+1) from a_n and the Original estimate of a_n. But maybe a refinement of the previous post helps. Am I wrong in assuming we prefer to solve for one variable ? regards tommy1729 « Next Oldest \| Next Newest » Messages In This Thread Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/07/2014, 12:22 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/08/2014, 04:25 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/10/2014, 12:14 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/10/2014, 11:31 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/10/2014, 11:48 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/10/2014, 11:58 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/09/2014, 11:19 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/10/2014, 11:56 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/13/2014, 04:23 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/14/2014, 05:54 AM RE: Searching for an asymptotic to exp[0.5] - by JmsNxn - 05/12/2014, 03:48 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/12/2014, 03:56 PM RE: Searching for an asymptotic to exp[0.5] - by JmsNxn - 05/12/2014, 05:06 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/12/2014, 11:35 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/13/2014, 11:44 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/14/2014, 11:42 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/15/2014, 06:15 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/15/2014, 09:49 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/16/2014, 07:27 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/14/2014, 12:20 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 07/17/2014, 05:46 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/15/2014, 08:53 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/16/2014, 09:36 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/16/2014, 10:16 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/18/2014, 06:14 PM RE: Searching for an asymptotic to exp[0.5] - by JmsNxn - 05/22/2014, 12:16 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/22/2014, 07:08 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/22/2014, 08:31 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/22/2014, 10:16 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/23/2014, 10:53 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/25/2014, 03:00 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/29/2014, 11:09 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 07/12/2014, 07:46 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/29/2014, 11:32 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/29/2014, 11:54 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/30/2014, 09:41 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 06/28/2014, 11:16 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/28/2014, 09:52 PM RE: Searching for an asymptotic to exp[0.5] - by JmsNxn - 06/29/2014, 01:40 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 06/30/2014, 12:56 AM RE: Searching for an asymptotic to exp[0.5] - by JmsNxn - 06/30/2014, 03:21 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 06/30/2014, 11:56 PM RE: Searching for an asymptotic to exp[0.5] - by JmsNxn - 07/01/2014, 12:35 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 06/30/2014, 01:27 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/01/2014, 10:10 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 07/01/2014, 11:41 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/02/2014, 09:53 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/10/2014, 11:48 PM RE: Searching for an asymptotic to exp[0.5] - by MorgothV8 - 07/13/2014, 06:48 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/14/2014, 12:27 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/14/2014, 11:16 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/15/2014, 08:22 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/15/2014, 09:43 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/15/2014, 09:48 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/24/2014, 12:10 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/24/2014, 10:47 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 07/25/2014, 02:46 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/24/2014, 10:54 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/26/2014, 12:21 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/27/2014, 08:37 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/27/2014, 05:01 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/28/2014, 12:17 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/28/2014, 10:30 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 07/30/2014, 04:07 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/01/2014, 11:20 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/01/2014, 11:36 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/02/2014, 12:26 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 08/02/2014, 03:44 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/02/2014, 11:02 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/02/2014, 11:48 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 08/03/2014, 04:54 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/03/2014, 08:46 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 08/03/2014, 12:06 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/03/2014, 12:10 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/05/2014, 11:31 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/08/2014, 10:28 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/09/2014, 12:24 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 08/10/2014, 06:08 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/01/2014, 10:24 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/03/2014, 01:04 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/02/2014, 07:46 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/02/2014, 07:53 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/08/2014, 12:56 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/08/2014, 04:15 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/08/2014, 11:03 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/09/2014, 04:33 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/09/2014, 06:26 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/10/2014, 11:02 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/11/2014, 08:02 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/11/2014, 02:13 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/12/2014, 07:49 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/12/2014, 06:35 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/13/2014, 07:15 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/13/2014, 11:25 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/13/2014, 11:45 PM RE: Searching for an asymptotic to exp[0.5] - by jaydfox - 09/13/2014, 11:49 PM RE: Searching for an asymptotic to exp[0.5] - by jaydfox - 09/14/2014, 12:00 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/14/2014, 05:07 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/15/2014, 03:53 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/14/2014, 09:34 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/16/2014, 12:14 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/16/2014, 12:27 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/18/2014, 10:20 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/18/2014, 11:07 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/19/2014, 12:23 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/29/2014, 11:40 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/19/2014, 04:02 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/03/2014, 01:26 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/03/2014, 10:49 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/03/2014, 11:34 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/03/2014, 11:39 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/04/2014, 09:41 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/04/2014, 10:38 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/05/2014, 11:58 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 11/07/2014, 12:27 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 03/28/2015, 11:11 PM RE: Searching for an asymptotic to exp[0.5] - by marraco - 03/29/2015, 12:59 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/25/2015, 10:24 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/25/2015, 10:52 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/15/2015, 06:45 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/15/2015, 06:55 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/17/2015, 01:45 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/18/2015, 09:34 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/18/2015, 09:56 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/18/2015, 10:09 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/31/2015, 04:57 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/31/2015, 05:12 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/15/2015, 10:22 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/16/2015, 02:49 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/16/2015, 03:23 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/26/2015, 07:36 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/03/2015, 10:31 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/05/2015, 08:16 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/09/2015, 12:17 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/12/2015, 01:14 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/14/2015, 01:30 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/18/2015, 11:31 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/21/2015, 10:53 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/21/2015, 05:58 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/24/2015, 08:10 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/25/2015, 12:59 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 09/25/2015, 08:26 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/26/2015, 12:24 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/29/2015, 12:28 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 10/01/2015, 07:56 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/30/2015, 12:25 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/30/2015, 09:27 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/01/2015, 11:25 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/02/2015, 02:56 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/03/2015, 10:42 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/06/2015, 12:11 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/06/2015, 12:26 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/08/2015, 07:52 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/08/2015, 12:26 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/08/2015, 10:43 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/08/2015, 11:08 PM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 10/09/2015, 08:15 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/09/2015, 11:56 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 10/10/2015, 03:08 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/10/2015, 07:40 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/08/2015, 11:12 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/09/2015, 07:18 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/10/2015, 08:15 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/10/2015, 08:26 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/11/2015, 07:17 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/17/2015, 11:59 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/18/2015, 11:07 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/18/2015, 11:22 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/19/2015, 12:20 AM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 10/27/2015, 01:27 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 02/16/2016, 03:17 AM RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 02/18/2016, 06:51 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 02/17/2016, 01:25 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 02/18/2016, 12:53 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 02/18/2016, 01:11 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 02/23/2016, 01:01 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 02/23/2016, 01:23 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 03/21/2016, 01:26 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 04/05/2016, 01:29 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/04/2016, 07:21 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/04/2016, 08:16 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/04/2016, 08:29 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/06/2016, 03:12 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/06/2016, 03:47 PM RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 03/15/2018, 01:23 PM Possibly Related Threads... Thread Author Replies Views Last Post Merged fixpoints of 2 iterates ? Asymptotic ? [2019] tommy1729 1 586 09/10/2019, 11:28 AM Last Post: sheldonison Another asymptotic development, similar to 2sinh method JmsNxn 0 2,559 07/05/2011, 06:34 PM Last Post: JmsNxn Users browsing this thread: 1 Guest(s)	{}
Some aspects of fluctuations of random walks on R and applications to random walks on R+ with non-elastic reflection at 0 Abstract : In this article we refine well-known results concerning the fluctuations of one-dimensional random walks. More precisely, if $(S_n)_{n \geq 0}$ is a random walk starting from $0$ and $r\geq 0$, we obtain the precise asymptotic behavior as $n\to\infty$ of $\mathbb P[\tau^{>r}=n, S_n\in K]$ and $\mathbb P[\tau^{>r}>n, S_n\in K]$, where $\tau^{>r}$ is the first time that the random walk reaches the set $]r,\infty[$, and $K$ is a compact set. Our assumptions on the jumps of the random walks are optimal. Our results give an answer to a question of Lalley stated in \cite{L}, and are applied to obtain the asymptotic behavior of the return probabilities for random walks on $\mathbb R^+$ with non-elastic reflection at $0$. Keywords : Type de document : Article dans une revue ALEA : Latin American Journal of Probability and Mathematical Statistics, Instituto Nacional de Matemática Pura e Aplicada, 2013, 10 (2), pp.591-607 Domaine : Littérature citée [19 références] https://hal.archives-ouvertes.fr/hal-00780453 Contributeur : Kilian Raschel <> Soumis le : vendredi 28 juin 2013 - 11:07:45 Dernière modification le : jeudi 7 février 2019 - 16:52:08 Document(s) archivé(s) le : dimanche 29 septembre 2013 - 04:36:30 Fichier Fluctuations.pdf Fichiers produits par l'(les) auteur(s) Identifiants • HAL Id : hal-00780453, version 2 Citation Rim Essifi, Marc Peigné, Kilian Raschel. Some aspects of fluctuations of random walks on R and applications to random walks on R+ with non-elastic reflection at 0. ALEA : Latin American Journal of Probability and Mathematical Statistics, Instituto Nacional de Matemática Pura e Aplicada, 2013, 10 (2), pp.591-607. 〈hal-00780453v2〉 Métriques Consultations de la notice 207 Téléchargements de fichiers	{}
# Prove that $\text{exp}(tX)=\alpha_X(t)$ for all $t\in\mathbb{R}$? Let $G$ be a Lie group and $\mathfrak{g}$ be the lie algebra of $G$. We know that for any $X\in\mathfrak{g}$ there exists an unique $\alpha_X:(\mathbb{R},+)\longrightarrow (G,\cdot)$ one-parameter subgroup such that $X_{\alpha_X(t)}=\alpha_X'(t)$ for all $t\in\mathbb{R}$. If the exponential map is defined by: $$\begin{array}{rcll} \text{exp}:&\mathfrak{g}&\longrightarrow & G\\ &X&\longmapsto& \text{exp}(X)=\alpha_X(1) \end{array}$$ How can I prove the following statement? If $X\in\mathfrak{g}$ then $\text{exp}(tX)=\alpha_X(t)$ for all $t\in\mathbb{R}$? Fix $s\in\Bbb R$. By uniqueness of solutions to ODEs we find $\alpha_{sX}(t)=\alpha_X(st)$ for all $t\in\Bbb R$. In particular, taking $t=1$, we find $\alpha_{sX}(1)=\alpha_X(s)$. By the definition of $\exp$, this means $$\exp(sX)=\alpha_X(s),\quad\forall s\in\Bbb R.$$	{}
Chronoamperometry is a technique where the potential of the working electrode is stepped for a specified period of time. Current is plotted as a function of time. Chronoamperometry is also known as potential step amperometry. ### Detailed Description Like most of the other electrochemical techniques offered by the AfterMath software, this experiment begins with an induction period. During the induction period, a set of initial conditions is applied to the electrochemical cell and the cell is allowed to equilibrate to these conditions. The default initial condition involves holding the working electrode potential at the initial potential for a brief period of time (i.e., 3 seconds). After the induction period, the potential of the working electrode is stepped to a specified potential for a period of time. After the step has finished, the experiment concludes with a relaxation period. The default condition during the relaxation period involves holding the working electrode potential at the initial potential for an additional brief period of time (i.e., 1 seconds). At the end of the relaxation period, the post experiment idle conditions are applied to the cell and the instrument returns to the idle state. Current is plotted as a function of time, resulting in a chronoamperogram. You may also choose to do some post experiment processing in order to generate a Cottrell plot. ### Parameter Setup The parameters for this method are arranged on various tabs on the setup panel. The most commonly used parameters are on the Basic tab. Additional tabs for Ranges and post experiment idle conditions are common to all of the electrochemical techniques supported by the AfterMath software. Finally, a Post Experiment Processing tab deals with manipulating the data automatically when the experiment is finished. ### Basic Tab You can click on the “I Feel Lucky” button (located at the top of the setup) to fill in all the parameters with typical default values (see Figure 1). You will no doubt need to change the Potential and Hold time in the Forward step period box to values which are appropriate for the electrochemical system being studied. You may also want to change the Number of intervals in the Sampling Control box. Figure 1: Basic Chronoamperometry Setup The Electrode Range on the Basic tab is used to specify the expected range of current. If the choice of electrode range is too small, actual current may go off scale and be truncated. If the electrode range is too large, the chronoamperogram may have a noisy, choppy, or quantized appearance. Some Pine potentiostats (such as the WaveNow and WaveNano portable USB potentiostats) have current autoranging capability. To take advantage of this feature, set the electrode range parameter to “Auto”. This allows the potentiostat to choose the current range “on-the-fly” while the chonoamperogram is being acquired. The waveform that is applied to the electrode is a simple pulse to the Potential listed in the Forward step period box. Note that the actual waveform that is measured (see Figure 2, red trace) fluctuates slightly compared to the applied potential (see Figure 2, black trace). Figure 2: Waveform for CA. Black trace = applied potential, Red trace = measured potential. ### Ranges Tab AfterMath has the ability to automatically select the appropriate ranges for voltage and current during an experiment. However, you can also choose to enter the voltage and current ranges for an experiment. Please see the separate discussions on autoranging and the Ranges Tab for more information. ### Post Experiment Conditions Tab After the Relaxation Period, the Post Experiment Conditions are applied to the cell. Typically, the cell is disconnected but you may also specify the conditions applied to the cell. Please see the separate discussion on post experiment conditions for more information. ### Post Experiment Processing Tab The Post Experiment Processing Tab (see Figure 3) allows you to automatically generate Cottrell current or Cottrell charge plots. Please see the Typical Results and Theory sections of this wiki for more information regarding Cottrell plots. Figure 3: Post Experiment Processing Options. ### Typical Results The results for a $2 mM$ solution of ferrocene in $0.1 M Bu_4NClO_4/MeCN$are shown in Figure 4. Figure 4: Chronoamperogram of a ferrocene solution using a potential = $0.75 V vs. Ag/AgCl (aq)$ and a $2 mm$ Pt WE. If you selected to automatically generate Cottrell plots, the plots are under the other plots folder in the Archive navigation panel. Choosing Cottrell current displays a plot of $i\;vs. t^{-1/2}$ (see Figure 5A). Choosing Cottrell charge displays a plot of $Q \;vs. t^{-1/2}$ (see Figure 5B). Note that for the Cottrell Current plot, the level portion in the plot is actually the time prior to the current spike shown in Figure 4. That is, earlier time points are to the right in a Cottrell Current Plot. This is not the case for a Cottrell Charge plot because integrating the current with respect to time gives charge. The endpoint in a Cottrell Charge plot is the total amount of charge passed during the experiment. Monitoring the charge passed during an experiment is a variant of chronoamperometry called Chronocoulometry. See the Applications section for an example of Chronocoulometry. Figure 5 : Post experiment processing plots. A – Cottrell Current $(i)$, B – Cottrell Charge $(Q)$. Conditions as listed in Figure 4. The Cottrell current plot is a useful diagnostic tool to examine that your species of interest is freely diffusion in solution. Upon applying the potential step, the current initially spikes, then begins trail off. As will be explained in more detail in the Theory section, the current during the trailing off period is a diffusion-limited current dictated by the Cottrell equation. Examining the Cottrell Current plot in this region reveals that the current is linear with respect to $t^{-1/2}$ (see black ellipse in Figure 6). Figure 6: Highlight of Diffusion-limited Current in Cottrell Current Plot. The addition of a Baseline tool to the diffusion limited current region allows for calculation of the diffusion coefficient for the species of interest if the concentration and electrode area are known (see Figure 7). First you will have to delete the region of the Cottrell Plot where current is not diffusion limited. Use the Point Selection Tool to delete the unwanted points of the Cottrell Plot. Next, add a baseline tool by right clicking on the leftover point and select Add Tool » Baseline. Manipulate the control points to provide an adequate fit. AfterMath automatically provides the slope and intercept for the Baseline tool. Figure 7: Addition of Baseline Tool in Diffusion-limited Current Region. The slope of the line in the plot is given by the equation $slope={\frac{nFAD_O^{1/2}C_O^}{{\pi}^{1/2}}}$ where $n$ is the number of electrons, $F$ is Faraday’s Constant $(96485 C/mol)$, $A$ is the electrode area $(cm^2)$, $D_O$ is the diffusion coefficient $(cm^2/s)$, and $C$ is the concentration $(mol/cm^3)$. In the example above, $D$ is calculated to be $1.6{\times}10^{-5}cm^2/s$. ### Theory The theory section is split into two segments. The first segment deals with Chronoamperometry (CA) and the second section deals with Chronocoulometry (CC). Chronoamperometry leads to Chronocoulometry natually since charge is obtained by integrating current with respect to time. ### Chronoamperometry The following is a basic description of the theory behind Chronoamperometry. Please see the literature1 for a more detailed description of the technique. Consider a reaction $O+e\;{\rightleftharpoons}\;R$ with a formal potential $E^{{\circ}'}$. In general, the potential step applied to the working electrode should be sufficiently more negative than $E^{{\circ}'}$ such that reduction of $O$ to $R$ is complete at the surface of the electrode (i.e. surface concentration of $O$ at the electrode surface is $0$). When this occurs, the current is diffusion-limited, much like the current that flows in CV after the potential of the electrode sweeps past $E_p$. When the current is diffusion-limited in CA, the current-time response is described by the Cottrell2 equation $i=\frac{nFAD_O^{1/2}C_O^}{({\pi}t)^{1/2}}&s=3$ where $n$ is the number of electrons, $F$ is Faraday’s Constant $(96485 \:C/mol)$, $A$ is the electrode area $(cm^2)$, $D_O$ is the diffusion coefficient $(cm^{2}/s)$, and $C_O^$ is the concentration $(mol/cm^3)$. As described in the Typical Results section, plotting $i \;vs. t^{-1/2}$ gives a slope that contains the diffusion coefficient, $D$. ### Chronocoulometry The following is a brief description of chronocoulometry. Please see the literature1 for a more detailed description of the technique. Chronocoulometry is advantages in some ways over chronoamperometry. First is that the signal grows with time, meaning that the later portion of the experiment are least distorted by the nonideal potential rise, giving better signal-to-noise ratios. Second, integrating smooths random noise making the data cleaner. Third, contributions from double-layer charging and reactions of adsorbed species can be separating from those of freely diffusion species. The cumulative charge passed during the experiment is given by the equation $Q_d={\frac{2nFAD_O^{1/2}C_O^t^{1/2}}{\pi^{1/2}}}$ where the parameters are as described in the above section. Typically, a plot of $Q \;vs.\; t^{-1/2}$ has a non-zero y-intercept. This intercept is related to double-layer charging and reduction of adsorbed $O$. The overall equation describing the charge is then given by $Q_d={\frac{2nFAD_O^{1/2}C_O^t^{1/2}}{\pi^{1/2}}}+Q_{dl}+nFA\Gamma_O$ where $Q_{dl}$ is the charge due to double-layer charging and $nFA\Gamma_O$ is the charge due to reduction of adsorbed $O$. $Q_{dl}$ and $nFA\Gamma_O$are not easily separated and usually require additional experiments such as double potential step chronoamperometry. However, approximations for $Q_{dl}$ can be made by performing the same potential step in a solution of only electrolyte. This is only an approximation because the true double-layer charging will be influenced by adsorbed $O$. The Application section in the next tab contains examples of both Chronoamperometry and Chronocoulometry. ### Application In the first example, Tennyson et al.3 used chronoampometry to determine the diffusion coefficients (D) and number of electrons (n) for a series of indirectly connected bimetallic complexes. In this instance, the researchers used microelectrodes to generate steady-state currents. Plotting of the function ${\frac{i(t)}{i_{ss}}}$ $vs.$ $t^{-1/2}$ allows for the calculation of $D_O$ without knowledge of $C_O^$ or $n$. Though the experiments are relatively simple and straightforward, the researchers then used these values for more complicated electrochemical experiments that enabled them to determine the degree of electronic coupling between the two metallic centers. Understanding the degree of electronic coupling could allow for facilitation to otherwise inaccessible catalysts and materials. The next example uses chronoamperometry to measure extremely small diffusion coefficients of redox polyether hybrid cobalt bipyridine molten salts. Crooker and Murray4 performed chronoamperometry on a series of undiluted molten salts and were able to obtain diffusion coefficients as low as $10^{-17} cm^2/s$. The researchers went on to obtain heterogeneous rate constants $(k^0)$ using CV and were able to show that the reaction remains quasireversible despite small $k^0$ rates because $D_O$ values are so low. The next example uses chronoamperometry to directly measure rate constants. Smalley et al.5 examined a series of monolayers terminated with either a ferrocene or ruthenium redox moiety. For monolayers containing more than 11 methylene units, chronoamperometry was used to determine the heterogeneous rate constants $(k^0)$. The researchers also determined preexponential factors and found them to be much lower that expected based aqueous solvent dynamics. The fourth example uses chronocoulometry. Wolfe et al.6 prepared a series of ferrocenated hexanethiolate protected Au nanoparticles. Normallly, TGA would be used to calculate the number of thiolates per nanoparticle. However, in this case, TGA was inconclusive due to a slow loss of mass over the applied temperature range. The researchers instead monitored the charge pass over time for solution of nanoparticles with known concentrations. Knowing the total charge passed and the particle concentration allowed them to determine the number of ligands per nanoparticle. This is a good example of how electrochemistry can be used in place of a more traditional characterization technique. The final example also uses chronocoulometry, but rather than using it for quantification purposes, the researchers use it to evoke a change in the system. Zhang et al.7 use chronocoulometry with predefined endpoints to produce Ag nanoparticles, confined on a electrode surface, with varying shapes and sizes. By varying the amount of charge passed the researchers were able to show that triangular nanoparticles oxidize first on the bottom edge, followed by triangular tips, followed by out-of-plane height. Since the researchers also coupled spectroscopy to their electrochemical experiments they were also able to show how the SPR band changes with shape evolution. This is a nice example of how electrochemistry can be used to evoke a change in a system rather than simply to monitor changes or determine physical parameters of the system.	{}
Tags: forensics Rating: --- title: "Write-up - Sunshine CTF 2016 - That's No Moon" date: 2016-03-14 00:00:00 tags: [Write-up, Sunshine CTF 2016] summary: "Write-up about Sunshine CTF 2016 - That's No Moon" --- For this Forensics challenge, we need to download a picture of a moon. The file is available [here](https://raw.githubusercontent.com/d0tslashpwn/ctf-files/master/Sunshine-CTF-2016/forensics/thats-no-moon/moon.png). After trying to play with colors with [Steganabara](https://www.wechall.net/downloads/by/user_name/ASC/page-1) or [Stegsolve](https://www.wechall.net/forum/show/thread/527/Stegsolve_1.3/page-1) to find a hidden text, we change of strategy. We use the strings command. By the end of the output, we can see a flag.txt is hidden inside the file. % strings moon.png ... flag.txtUT *nAb( flag.txtUT) Let is change the extension of the file to a zip archive and extract it. % mv moon.png moon.zip % unzip moon.zip Archive: moon.zip warning [moon.zip]: 411781 extra bytes at beginning or within zipfile (attempting to process anyway) A password is required. After some guessing, we notice that the challenge is titled _That's No Moon_. So let is try moon as a password. Success! The flag is sun{0kay_it_is_a_m00n}.	{}
# Eavesdropping! (Physics-Sound) You are trying to overhear a juicy conversation, but from your distance of 21.0m , it sounds like only an average whisper of 25.0dB . So you decide to move closer to give the conversation a sound level of 75.0dB instead. How close should you come? $dB = 10log(I / I_0)$ $25 =10log(I_1/I_0)$ $I_1/I_0 = 10^2.5$ $75=10*log(I_2/I_0)$ $I_2/I_0 =10^7.5$ $I_2/I_1 =10^7.5-2.5 =10^5$ We know that $I = C/R^2$ (because the sound wave front is spherical) where $C$ is a constant $I_2/I_1= (R_1/R_2)^2 =10^5$ $(21/R_2)^2 =100000$ $R_2^2 =21^2/100000 =0.00441 m$ $R_2 = \sqrt{R_2^2} =6.6 cm$	{}
# If a cold front is approaching, a weather forecaster might predict _____. A. hurricanes B. clear skies C. steady drizzle D. thunderstorms ###### Question: If a cold front is approaching, a weather forecaster might predict _____. A. hurricanes B. clear skies D. thunderstorms ### A 24g sample of carbon combines with 64g of oxygen to form CO2. What is the mass of the reactants? What is the mass of the product? Which law do these data support? A 24g sample of carbon combines with 64g of oxygen to form CO2. What is the mass of the reactants? What is the mass of the product? Which law do these data support?... ### Which best describes the benefits of renting a home? Which best describes the benefits of renting a home?... ### How many Olmec colossal heads have been discovered? How many Olmec colossal heads have been discovered?... ### Assume you have an XML Tree variable called quiz initialized with the following valid XML document. You are not required to draw the tree, but you might find it helpful. <?xml version="1.0" encoding="UTF-8"?> 155 What is the size of this tree? What would be the value returned by quiz. child(0).child(1).child(0).label() What would be the value returned by quiz. child(1).attributeValue("number") Assume you have an XML Tree variable called quiz initialized with the following valid XML document. You are not required to draw the tree, but you might find it helpful. <?xml version="1.0" encoding="UTF-8"?> 155 What is the size of this tree? What would be the value returned by quiz. child(0)... ### Can anybody help me with this please hurry Can anybody help me with this please hurry... ### You want crown? hvvvvvvvvvvvvvvvvvvvvvv you want crown? hvvvvvvvvvvvvvvvvvvvvvv... ### Which plant and animal species are characteristics of the Sahara desert in Africa which plant and animal species are characteristics of the Sahara desert in Africa... ### 1 2 ( − 1 . 4 + 0 . 4 ) = + 0 . 2 1 2 ( − 1 . 4 + 0 . 4 ) = + 0 . 2... ### Match each art movement with one of its defining concerns. Romanticism ? Representation of virtue Impressionism 2 Opposition to the violence of war Post- Impressionism 2 Depiction of gritty street life Neoclassicism ? Focus on the joy of modern life Realism Individualistic approach to style Match each art movement with one of its defining concerns. Romanticism ? Representation of virtue Impressionism 2 Opposition to the violence of war Post- Impressionism 2 Depiction of gritty street life Neoclassicism ? Focus on the joy of modern life Realism Individualistic approach to style... ### Helppp I need to pass critical! Oh um will give brainliest if requested! Helppp I need to pass critical! Oh um will give brainliest if requested!... ### How can teeth be used in a forensics investigation? Explain How can teeth be used in a forensics investigation? Explain... ### If a star appears to be very red (like Betelgeuse), it is likely to be in which spectral class? If a star appears to be very red (like Betelgeuse), it is likely to be in which spectral class?... ### Which expression represents the area of the rectangle? A) 4x2 + 19x + 17.5 B) 6x2 + 38x + 17.5 C) 8x2 + 38x + 35 D) 12x + 24 The expressions are (4x+5) and (2x+7) Which expression represents the area of the rectangle? A) 4x2 + 19x + 17.5 B) 6x2 + 38x + 17.5 C) 8x2 + 38x + 35 D) 12x + 24 The expressions are (4x+5) and (2x+7)... ### I can't answer that question. Someone help me please :] The gene for albinism is expressed only when it is in pair, and it is located at the same locus of chromosomes that have similar genetic charges, skin color being normal, transmitted by a gene that is expressed, even in a single dose. According to the highlighted words, mark the concept that seems correct, respectively: a) homologous, alleles, recessive, dominant. b) alleles, recessive, alleles, dominant. c) recessive, dominant, alle I can't answer that question. Someone help me please :] The gene for albinism is expressed only when it is in pair, and it is located at the same locus of chromosomes that have similar genetic charges, skin color being normal, transmitted by a gene that is expressed, even in a single dose. Acc...	{}
# Lecture Notes math214:04-03 ## 2020-04-03, Friday $$\gdef\End{\text{ End}}$$ ## Parallel Transport. Let $\gamma: [0,1] \to M$ be an embedded smooth curve. (If you worry about the boundary, think of an embedded curve $(-\epsilon, 1+\epsilon) \to M$.) Let $E \to M$ be a vector bundle, $\nabla$ be a connection. Our goal is to define the isomorphism $$P_\gamma: E_{\gamma(0)} \to E_{\gamma(1)}.$$ Suppose $u_0 \in E_{\gamma(0)}$, we want to find a section $u_t \in E_{\gamma(t)}$, such that $$\nabla_{\dot \gamma(t)} u_t = 0.$$ Namely, we should have a 'constant' (or flat) section $t \mapsto u_t$, living over the image of $\gamma$. The above statement is correct morally, however, $\nabla_{\dot \gamma(t)} u_t$ notation is problematic since $u_t$ is only a section living on a line, not on an open set of $M$. There are two ways to make this rigorous. 1. One is to go to a coordinate patch. Say image of $\gamma$ is contained in a trivializing patch $U$ of $E$, and we have $\{e_\alpha\}$ a frame of $E$, and $x^1, \cdots, x^n$ are base coordinate, then we may express a section using the coefficients $$u_t = \sum_{\alpha} u^\alpha(t) e_\alpha(\gamma(t)) \in E_{\gamma(t)}.$$ The collection of coefficients $u^\alpha(t)$ should satisfy a system of ODE $$\frac{d u^\alpha(t)}{dt} + \Gamma^\alpha_{i \beta}(\gamma(t)) \dot \gamma^i(t) u^{\beta}(t) = 0.$$ 2. The second way, is to define the pullback bundle $\gamma^E$ and the pull-back connection $\gamma^\nabla$. In fact, this can be defined more generally. Let $(M, E, \nabla)$ be a bundle with connection, and $F : N \to M$ be a smooth map. We can define $F^E$ the pull-back bundle on $N$, by setting $(F^E)_p = E_{F(p)}$ for any $p \in N$, and we can define the connection on $F^E$ by setting $$(F^\nabla)_{X_p} (F^* s) = (\nabla_{F_(X_p)} s)\|_p$$ where $s$ is a section of $E$ defined near $F(p)$, and $X_p$ is a tangent vector in $T_p N$. see this mathoverflow discussion for why this defines the pullback connection. ## Curvature Prop 3.3.8 [Ni] We may extend $\nabla: \Omega^0(M, E) \to \Omega^1(M, E)$ to $\nabla: \Omega^k(M, E) \to \Omega^{k+1}(M, E)$, such that it satisfies the Leibniz rule. If $\omega \in \Omega^r(M)$ and $u \in \Omega^s(M, E)$, then $$d^\nabla( \omega \wedge u) = d(\omega) \wedge u + (-1)^{\|\omega\|}\omega \wedge \nabla(u).$$ ([Ni] uses $d^\nabla$ for this extension, where I still use $\nabla$.) Prop For any smooth function $f \in C^\infty(M)$ and $\omega \in \Omega^r(M, E)$, we have $$(\nabla^2) (f \omega ) = f \nabla^2(\omega)$$ Proof: This is a calculation worth doing, $$\nabla( df \omega + f \nabla(\omega)) = dd(f) \omega - df \nabla(\omega) + df \nabla(\omega) + f \nabla^2(\omega) = f \nabla^2(\omega).$$ Recall that, if a map $a: \Omega^k(M, E) \to \Omega^{k+s}(M, E)$ is a $C^\infty(M)$-linear map, then the action of $a$ is point-wise (no derivative of section of $E$ is neede). In other word, we may view $a \in \Omega^s(M, \End(E))$. Curvature We define the curvature $F_\nabla = \nabla^2 \in \Omega^2(M, \End(E))$ #### Local connection 1-form and Curvature 2-form Suppose we have a local trivialization $\{e_\alpha\}$ over $U \In M$. Then, we have an induced trivial connection $d_U$ on $E\|_U$, and we may write $\nabla\|_U = d_U + A$, for some $A \in \Omega^1(U, \End(E))$. We then have $$F_\nabla\|_U = (\nabla\|_U)^2 = (d_U + A)^2 = d_U A + A \wedge A$$ where we define $\wedge: \Omega^r(M, \End(E)) \times \Omega^s(M, \End(E)) \to \Omega^{r+s}(M, \End(E))$ by $\wedge$ on the form factor, and compose on the associative algebra factor $\End(E)$ $$(\eta \ot a) \wedge (\xi \ot b) = (\eta \wedge \xi) \ot (ab), \quad \eta, \xi \in \Omega^(M), a,b \in\Omega^0(\End(E)).$$ ### Useful Formula In the following, $\omega$ denote $k$-form, $u$ denote section of $E$. $X, Y, Z$ are vector fields. We have $$\nabla_X(\omega \ot u) = L_X(\omega) \ot u + \omega \ot \nabla_X(u)$$ $$i_X \nabla + \nabla i_X = \nabla_X$$ $$i_X i_Y + i_Y i_X = 0$$ $$\nabla_X i_Y - i_Y \nabla_X = i_{[X,Y]}$$ For example, we test the last formula on $\Omega^1(M, E)$, $\eta \in \Omega^1(M)$, $u$ a section of $E$, $$(\nabla_X i_Y - i_Y \nabla_X) (\eta \ot u) = \nabla_X( \eta(Y) \ot u) - i_Y (L_X(\eta)\ot u + \eta \nabla_X u)$$ $$= L_X( \eta(Y)) \ot u + \eta(Y) \ot \nabla_X(u) - [Y (\eta(X)) + d\eta(X, Y)] \ot u - \eta(Y) \ot \nabla_X(u)$$ $$= \iota_{[X,Y]}(\eta) \ot u$$ Some formula about curvature $$F(X, Y) = [\nabla_X, \nabla_Y] - \nabla_{[X,Y]}$$ In local coordinates $x_i$ on $U$, we have $$F_{ij} = - F_{ji} = [\nabla_i, \nabla_j]$$ where $\nabla_i = \nabla_{\d_i}$. math214/04-03.txt · Last modified: 2020/04/03 00:42 by pzhou	{}
# Problem with Union and Intersection Consider the following: list={{a,b,c},{c,d,e},{d,e,f},...{x_,y_,z_}}; I would like to apply Union on the elements of the list in the following way Union[{a,b,c},{c,d,e},{d,e,f},...{x_,y_,z_}]; The problem Union[list] does not return the desired result when applied on list. Please consider the following example: list={{10,2,3},{2,3,4},{2,3,50}}; Union[list] The same problem occures with Intersection. (* Out={{10,2,3},{2,3,4},{2,3,50}} ) Union[{10,2,3},{2,3,4},{2,3,50}] ( Out={2,3,4,10,50} *) - You're looking for Apply? – Rojolalalalalalalalalalalalala Apr 19 '12 at 21:50 Union@Flatten@list – F'x Apr 19 '12 at 21:51 Applyworks. Many thanks – John Apr 19 '12 at 21:52 You're looking for Apply Apply[Union, list] which can be written in short form as Union@@list - Union eliminates duplicate elements in a list, or duplicate sublists in a list of lists. In[9]:= list = {{a, a, a}, {a, a, a}, {d, e, f}}; Union[list] Out[10]= {{a, a, a}, {d, e, f}} list={{10,2,3},{2,3,4},{2,3,50}}; Union[list] even though parts of the sublists are the same, no sublist was a complete duplicate of any other sublist. So there was nothing for Union to do. - +1 for the explanation why the OP's example does not behave as expected. – István Zachar Jun 1 '12 at 9:46 Welcome to Mathematica.SE and thanks for taking the time to explain this. Please consider registering your account so your reputation score will be preserved and you can access to more features of the site. – Szabolcs Jun 1 '12 at 9:55 Just stir things up, you can also do the following: Union[Sequence@@list] It uses the same function referenced (Map), but makes forces the problem to look like you were expecting: Union[{a,b,c},{c,d,e},{d,e,f},...{x_,y_,z_}]; -	{}
## Abstract and Applied Analysis ### Boundedness for a Class of Singular Integral Operators on Both Classical and Product Hardy Spaces Chaoqiang Tan #### Abstract We found that the classical Calderón-Zygmund singular integral operators are bounded on both the classical Hardy spaces and the product Hardy spaces. The purpose of this paper is to extend this result to a more general class. More precisely, we introduce a class of singular integral operators including the classical Calderón-Zygmund singular integral operators and show that they are bounded on both the classical Hardy spaces and the product Hardy spaces. #### Article information Source Abstr. Appl. Anal., Volume 2014, Special Issue (2013), Article ID 987214, 7 pages. Dates First available in Project Euclid: 7 October 2014 https://projecteuclid.org/euclid.aaa/1412687020 Digital Object Identifier doi:10.1155/2014/987214 Mathematical Reviews number (MathSciNet) MR3176784 #### Citation Tan, Chaoqiang. Boundedness for a Class of Singular Integral Operators on Both Classical and Product Hardy Spaces. Abstr. Appl. Anal. 2014, Special Issue (2013), Article ID 987214, 7 pages. doi:10.1155/2014/987214. https://projecteuclid.org/euclid.aaa/1412687020 #### References • C. Fefferman and E. M. Stein, “${H}^{p}$ spaces of several variables,” Acta Mathematica, vol. 129, no. 3-4, pp. 137–193, 1972. • R. F. Gundy and E. M. Stein, “${H}^{p}$ theory for the poly-disc,” Proceedings of the National Academy of Sciences of the United States of America, vol. 76, no. 3, pp. 1026–1029, 1979. • L. Carleson, “A counterexample for measures bounded on Hp for the bi-disc,” Mittag Leffler Report 7, 1974. • S. Y. Chang and R. Fefferman, “A continuous version of duality of ${H}^{1}$ with BMO on the bidisc,” Annals of Mathematics, vol. 112, no. 1, pp. 179–201, 1980. • S. Y. Chang and R. Fefferman, “Some recent developments in Fourier analysis and ${H}^{p}$-theory on product domains,” Bulletin of the American Mathematical Society, vol. 12, no. 1, pp. 1–43, 1985. • C. Q. Tan, “Boundedness of classical Calderón-Zygmund convolution operators on product Hardy space,” Mathematical Research Letters, vol. 20, no. 3, pp. 591–599, 2013. • R. Fefferman, “Calderón-Zygmund theory for product domains: ${H}^{p}$ spaces,” Proceedings of the National Academy of Sciences of the United States of America, vol. 83, no. 4, pp. 840–843, 1986. • C. Q. Tan, “Singular integral operators between the classical type and the product type,” Acta Scientiarum Naturalium Universitatis Sunyatseni, vol. 51, no. 5, pp. 58–62, 2012. • Y. S. Han, G. Z. Lu, and K. Zhao, “Discrete Calderón's identity, atomic decomposition and boundedness criterion of operators on multiparameter hardy spaces,” Journal of Geometric Analysis, vol. 20, no. 3, pp. 670–689, 2010. • Y.-S. Han, “Plancherel-Pólya type inequality on spaces of homogeneous type and its applications,” Proceedings of the American Mathematical Society, vol. 126, no. 11, pp. 3315–3327, 1998. • E. M. Stein, “Note on singular integrals,” Proceedings of the American Mathematical Society, vol. 8, pp. 250–254, 1957. \endinput	{}
# Behavior of $\sum_{k=0}^n (-1)^k {2n \choose k} (1-\frac{k}{n})^2 \ln(1-\frac{k}{n})$ I came across the sum $\sum_{k=0}^n (-1)^k {2n \choose k} (1-\frac{k}{n})^2 \ln(1-\frac{k}{n})$ while working through some integrals, and I believe that it tends as $\frac{4^{n}}{n^{5/2}}$. I'm particularly interested in the coefficient on this putative dominant term in an asymptotic expansion as n tends to infinity. I tried Taylor expanding the piece $(1-\frac{k}{n})^2 \ln(1-\frac{k}{n})$, which yields a relatively friendly series. Then my next step would be to evaluate sums of the form $\sum_{k=0}^n (-1)^k {2n \choose k} k^m$ for some arbitrary natural number m. I saw a more specialized sum with more complete bounds here. However, I'm having difficulty generalizing the accepted answer's technique with my bounds. I found, via derivatives of the trigonometric power formula for sin^2n, that $\sum_{k=0}^n (-1)^k {2n \choose k} (1-\frac{k}{n})^{2m}=0$ for natural m < n. That might help if one expands just the logarithm. Note that I take $x^2 \ln(x)$ to vanish at x=0 so that I can write my upper bound to n for aesthetic reasons alone. • Good question but one thing missed: is $n$ tending to infinity? – Szeto Jun 19 '18 at 5:42 • Yep, n is tending to infinity. I've added that in. – user196574 Jun 19 '18 at 5:44 As J. D'Aurizio notes, the proposer's question is answered if the asymptotics of his eq (1) can be determined. I'll outline a proof of the following: $$\sum_{k=1}^n (-1)^k \binom{2n}{n+k} k^2 \log{k} \sim \binom{2n}{n} \frac{\pi}{4}\Big( 7\frac{\zeta(3)}{\pi^3} + \frac{93}{n} \frac{\zeta(5)}{\pi^5} + ...\Big).$$ The $\zeta(n)$ are Riemann's zeta. The OP conjectured an asymptotic form, which is correct up to an alternating factor, and asked what the value of the leading constant is. By using the asymptotics for the central binomial, $$\sum_{k=0}^n (-1)^k \binom{2n}{k} (1-k/n)^2 \log{(1-k/n)} \sim \frac{4^n}{n^{5/2}}\, (-1)^n \, \frac{7\zeta(3)}{4\pi^2\sqrt{\pi}}.$$ My top equation is not so difficult to derive if one should have the identity available, $$(A)\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = \binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}.$$ Differentiate with respect to $s$ and set it afterwards to 2. Use the asymptotic expansion for the ratio of gamma functions to get $$\sum_{k=1}^n (-1)^{k} \binom{2n}{n+k} k^2 \log{k} = \binom{2n}{n} \frac{\pi}{2} \int_0^\infty \frac{dx \, \,x^2}{\sinh{\pi x}} \big(1+\frac{x^2}{n} + ...\big).$$ Integrate term-by-term and the integrals are known by Mathematica in terms of $\zeta(n)$. I don't know of a reference that has (A). My proof is long and I'll skip it unless there is suitable interest. I was really surprised at the final result because I was expecting a $\log(n)$ to be in the mix. • "A" is quite the striking identity! It seems lucky that s=2 causes the sine term to vanish on product rule, was the identity designed for such? I'll be posting a whole host of curious and surprising sums over the next few days. – user196574 Jun 20 '18 at 20:28 • proof for $A$? thx :-) – tired Jun 20 '18 at 23:25 • tired and user196574: posted my proof as MSE 2827591. Question asks for a generalization or simplification. – skbmoore Jun 21 '18 at 18:37 A partial answer. $$\frac{1}{n^2}\sum_{k=0}^{n}(-1)^k \binom{2n}{k}(n-k)^2 = \delta(n-1)$$ hence it is enough to find the asymptotic behaviour of $$\frac{(-1)^n}{n^2}\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^2\log k\tag{1}$$ where for any $n\geq 2$ we have $$\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k = \frac{1}{2}\binom{2n}{n} \tag{O}$$ $$\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k = -\frac{n}{4n-2}\binom{2n}{n} \tag{A}$$ $$\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^2 = 0 \tag{B}$$ $$\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^3 = \frac{n^2}{2(2n-1)(2n-3)}\binom{2n}{n}. \tag{C}$$ We have $H_k=\log k+\gamma+O\left(\frac{1}{k}\right)$ and $$-\frac{\log(1-z)}{1-z}=\sum_{n\geq 1} H_n z^n\tag{D}$$ $$\sin^{2n}(x)=\frac{1}{4^n}\binom{2n}{n}+\frac{(-1)^n}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^k\binom{2n}{k}\cos(2(n-k)x)\tag{E}$$ hence the asymptotic behaviour of the sum in which $\log u$ is replaced by $H_u$ can be found through the following approach: 1. Apply $\frac{d^2}{dx^2}$ to $(E)$; 2. Evaluate $(D)$ at $z=e^{2ix}$ and $z=e^{-2ix}$; 3. Write the wanted combinatorial sum, where $\log k$ has been replaced by $H_k$, as a contour integral, by exploiting Parseval's theorem and the previous points; 4. Estimate such integral through the saddle point/Laplace method. On the other hand, what we really need to find is the derivative at $m=2$ of $\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^m$. • Could you explain further the replacement of log(k) by the harmonic number? It seems to me that the 1/k^3 and higher corrections might have a non-negligible contribution to the sums. For example, for very large n it seems that we have approximately $\sum_{k=1}^n {2n \choose n+k} (-1)^k \frac{1}{k} = -\ln(2) {2n \choose n}$ which would make such a term of order $4^n / \sqrt{n}$ after Stirling's approx and would arise from an O(1/k^3) correction to the harmonic number approximation. – user196574 Jun 19 '18 at 20:29 • I've made another question addressing sums of the form $\sum_{k=0}^n {2n \choose n+k} (-1)^k k^m$ and their large-n form. I've quoted some of your results O through C there since I find them really interesting! Specifically how all seem to depend on n similarly in the large-n limit. – user196574 Jun 19 '18 at 22:05 • @user196574: you are right, I assumed that the contribution provided by $\frac{1}{k},\frac{1}{k^3}$ were negligible, but that might not be the case. I am going to fix the answer and clarify it is just a partial one. – Jack D'Aurizio Jun 19 '18 at 22:10	{}
# Math Help - tangent line equation of curve.. 1. ## tangent line equation of curve.. This is im sure very easy, but it was a busy summer. I am going through some questions early in my calc book, and im not sure of the best/easiest way to tackle this question. (please explain steps) The point P(2, ln 2) lies on the curve y = ln x. Using the point Q(2.001, ln 2.001) on the curve, a reasonable equation of the tangent line to the curve at P(2, ln 2) is? Cheers 2. Originally Posted by captain_jonno This is im sure very easy, but it was a busy summer. I am going through some questions early in my calc book, and im not sure of the best/easiest way to tackle this question. (please explain steps) The point P(2, ln 2) lies on the curve y = ln x. Using the point Q(2.001, ln 2.001) on the curve, a reasonable equation of the tangent line to the curve at P(2, ln 2) is? Cheers I suppose that by "reasonable" they mean you calculate the equation of the line passing through the points P and Q . Use the folllowing, which is Junior H.S. stuff: Equation of the line passing through $(x_1,y_1)$ and with slope equal to $m$ : $y-y_1=m(x-x_1)$ Slope of the line passing through the points $(x_1,y_1)\,,\,(x_2,y_2)\,:\;\frac{y_2-y_1}{x_2-x_1}$ , unless $x_1=x_2$ and then the line passing through these points is a vertical one and its slope is undefined. Tonio 3. Ok super, thanks. so y= x/2 -lnx +1 or what do you think 4. Originally Posted by captain_jonno Ok super, thanks. so y= x/2 -lnx +1 or what do you think $m=\frac{\ln(2.001)-\ln 2}{2.001-2}\cong \frac{1}{2}$ , so the wanted line is: $y - \ln 2=\frac{1}{2}(x-2)\Longrightarrow y = \frac{x}{2}+\ln 2 - 1$	{}
Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. # Localized electronic vacancy level and its effect on the properties of doped manganites ## Abstract Oxygen vacancies are common to most metal oxides and usually play a crucial role in determining the properties of the host material. In this work, we perform ab initio calculations to study the influence of vacancies in doped manganites $$\text {La}_{(1-\text {x})} \text {Sr}_{\text {x}} \text {MnO}_{3}$$, varying both the vacancy concentration and the chemical composition within the ferromagnetic-metallic range ($$0.2\,<\,\text {x}\,<\,0.5$$). We find that oxygen vacancies give rise to a localized electronic level and analyse the effects that the possible occupation of this defect state can have on the physical properties of the host. In particular, we observe a substantial reduction of the exchange energy that favors spin-flipped configurations (local antiferromagnetism), which correlate with the weakening of the double-exchange interaction, the deterioration of the metallicity, and the degradation of ferromagnetism in reduced samples. In agreement with previous studies, vacancies give rise to a lattice expansion when the defect level is unoccupied. However, our calculations suggest that under low Sr concentrations the defect level can be populated, which conversely results in a local reduction of the lattice parameter. Although the exact energy position of this defect level is sensitive to the details of the electronic interactions, we argue that it is not far from the Fermi energy for optimally doped manganites ($$\text {x}\,\sim \,1/3$$), and thus its occupation could be tuned by controlling the number of available electrons, either with chemical doping or gating. Our results could have important implications for engineering the electronic properties of thin films in oxide compounds. ## Introduction It is well known that oxygen vacancies in perovskite oxides have a great impact on the functional properties of these materials, modifying the interplay among lattice, charge, spin, and orbital degrees of freedom. In mixed-valence manganites, the coexistence of ferromagnetism and metallic conduction has been traditionally described via a combination of a strong electron-phonon interaction, and the double-exchange mechanism in which the Mn 3d electrons hop through the O 2p orbitals. An oxygen vacancy disrupts this mechanism, inducing structural, electronic and magnetic changes that could have dramatic consequences. A considerable number of studies, both experimental and theoretical works, have tried to identify these effects, mitigating the negative consequences that they could pose for applications, for example in spintronics1,2. In other cases, understanding the effect of oxygen vacancies is pursued to improve their positive consequences, such as for applications in catalysis, or solid oxide fuel cells3. X-ray photoemission (XPS) and photoabsorption spectroscopies (XAS) on oxygen or manganese core-levels have been used to identify the presence of oxygen vacancies in samples grown under different oxygen environments, or subject to annealing treatments. Although O-1s photoemission and O-K edge XAS spectra show features that have been correlated to oxygen vacancy states, probably the most clear evidence for their presence comes from the Mn-L edge XAS spectra. In particular, a peak or shoulder at $$\sim \,640\,\text {eV}$$ has been attributed to the presence of $$\text {Mn}^{+2}$$, that is a deviation from the characteristic $$\text {Mn}^{+3}/\text {Mn}^{+4}$$ valence ratio4. It can be argued that the oxygen vacancy acts as electron donor, and the excess electron is localized in the proximity of the defect, giving rise to such $$\text {Mn}^{+2}$$ peak. Indeed, first principles DFT calculations have demonstrated the existence of a defect level in the minority spin gap2,5,6,7. The partial occupation of this level results in the lost of half-metallicity and has been linked to the degradation of the magnetic properties5,6. However, calculations show that the position of the defect level is very sensitive to structural relaxations2,7 and it is usually located above the Fermi level, close to the bottom of the conduction band, hence decreasing the minority spin-gap. Ultraviolet photoemission measurements also reveal a significant reduction of the work function in samples where the $$\text {Mn}^{+2}$$ peak is observed in XAS8, evidencing that the extra electrons left behind by the oxygen vacancy play an important role on the physical properties of the material. The splitting of the Mn 3s XPS spectra is commonly used to estimate the oxidation state (and magnetic moment) of Mn atoms. In samples with significant presence of oxygen vacancies, an increase in the splitting was reported7,9 corresponding to a decrease in the formal valence of Mn ions. However, the opposite trend was found in samples annealed in UHV for an extended period of time (where oxygen vacancies are expected to develop), and not exposed to air5. It can be argued that under these conditions, the $$\text {Mn}^{+2}$$ state was not formed at the surface5,10, and that the XPS spectra reveals an increase of the bonding covalency. Interestingly, depopulation of the defect level in previous DFT calculations seems to be related to a weakening of the Mn bonding covalency and an increase of the Mn-Mn distance, which correlates nicely with a slight expansion of the lattice observed in X-ray diffraction experiments7,11. In any case, these contradictory results might indicate that under certain circumstances the electronic defect level could be occupied, and under other conditions, the level is empty. The aim of this paper is to determine under what conditions could the defect state be occupied and to clarify what are the consequences in the materials properties. Using first principles calculations, we provide a description at the atomic level of the electronic, structural and magnetic properties of bulk $$\text {La}_{(1-\text {x})} \text {Sr}_{\text {x}} \text {MnO}_{3}$$ (LSMO) manganite with oxygen vacancies. We focus on the most studied and technologically relevant doping concentration, namely $$\text {La}_{2/3} \text {Sr}_{1/3} \text {MnO}_{3}$$, usually referred to as optimal-doping, mainly due to the fact that it has the highest Curie temperature $$\sim \,370\,\text {K}$$12. To study the influence of both global and local Sr doping, we work in the composition range $$0.2\,<\,\text {x}\,<\,0.5$$, where the metallic-ferromagnetic character is preserved, and we address the effect of vacancy concentration considering different sizes of supercell. In this way, we can tune the position of the Fermi energy and modify the occupation of the defect state, which could form a dispersive band for large concentration of vacancies due to defect-defect interactions. The role of electronic correlations is also reviewed, and a comparison with previous works is discussed. By analysing the character of the vacancy electronic level along with the effect of structural parameters and the magnetic interactions, we connect our results with experimental evidences published in the literature. Our main findings can be summarized as: (i) oxygen vacancies give rise to a new electronic defect level that has its main contribution mostly from neighbouring Mn d orbitals pointing towards the vacant site; (ii) this defect level acts as a scattering center that might be partially responsible for the diminished transport properties observed in reduced LSMO samples; (iii) the occupation of this bonding level reduces the interatomic distance of the surrounding Mn leading to a local shrinking of the cell volume; (iv) conversely, depopulation of this defect level results in an increased interatomic distance leading to an overall volume expansion; (v) oxygen vacancies produce a weakening of the ferromagnetic interaction, as revealed by the substantial reduction of the exchange energy for the Mn ions close to the vacancy, regardless of the defect state occupation, and in agreement with the observed degradation of the magnetic properties of reduced LSMO. ## Results ### Pristine LSMO The $$\text {La}_{(1-\text {x})} \text {Sr}_{\text {x}} \text {MnO}_{3}$$ ($$0.2\,<\,\text {x}\,<\,0.5$$) crystal structure belongs to the $$a^{-}a^{-}a^{-}$$ Glazer’s tilt system13, which presents equal counterphase rotations of the octahedra around the three pseudocubic crystallographic axes. All the stoichiometric (pristine) supercells used in this study display this distortion pattern, with minimal dispersion of the octahedra rotations due to the direct substitution of the La cations for Sr dopant. Details can be found in the Supplementary Information (SI). The octahedral crystal field gives rise to the Mn 3d levels splitting into a lower energy triplet $$\text {t}_{\text {2g}}$$ and a higher energy $$\text {e}_{\text {g}}$$ doublet. Figure 1 plots the projected density of states (PDOS) and the orbital-projected Mn 3d and O 2p band structure for each spin channel. For the majority spin (up channel), the $$\text {t}_{\text {2g}}$$ level is completely filled while the $$\text {e}_{\text {g}}$$ level is just partially filled due to the holes introduced by the Sr doping. Therefore, the Mn 3d $$\text {e}_{\text {g}}$$ and O 2p levels extend around the Fermi level vicinity. This agrees with the fact that we do not observe any appreciable Jahn-Teller distortion of the octahedra14. For the minority spin (down channel), there is a pseudo-bandgap that decreases when increasing the Sr doping (see also Fig. S2 of SI online). The valence band maximum (VBM) is composed of O 2p orbitals while the conduction band minimum (CBM) has mainly empty Mn 3d $$\text {t}_{\text {2g}}$$ orbitals. Sr and La levels are very high in energy, well above the Fermi level. Consequently, within the range of doping considered in this work, the material presents half-metallicity with a 100% spin-polarization at the Fermi level2,15. As the amount of Sr doping is increased (i.e. more holes are added), there is a rigid shift of the levels due to the Fermi energy moving downwards, but the half-metallic character of the system remains. Addition of a Hubbard term shifts the $$\text {t}_{\text {2g}}^{\uparrow }$$ levels towards lower energies and the $$\text {t}_{\text {2g}}^{\downarrow }$$ ones to higher energies. This results in an increased oxygen hybridization and delocalization of electronic levels, and a widening of the pseudo-bandgap with GGA + U, as compared to the bare GGA results. As expected, there is a minor underestimation of the pseudo-bandgap value ($$\text {E}_{\text {g}}$$) for bare GGA, and a better agreement with the experimentally reported gap16 when a value of $$\text {U}\,\sim \,1\,\text {eV}$$ is used (Fig. S2 in SI online). Our results also reproduce the strong ferromagnetic coupling resulting from the double-exchange Mn interaction mediated by oxygen atoms, as flipping the magnetization for any single Mn atom in this scenario has an energy cost of roughly 0.5 eV. ### Reduced LSMO: oxygen vacancies in the bulk In this section we focus our study on the reduced host, namely $$\text {La}_{(1-\text {x})} \text {Sr}_{\text {x}} \text {MnO}_{3-\delta }$$, by introducing oxygen vacancies in a supercell structure. The size of the supercell determines the defect concentration, or the degree of reduction in the system ($$\delta$$). The diluted limit is studied by using a $$2\times 2\times 6$$ supercell with a low defect concentration ($$\delta \,=\,1.4\,\%$$), and the vacancy-rich limit is addressed with a $$2\times 2\times 2$$ supercell corresponding to a larger defect concentration ($$\delta \,=\,4.2\,\%$$). The oxygen vacancy formation energy ($$\text {E}_{\text {for}}$$) is calculated by evaluating the Kohn-Sham energy of the pristine system $$\text {E}_{\text {pristine}}$$, the reduced system $$\text {E}_{\text {defect}}$$ and the chemical potential of the removed oxygen atom, $$\mu _{\text{ O }}$$: \begin{aligned} E_{\text {for}}\,=\,E_{\text {defect}}-E_{\text {pristine}}+\mu _{\text {O}} \end{aligned} (1) where the chemical potential for an oxygen atom is determined from the energy of the oxygen molecule in the gaseous phase, $$\mu _{\text {O}}=1/2E_{\text {O}_2}$$, thus assuming an oxygen-rich environment. Equation (1) is a simplification of a general expression extensively used in DFT approaches17,18 which is restricted in our case for neutral defects, because our system is metallic and the electronic reservoir is determined by the position of the Fermi level. The lack of an oxygen atom bridging two neighbors Mn ions reduces the electronic hopping behind the double-exchange ferromagnetic mechanism, and favors superexchange interactions that can result in anti-parallel magnetic configurations for the two Mn atoms linked to the vacancy. To explore this hypothesis, in the following we analyze two different magnetic states illustrated in Fig. 2: (i) the ferromagnetic configuration (which we name FM), where all Mn have parallel magnetic moments; and (ii) a local antiferromagnetic arrangement that involves a spin-flip process for one of the Mn atoms that is nearest neighbour to the vacancy (which we label as SF configuration). #### Diluted limit To explore the limit corresponding to a diluted vacancy concentration, where vacancies are considered as isolated defects within the bulk of LSMO, we take a $$2\times 2\times 6$$ supercell with 120 atoms. This supercell is large enough to simulate a variety of Sr compositions and in addition, allows us to explore the effect of different local distributions of dopants (in particular, the effect of Sr clustering). With this purpose, we take two possible arrangements for the Sr atoms within the cell: (a) an ordered distribution were the Sr cations are uniformly dispersed across the bulk and local effects are minimized; and (b) a non-uniform distribution inspired on the concept of a random substitutional binary alloy, where the occupation probability of each A-site is given by the composition weight of the element. We refer to the latter as random distribution. Figure 3 illustrates two examples of the supercells used. Regarding oxygen vacancy concentration, we study the diluted limit with a reduction of $$\delta \,=\,1.4\,\%$$ ($$\text {La}_{(1-\text {x})} \text {Sr}_{\text {x}} \text {MnO}_{2.958}$$). We verified, by placing one oxygen vacancy on the LaSrO ab-planes of the supercell (highlighted in Fig. 3), that the interaction between periodic image replicas is negligible (this is not the case when the vacancy is introduced in a $$\text {MnO}_2$$ plane). Besides, it is also reasonable to assume that a very low concentration of defects (diluted limit) will only affect the structure locally, and no significant changes in the lattice constants should happen. Therefore, in this limit we allow relaxation of the internal atomic coordinates, without modification on the lattice cell parameters (our relaxed geometries have values of internal stress lower than 0.8 GPa). When an oxygen vacancy is formed, the two electrons left behind are available for the bulk states. Consequently, there are visible modifications in the electronic structure, particularly in the levels of the Mn atoms surrounding the defect. This can be seen in Fig. 4 which compares the projected density of states (PDOS) for a reference bulk-like Mn atom (that is, a Mn atom located far from the defect, named as $$\text {Mn}_{\text {far}}$$) with the PDOS corresponding to the Mn atoms which are nearest neighbors (NN) to the oxygen vacancy, $$\text {Mn}_{\text {NN}}$$. The location of the oxygen vacancy is such that the Mn-$$\text {V}_{\text{ O }}$$-Mn chain is aligned along the z axis. Figure 4b corresponds to the FM case, where both $$\text {Mn}_{\text {NN}}$$ atoms are equivalent, thus only the PDOS corresponding to one of them is shown. The energy of the $$\text {d}_{\text {z}^{2}-\text {r}^{2}}$$ level, which in the pristine system overlaps with the oxygen 2p orbitals at − 6 eV, is now shifted upwards in energy by $$\sim$$ 5.5 eV, as schematically highlighted by the black arrow. Moreover, the contribution coming from the $$\text {d}_{\text {z}^{2}-\text {r}^{2}}^{\uparrow }$$ to the broad $$\text {e}_\text {g}$$ conduction band characteristic of LSMO half-metallicity in Fig. 4a, becomes more localized and forms a peak at about − 2 eV, with just a little tail crossing the Fermi energy in 4b. The $$\text {d}_{\text {x}^{2}-\text {y}^{2}}$$ on the other hand, remains mostly unaffected by the vacancy. Notice also that the removal of the oxygen atom transforms the symmetry of the system from octahedral to square pyramidal, thus lifting the degeneracy of the $$\text {t}_{\text {2g}}$$ levels. There is, however, a feature that stands out when the vacancy is formed: a defect level in the minority spin channel. We argue that the position of this level relative to the Fermi energy critically determines the behaviour of the system. This defect state, can be either above or below the Fermi energy, thus changing its occupation depending on the Sr doping (i.e. the amount of holes added into the system). As a result, a slope change is observed in the formation energy as a function of Sr doping (central panel in Fig. 4), and three regions can be defined within the composition range analyzed: low, intermediate and high doping respectively, which we discuss in the following. Let us first address the case where the Sr distribution is homogeneous (ordered), and there is no local clustering. At low doping concentration (equivalent to a few number of holes in the system) the defect level is occupied (OL) and the FM configuration is the ground state, as can be seen in the central panel of Fig. 4. The total magnetic moment in the unit cell does not change upon the formation of an oxygen vacancy, which is consistent with a picture where the two electrons left behind by the vacancy have opposite spins. The defect state has mostly s-type character, with bonding contribution from the neighbouring Mn $$\text {d}_{\text {z}^{2}-\text {r}^{2}}$$ orbitals. Consequently, these $$\text {Mn}_{\text {NN}}$$ ions shorten their interatomic distance and reduce their valence charge, as compared to the reference bulk values (see Tables 1 and 2, respectively). Although the reduced valence correlates well with the image of the $$\text {Mn}^{+2}$$ peak observed in XAS experiments and relates to the formation of vacancies close to the surface of the manganite7,20, the reduced Mn-Mn distance goes in the opposite direction to a lattice expansion reported for vacancy-rich samples11,21. At high doping concentration, the increased number of holes introduced by the Sr dopant lowers the Fermi energy and the defect level becomes depopulated (UL). The electron that was localized in this state, is now distributed in the available electronic levels, which in this case are the $$\text {e}_{\text {g}}^\uparrow$$ ones. Hence, the extra spin is distributed among the bulk Mn ions, resulting in a slight increase (by about $$0.13\,\mu _{B}$$) of their magnetic moment, as can be seen in Table 2. As a consequence, the total magnetic moment of the system is increased by $$+2\,\mu _{B}$$ with respect to the pristine reference value. From a structural viewpoint, the depopulation of the (bonding) defect level increases the $$\text {Mn}_{\text {NN}}$$ interatomic distance, which is in agreement with the aforementioned volume expansion. The above mentioned increase in the system magnetization induced by the defect seems counterintuitive with the degradation of the reported saturation magnetization11. To elucidate this issue, we explore spin-flip (SF) configurations for the neighbouring Mn atoms (Fig. 2b) to have an estimation of the strength of the ferromagnetic interactions. Although this SF configuration is less stable than the FM solution, their energy difference ($$\lesssim \,0.2\,\text{eV}$$) is significantly diminished with respect to the pristine LSMO ($$\sim \,0.5\,\text {eV}$$). Analysis of the PDOS for both $$\text {Mn}_{\text {NN}}$$ is shown in the shaded lateral panels in Fig. 4. The most remarkable feature is that now the defect level shifts to higher energies, falling above the Fermi level (both in the low and high doping regions). Otherwise, the PDOS for $$\text {Mn}_{\text {NN}}^{\uparrow }$$ resembles the FM state, with $$\text {d}_{\text {z}^{2}-\text {r}^{2}}$$ state shifting by $$\sim \,6.5\,\text {eV}$$ for the spin down towards higher energies. For $$\text {Mn}_{\text {NN}}^{\downarrow }$$ the image is very similar, except that the peaks are sharper possibly due to a reduced interaction with the host FM matrix. Interestingly, as a function of Sr doping, the energy difference between the SF and the FM configurations decreases in the low doping region while increases in the high doping one. This defines an intermediate doping region, where the FM and SF configurations are almost degenerate in energy (central panel of Fig. 4). This region can also be identified by a change in the slope of the defect formation energy versus the Sr doping. It is worth mentioning that the so called optimally-doped manganite composition, $$\text {La}_{2/3} \text {Sr}_{1/3} \text {MnO}_{3}$$, falls within this intermediate region. Electronically, the Fermi level falls on top (or very close) to the defect state in the FM configuration and this could drive an electronic instability, favouring the spin flipping in one of the neighbour Mn atoms. To tune the occupation of the defect level, we perform a Fix Spin (FS) calculation, by forcing a constrain over the total magnetic moment of the simulation cell. Taking the pristine LSMO magnetic moment as a reference, we impose a $$+2\,\mu _{B}$$ increase in the FM configuration ($$\Delta \text {M}$$ in Table 1), which forces the extra electrons from the vacancy to occupy majority spin states. In the low doping regime, this procedure artificially pushes the defect level in the minority spin above the Fermi level. The vacancy formation energy in this situation is shown in Fig. 4 with blue symbols, and is higher than the FM case. This excited state is metastable, and releasing the FS constrain on the magnetic moment provokes the system to come back to the FM ground state. We use the same approach in the high doping regime to induce the occupation of the defect level by fixing the total magnetic moment to the pristine value. Again, this results in higher metastable energy solutions (shown by blue crosses in the central panel of Fig. 4). In this way, using constrained and unconstrained calculations, we can set the defect level to be either occupied (OL) or unoccupied (UL) across the whole composition range as highlighted by the visual guidelines in Fig. 4 (dotted black lines). In the following, we discuss the effect of Sr clustering in the system. To address how different local Sr environments influence the electronic and magnetic properties of bulk LSMO in the presence of oxygen vacancies, we take $$2\times 2\times 6$$ supercells with different disordered Sr distributions with an overall global composition ($$\text {x}\,=\,0.250, 0.375$$ and 0.458) corresponding to the low, intermediate and high doping regimes. For each of these supercells, we placed an isolated oxygen vacancy with different local Sr concentrations. For low doping, the defect level is always occupied and the local environment introduces a $$\sim \,0.05\,\text {eV}$$ dispersion in the formation energy (green squares in Fig. 4). For high doping, the defect level is always unoccupied, with slightly larger dispersion in the energies depending on the local Sr environment ($$\sim \,0.07\,\text {eV}$$). The intermediate doping regime shows a richer landscape, with the defect level being occupied when the vacancy is placed in a Sr-poor environment while it is unoccupied when located in a Sr-rich region. As a side note, the formation energies for these two scenarios match nicely the dotted lines in Fig. 4. These findings are threefold relevant because the Sr doping mainly used for applications in devices based on LSMO, matches our intermediate composition, where nearly degenerate magnetic configurations (FM and SF) arise in the defective system. Besides, this region constitutes a narrow frontier between occupied/unoccupied defect state and a slight composition variation might lean the scale towards one state or the other. In this respect, segregation of Sr at the surface of LSMO in thin films has been broadly reported22,23,24 and our results evidence how local cation distribution strongly influences both the electronic and magnetic properties. At this point, we stress that our GGA treatment shows the presence of a defect level in the gap of the minority spin, which can be occupied or empty (below or above the Fermi level respectively), depending on the number of available electrons in the system. Although the presence of this defect level has been reported in previous $$\textit{ab initio}$$ calculations2,5, to our knowledge it has not been observed below the Fermi level and therefore, occupied. Most of the prior calculations include a Hubbard U term to account for strong electronic correlations. Even though our GGA results seem to give electronic features, particularly the energy position of Mn $$\text {t}_{\text {2g}}$$ peaks for the pristine system, that are in good agreement with experiments25, we also performed GGA + U calculations for the reduced case. Results obtained with a standard $$\text {U}\,=\,4.5\,\text {eV}$$ for all Mn 3d orbitals26 are presented in Fig. 5. In this case, the shift of the $$\text {d}^{\downarrow }_{\text {z}^{2}-\text {r}^{2}}$$ level increases to 6 eV, placing the defect state above the Fermi level for all the studied compositions. The $$\text {3d}^{\uparrow \downarrow }$$ exchange energy and the pseudo-bandgap, are also increased. The ground state is FM across the entire Sr range, and the SF configuration is consistently higher by $$\sim \,0.25\,\text {eV}$$. The electronic structure sketched by the PDOS on Mn nearest neighbors to the oxygen vacancy shares the same features described for the high doping regime using bare GGA, where the defect level is empty. The local density of states (LDOS) due to the defect state wavefunction is shown in the right panel of Fig. 5 where the bonding (antibonding) character for the FM (SF) configuration is evident. #### Concentrated limit When the number of oxygen vacancies increases, the interaction among the defect states can give rise to a dispersive band which can have an impact on the magnetic properties of the system. We explore this high vacancy concentration limit using a pseudocubic $$2\times 2\times 2$$ supercell with $$R{{\overline{3}}}c$$ symmetry that allows an adequate description of the octahedral distortions (Fig. 6). As before, Sr doping is modelled by direct substitution of La atoms by Sr ones. In this rather small supercell, clustering of the Sr atoms could result in artificially stacked heterostructures. Therefore, we adopt the so called anticlustering distribution, where the Sr atoms are disposed so that their mutual distance is maximized. The removal of a single oxygen atom in this supercell corresponds to a $$\delta \,=\,4.2\,\%$$ reduction ($$\text {La}_{0.825} \text {Sr}_{0.175} \text {MnO}_{2.875}$$), where significant modifications to the lattice are expected to take place. Previous studies report a lattice expansion induced by the presence of oxygen vacancies in the system11,27. In particular, Guo et al.7 ascribed this expansion to the repulsion between the $$\text {Mn}^{+3/+4}$$ and $$\text {La}^{+3}/\text {Sr}^{+2}$$ cations which are nearest neighbors of the vacancy and the donor defect with a +2 formal charge. Here, we evaluate the effect on the geometry due to the Sr doping and the oxygen vacancies by allowing a full relaxation of both the atomic coordinates (with forces below 0.04 eV/Å), and the lattice parameters (with cell pressures smaller than 0.1 GPa). Our results agree with an expansion of the lattice parameter, except for the FM solution in the low Sr concentration regime, where GGA predicts an occupied defect state in the minority spin, and a contraction of the cell volume (see Fig. S2 of SI for more details). Notice that a local contraction of the Mn-Mn distance (Table 1) does not result in an overall volume contraction for the SF configuration for the same Sr concentration, which is slightly lower in energy than the FM phase. As we will discuss later, we suggest that this apparent dichotomy could actually explain some recent experimental observations. The formation energy for oxygen vacancies in the concentrated limit is reported in Table 1 for both magnetic configurations analysed (FM and SF) and the corresponding doping concentration considered. We also include results obtained with the constrained-DFT approach (FS). The local distortion due to the vacancy defect is characterized by $$\text {d}_{\text {Mn-Mn}}$$, the interatomic distance between the two Mn atoms which are nearest neighbors to the vacant site. The difference in the total magnetic moment between the pristine and the defective system ($$\Delta \text {M}$$) is also reported in the table. As in the case of the diluted vacancy limit, the formation energy decreases as the Sr content increases. Furthermore, our data reveals that throughout the whole range of Sr compositions analyzed, the GGA ground state is the local spin-flip arrangement (SF). To the best of our knowledge, such a clear evidence of the weakening of the double-exchange mechanism in favor of superexchange antiferromagnetic interactions has not been reported before. This SF solution also results in a decreased magnetization of the system, and a slight shortening of $$\text {d}_{\text {Mn-Mn}}$$. On the contrary, under the GGA + U approximation, the FM becomes the most stable configuration with an unoccupied defect level, an augmented total magnetization in the defective system and an increased $$\text {d}_{\text {Mn-Mn}}$$. Mulliken population analysis reinforces our interpretation of charge distribution throughout the studied cases. Table 2 collects the data of characteristic Mn atoms in the defective system for two specific Sr concentrations illustrating either the low and high doping regimes. For reference, Mulliken charges for the pristine ferromagnetic system are also presented. When the defect level in the FM state is occupied, the magnetic moment of neighbouring Mn atoms to the vacant site remains unaltered with respect to the pristine LSMO value, but the valence is reduced approaching the formal $$\text {Mn}^{+3}$$ value. This is a consequence of the localization of the extra electronic charge in a singlet state. The SF configuration shows a preferential increase in the net spin for the $$\text {Mn}_{\text {NN}}^{\uparrow }$$ atom that is aligned with the ferromagnetic host, and a decrease in the magnetic moment of the inverted $$\text {Mn}_{\text {NN}}^{\downarrow }$$. In the high doping regime (within bare GGA) or through the whole composition range upon the addition of the Hubbard term, the defect level becomes depopulated for the FM case and there is an increase in the magnetic moment of $$0.13\,\mu _{B}$$ in the $$\text {Mn}_{\text {far}}$$ and of $$0.3{-}0.4\,\mu _{B}$$ in both $$\text {Mn}_{\text {NN}}^{\uparrow }$$ with respect to the pristine system. Therefore in this case, the extra electronic charge left behind by the vacancy spreads over a wider region around the defect, and not just in the nearby Mn atoms. ## Discussion and conclusions In this work, first-principles calculations have been used to study the influence of oxygen vacancies in LSMO within the most typical range of compositions. In agreement with previous works7, we show that the creation of oxygen vacancies pushes the $$\text {d}_{\text {z}^{2}-\text {r}^{2}}$$ states from neighboring Mn atoms, towards higher energies. This results in a localized defect state whose energy falls within the gap of the minority spin component. We found that this defect level can be occupied, or at least partially occupied, under particular circumstances. This bonding state leads to: (i) a local accumulation of charge that could be linked to the $$\text {Mn}^{+2}$$ peak observed in XAS spectra4; (ii) a degradation of the average magnetization around the defect; (iii) an overall weakening of ferromagnetic interactions in the system and (iv) a scattering center that can be responsible for the degradation of the electronic conductivity in reduced systems. Previous calculations were mostly based on a DFT + U approach, and showed no evidence that the defect state could be occupied7,28. When the Hubbard term is included in the Hamiltonian (with typical U values in the $$3{-}5\,\text {eV}$$ range, for all Mn 3d orbitals) the defect level is pushed towards higher energies and falls well above the Fermi level, for any Sr doping concentration. The extra charge left behind by the removed oxygen is distributed over the $$\text {e}_{\text {g}}$$ states, resulting in an increased net magnetization at odds with the degradation of the magnetization observed in experiments. The above results are based on the assumption that the U term is the same for all the Mn 3d orbitals, but the chemical environment close to the vacancy is different from that in the bulk, so there is no reason to support this approximation, and to assume that the defect state is always empty. Likewise, as we will discuss next, a few experimental observations in reduced LSMO samples can be better understood within an occupied defect level scenario. The volume expansion measured by X-ray diffraction experiments11 is also found in our FM calculations (along with an increase of the Mn-Mn interatomic distance) when the defect level is empty, both in the GGA high doping limit and in GGA + U calculations. On the other hand, the bonding character of the defect state gives a reduced Mn-Mn distance when the level is occupied, leading to a small volume contraction, in opposition with the above-mentioned experimental findings. Although this might seem contradictory with our low doping bare GGA results, the DOS spectral features (including the pseudo-bandgap value) obtained with GGA are in better agreement with experimental measurements compared to GGA + U. Furthermore, LSMO thin films grown under low oxygen pressure (on top of $$\text {SrTiO}_3$$) with a Sr compositional gradient, recently revealed Sr-poor regions displaying a compression of the out of plane lattice parameter along with a complex magnetic behaviour29. Under these growth conditions, oxygen vacancies are expected throughout the sample and according to our predictions, the occupation of the bonding defect level is favoured, thus locally reducing the lattice parameter and inducing a softening of the ferromagnetic properties. In this way, our results might contribute to shed light on the mentioned contrasting observations in these monolithic films, which could not be explained when the defect level is empty. A ferromagnetic metal to insulating transition was reported and related to the electromigration of mobile oxygen ions when electrons were added to doped manganites by means of electrolyte-gating30. In this regard, we found a weakening of the double-exchange mechanism revealed by the substantial reduction of the spin-flip energy configuration with respect to the pristine LSMO value. In the vacancy-diluted limit, the ground state remains ferromagnetic, but when the vacancy concentration increases, the superexchange interaction favors the SF arrangement over the FM one. It is expected that under a high vacancy concentration the interaction among defect levels gives rise to a dispersive defect band, increasing the range of Sr compositions for which the defect state could be occupied, thus destabilizing the ferromagnetic coupling. Previous calculations28 have shown that a gap is opened for an antiferromagnetic phase in bulk LSMO within the range of chemical compositions addressed here. All the above pieces could fit together considering a large concentration of vacancies along with the addition of electrons to populate the defect level. This procedure would weaken the ferromagnetic interactions, favoring antiferromagnetism and a bandgap opening. Electrostatic gating in field-effect transistors30,31 could be used to tune the electronic chemical potential, thus changing the occupation of the defect state. In this regard, our GGA calculations predict that the optimal Sr doping concentration ($$\text {x}\,=\,1/3$$) belongs to the region where the Fermi level falls near the defect level, driving the electronic instability that favours spin flip arrangements. In this manner, small variations in the gate voltage could have dramatic effects on the electronic properties of reduced manganites, and we propose this experimental approach as a mean to validate the predictions of our calculations. We anticipate that the important insights revealed in this paper regarding the effects arising from an electronic defect state due to oxygen vacancies will improve our understanding of the LSMO physical properties and trigger further development of devices and applications. ## Methods To perform total energy calculations we employ the SIESTA code32 within the GGA scheme, using the exchange-correlation functional WC33 (Wu-Cohen, modification of the PBE functional). Non local Troulliers-Martins pseudopotentials34 are employed, which conserve the norm and are factorized by the Kleinman-Bylander method35. All pseudopotentials and bases used in this work were developed and optimized in Ref.25. The pseudopotential of manganese was generated using the atomic configuration $$3\text {s}^{2}3\text {p}^{6}3\text {d}^{5}5\text {f}^{0}$$ with cutoff radii of 1.39, 1.88, 1.48 and 1.88 for the orbitals s, p, d and f respectively. In the case of oxygen, a configuration $$2\text {s}^{2}2\text {p}^{4}3\text {d}^{0}4\text {f}^{0}$$ was used with cutoff radii equal to 1.14 for each of the channels s, p, d and f. For strontium, the valence configuration for the pseudopotential was $$4\text {s}^{2}4\text {p}^{6}$$ (cutoff radius 1.49 for both cases) $$4\text {d}^{0}4\text {f}^{0}$$ (cutoff radius 1.99). The pseudopotential of lanthanum was generated using the atomic configuration $$5\text {s}^{2}5\text {p}^{6}5\text {d}^{0}4\text {f}^{0}$$ with cutoff radii 1.83, 2.19, 3.06 and 1.39 for the orbitals s, p, d and f respectively. For Mn, La and Sr relativistic effects were considered, but spin-orbit interactions were discarded. In these three elements, semicore states were also included. Finally, to model oxygen vacancies we employ the ghost atom feature included in SIESTA, considering the same pseudopotential and basis used for the oxygen atom. For the $$2\times 2\times 2$$ supercell, we take a 1600 Ry energy cutoff to perform the numerical integration in the real space and a $$18\times 18\times 18$$ k-points mesh in the Monkhorst-Pack scheme36, for the structural optimization. K-points mesh is reduced accordingly to $$18\times 18\times 6$$ for the $$2\times 2\times 6$$ supercell (with 120 atoms). For the density of states (DOS) calculation, we take a denser grid of k-points to obtain an accurate description of the energy levels. Convergence analysis were performed indicating less than 1.25 meV/u.f. energy error. The use of a computational methodology that accentuate Coulomb interactions among highly localized orbitals, such as DFT + U37,38 or hybrid functionals39 (HSE40, B3LYP41, PBE042, etc) has been shown not necessary by previous studies in the composition range analyzed of pristine LSMO, characterized by a metallic state and Jahn-Teller dynamic distortions25,43,44,45. Regarding reduced LSMO and its defect state, it is fair to assume that bare DFT might underestimate the position of that level. Although a more complete description of the electronic interactions (such as hybrid functionals or beyond) might be needed to precisely determine the energy position of the defect level, these calculations are computationally challenging and exceed the scope of this work. ## Data availability All data generated or analysed during this study are included in this published article and its Supplementary Information files. ## References 1. de Jong, M. P., Dediu, V. A., Taliani, C. & Salaneck, W. R. Electronic structure of $${\text{ La}_{0.7}\text{ Sr}_{0.3}\text{ MnO}_{3}}$$ thin films for hybrid organic/inorganic spintronics applications. J. Appl. Phys. 94, 7292–7296. https://doi.org/10.1063/1.1625081 (2003). 2. Wang, K., Ma, Y. & Betzler, K. Defect-induced spin deterioration of $${\text{ La}_{0.64}\text{ Sr}_{0.36}\text{ MnO}_{3}}$$: Ab initio study. Phys. Rev. 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Transformation of spin information into large electrical signals using carbon nanotubes. Nature 445, 410–413. https://doi.org/10.1038/nature05507 (2007). 45. Pruneda, J. M. et al. Ferrodistortive instability at the (001) surface of half-metallic manganites. Phys. Rev. Lett. https://doi.org/10.1103/PhysRevLett.99.226101 (2007). ## Acknowledgements Work at ICN2 was funded by FEDER/Ministerio de Ciencia e Innovación – Agencia Estatal de Investigación (grant PGC2018-096955-B-C43), Generalitat de Catalunya (2017SGR1506), and the EU-H2020 programme (654360 and 824143) under the NFFA-Europe Transnational Access Activity and the EU MaX Center of Excellence. ICN2 is also supported by the Severo Ochoa Program (SEV--2013--0295) and the CERCA Program of Generalitat de Catalunya. D. J. and V. F. acknowledge support from CONICET (PIP 00020-2019) and ANPCyT (PICT1857), Argentina. D. J. acknowledges CONICET for a doctoral fellowship. The authors thank computational time in the HPC Cluster at CNEA, Argentina and the Spanish Supercomputing Network (RES). ## Author information Authors ### Contributions D.J. performed DFT calculations under the supervision of M.P. and V.F. All authors analysed the results, discussed the ideas and wrote the manuscript. ### Corresponding author Correspondence to Dilson Juan. ## Ethics declarations ### Competing interests The authors declare no competing interests. ### Publisher's note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Juan, D., Pruneda, M. & Ferrari, V. Localized electronic vacancy level and its effect on the properties of doped manganites. Sci Rep 11, 6706 (2021). https://doi.org/10.1038/s41598-021-85945-5 • Accepted: • Published: • DOI: https://doi.org/10.1038/s41598-021-85945-5	{}
Fried Bean Curd Skin Roll With Shrimp Recipe, Masterbuilt 40'' Smoker Cover, Spiked Club Weapon, Thatchers Gold Cider For Sale, Samsung Rf28r7201sr Reviews, Image Distance Formula, Black Woman Silhouette Png, " /> Fried Bean Curd Skin Roll With Shrimp Recipe, Masterbuilt 40'' Smoker Cover, Spiked Club Weapon, Thatchers Gold Cider For Sale, Samsung Rf28r7201sr Reviews, Image Distance Formula, Black Woman Silhouette Png, " /> Fried Bean Curd Skin Roll With Shrimp Recipe, Masterbuilt 40'' Smoker Cover, Spiked Club Weapon, Thatchers Gold Cider For Sale, Samsung Rf28r7201sr Reviews, Image Distance Formula, Black Woman Silhouette Png, " /> Let us consider a unit cell of ccp or fcc lattice [Fig. What is the disadvantage of using impact sockets on a hand wrench? This happens when the radius ratio exceeds 0.732. This situation is relatively unstable. All tetrahedral sites are filled with cations. The ccp structure has 4 atoms per unit cell. In this case the radius ratio model fails rather badly. Complexes that adapt this structure include.. CdF 2, HgF 2, SrF 2, PbF 2, BaF 2 and HfO 2. The vacant space left in the closest pack arrangement of constituent particles is called an interstitial void or interstitial site. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. What would result from not adding fat to pastry dough. Effect of Radius Ratio on Coordination Number: A cation would fit exactly into the octahedral void and would have a coordination number of six if the radius ratio were exactly 0.414. would fit exactly into the octahedral void and would have a coordination number the wurtize structures (zns): caesium chloride (cscl) structure: calcium floride (florite) structure: antifluorite structure (revere of fluorite) na2o: spinel structure : close packing in crystals (2d and 3d): study of interstitial voids: locating tetrahedral and octahedral voids: radius ratio rules in ionic solids: simple cubic cell(scc): The radius ratio at which anions just touch each other, as well as central cation, is called the critical radius ratio. Let a be each side of the cube. It is clear that since the antifluorite structure has the smaller ion filled into the tetrahedral voids, radius ratio can range from 0.225 − 0.414, with 0.225 being the best fit for this type of void. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Chemistry Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Determining cation-to-anion radius ratio for “ideal” Anti-fluorite structure, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…, “Question closed” notifications experiment results and graduation. Thanks. A D. 1 2. It would be great if this is clarified. the ccp structure, each unit cell has 4 atoms. tetrahedral void and coordination number decreases from 6 to Thanks for contributing an answer to Chemistry Stack Exchange! When the radius ratio is greater than this, then the anions move apart to accommodate larger cation. In case of Were any IBM mainframes ever run multiuser? If the radius ratio is further increased the anions will move farther and farther apart till to reach a stage at which more anions can be accommodated. Your email address will not be published. of six if the radius ratio were exactly 0.414. 4. Let us How do I legally resign in Germany when no one is at the office? Each small cube has atoms at alternate corners as shown. How should this half-diminished seventh chord from "Christmas Time Is Here" be analyzed in terms of its harmonic function? How to find individual probabilities of all numbers from a list? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. C. 8. toppr. Use MathJax to format equations. C.N. Besides the In crystals, cations tend to get surrounded by the largest possible number of anions around it. cube, C is not occupied but it is surrounded by six atoms on face centres. Thus, the number of tetrahedral voids is twice the number of atoms. each other, which is an unstable situation. 4. Hence the structure changes to some suitable stable arrangement. We have, Thus radius ratio for the octahedral void is 0.414. B. Thus, the number of Therefore the ratio between cationic and anionic radii in zinc blend is 0.39 (74pm/190 pm) .This suggests a tetrahedral ion arrangement and four nearest neighbors from standard crystal structure prediction tables. 9000 ft.) is 15,000 feet high? Each small cube has 4 atoms. It only takes a minute to sign up. To learn more, see our tips on writing great answers. of close pack anions and have the coordination number of six, in this case, the there are many exceptions to it. Since each What is the proper etiquette with regards to reciprocating Thanksgiving dinner invitations? The ratio of Each of the eight small cubes has one void in one unit cell of ccp structure. Only 1/4 th of each void belongs to a particular Can we omit "with" in the expression glow with (something)? For cubic close-packed structure: I do not understand why the ideal radius ratio should be $0.228$. Determining structure factors for crystals from lattice and motif, Cation-anion radius ratio of manganese difluoride, Is Firefox so insecure it's worth blocking. Attempts have been made to investigate the factors governing the relative stability of the various structure types found by mixing different kinds of atoms, A or B, in a large number of AB 3 intermetallic compounds related to Cu 3 Au. A octahedral site in a cube having octahedral void of radius r is as shown at the centre of the cube. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. When joined to each other, they make a regular tetrahedron as shown in the figure. Video Explanation. has one octahedral void at the body centre of the cube. Effect of Radius Ratio on Coordination Number: A cation would fit exactly into the octahedral void and would have a coordination number of six if the radius ratio were exactly 0.414. The radius ratio at which anions just touch each other, as well as central cation, is called the critical radius ratio. Hence smaller cation moves to In antifluorite structure, co-ordination number of anion is: A. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Examples of Radius Ratio Rule. Moreover, the limiting ratio has to be greater than 0.414, (radius ratio greater than 0.414) to fit an octahedral arrangement of anions. edges. Quantifier elimination of the complex field in a restricted language. radius ratio is exactly 0.414. If ANTIFLUORITE STRUCTURE (REVERE OF FLUORITE) Na2O: ... radius ratio , Packing efficiency, density and void %: Other Examples: – Na 2 O, K 2 O, K 2 S, Na 2 S, Rb 2 O, Rb 2 S. Note: Metals like Al, Ag, Au, Cu, Ni, and Pt have ccp structure and Be, Mg, Co and Zn have a hcp structure. Similarly, a cation would fit It is however unclear as to why there is a difference in the ideal radius ratio for antifluorite structure and for the best fit of the tetrahedral void. Consider face diagonal AB (Right-angled triangle ABC). Number of octahedral void at the body-centre of the cube = 1 Answer. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. the unit cell. In antifluorite, of formula A2X, the coordination numbers of cation and anion must be in the ratio of 1:2. if the radius ratio were exactly 0.225. The body centre of the In such case packing of anions is very close to each other and due to repulsion between anions, the system becomes unstable. (a)]. Generally, if the radius ratio r-/r+ exceeds ~1.37 then a rutile rather than fluorite structure is favoured. If a piece of software does not specify whether it is licenced under GPL 3.0 "only" or "or-later", which variant does it "default to"? Why does Lovecraft write that Mount Nansen (approx. The ionic radius for Zn 2+ is 74pm and for S 2-is 190pm. Since the number of atoms in a single unit cell of Zn and S is the same, it is consistent with the formula ZnS. edge of the cube is shared between four adjacent unit cells, so is the My book under the heading of Anti-fluorite structure makes the following remark, about the radius ratio of an ideal Anti-fluorite: The ideal antifluorite would have a cation-to-anion radius ratio of $0.228$. unit cell (2 on the corners and 2 on face centre) and two belonging to two consider a case in which a cation is fitting exactly into the octahedral void	{}
# How to account for speed of the vehicle when shooting shells from it? I'm developing a simple 3D ship game using libgdx and bullet. When a user taps the mouse I create a new shell object and send it in the direction of the mouse click. However, if the user has tapped the mouse in the direction where the ship is currently moving, the ship catches up to the shells very quickly and can sometimes even get hit by them - simply because the speed of shells and the ship are quite comparable. I think I need to account for ship speed when generating the initial impulse for the shells, and I tried doing that (see "new line added"), but I cannot figure out if what I'm doing is the proper way and if yes, how to calculate the correct coefficient. public void createShell(Vector3 origin, Vector3 direction, Vector3 platformVelocity, float velocity) { long shellId = System.currentTimeMillis(); // hack ShellState state = getState().createShellState(shellId, origin.x, origin.y, origin.z); ShellEntity entity = EntityFactory.getInstance().createShellEntity(shellId, state); entity.getBody().applyCentralImpulse(platformVelocity.mul(velocity * 0.02f)); // new line added, to compensate for the moving platform, no idea how to calculate proper coefficient entity.getBody().applyCentralImpulse(direction.nor().mul(velocity)); } private final Vector3 v3 = new Vector3(); public void shootGun(Vector3 direction) { Vector3 shipVelocity = world.getShipEntities().get(id).getBody().getLinearVelocity(); world.getState().getShipStates().get(id).transform.getTranslation(v3); // current location of our ship v3.add(direction.nor().mul(10.0f)); // hack; this is to avoid shell immediately impacting the ship that it got shot out from world.createShell(v3, direction, shipVelocity, 500); } Edit - I switched to: v3.set(direction); v3.nor().mul(velocity); if (platformVelocity != null) entity.getBody().setLinearVelocity(v3); And it seems to work (so adding the velocities and setting them on the body). However, I'm not sure what the drawbacks are and why the tutorial application used impulses instead of setting velocities. I also think that while this is physically correct it leads to strange results (gun range is two times further when shooting forward than when shooting sideways). The root cause is because really the shells are indeed too slow, but I like them slow as otherwise the gameplay is too fast-paced, and they travel too far. I think I need some compromise between physics-realistic and ignore-platform-velocity. • What I feel from reading this is that either the ship is too fast or your bullet is too slow. You could account for the velocity of the ship when firing bullets, but does that mean the bullets will have slower velocity when shot in a direction opposite to the ship's movement? – Vite Falcon Nov 28 '12 at 15:24 The shell's speed was the same as the ship's speed before it got fired, if you think about it. It was sitting on the ship. So, if you add the velocity from being fired to the velocity of the ship, you should get the correct result. • I still feel his bullet is too slow or his ship is too fast. If he feels that his ship catches up with the bullet, if he bullet is fired against the direction of motion, adding the ship's velocity will either make the bullet stand-still or move in the direction of ship (and may even hit it :O). – Vite Falcon Nov 28 '12 at 15:28 • That doesn't make conservation of momentum the wrong thing to do, it just means his ship's gun is embarassingly ineffective. Maybe the sailors should consider using a little more gun powder. If the bullet catches up with the ship when fired AGAINST the direction of motion, that means the ship is either slowing down incredibly fast, or the bullet was fired with negative velocity. In the first case, the boys in the engine room are really working hard! Maybe too hard. In the second case, I'd say that gun needs serious repairs. – Eric B Nov 28 '12 at 15:29 • LOL, or programmatically, the Bullet-to-Ship velocity ratio should be higher than what it is. – Vite Falcon Nov 28 '12 at 15:34 • From a gameplay perspective, conservation of momentum may very well be wrong. So few games use it that it can (and often does) feel wrong in games that use it. Whether that holds true for this particular genre of game, I don't know. But in general, do what's fun, not what's most realistic. – Sean Middleditch Nov 28 '12 at 17:31 • Tribes conserves momentum, making it incredibly hard to hit targets, which is exactly why some players LIKE it. – Eric B Nov 28 '12 at 19:36 entity.getBody().setLinearVelocity(platformVelocity);	{}
# Documentation ### This is machine translation Translated by Mouseover text to see original. Click the button below to return to the English verison of the page. # double Double-precision floating-point real-world value of `fi` object ## Syntax `double(a)` ## Description `double(a)` returns the real-world value of a `fi` object in double-precision floating point. `double(a)` is equivalent to `a.double`. Fixed-point numbers can be represented as `$real\text{-}worldvalue={2}^{-fractionlength}×storedinteger$` or, equivalently as `$real\text{-}worldvalue=\left(slope×storedinteger\right)+bias$` ## Examples The code ```a = fi([-1 1],1,8,7); y = double(a) z = a.double``` returns ```y = -1 0.9922 z = -1 0.9922``` Get trial now	{}
Thesis ### The bounded H^{infinity} calculus for sectorial, strip-type and half-plane operators Abstract: The main study of this thesis is the holomorphic functional calculi for three classes of unbounded operators: sectorial, strip-type and half-plane. The functional calculus for sectorial operators was introduced by McIntosh as an extension of the Riesz-Dunford model for bounded operators. More recently Haase has developed an abstract framework which incorporates analogous constructions for strip-type and half-plane operators. These operators are of interest since they arise naturally as gen... ### Access Document Files: • (pdf, 749.8KB) ### Authors More by this author Division: MPLS Department: Mathematical Institute Role: Author #### Contributors Role: Supervisor Type of award: DPhil Level of award: Doctoral Awarding institution: University of Oxford Keywords: Subjects: UUID: uuid:6bff352d-f858-492a-a00b-3a3dd2049b5c Deposit date: 2017-07-10	{}
# Can anyone express the number 215 in the 32-bit floating point format IEEE-754 [closed] i would be most grateful if you could solve this question Thank you - 1000011010101110000000000000000, proof techniques? – Pål GD Jan 28 at 21:21 Would you please explain how you solved – Dionysus Jan 28 at 21:22 Java: Integer.toBinaryString(Float.floatToIntBits(215)). Obviously, you'll have to check it for yourself. – Pål GD Jan 28 at 21:41 This can be solved by hand looking at the IEEE floating point standards, or by writing a small program, or... No reason to ask here. – vonbrand Jan 28 at 21:44 ## closed as off topic by Pål GD, Raphael♦Jan 28 at 23:49 Questions on Computer Science Stack Exchange are expected to relate to computer science within the scope defined in the FAQ. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about closed questions here. #include <stdio.h>	{}
# -----CodeForces 3B Lorry-----贪心------- 296人阅读评论(0) B - Lorry Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Description A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres). Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body. Input The first line contains a pair of integer numbers n and v (1 ≤ n ≤ 1051 ≤ v ≤ 109), where n is the number of waterborne vehicles in the boat depot, and v is the truck body volume of the lorry in cubic metres. The following n lines contain the information about the waterborne vehicles, that is a pair of numbers ti, pi (1 ≤ ti ≤ 21 ≤ pi ≤ 104), where ti is the vehicle type (1 – a kayak, 2 – a catamaran), and pi is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file. Output In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them. Sample Input Input 3 2 1 2 2 7 1 3 Output 7 2 #include <iostream> #include <algorithm> #include <stdio.h> using namespace std; struct sa { int id; int val; } one[100005],two[100005]; int cmp(const sa &a,const sa &b) { return a.val>b.val; } int sum1[100005],sum2[100005]; int main() { int n,v; while(scanf("%d%d", &n, &v) != EOF) { int kind,val; int one_num=0,two_num=0; for(int i=1; i<=n; i++) { scanf("%d%d",&kind,&val); if(kind==1) { one[++one_num].val=val; one[one_num].id=i; } else { two[++two_num].val=val; two[two_num].id=i; } } sort(one+1,one+one_num+1,cmp); sort(two+1,two+two_num+1,cmp); //int sum1[100005],sum2[100005]; sum1[0]=0; sum2[0]=0; for(int i=1; i<=one_num; i++) { sum1[i]=sum1[i-1]+one[i].val; } for(int i=1; i<=two_num; i++) { sum2[i]=sum2[i-1]+two[i].val; } int maxv=-1,io=0,it=0; for(int i=0; i<=one_num; i++) { if(i>v) break; int d=sum1[i]+sum2[min((v-i)/2,two_num)]; if(d>maxv) { maxv=d; io=i; it=min((v-i)/2,two_num); } } if(maxv==-1) { cout<<"0"<<endl; continue; } cout<<maxv<<endl; int num[100005]; int cnt=0; for(int i=1; i<=io; i++) num[++cnt]=one[i].id; for(int i=1; i<=it; i++) num[++cnt]=two[i].id; for(int i=1; i<=cnt; i++) printf("%d%c", num[i], i == cnt? '\n' : ' '); } return 0; } 0 0 * 以上用户言论只代表其个人观点，不代表CSDN网站的观点或立场个人资料 • 访问：12604次 • 积分：1003 • 等级： • 排名：千里之外 • 原创：93篇 • 转载：1篇 • 译文：0篇 • 评论：0条阅读排行评论排行	{}
(i) I do not want e-mail notification, my inbox is busy enough! (ii) I would prefer not to have to keep clicking on the "Reputation" link to check for the possibility that someone might have a query. • Doesn't the StackExchange logo on the top left work for you? It is not entirely reliable but most of the time someone left a comment for me I see a red bullet (conatining the number of messages I have on the top left). If you leave a reply to this comment I'll make a screen shot and post it, so you can see how it looks for me. – t.b. Aug 30 '11 at 0:01 • Here's a picture. The white $1$ in the red bullet on the top left shows that I received one comment notification. – t.b. Aug 30 '11 at 0:10 • @Theo: Thank you, I was unaware of it! There is a slight complication, in that I have difficulty noticing red (blue would be useful). But in principle, and, with some effort, probably in practice, your comment solves my problem. – André Nicolas Aug 30 '11 at 0:18 • I suggest that you make this a feature request. Accessibility should be one of the highest priorities of such a site and red may indeed be problematic for many enough. Unfortunately, on the usual sister sites I couldn't find anything allowing you to customize colors. However, here are two links for you that you may want to bookmark for easier accessibility of your inbox: 1. direct link to your responses tab and 2. raw inbox. – t.b. Aug 30 '11 at 0:37 • @Theo: Thanks, that too will help. – André Nicolas Aug 30 '11 at 1:08 • Another color blind? Hooray. I am not alone! – Asaf Karagila Aug 30 '11 at 6:37	{}
# Function defined by infinite series A function $f$ is defined as follow: $$f(x)=\sum_{n=1}^{\infty}\frac{b_{n}}{(x-a_{n})^{2}+b_{n}^{2}}\;\;, x\in \mathbb R$$ where $(a_{n}, b_{n})$ are points in the $xy$-plane, $b_{n}>0$ for all $n$. When is the function $f(x)$ bounded away from zero? that is $f(x)\geq a>0$, for some $a>0$, for all $x\in \mathbb R$. I believe that this would somehow depend on the points $(a_{n}, b_{n})$, for example if $\{b_{n}\}$ converges to zero or not, but I cannot find out how! Thanks EDIT: I still don't know when $f(x)$ will be bownded away from zero! Edit: $f(x)$ is bounded above by some constant, say $C>0$. - Instead of "strictly positive" you can say: "bounded away from zero". – GEdgar Feb 20 '12 at 0:51 Yes, thanks! I fixed it. – Chelsea Feb 20 '12 at 1:10 All summands are strictly positive, so if the series converges (or if you accept $+\infty$ as strictly positive) the sum is strictly positive. - So this means that the case $\lim_{x\rightarrow x_{0}} f(x)=0$ is not possible for any $x_{0}\in \mathbb R$?! – Chelsea Feb 20 '12 at 0:42 Of course; again, for any summand the limit is strictly positive. – Robert Israel Feb 20 '12 at 0:53 Changing "strictly positive" to "bounded away from zero" changes the question completely. – Robert Israel Feb 20 '12 at 4:41 My fault! I meant bounded away from zero. – Chelsea Feb 20 '12 at 17:23 Here's one in the opposite direction. Note that for any $c > 0$ the curve $y/(x^2 + y^2) = c$ is a circle in the upper half plane, tangent to the $x$ axis at the origin. If there is some $r > 0$ such that every circle of radius $r$ with centre on the line $y=r$ contains at least one $(a_n, b_n)$, then $f(x)$ is bounded below. - How about $$\lim_{x\to-\infty}\sum_{n = 0}^{\infty} \frac{1}{(x-n)^2 + 1} = 0$$ - And somewhat more generally, if there is $x_0$ such that all $a_n \ge x_0$ and $\sum_{n=1}^\infty \frac{b_n}{(x_0 - a_n)^2 + b_n^2} < \infty$, then $\lim_{x \to -\infty} \sum_{n=1}^\infty \frac{b_n}{(x-a_n)^2 + b_n^2} = 0$ so in that case your sum is not bounded away from $0$. – Robert Israel Feb 20 '12 at 5:42 @Robert Israel: We can assume that $f(x)$ is bounded above, but does the $b_{n}$'s play any role here? – Chelsea Feb 20 '12 at 13:27 How do you see that the limit is zero? – Chelsea Feb 21 '12 at 19:57 @Chelsea: take $x<-k$, $k\in\mathbb{N}$. Then $\sum_{n = 0}^{\infty} \frac{1}{(x-n)^2 + 1}\leq\sum_{n = k}^{\infty} \frac{1}{n^2 + 1}$. – Martin Argerami Feb 25 '12 at 16:29 @Chelsea: Note that if $x<x_0$, then the smaller $x$ the smaller $f(x)$ (because every summand gets smaller). If you want it $f(x)$ to be smaller than $\varepsilon$, then you find $N$ such that $\sum_{n=N}^\infty \frac{b_n}{(x_0-a_n)^2 + b_n^2} < \varepsilon/2$ and look for $x$ so close to $-\infty$, that the beginning of the series gets smaller than $\varepsilon/2$. – savick01 Feb 25 '12 at 16:43 show 1 more comment	{}
Browse Questions # Events A and B are such that $P (A) = \frac{1}{2} , P(B) = \frac{7}{12}$ and P(not A or not B) = $\frac{1}{4}$ . State whether A and B are independent? $\begin{array}{1 1} \text{A and B are independent} \\\text{A and B are not independent} \end{array}$ Toolbox: • If A and B are independent events, $P(A\cap\;B)=P(A)\;P(B)$ • P (not A and not B) = $P(\bar{A}\cap\;\bar{B})$ = $P(\overline{A \cup B})$ = 1 - P (A $\cup$ B) • P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B) • P ($\bar{A})$ = 1 - P(A) Given $\;P(A)=\large \frac{1}{2},$ $\;;P(B)=\large \frac{7}{12},$$\;$ and $\;P(\bar{A}\cup\;\bar{B})=\large \frac{1}{4}$ If A and B are independent events, $P(A\cap\;B)=P(A)\;P(B)$ We know that P (not A and not B) = $P(\bar{A}\cap\;\bar{B})$ = $P(\overline{A \cup B})$ = 1 - P (A $\cup$ B) $\Rightarrow \large \frac{1}{4} $$= 1 - P (A \cup B) \rightarrow P (A \cup B) = 1 - \large \frac{1}{4} = \frac{3}{4} \Rightarrow P (A \cap B) = P (A) + P(B) - P (A \cup B) = \large \frac{1}{2} + \frac{7}{12} - \frac{3}{4} = \large \frac{6 + 7 - 4}{12} = \frac{9}{12} = \frac{3}{4} However, P(A) \times P(B) = \large \frac{1}{2} \times \large\frac{7}{12} = \frac{7}{24}$$\;\neq P (A \cap B)$ Therefore A and B are not independent.	{}
# Kyle solved 18 of 24 puzzles ina puzzle book. he says that he can use an equivalent fraction ###### Question: Kyle solved 18 of 24 puzzles ina puzzle book. he says that he can use an equivalent fraction to find the percent of puzzles in the book that he solved. how can he do that? what is the percent? ### Of three positive integers, the second is twice the first, and the third is twice the second. one of Of three positive integers, the second is twice the first, and the third is twice the second. one of these integers is 17 more than another. what is the sum of the three integers?... ### By what year did mecca have control By what year did mecca have control... ### How do you convert 4680 weeks into years? Need answer now please How do you convert 4680 weeks into years? Need answer now please... ### Importance Of Drawing Study GuideThe act if drawing is important because it develops the following Importance Of Drawing Study Guide The act if drawing is important because it develops the following skills.1. Using skills teach us how to see.... ### Many cars are filled with an air bag. In a collision, the air bag inflates and reduces the effect of the impact between the passenger Many cars are filled with an air bag. In a collision, the air bag inflates and reduces the effect of the impact between the passenger and the dashboard. In a test of the air bag, a heavy ball is used instead of the passenger. The car is travelling at 14m/s when it hits a wall. The air bag inflates a... ### Select the sentence that is written informally. it is almost baseball season, and i am getting excited. Select the sentence that is written informally. it is almost baseball season, and i am getting excited. jane loves to go to the library to check out books. it's always awesome to check out new video games. the ground was frozen solid... ### Can somebody help me with this. Will mark brainliest! Can somebody help me with this. Will mark brainliest! $Can somebody help me with this. Will mark brainliest!$... ### Water makes up approximately how much of a human's body weight? Water makes up approximately how much of a human's body weight?... ### Through what activity can you achieve the goal of a peaceful retirement and reaching your milestones Through what activity can you achieve the goal of a peaceful retirement and reaching your milestones in life?... ### ( What are the challenges of keeping historical sites?) ( What are the challenges of keeping historical sites?)... ### The battle between the Merrimack and the Monitor showed that new technology gave the Union an advantage. surprise attacks were The battle between the Merrimack and the Monitor showed that new technology gave the Union an advantage. surprise attacks were effective. O the South had better ships than the North. the North and South were equally matched at sea.... ### On the last day of a Shakespeare class, an English teacher asked her students which play they liked On the last day of a Shakespeare class, an English teacher asked her students which play they liked most. Out of the 18 students who have submitted responses so far, 6 liked Much Ado About Nothing best. What is the experimental probability that the next student to respond likes Much Ado About Nothin... ### Explain the differnce between a condordant and discordant coastline. (4 marks) Explain the differnce between a condordant and discordant coastline. (4 marks)... ### Spanish conquistador Hernán Cortés wrote the following excerpt in a letter to the Spanish king:The Spanish conquistador Hernán Cortés wrote the following excerpt in a letter to the Spanish king: The inhabitants of this city pay a greater regard to style in their mode of dress and politeness of manners than those of the other provinces and cities; since, as the Cacique Moctezuma has his residen... ### Match the leaders to there goals at the paris peace conference Match the leaders to there goals at the paris peace conference... ### The rule of a certain sequence is k = 3n. Find the first four terms of the sequence The rule of a certain sequence is k = 3n. Find the first four terms of the sequence... 23. Find the value of B. 24.Find the value of x. $23. Find the value of B. 24.Find the value of x.$... Daniel plays two games in the casino. The first game he believes he has 60% chance to win. If he wins the first game, he will win the second game with 35% chance, if he loses the first game, he will win the second game with 65% chance. The winning prices for the first and second game are $1 and$2 r...	{}
May 23 – 27, 2017 Seoul Asia/Seoul timezone ## The BetaCage: An Ultra-sensitive Screener for Surface Contamination May 25, 2017, 2:30 PM 1h Seoul Board: 12 Poster ### Speaker Dr Eric Miller (South Dakota School of Mines) ### Description Material screening for identifying low-energy electron emitters and alpha-decaying isotopes is now a prerequisite for rare-event searches (e.g., dark-matter direct detection and neutrinoless double-beta decay) for which surface radiocontamination has become an increasingly important background. The BetaCage, a gaseous neon time-projection chamber, is a proposed ultra-sensitive (and nondestructive) screener for alpha and beta-emitting surface contaminants to which existing screening facilities are insufficiently sensitive. The expected sensitivity is 0.1 betas (per keV-m$^2$-day) and 0.1 alphas (per m$^2$-day), where the former will be limited by Compton scattering of external photons in the screening samples and (thanks to tracking) the latter is expected to be signal-limited; radioassays and simulations indicate backgrounds from detector materials and radon daughters should be subdominant. We will report on details of the background simulations and detector design that provide the discrimination, shielding, and radiopurity necessary to reach our sensitivity goals for a chamber with a 95x75 cm^2 sample area positioned below a 40 cm drift region and monitored by crisscrossed anode and cathode planes consisting of 151 wires and 112 wires, respectively. ### Primary author Dr Eric Miller (South Dakota School of Mines) Poster	{}
# Puzzle out the Cryptogram Patterns So I haven't posted very many questions and I'm not yet very good at figuring out how many clues to include. In addition, I realize that the key behind the third cryptogram is not at all obvious so it might need some more help but I don't want to give the game away. I'll give it until Monday and then I'll post some more obvious hints. Can you decode each of the cryptograms below and then answer the challenge they pose? ISAEBZMEIYIQWXXOIYSCIZSJCOTFSYJALIAJQEICJGVMEIQOSJGMIPMJYGBMMEIAESOOIGVISYFSYJAMBBOYASGLBMESMRBFDICTKWJAUOTMEIQOSJGMIPMJYHICIOTMEIIPQOSGSMJBGBZMEIMCWIQWXXOI XKEJGBVUUCMZCFSAKNFZEJCKEAABCRVAACJMICBFMYCVGBGFRBCJAJXAKCOVQFMCCVGBIXFASCUNVSABCEMYCJUXFMZRVAACJMSVJCEMFDECABCXVJCSFQRUCFYCVSIEAQVXICAJFGHXAKFYCMAFNXICGVESCABCXJCDEFJCVMCRFRBVMX CIZQYNDRFRYAKPQLRNDRRQQKDZDPYEPQLRJDGHSZQOVYBERQAQLAGYFCQLFVYFRKDDENQCLNHLPMHFMCRQYRRMDJDFRHLFTDOTCGGMHWDHGGRMODDEGHCLRDXRFRMDYLPDOGZCLNEHRRDOLIQODHAMHLPRMDOYGDRMHRRMDICOFRHLPRMCOPFMHODRQDLFYODHYLCUYDFYJFRCRYRCQLNQQPGYAKEYSSGCLN There may be solutions to the puzzle-within-the-puzzle that I haven't thought of yet. Any such solutions will receive an upvote but will not be accepted as the answer unless the answer I was thinking of is never found (doubtful) in which case the most upvoted answer will be accepted. Clarification on the hidden challenge: NAFJREVATGURUVQQRAPUNYYRATRJVGUFBZRGUVATYVXRNRAPELCGFGBAJVYYORPBAFVQRERQVAFHSSVPVRAGLBHZHFGRKCYNVAGURCNGGREAGUNGYRQGBNRAPELCGVATNFA End of the first day hint: As @JasonPatterson hinted at on the answer by @GentlePurpleRain there are clues in the third cryptogram. In fact, the first three sentences of the cryptogram each contain a clue. Three sentences, three clues, three cryptograms. That nae be a coincidence, laddy. End of the second day hint: For the first cryptogram: Don't think of the key as starting with "A encodes to S". It starts with "Q encodes to K". Now, why does it start with "Q" and why does that become "K"? For the second cryptogram: The words from the first clue with the Z and J are there for a reason. Those are rare letters. Where have you seen those words together before? (@GentlePurpleRain has already solved this one) For the third cryptogram: @JasonPatterson already found the key word in the encoded clue. The method of applying it may be fremd for individuals vacant of experience. How can you go from the resulting mess back to a letter? End of the third day hint: (not optimized because kids just started screaming) 1) Your clues are: "IF YOU GET STUCK DONT GET TOO KEYED UP". It starts with "Q encodes to K". Try looking down. Recall that Q is the 17th letter of the alphabet. But what if two letters encode to the same result? That's not allowed. 2) (solved) 3) Your clues are: "JUST KEEP GOING AND HASH IT OUT". The method of applying it may be fremd for individuals vacant of experience. After you hash it out, you have to convert it back to a number. Try to use as little of the mess as possible to get a unique encoding. But what if two letters encode to the same result? That's not allowed. Beginning of the second week hint: 1) If you're viewing this on the desktop site - I.E. not the mobile site - and you look down, what do you see with keys on it? Since Q is the 17th letter of the alphabet, what comes 17 keys after Q? What comes 23 keys after W? 2) (Solved) 3) After you hash it out, you need to get back to a letter. How can you take the first tiny bit of the hex mess and modify it to get a decimal equivalent to a letter? How little a bit must you take to get a unique result? • MEIQOSJGMIPMJY appears twice in the first cryptogram.. Also, MEI alone appears very many times. – Ben Frankel Apr 20 '15 at 14:58 Cryptogram 1 - Based on the hints, here is the encryption method: On a QWERTY keyboard, the keys are numbered from L->R and T->B, as shown below. The plaintext key number plus its alphabet number determine its associated ciphertext letter. For Q, 1+17=18, so the CT letter is K. If the CT letter has already been used, then the next available key becomes the CT letter. For T, 5+20=25, but N has been used so the next available letter is M. The encoding table is shown below. Cryptogram 3 - Here is the encryption method: Step 1 - make MD5 hash for each letter of the alphabet, individually Step 2 - start by taking the first digit from each hash, then convert to decimal Step 3 - use mod-26, add 1 and select the equivalent letter of the alphabet Example for plaintext D - the first digit of the hash is f which is decimal 15. Add 1, then select P for the ciphertext. Step 4 - if the CT letter has already been taken, then repeat steps 2 and 3 with the first 2 digits of the hash. Keep adding a digit from the hash until a unique letter can be selected for the CT. Example for PT L - a unique letter is not selected until d20 (3 digits) is used. Convert to 3360 (decimal) then mod-26 plus 1 is 7, so the CT character is G. The encoding table is shown below. • +1 the method behind Cryptogram 1 is spot on. For Cryptogram 3, you're missing two key points. 1) You need a step 2.5 to modify the decimal version. The need for this is not obvious because you don't have Step 4 yet. 2) Step 4 has a little different rule that Cryptogram 1. Where else can you make adjustments to get a different CT? Here's a hint: I is the first PT to need adjustment and PT X needed the most adjustment. – Engineer Toast Apr 28 '15 at 12:52 • Instead of going to the next available character, what if Step 4 was to loop back to Step 2? The only way to get something different would be to either pull out a different character or pull out more characters. – Engineer Toast Apr 29 '15 at 12:34 • That's the ticket! Encoding X was a pain. it far exceeded my standard calculator's ability to convert hex to decimal and to do mod. – Engineer Toast Apr 29 '15 at 20:18 • And, after all that effort, it just encodes to itself. X is a jerk letter. – Engineer Toast Apr 30 '15 at 12:33 ## -- Partial solution -- The cryptograms prove relatively easy to decrypt. Their solutions are as follows: ### Cryptogram 1 EACH OF THESE PUZZLES ARE FAIRLY BASIC DECIPHERING THE PLAINTEXT IS NOT THE CHALLENGE AS BASIC TOOLS CAN DO THAT JOB VERY QUICKLY THE PLAINTEXT IS MERELY THE EXPLANATION OF THE TRUE PUZZLE Key: ABCDEFGHIJKLMNOPQRSTUVWXYZ CORVHBNMEIQDTWLXPJAYKGUZSF ### Cryptogram 2 YOUR CHALLENGE IS TO FIGURE OUT THE PATTERN BEHIND EACH CIPHER TRY TO EXAMINE EACH BY ITSELF AS THE UNDERLYING PATTERNS ARE UNIQUE THEY ARE SIMPLE IDEAS BUT MAYBE TRICKY TO IDENTIFY BECAUSE THEY REQUIRE AN EPIPHANY Key: ABCDEFGHIJKLMNOPQRSTUVWXYZ THEQUICKBROWNFXJMPSVLAZYDG ### Cryptogram 3 IF YOU GET STUCK DONT GET TOO KEYED UP DONT BE LAZY OR JUMP TO CONCLUSIONS JUST KEEP GOING AND HASH IT OUT THE BEST ANSWER WILL HAVE ALL THREE PLAINTEXTS THE UNDERLYING PATTERN FOR EACH AND THE RULE THAT THE FIRST AND THIRD SHARE TO ENSURE A UNIQUE SUBSTITUTION GOOD LUCK PUZZLING Key: ABCDEFGHIJKLMNOPQRSTUVWXYZ CMIEPSLAFBKNHGRDOTZWQJVXUY Right away, the key for the second cryptogram jumps out. It is obviously the common pangram "The quick brown fox jumps over the lazy dog" with duplicate letters removed. The keys for the first and third cryptograms continue to elude me, though. I suspect that is the part of the puzzle that is intended to be difficult. It doesn't seem like they can be following a similar pattern to the second cryptogram, because any phrase with repeated letters removed would likely still look like English for the first few characters (until there is a repeated letter), and would probably have most of the more common letters (ETAOIN SHRDLU) near the beginning of the key. • In the solution to cryptogram 3, the phrase after "keep going and - - -" sticks out to me. I'm thinking of that particular verb's application to passwords, but I'm not sure how to apply it here and conserve length and keep everything in letters rather than hexadecimal. – Jason Patterson Apr 20 '15 at 18:27 • I noticed something weird and I don't want time lost on it: There are some cases where the letter encodes to itself. That's a coincidence. I only noticed one of them before I posted because [REDACTED]. – Engineer Toast Apr 21 '15 at 19:36	{}
2 added 37 characters in body Zariski introduced an abstract notion of Riemann surface associated to, for example, a finitely generated field extension $K/k$. It's a topological space whose points are equivalence classes of valuations of $K$ that are trivial on $k$, or equivalently valuation rings satisfying $k\subset R_v\subset K$. If $A$ is a finitely generated $k$-algebra inside $K$ then those $R_v$ which contain $A$ form an open set. In the case of a (finitely generated and) transcendence degree 1 extension all of these valuation rings are the familiar DVRs -- local Dedekind domains -- and they serve to identify the points in the unique complete nonsingular curve with this function field. (There is also the trivial valuation with $R_v=K$, which corresponds to the generic point of that curve.) In higher dimensions there are lots of complete varieties to contend with -- you can keep blowing up. Also there are more possibilities for valuations. Most of the valuation rings are not Noetherian. A curve in a surface gives you a discrete valuation ring, consisting of those rational functions which can meaningfully be restricted to rational functions on the curve: those which do not have a pole there. A point on a curve on a surface gives you a valuation whose ring consists of those functions which do not have a pole all along the curve, and which when restricted to the curve do not have a pole at the given point. The value group is $\mathbb Z\times \mathbb Z$ lexicographically ordered. A point on a transcendental curve in a complex surface, or more generally a formal (power series) curve in a surface gives you a valuation by looking at the order of vanishing; the value group is a subgroup of $\mathbb R$. This space of valuations has something of the flavor of Zariski's space of prime ideals in a ring: it is compact but not Hausdorff, for example. It can be thought of as the inverse limit, over all complete surfaces $S$ with this function field, of the space (Zariski topology) of points $S$. Zariski introduced an abstract notion of Riemann surface associated to, for example, a finitely generated field extension $K/k$. It's a topological space whose points are equivalence classes of valuations of $K$ that are trivial on $k$, or equivalently valuation rings satisfying $k\subset R_v\subset K$. If $A$ is a finitely generated $k$-algebra inside $K$ then those $R_v$ which contain $A$ form an open set. In the case of transcendence degree 1 all of these valuation rings are the familiar DVRs -- local Dedekind domains -- and they serve to identify the points in the unique complete nonsingular curve with this function field. (There is also the trivial valuation with $R_v=K$, which corresponds to the generic point of that curve.) In higher dimensions there are lots of complete varieties to contend with -- you can keep blowing up. Also there are more possibilities for valuations. Most of the valuation rings are not Noetherian. A curve in a surface gives you a discrete valuation ring, consisting of those rational functions which can meaningfully be restricted to rational functions on the curve: those which do not have a pole there. A point on a curve on a surface gives you a valuation whose ring consists of those functions which do not have a pole all along the curve, and which when restricted to the curve do not have a pole at the given point. The value group is $\mathbb Z\times \mathbb Z$ lexicographically ordered. A point on a transcendental curve in a complex surface, or more generally a formal (power series) curve in a surface gives you a valuation by looking at the order of vanishing; the value group is a subgroup of $\mathbb R$. This space of valuations has something of the flavor of Zariski's space of prime ideals in a ring: it is compact but not Hausdorff, for example. It can be thought of as the inverse limit, over all complete surfaces $S$ with this function field, of the space (Zariski topology) of points $S$.	{}
Distance and Displacement from a velocity vs. time graph My teacher is saying that the distance covered will be equal to the area of the trapezium in the graph, but the displacement will be equal to the area of the triangle (with purple hypotenuse). I studied in the internet, and it seemed to me that in this case, the distance will be equal to the displacement (as the velocity was not negative at any point of time). But my teacher is saying otherwise. He is not explaining the reason either. Can anyone please explain it to me? I really need to understand it. *unit of time is second and unit of velocity is metre per second • How do you define distance and displacement? – Steeven Feb 10 '18 at 11:15 • There are four different colored lines. What is each one supposed to represent? – Chris Feb 10 '18 at 11:23 • Nothing. I just wanted to provide a clear picture. So I recreated the graph in my question paper in my phone. – Hoque Feb 10 '18 at 11:25 Position is the time-integral to velocity: $$s=\int v\; dt$$ And the velocity is your red-and-then-green curve according to your text. The purple line is describing another motion. The total length/distance/displacement (depending of how you define each word) covered over some duration is simply the definite integral (with $t_1<t_2$): $$s=\int_{t_1}^{t_2} v\; dt$$ And such an integral is the area under the $v$-curve (still the red-and-green curve, not the purple). If you define one of these words as sign-independent, then we must at all times keep a positive speed, and we simply add: $$s=\int_{t_1}^{t_2} \|v\|\; dt$$ But, regardless of which version you are actually looking for, we can see that there are no situations of negative $v$-values on this graph. Both formulae are completely equal in this case. Now, that your teacher should claim the purple line to be the real motion does not make much sense. I am guessing that you must have misunderstood, or the teacher has misspoken. Imagine two race cars with the same max speed, but one with better acceleration. One will quickly reach it's max speed and thus move further-per-second at every moment than the other one. It will move far ahead. When the other one reaches the same max speed, the first one is already much further ahead. If you stop the race now, then of course it will end up having covered a longer distance, intuitively. The graphs below show that there can be a difference between the distance travelled and the magnitude of the displacement. For the motion depicted by the left hand graph the displacement and the distance travelled are both equal to the shaded area $A$. They are the same because the object is always travelling in the same direction. For the motion depicted by the centre and right hand graphs the direction of motion of the object changes - its velocity becomes negative after a certain time. For the centre graph you need to find the difference between the two area $B$ and $C$ to obtain the displacement whereas with the right hand graph you need to add the two areas to obtain the distance travelled. If areas $B$ and $C$ are the same then the displacement is zero whereas the distance travelled is $B+C = 2B = 2C$ .	{}
## Algebra 2 Common Core Any integer $x$ can expressed as $\frac{x}{1}$. A rational number must be able to be expressed by $\frac{a}{b}$, where a and b are both integers, and b is not 0.	{}
# NCERT Solutions for Class 11 Maths Chapter 16 Probability NCERT Solutions for Class 11 Maths Chapter 16 Probability: In our previous classes, you have studied the classical theory of probability as a measure of uncertainty of the various phenomenon in the random experiment. In the previous chapter, you have studied statistics. In this chapter, you will learn the statistical approach of probability and the axiomatic approach of probability. In NCERT solutions for class 11 maths chapter 16 probability, you will learn to find the probability on the basis of collected data and observations which is known as the statistical approach of probability. There are 44 questions in 3 exercises in the NCERT textbook. First, try to solve it on your own, if you are not able to do so, you can take help of CBSE NCERT solutions for class 11 maths chapter 16 probability. This chapter is very important for CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT etc. Solutions of NCERT for class 11 maths chapter 16 probability is useful to study advanced topics like probability distribution, stochastic process, mathematics statics and probability(MSP). Check all NCERT solutions to learn CBSE science and maths. ## Topics of NCERT Grade 11 Maths Chapter-16 Probability 16.1 Introduction 16.2 Random Experiments 16.3 Event 16.4 Axiomatic Approach to Probability The complete Solutions of NCERT Class 11 Mathematics Chapter 16 is provided below: ## NCERT solutions for class 11 maths chapter 16 probability-Exercise: 16.1 A coin is tossed three times. Let H denote Heads and T denote Tails. For each toss, there are two possible outcomes = H or T The required sample space is: S = {HHH, HHT, HTH, THH, TTH, HTT, THT, TTT} A die is thrown two times. When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6} The required sample space is: S = { $\dpi{100} (x, y) : x, y$ = 1,2,3,4,5,6} or S = {(1,1), (1,2), (1,3), ..., (1,6), (2,1), (2,2), ..., (2,6), ..., (6, 1), (6, 2), ..., (6,6)} A coin is tossed four times. Let H denote Heads and T denote Tails. For each toss, there are two possible outcomes = H or T The required sample space is: S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT} A coin is tossed and a die is thrown. Let H denote Heads and T denote Tails. For each toss, there are two possible outcomes = H or T And, When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6} The required sample space is: S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} A coin is tossed and then a die is rolled only in case a head is shown on the coin. Let H denote Heads and T denote Tails. For each toss, there are two possible outcomes = H or T For H, when a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6} The required sample space is: S = {H1, H2, H3, H4, H5, H6, T} 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person. Let X denote the event Room X is selected, Y denote the event Room Y is selected. B1, B2 denote the event a boy is selected and G1, G2 denote the event a girl is selected from room X. B3 denotes the event a boy is selected and G3, G4, G5 denote the event a girl is selected from Room Y. The required sample space is: S = {XB1 , XB2 , XG1 , XG2 , YB3 , YG3 , YG4 , YG5 } One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space. Let, R denote the event the red die comes out, W denote the event the white die comes out, B denote the event the Blue die is chosen When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6} The required sample space is: S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6} What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births? Let B denote the event a boy is born, G denote the event a girl is born The required sample space with a boy or girl in the order of their births is: S = {BB, BG, GB, GG} What is the sample space if we are interested in the number of girls in the family? (ii) In a family with two child, there can be only three possible cases: no girl child, 1 girl child or 2 girl child The required sample is: S = {0, 1, 2} Given, Number of red balls =1 Number of white balls = 3 Let R denote the event that the red ball is drawn. And W denotes the event that a white ball is drawn. Since two balls are drawn at random in succession without replacement, if the first ball is red, the second ball will be white. And if the first ball is white, second can be either of red and white The required sample space is: S = {RW, WR, WW} Let H denote the event that Head occurs and T denote the event that Tail occurs. For T in first toss, the possible outcomes when a die is thrown = {1, 2, 3 ,4 ,5 ,6} The required sample space is : S = {HH, HT, T1, T2, T3, T4, T5, T6} Let D denote the event the bulb is defective and N denote the event the bulb is non-defective The required sample space is: S = {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN} Possible outcomes when a coin is tossed = {H,T} Possible outcomes when a die is thrown = {1,2,3,4,5,6} When T occurs, experiment is finished. S1 = {T} When H occurs, a die is thrown. If the outcome is odd ({1,3,5}), S2 = {H1, H3, H5} If the outcome is even({2,4,6}), the die is thrown again., S3 = {H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66} The required sample space is: S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66} Given, two slips are drawn from the box, one after the other, without replacement. Let 1, 2, 3, 4 denote the event that 1, 2, 3, 4 numbered slip is drawn respectively. When two slips are drawn without replacement, the first event has 4 possible outcomes and the second event has 3 possible outcomes S = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)} Possible outcomes when a die is thrown = {1,2,3,4,5,6} Possible outcomes when a coin is tossed = {H,T} If the number on the die is even {2,4,6}, the coin is tossed once. S1 = {2H, 2T, 4H, 4T, 6H, 6T} If the number on the die is odd {1,3,5}, the coin is tossed twice. S2 = {1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT} The required sample space is: S = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T} Possible outcomes when a coin is tossed = {H,T} Possible outcomes when a die is thrown = {1,2,3,4,5,6} Let R1 and R2 denote the event that a red ball is drawn and B1, B2, B3 denote the event that a blue ball is drawn If H occurs, a die is thrown. S1 = {H1, H2, H3, H4, H5, H6} If T occurs, a ball from a box which contains $2$ red and $3$ black balls is drawn. S2 = {TR1 , TR2 , TB1 , TB2 , TB3} The required sample space is: S = {TR1 , TR2 , TB1 , TB2 , TB3 , H1, H2, H3, H4, H5, H6} Given, a die is thrown repeatedly untill a six comes up. Possible outcomes when a die is thrown = {1,2,3,4,5,6} In the experiment 6 may come up on the first throw, or the 2nd throw, or the 3rd throw and so on till 6 is obtained. The required sample space is: S = {6, (1,6), (2,6), (3,6), (4,6), (5,6), (1,1,6), (1,2,6), ..., (1,5,6), (2,1,6). (2,2,6), ..., (2,5,6), ..., (5,1,6), (5,2,6), ... } NCERT solutions for class 11 maths chapter 16 probability-Exercise: 16.2 When a die is rolled, the sample space of possible outcomes: S = {1, 2, 3, 4, 5, 6} Now, E = event that the die shows 4 = {4} F = event that the die shows even number = {2, 4, 6} $\dpi{100} \cap$ F = {4} $\dpi{100} \cap$ {2, 4, 6} = {4} $\dpi{100} \neq \phi$ Hence E and F are not mutually exclusive event. Question:2(i) A die is thrown. Describe the following events: A: a number less than 7 When a die is rolled, the sample space of possible outcomes: S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7} Given, A : a number less than 7 As every number on a die is less than 7 A = {1, 2, 3, 4, 5, 6} = S Question:2(ii) A die is thrown. Describe the following events: B: a number greater than 7 When a die is rolled, the sample space of possible outcomes: S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7} Given, B: a number greater than 7 As no number on the die is greater than 7 B = $\dpi{100} \phi$ Question:2(iii) A die is thrown. Describe the following events: C: a multiple of 3. When a die is rolled, the sample space of possible outcomes: S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7} Given, C : a multiple of 3 C = {3, 6} Question:2(iv) A die is thrown. Describe the following events: When a die is rolled, the sample space of possible outcomes: S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7} Given, D : a number less than 4 D = {1, 2, 3} Question:2(v) A die is thrown. Describe the following events: E: an even multiple greater than 4 When a die is rolled, the sample space of possible outcomes: S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7} Given, E : an even number greater than 4 S1 = Subset of S containing even numbers = {2,4,6} Therefore , E = {6} Question:2(vi). A die is thrown. Describe the following events: F: a number not less than 3 When a die is rolled, the sample space of possible outcomes: S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7} Given, F : a number not less than 3 F = {x: x $\dpi{100} \in$ S, x $\dpi{80} \geq$ 3 } = {3, 4, 5, 6} Question:2.(vi) A die is thrown. Describe the following events: Also find (a) $A\cup B$ A = {1, 2, 3, 4, 5, 6} B= $\dpi{100} \phi$ $\dpi{100} \therefore$ A $\dpi{80} \cup$ B = {1, 2, 3, 4, 5, 6} $\dpi{80} \cup$ $\dpi{100} \phi$ = {1, 2, 3, 4, 5, 6} Question:2.(vi) A die is thrown. Describe the following events: Also find (b) $A\cap B$. A = {1, 2, 3, 4, 5, 6} B= $\dpi{100} \phi$ $\dpi{100} \therefore$ A $\dpi{80} \cap$ B = {1, 2, 3, 4, 5, 6} $\dpi{80} \cap$ $\dpi{100} \phi$ = $\dpi{100} \phi$ Question:2.(vi) A die is thrown. Describe the following events: Also find (c) $B\cup C$ B= $\dpi{100} \phi$ C= {3, 6} $\dpi{100} \therefore$ B $\dpi{80} \cup$ C = $\dpi{100} \phi$ $\dpi{80} \cup$ {3, 6} = {3, 6} Question:2.(vi) A die is thrown. Describe the following events: (d) Also find $E\cap F$ E = {6} F = {3, 4, 5, 6} $\dpi{100} \therefore$ E $\dpi{80} \cap$ F = {6} $\dpi{80} \cap$ {3, 4, 5, 6} = {6} Question:2.(vi) A die is thrown. Describe the following events: Also find (e) $D\cap E$ D = {1, 2, 3} E = {6} $\dpi{100} \therefore$ D $\dpi{80} \cap$ E = {1, 2, 3} $\dpi{80} \cap$ {6} = $\dpi{100} \phi$ (As nothing is common in these sets) Question:2.(vi) A die is thrown. Describe the following events: Also find (f) $A-C$ A = {1, 2, 3, 4, 5, 6} C = {3, 6} $\dpi{100} \therefore$ A - C = {1, 2, 3, 4, 5, 6} - {3, 6} = {1, 2, 4, 5} Question:2.(vi) A die is thrown. Describe the following events: Also find (g) $D-E$ D = {1, 2, 3} E = {6} $\dpi{100} \therefore$ D - E = {1, 2, 3} - {6} = {1, 2, 3} Question:2.(vi) A die is thrown. Describe the following events: Also find (h) $E\cap F'$ E = {6} F = {3, 4, 5, 6} $\dpi{100} \therefore$ F' = {3, 4, 5, 6}' = S - F = {1, 2} $\dpi{100} \therefore$ E $\dpi{80} \cap$ F' = {6} $\dpi{80} \cap$ {1, 2} = $\dpi{100} \phi$ Question:2.(vi) A die is thrown. Describe the following events: Also find (i) ${F}'$ F = {3, 4, 5, 6} $\dpi{100} \therefore$ F' = {3, 4, 5, 6}' = S - F = {1, 2} the sum is greater than $8$ Sample space when a die is rolled: S = {1, 2, 3, 4, 5, 6} Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes] E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)} Now, A : the sum is greater than 8 Possible sum greater than 8 are 9, 10, 11 and 12 A = [ {(a,b): (a,b) $\dpi{100} \in$ E, a+b>8 } ]= {(3,6), (4,5), (5, 4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)} $2$ occurs on either die Sample space when a die is rolled: S = {1, 2, 3, 4, 5, 6} Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes] E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)} Now, B: 2 occurs on either die Hence the number 2 can come on first die or second die or on both the die simultaneously. B = [ {(a,b): (a,b) $\dpi{100} \in$ E, a or b = 2 } ]= {(1,2), (2,2), (3, 2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)} the sum is at least $7$ and a multiple of $3$ Sample space when a die is rolled: S = {1, 2, 3, 4, 5, 6} Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes] E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)} Now, C: the sum is at least 7 and a multiple of 3 The sum can only be 9 or 12. C = [ {(a,b): (a,b) $\dpi{100} \in$ E, a+b>6 & a+b = 3k, k $\dpi{100} \in$ I} ]= {(3,6), (6,3), (5, 4), (4,5), (6,6)} Which pairs of these events are mutually exclusive? For two elements to be mutually exclusive, there should not be any common element amongst them. Also, A = {(3,6), (4,5), (5, 4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)} B = {(1,2), (2,2), (3, 2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)} C = {(3,6), (6,3), (5, 4), (4,5), (6,6) Now, A $\cap$ B = $\phi$ (no common element in A and B) Hence, A and B are mutually exclusive Again, B $\cap$ C = $\phi$ (no common element in B and C) Hence, B and C are mutually exclusive Again, C $\cap$ A = {(3,6), (6,3), (5, 4), (4,5), (6,6)} Therefore, A and B, B and C are mutually exclusive. mutually exclusive? Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Now, A = Event that three heads show up = {HHH} B = Event that two heads and one tail show up = {HHT, HTH, THH} C = Event that three tails show up = {TTT} D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT} (i). For two elements X and Y to be mutually exclusive, X $\cap$ Y = $\phi$ $\cap$ B = {HHH} $\cap$ {HHT, HTH, THH} = $\phi$ ; Hence A and B are mutually exclusive. $\cap$ C = {HHT, HTH, THH} $\cap$ {TTT} = $\phi$ ; Hence B and C are mutually exclusive. $\cap$ D = {TTT} $\cap$ {HHH, HHT, HTH, HTT} = $\phi$ ; Hence C and D are mutually exclusive. $\cap$ A = {HHH, HHT, HTH, HTT} $\cap$ {HHH} = {HHH} ; Hence D and A are not mutually exclusive. $\cap$ C = {HHH} $\cap$ {TTT} = $\phi$ ; Hence A and C are mutually exclusive. $\cap$ D = {HHT, HTH, THH} $\cap$ {HHH, HHT, HTH, HTT} = {HHT, HTH} ; Hence B and D are not mutually exclusive. simple? Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Now, A = Event that three heads show up = {HHH} B = Event that two heads and one tail show up = {HHT, HTH, THH} C = Event that three tails show up = {TTT} D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT} (ii).If an event X has only one sample point of a sample space, it is called a simple event. A = {HHH} and C = {TTT} Hence, A and C are simple events. Compound? Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, THH, TTH, TTT} Now, A = Event that three heads show up = {HHH} B = Event that two heads and one tail show up = {HHT, HTH, THH} C = Event that three tails show up = {TTT} D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT} (iv). If an event has more than one sample point, it is called a Compound event. B = {HHT, HTH, THH} and D = {HHH, HHT, HTH, HTT} Hence, B and D are compound events. Question:5(i) Three coins are tossed. Describe Two events which are mutually exclusive. Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, THH, TTH, TTT} (i) A = Event that three heads show up = {HHH} B = Event that three tails show up = {TTT} $\cap$ B = {HHH} $\cap$ {TTT} = $\phi$ ; Hence A and B are mutually exclusive. Question:5(ii) Three coins are tossed. Describe Three events which are mutually exclusive and exhaustive. Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Let , A = Getting no tails = {HHH} B = Getting exactly one tail = {HHT, HTH, THH} C = Getting at least two tails = {HTT, THT, TTH} Clearly, A $\cap$ B = $\phi$ ; B $\cap$ C = $\phi$ ; C $\cap$ A = $\phi$ Since (A and B), (B and C) and (A and C) are mutually exclusive Therefore A, B and C are mutually exclusive. Also, $\cup$ B $\cup$ C = S Hence A, B and C are exhaustive events. Hence, A, B and C are three events which are mutually exclusive and exhaustive. Question:5(iii). Three coins are tossed. Describe Two events, which are not mutually exclusive. Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Let , A = Getting at least one head = {HHH, HHT, HTH, THH, TTH} B = Getting at most one head = {TTH, TTT} Clearly, A $\cap$ B = {TTH} $\neq$ $\phi$ Hence, A and B are two events which are not mutually exclusive. Question:5.(iv) Three coins are tossed. Describe Two events which are mutually exclusive but not exhaustive. Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Let , A = Getting exactly one head = {HTT, THT, TTH} B = Getting exactly one tail = {HHT, HTH, THH} Clearly, A $\cap$ B = $\phi$ Hence, A and B are mutually exclusive. Also, A $\cup$ B $\neq$ S Hence, A and B are not exhaustive. Question:5.(v) Three coins are tossed. Describe Three events which are mutually exclusive but not exhaustive Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Let , A = Getting exactly one tail = {HHT, HTH, THH} B = Getting exactly two tails = {HTT, TTH, THT} C = Getting exactly three tails = {TTT} Clearly, A $\cap$ B = $\phi$ ; B $\cap$ C = $\phi$ ; C $\cap$ A = $\phi$ Since (A and B), (B and C) and (A and C) are mutually exclusive Therefore A, B and C are mutually exclusive. Also, $\cup$ B $\cup$ C = {HHT, HTH, THH, HTT, TTH, THT, TTT} $\neq$ S Hence A, B and C are not exhaustive events. A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$. Describe the events $A{}'$ Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} (i) Therefore, A'= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} = B : getting an odd number on the first die. Question:6.(ii) Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$. Describe the events not B Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} (ii) Therefore, B'= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} = A : getting an even number on the first die. Question:6.(iii) Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$. Describe the events A or B Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} (iii) A or B = A $\cup$ B = {(1,1), (1,2) .... (1,6), (3,1), (3,2).... (3,6), (5,1), (5,2)..... (5,6), (2,1), (2,2)..... (2,6), (4,1), (4,2)..... (4,6), (6,1), (6,2)..... (6,6)} = S Question:6.(iv) Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$ Describe the events A and B Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} (iii) A and B = A $\cap$ B = A $\cap$ A' = $\phi$ (From (ii)) A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$ Describe the events A but not C Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5 The possible sum are 2,3,4,5 C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} (v) A but not C = A - C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Question:6.(vi) Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$ Describe the events B or C Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5 The possible sum are 2,3,4,5 C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} (vi) B or C = B $\cup$ C = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} Question:6.(vii) Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$ Describe the events B and C Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5 The possible sum are 2,3,4,5 C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} (vi) B and C = B $\cap$ C = {(1, 1), (1,2), (1,3), (1,4), (3,1), (3,2)} Question:6.(viii) Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice $\leq 5$ Describe the events $A\cap {B}'\cap {C}'$ Sample space when two dice are thrown: S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5 The possible sum are 2,3,4,5 C = {(a,b): 2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} (viii) A $\cap$ B' $\cap$ C' = A $\cap$ A $\cap$ C' (from (ii)) = A $\cap$ C' = A - C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Question:7.(i) Refer to question 6 above, state true or false: (give reason for your answer) A and B are mutually exclusive Here, A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} (i) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$ $\cap$ B = $\phi$ , since A and B have no common element amongst them. Hence, A and B are mutually exclusive. TRUE A and B are mutually exclusive and exhaustive Here, A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} (ii) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$ $\cap$ B = $\phi$ , since A and B have no common element amongst them. Hence, A and B are mutually exclusive. Also, $\cup$ B = {(2,1), (2,2).... (2,6), (4,1), (4,2).....(4,6), (6,1), (6,2)..... (6,6), (1,1), (1,2).... (1,6), (3,1), (3,2)..... (3,6), (5,1), (5,2).... (5,6)} = S Hence, A and B are exhaustive. TRUE Question:7.(iii) Refer to question 6 above, state true or false: (give reason for your answer) $A=B{}'$ Here, S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} (iii) Therefore, B' = S -B = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} = A TRUE Question:7.(iv) Refer to question 6 above, state true or false: (give reason for your answer) A and C are mutually exclusive Here, S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} C = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} (iv) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$ $\cap$ C = {(2,1), (2,2), (2,3), (4,1)} , Hence, A and B are not mutually exclusive. FALSE $A$ and ${B}'$ are mutually exclusive. X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$ $\cap$ B' = A $\cap$ A = A (From (iii)) $\dpi{100} \therefore$ A $\cap$ B’ $\dpi{100} \neq \phi$ Hence A and B' not mutually exclusive. FALSE Question:7.(vi) Refer to question 6 above, state true or false: (give reason for your answer) ${A}',{B}',C$ are mutually exclusive and exhaustive. Here, S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6} A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} C = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} (vi) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$ $\dpi{100} \therefore$ A' $\cap$ B' = B $\cap$ A = $\phi$ (from (iii) and (i)) Hence A' and B' are mutually exclusive. Again, $\dpi{100} \therefore$ B' $\cap$ C = A $\cap$ C $\dpi{80} \neq$ $\phi$ (from (iv)) Hence B' and C are not mutually exclusive. Hence, A', B' and C are not mutually exclusive and exhaustive. FALSE CBSE NCERT solutions for class 11 maths chapter 16 probability-Exercise: 16.3 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (a) $0.1$ $0.01$ $0.05$ $0.03$ $0.01$ $0.2$ $0.6$ (a) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one. Condition (ii): Sum of probabilities = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1 Therefore, the assignment is valid Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (b) $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ (b) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one. Condition (ii): Sum of probabilities = $\dpi{100} \frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7} = 1$ Therefore, the assignment is valid Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (c) $0.1$ $0.2$ $0.3$ $0.4$ $0.5$ $0.6$ $0.7$ (c) Since sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1 Hence, Condition (ii) is not satisfied. Therefore, the assignment is not valid Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (d) $-0.1$ $0.2$ $0.3$ $0.4$ $-0.2$ $0.1$ $0.3$ (d) Two of the probabilities p( $\omega_1$ ) and p( $\omega_5$ ) are negative, hence condition(i) is not satisfied. Therefore, the assignment is not valid. Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (e) $\frac{1}{14}$ $\frac{2}{14}$ $\frac{3}{14}$ $\frac{4}{14}$ $\frac{5}{14}$ $\frac{6}{14}$ $\frac{15}{14}$ (e) Each of the number p( $\dpi{100} \omega_i$) is positive but p( $\dpi{100} \omega_7$) is not less than one. Hence the condition is not satisfied. Therefore, the assignment is not valid. Sample space when a coin is tossed twice, S = {HH, HT, TH, TT} [Note: A coin tossed twice is same as two coins tossed at once] $\therefore$ Number of possible outcomes n(S) = 4 Let E be the event of getting at least one tail = {HT, TH, TT} $\therefore$ n(E) = 3 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{4}$ = 0.75 A prime number will appear Sample space when a die is thrown, S = {1,2,3,4,5,6} $\therefore$ Number of possible outcomes n(S) = 6 Let E be the event of getting a prime number = {2,3,5} $\therefore$ n(E) = 3 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{6}$ = 0.5 Question:3.(ii) A die is thrown, find the probability of following events: A number greater than or equal to $3$ will appear Sample space when a die is thrown, S = {1,2,3,4,5,6} $\therefore$ Number of possible outcomes n(S) = 6 Let E be the event of getting a number greater than or equal to 3 = {3,4,5,6} $\therefore$ n(E) = 4 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{6} = \frac{2}{3}$ = 0.67 Question:3.(iii) A die is thrown, find the probability of following events: A number less than or equal to one will appear Sample space when a die is thrown, S = {1,2,3,4,5,6} $\therefore$ Number of possible outcomes n(S) = 6 Let E be the event of getting a number less than or equal to one = {1} $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{6}$ = 0.167 Question:3.(iv) A die is thrown, find the probability of following events: A number more than $\small 6$ will appear Sample space when a die is thrown, S = {1,2,3,4,5,6} $\therefore$ Number of possible outcomes n(S) = 6 Let E be the event of getting a number more than 6 will appear = $\phi$ $\therefore$ n(E) = 0 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{0}{6}$ = 0 A number less than $\small 6$ will appear. Sample space when a die is thrown, S = {1,2,3,4,5,6} $\therefore$ Number of possible outcomes n(S) = 6 Let E be the event of getting a number less than 6 will appear = {1,2,3,4,5} $\therefore$ n(E) = 5 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{5}{6}$ = 0.83 (a) Number of points(events) in the sample space = Number of cards in the pack = 52 Calculate the probability that the card is an ace of spades. Number of possible outcomes, n(S) = 52 Let E be the event that the card is an ace of spades $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{52}$ The required probability that the card is an ace of spades is $\dpi{80} \frac{1}{52}$. Question:4(c)(i) A card is selected from a pack of 52 cards. Calculate the probability that the card is an ace Number of possible outcomes, n(S) = 52 Let E be the event that the card is an ace. There are 4 aces. $\therefore$ n(E) = 4 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{52} = \frac{1}{13}$ The required probability that the card is an ace is $\dpi{80} \frac{1}{13}$. Calculate the probability that the card is black card. Number of possible outcomes, n(S) = 52 Let E be the event that the card is a black card. There are 26 black cards. (Diamonds and Clubs) $\therefore$ n(E) = 26 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{26}{52} = \frac{1}{2}$ The required probability that the card is an ace is $\dpi{80} \frac{1}{2}$. The coin and die are tossed together. The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Number of possible outcomes, n(S) = 12 (i) Let E be the event having sum of numbers as 3 = {(1, 2)} $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{12}$ The required probability of having 3 as sum of numbers is $\dpi{80} \frac{1}{12}$. The coin and die are tossed together. The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}} = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Number of possible outcomes, n(S) = 12 (ii) Let E be the event having sum of numbers as 12 = {(6, 6)} $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{12}$ The required probability of having 12 as sum of numbers is $\dpi{80} \frac{1}{12}$. There are four men and six women on the city council $\therefore$ n(S) = n(men) + n(women) = 4 + 6 = 10 Let E be the event of selecting a woman $\therefore$ n(E) = 6 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{6}{10} = \frac{3}{5}$ Therefore, the required probability of selecting a woman is 0.6 Here the sample space is, S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT} According to question, 1.) 4 heads = 1 + 1 + 1 + 1 = Rs. 4 2.) 3 heads and 1 tail = 1 + 1 + 1 - 1.50 = Rs. 1.50 3.) 2 heads and 2 tails = 1 + 1 - 1.50 - 1.50 = - Rs. 1 : he will lose Re. 1 4.) 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50 = - Rs. 3.50 : he will lose Rs. 3.50 5.) 4 tails = – 1.50 – 1.50 – 1.50 – 1.50 = - Rs. 6 = he will lose Rs. 6 Now, sample space of amounts corresponding to S: S' = {4, 1.50, 1.50, 1.50, 1.50, - 1, - 1, - 1, - 1, - 1, - 1, - 3.50, - 3.50, - 3.50, - 3.50, - 6} $\therefore$ n(S') = 12 $\therefore$ Required Probabilities are: $\dpi{100} P(Winning\ Rs.\ 4) = \frac{n(Winning\ Rs.\ 4)}{n(S')}$ $= \frac{1}{16}$ $\dpi{100} P(Winning\ Rs.\ 1.50) = \frac{n(Winning\ Rs.\ 1.50)}{n(S')}$ $= \frac{4}{16} = \frac{1}{4}$ $\dpi{100} P(Losing\ Re.\ 1) = \frac{n(Losing\ Re.\ 1)}{n(S')}$ $= \frac{6}{16} = \frac{3}{8}$ $\dpi{100} P(Losing\ Rs.\ 3.50) = \frac{n(Losing\ Rs.\ 3.50)}{n(S')}$ $= \frac{4}{16} = \frac{1}{4}$ $\dpi{100} P(Losing\ Rs.\ 6) = \frac{n(Losing\ Rs.\ 6)}{n(S')}$ $= \frac{1}{16}$ $\small 3$ heads Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting 3 heads = {HHH} $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$ The required probability of getting 3 heads is $\dpi{80} \frac{1}{8}$. Question:8.(ii) Three coins are tossed once. Find the probability of getting $\small 2$ heads Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting 2 heads = {HHT, HTH, THH} $\therefore$ n(E) = 3 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{8}$ The required probability of getting 2 heads is $\dpi{80} \frac{3}{8}$. Question:8.(iii) Three coins are tossed once. Find the probability of getting atleast $2$ heads Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting atleast 2 heads = Event of getting 2 or more heads = {HHH, HHT, HTH, THH} $\therefore$ n(E) = 4 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{4}{8} = \frac{1}{2}$ The required probability of getting atleast 2 heads is $\dpi{80} \frac{1}{2}$. Question:8.(iv) Three coins are tossed once. Find the probability of getting atmost $\small 2$ heads Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting atmost 2 heads = Event of getting 2 or less heads = {HHT, HTH, THH, TTH, HTT, THT} $\therefore$ n(E) = 6 $= \frac{6}{8} = \frac{3}{4}$$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ The required probability of getting almost 2 heads is $\dpi{80} \frac{3}{4}$. Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting no head = Event of getting only tails = {TTT} $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$ The required probability of getting no head is $\dpi{80} \frac{1}{8}$. Question:8.(vi) Three coins are tossed once. Find the probability of getting $\small 3$ tails Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting 3 tails = {TTT} $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$ The required probability of getting 3 tails is $\dpi{80} \frac{1}{8}$. Question:8(vii) Three coins are tossed once. Find the probability of getting exactly two tails Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting exactly 2 tails = {TTH, HTT, THT} $\therefore$ n(E) = 3 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{3}{8}$ The required probability of getting exactly 2 tails is $\dpi{80} \frac{3}{8}$. Question:8.(viii) Three coins are tossed once. Find the probability of getting no tail Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting no tail = Event of getting only heads = {HHH} $\therefore$ n(E) = 1 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{1}{8}$ The required probability of getting no tail is $\dpi{80} \frac{1}{8}$. Question:8.(ix) Three coins are tossed once. Find the probability of getting atmost two tails Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8] Let E be the event of getting atmost 2 tails = Event of getting 2 or less tails = {HHT, HTH, THH, TTH, HTT, THT} $\therefore$ n(E) = 6 $\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$ $= \frac{6}{8} = \frac{3}{4}$ The required probability of getting atmost 2 tails is $\dpi{80} \frac{3}{4}$. Given, P(E) = $\small \frac{2}{11}$ We know, P(not E) = P(E') = 1 - P(E) = $\dpi{80} 1 - \small \frac{2}{11}$ $\dpi{80} \frac{9}{11}$ Given, ‘ASSASSINATION’ No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1 No. of letters = 13 No. of vowels = {3 A's,2 I's,O} = 6 One letter is selected: n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13 Let E be the event of getting a vowel. n(E) = $\dpi{100} ^{6}\textrm{C}_{1}$ = 6 $\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{6}{13}$ Given, ‘ASSASSINATION’ No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1 No. of letters = 13 No. of consonants = {4 S's,2 N's,T} = 7 One letter is selected: n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13 Let E be the event of getting a consonant. n(E) = $\dpi{100} ^{7}\textrm{C}_{1}$ = 7 $\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{7}{13}$ Total numbers of numbers in the draw = 20 Numbers to be selected = 6 $\dpi{100} \therefore$ n(S) = $\dpi{100} ^{20}\textrm{C}_{6}$ Let E be the event that six numbers match with the six numbers fixed by the lottery committee. n(E) = 1 (Since only one prize to be won.) $\dpi{100} \therefore$ Probability of winning = $\dpi{100} P(E) = \frac{n(E)}{n(S)}$$\dpi{100} = \frac{1}{^{20}\textrm{C}_{6}} = \frac{6!14!}{20!}$ $\dpi{100} = \frac{6.5.4.3.2.1.14!}{20.19.18.17.16.15.14!}$ $\dpi{100} = \frac{1}{38760}$ $P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$ (i) Given, $P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$ Now P(A $\cap$ B) > P(A) (Since $\cap$ B is a subset of A, P($\cap$ B) cannot be more than P(A)) Therefore, the given probabilities are not consistently defined. $P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$ (ii) Given, $P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$ We know, P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B) $\implies$ 0.8 = 0.5 + 0.4 - P(A $\cap$ B) $\implies$ P(A $\cap$ B) = 0.9 - 0.8 = 0.1 Therefore, P(A $\cap$ B) < P(A) and P(A $\cap$ B) < P(B) , which satisfies the condition. Hence, the probabilities are consistently defined Question:13 Fill in the blanks in following table: $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $...$ (ii) $0.35$ $...$ $0.25$ $0.6$ (iii) $0.5$ $0.35$ $...$ $0.7$ We know, $\dpi{100} P(A \cup B) = P(A)+ P(B) - P(A \cap B)$ (i) $\dpi{100} P(A \cup B)$ = $\dpi{100} \frac{1}{3}+\frac{1}{5}-\frac{1}{15}$ = $\dpi{100} \frac{5+3-1}{15} = \frac{7}{15}$ (ii) $\dpi{100} 0.6 = 0.35 + P(B) - 0.25$ $\implies$ $\dpi{100} P(B) = 0.6 - 0.1 = 0.5$ (iii) $\dpi{100} 0.7 = 0.5 + 0.35 - P(A \cap B)$ $\implies$ $\dpi{100} P(A \cap B) = 0.85 - 0.7 = 0.15$ $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $\dpi{100} \boldsymbol{\frac{7}{15}}$ (ii) $0.35$ 0.5 $0.25$ $0.6$ (iii) $0.5$ $0.35$ 0.15 $0.7$ Given, $\dpi{80} P(A)=\frac{3}{5}$ and $\dpi{80} P(B)=\frac{1}{5}$ To find : $\dpi{100} P(A or B) = P(A \cup B)$ We know, $P(A \cup B) = P(A)+ P(B) - P(A \cap B) = P(A)+ P(B)$ [Since A and B are mutually exclusive events.] $\implies$ $P(A \cup B) = \frac{3}{5}+\frac{1}{5} = \frac{4}{5}$ Therefore, $\dpi{100} P(A \cup B) = \frac{4}{5}$ Given, $P(E)=\frac{1}{4}$, $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$ To find : $P(E or F) = P(E \cup F)$ We know, $P(A \cup B) = P(A)+ P(B) - P(A \cap B)$ $\implies$ $P(E \cup F) =$ $\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$ $= \frac{5}{8}$ Therefore, $P(E \cup F) =$ $\frac{5}{8}$ Given, $P(E)=\frac{1}{4}$, $P(F)=\frac{1}{2}$ and $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$ To find :$P(not\ E\ and\ not\ F) = P(E' \cap F')$ We know, $P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)$ And $P(A\cup B) = P(A)+ P(B) - P(A \cap B)$ $\implies$ $P(E \cup F) =$$\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$ $= \frac{5}{8}$ $\implies$ $P(E' \cap F') = 1 - P(E \cup F)$ $= 1- \frac{5}{8}= \frac{3}{8}$ Therefore, $P(E' \cap F') =$ $\frac{3}{8}$ Given, $P(not\ E\ or\ not\ F) = 0.25$ For A and B to be mutually exclusive, $P(A \cap B) = 0$ Now, $P(not\ E\ or\ not\ F) = P(E' \cup F') = 0.25$ We know, $P(A' \cup B') = P(A \cap B)' = 1 - P(A \cap B)$ $\\ \implies 0.25 = 1 - P(E \cap F) \\ \implies P(E \cap F) = 1 - 0.25 = 0.75 \neq 0$ Hence, E and F are not mutually exclusive. Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16 (i) $P(not\ A) = P(A') = 1 - P(A)$ $\implies$ $P(not\ A) = 1 - 0.42 = 0.58$ Therefore, P(not A) = 0.58 Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16 (ii) $P(not\ B) = P(B') = 1 - P(B)$ $\implies$ $P(not\ B) = 1 - 0.48 = 0.52$ Therefore, P(not B) = 0.52 Given, P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16 (iii) We know, $P(A \cup B) = P(A)+ P(B) - P(A \cap B)$ $\implies$ $P(A \cup B) = 0.42 + 0.48 - 0.16 = 0.9 - 0.16$ = 0.74 Let M denote the event that the student is studying Mathematics and B denote the event that the student is studying Biology And total students in the class be 100. Given, n(M) = 40 $\implies$ P(M) = $\dpi{100} \frac{40}{100} = \frac{2}{5}$ n(B) = 30$\implies$ P(M) = $\dpi{100} \frac{30}{100} = \frac{3}{10}$ n(M $\cap$ B) = 10$\implies$ P(M) = $\dpi{100} \frac{10}{100} = \frac{1}{10}$ We know, P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B) $\implies$ P(M $\cup$ B) = 0.4 + 0.3 - 0.1 = 0.6 Hence, the probability that he will be studying Mathematics or Biology is 0.6 Let A be the event that the student passes the first examination and B be the event that the students passes the second examination. P(A $\cup$ B) is probability of passing at least one of the examination. Therefore, P(A $\cup$ B) = 0.95 , P(A)=0.8, P(B)=0.7 We know, P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B) $\implies$ P(A $\cap$ B) = 0.8 + 0.7 - 0.95 = 1.5 -0.95 = 0.55 Hence,the probability that the student will pass both the examinations is 0.55 Let A be the event that the student passes English examination and B be the event that the students pass Hindi examination. Given, P(A)=0.75, P(A $\cap$ B) = 0.5, P(A' $\cap$ B') =0.1 We know, P(A' $\cap$ B') = 1 - P(A $\cup$ B) $\implies$ P(A $\cup$ B) = 1 - 0.1 = 0.9 Also, P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B) $\implies$ P(B) = 0.9 - 0.75 + 0.5 = 0.65 Hence,the probability of passing the Hindi examination is 0.65 The student opted for NCC or NSS. Let A be the event that student opted for NCC and B be the event that the student opted for NSS. Given, n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24 Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$ P(B) = $\inline \frac{32}{60} = \frac{8}{15}$ P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$ (i) We know, P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)	{}
# The theory of simultaneous linear equations (Found in "Linear Algebra and Its Applications" by Gilbert Strang) "Find the value of c which makes it possible to solve: $$\left\{ \begin{array}{c} u+v+2w=2\\ 2u+3v-w=5\\ 3u+4v+w=c \end{array}\right.$$" - Use the appropriate extended matrix of coefficients: $$\left(\begin{array}{ccc\|c}1&1&2&2\\2&3&-1&5\\3&4&1&c \end{array}\right)$$ Row-reduce it, and find the value of $c$ for which you have no rows of the form $\begin{pmatrix}0&0&0&\|&\alpha\end{pmatrix}$ for $\alpha\neq0$. For that value there exists a solution. If there are no zero rows - then the solution is unique. If there is at least one zero-row then there are infinitely many solutions. -	{}
# Turbo Charger Jet? Discussion in 'Aircraft Design / Aerodynamics / New Technology' started by death31313, Mar 29, 2010. ### Help Support HomeBuiltAirplanes Forum by donating: 1. Mar 29, 2010 ### death31313 #### Member Joined: Mar 20, 2010 Messages: 11 0 Location: Colton, SD , United States I've been looking for jet engines to power an aircraft I've been trying to design and I stumbled accross multiple sets of plans for turning two turbochargers into a turbo jet engine. now im sure you have all probibly seen them and could probibly give me some answers. first I would like to ask, If constructed from quality parts, could a turbocharger jet work in an aircraft? Whats the reliability like? how much thrust do they put out? I'm Know they aren't ideal but would they work? thoughts and advice are welcome as always. Last edited: Mar 29, 2010 2. Mar 29, 2010 ### Bart #### Well-Known Member Joined: Apr 14, 2007 Messages: 299 2 Well, for sake of discussion, suppose one of those units would work fabulously, for $5 total cost. But, you'd still be stuck with the fact that fuel/air ratio needs to be something like ~13-14:1 to burn properly. The more air sucked & blown through the pipe, the more you gotta pay for gas. Such turbines suck a LOT of air, so to keep burning you gotta buy and carry aloft a LOT of gas, at ~ 6 lbs./gallon and ~$6/gallon. What you save in engine weight you'll generally wind up carrying in extra fuel, heavy and expensive. So, the uber-light \$5 engine might mean the plane is not so light and cheap, after all. I'm exaggerating for effect, obviously, to make a point about life-cycle cost and efficiency. Thoughts? 3. Mar 30, 2010 ### Lucrum #### Well-Known Member Joined: Jun 10, 2008 Messages: 956 189 Location: Canton, GA It's an interesting idea. But I'm inclined to think reality will (as usual) get in the way. While simple in theory, in practice a reliable and controllable turbine engine is fairly complex. It's one thing to have a novelty running on your driveway and a whole other to get one reliably working in the confines of an airframe. What about engine driven accessories? How do you drive them? What about over speed protection and fuel control? Truth be known I'm as big a dreamer as anyone here. And I'd love to see someone with the turbine knowledge, machining capabilities and of course MONEY successfully put something like this together. But I have my doubts. 4. Mar 30, 2010 ### Dan Thomas #### Well-Known Member Joined: Sep 18, 2008 Messages: 4,873 2,083 Turbochargers typically generate only low pressures, perhaps no more than 20 or 25 psi. The compressor of a jet generates as much as 350 psi. More squeeze means more power, so a turbo isn't going to give much at all. Turbos are small-diameter centrifugal compressors that have to spin at terrific speeds to do their jobs. Most turbine engines have considerable larger compressor diameters and still have to spin at terrific speeds. Dan 5. Mar 30, 2010 ### death31313 #### Member Joined: Mar 20, 2010 Messages: 11 0 Location: Colton, SD , United States the basics behind how a turbocharger Jet works is there are two turbos these are conected by piping wich serves as the combustion chamber. the blades of these turbos are attached by some sort of rod. One turbo takes in air compresses it in the combustion chamber and a spark plug ignites a fuel mixed with the compressed air. The ignition of this fuel causes a increse in pressure and sends that pressure out through the other turbo wich spins that turboes blades there by spining the blades of the intake turbo and the cycle repeats. they are throtialible by regulating the amount of fuel intake these jets have become somewhat popular with jet enthusiasts and has even been built on an episode of Junkyard Wars, but as far as I know, no one has put one on an airplane. I supose I should have described the turbocharger jet engine a little better for those who didn't know what i was talking about. (please excuse the spelling in this post, the computer im using has no spell check) 6. Mar 30, 2010 ### Sir Joab #### Well-Known Member Joined: Mar 28, 2010 Messages: 47 4 Location: Arlington, Wa, US I've seen these things too, but I think they generally only use 1 turbocharger. A turbocharger on a car has a turbine through which the engine's exhaust flows which in turn drives the intake turbine which compresses the fuel/air mixture and forces it into the engine. A turbocharger-jet simply replaces the automotive engine with a combustion chamber. My personal doubt about it's viability would be the engine's power-to-weight ratio. Here's a video of a very nice engine on youtube, but notice that the operator stops the engine from rolling on it's stand at full power by bracing his foot against it while sitting on a rolling chair... I personally have been intrigued at the idea of building a plane powered by a Pulsejet... but that's another story. 7. Mar 30, 2010 ### Dana #### Super ModeratorStaff Member Joined: Apr 4, 2007 Messages: 8,687 3,064 Location: CT, USA Turbocharger jets are popular ("popular" being relative, of course) among hobbyists, bur as a curiosity, nothing more. Oh, some may have stuck them on bicycles or go-karts, but that's about it. Most car turbochargers are quite small compared to even the smallest jet engines. About 30 years ago I read in Aviation Week that Garrett was working with a jet adapted from a large truck turbocharger for drone applications, but I don't know what came of it. Also they're heavy... turbos typically have bodies made of cast iron, not the stainless steel and Inconel of an aircraft turbine engine. If you want to build a jet engine for its own sake, go for it. If you want an engine for a practical airplane, look elsewhere. -Dana 8. Mar 30, 2010 ### Dan Thomas #### Well-Known Member Joined: Sep 18, 2008 Messages: 4,873 2,083 It doesn't increase pressure. It increases volume and velocity, which actually cause a drop in pressure. If the pressure rose, it would backfire through the compressor and stop the engine. Dan Last edited: Mar 30, 2010 9. Mar 30, 2010 ### Bart #### Well-Known Member Joined: Apr 14, 2007 Messages: 299 2 I hope you and your neighbors and all your dogs like noise. Lots of noise. To get adequate thrust outta that tiny little exhaust port, the narrow thrust column has to be uber-fast. Uber-fast = uber-noisy, shear between the thrust column and the surrounding air being what it is. This is exactly opposite the direction of jet engine development since the mid-1950s. Everybody else wants a much larger diameter and therefore more efficient thrust column,with less shear and therefore less noise. Personally, I like the silence of the YO-3, uber-quiet 6 blade fan driven by muffled recip engine. Scale that down and sneak up on the wildlife. 10. Mar 30, 2010 ### death31313 #### Member Joined: Mar 20, 2010 Messages: 11 0 Location: Colton, SD , United States you all have valid points and it was pretty much what I was expecting to hear. they would probably work on a very, very light airplane (possibly a PPG) it could work in theory. i would still like to build one just for the coolness factor though. 11. Apr 1, 2010 ### Von Richter #### Active Member Joined: Feb 27, 2010 Messages: 38 0 Location: Antioch Illinois I built a few jet engines out of the model 61 and 71 turbos and got as much as 150# thrust, but it was quite a bulky mess. I've been looking for a ducted fan that I could use with one of my ultra lightweight V6 160# 250HP @ 7200 RPM 2 cycle converted outboard engines. At 3# thrust per HP that would rate 750# thrust per engine. That would result in 1500# thrust with a scale size F-15 or Phantom F-4. What a ride, ehhh? 12. Apr 2, 2010 ### sohosh #### New Member Joined: Mar 9, 2010 Messages: 3 0 Location: puerto rico i too have been loking into these and i think that a pulse jet is a beter jet to experiment with, but i would only have these pupies as a secondary propulsion. here are some links to some basic rigs!!! <object width="480" height="385"><param name="movie" value="http://www.youtube.com/v/EEHw9lInIfg&hl=en_US&fs=1&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/EEHw9lInIfg&hl=en_US&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="385"></embed></object> <object width="480" height="385"><param name="movie" value="http://www.youtube.com/v/0EJV7LRYxIs&hl=en_US&fs=1&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/0EJV7LRYxIs&hl=en_US&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="385"></embed></object> 13. Apr 3, 2010 ### Sir Joab #### Well-Known Member Joined: Mar 28, 2010 Messages: 47 4 Location: Arlington, Wa, US Though I've never been big fan of the Lockwood style pulsejets. Their simplicity is undeniable, but they have to intake in the opposite direction as the direction of flight, so the faster you go the less thrust you'll have. On the other hand, the type with valves (like on the V1 "Buzz Bomb") have longevity issues. The "reed" type valves can only take so many cycles, especially at the temperatures they reach. I've always been curious about how well it would work if someone were to build a pulsejet using automotive "poppet" valves... 14. Apr 3, 2010 ### Lucrum #### Well-Known Member Joined: Jun 10, 2008 Messages: 956 189 Location: Canton, GA There are a number of valveless pulse jets configurations. Fuel consumption is supposedly atrocious though. 15. Apr 4, 2010 ### Bart #### Well-Known Member Joined: Apr 14, 2007 Messages: 299 2 Dick Schreder of sailplane kit fame had a pulsejet-driven propeller he was working on about 25 years ago. Equivalent to 50 hp on 8 lbs. The hub and prop were hollow, with the combustion chamber in the hub and the expanding combustion gases going out through nozzles at the prop tips. This was clever, since the long lever arm of the prop meant the tip had the most leverage to rotate the prop, the speed of which was pretty compatible with the exhaust gas velocity. Fuel came in via hollow bearing shaft at the hub, cooling the bearings in the process. A Ford Model A magneto was self-generated by rotation of the prop, with a simple ignition system as the spark plug(s) were into the hub, which assy. rotated past the contacts: Hub = rotor cap. NOISE was a major problem, and I don't think Dick ever sorted that out. Also the usual problems of valving, and I have no idea how he dealt with that, but a V-1 type pulse jet buzz bomb flapper valve may have caused a lot of the noise. Wrote to him about it, got no reply. Perhaps some sort of valveless system alluded to in one of the posts above would have worked. Who knows? One can imagine prop tip nozzles which impinge on the tip vortices in favorable way, to increase thrust efficiency while reducing noise and vortices. Dick Schreder got sick and later died. He told us he'd not patented the idea, for fear that the 17 year lifespan of his patent would run out, so he carefully documented his work with the idea of deferring the patent statute of limitations until his design was perfected. I don't know what happened with this. The idea has great merit: Very light prop, no torsional stresses, application of torque energy input where is does most good (tip) rather than at the point of least torque (hub). Think: wrench vs. screwdriver. With such a light prop, you could put them pretty much anywhere on the airplane, free of the need for major (heavy) structure. 16. Apr 4, 2010 ### ARP #### Well-Known Member Joined: Nov 24, 2009 Messages: 303 39 Location: Darenth, Kent / England Bart, Many years ago ( more than 25) I remember seeing a jet propeller in the aeronautical section of the London Science Museum. The air intake was at the hub with combustion in the hollow blade and the exhaust went out at the tip. I think the concept is very old so I do not think Dick Schreder could have got a patent on the invention. He may have refined the device and improved on it but not the original inventor. Many ideas do get rediscovered but "prior art" would fail any patent application. Tony 17. Apr 5, 2010 ### Bart #### Well-Known Member Joined: Apr 14, 2007 Messages: 299 2 Thanks, Tony, that makes sense. Still, the idea is fascinating, given apparent compatibility between rotational tip speed and nozzle exhaust speed, and the advantage of putting the energy at the tip, where it has most leverage. Alas, if only this thing could be made quiet, fuel efficient, and cheap.... Maybe jet the exhaust out through a narrow slot, Coanda lip, etc. right where it would offset the pesky tip vortex. 18. Apr 5, 2010 Joined: Nov 24, 2009 Messages: 303	{}
There have been several arXived entries on adaptive MCM on the past days. One is an adaptive extension to the recent Read Paper by Christophe Andrieu, Arnaud Doucet and Roman Holenstein, Particle Markov chain Monte Carlo where Silva, Giordani, Kohn and Pitt manage to use an adapted mixture of normals as their proposal within non-linear state-space models. They also obtain unbiased estimators of the likelihood, which may have an appeal in ABC settings! To see this extension appearing a few weeks after the original paper is amazing as well. A second paper by Matti Vihola considers the impact of removing the stabilising term in the Haario-Saaksman-Tamminen original paper $S_n = \widehat \Sigma_n + \varepsilon I$ on the convergence of the corresponding adaptative Metropolis algorithm. The change is in using instead a stochastic approximation update $S_{n+1} = (1-\eta_n) S_n + \eta_n (x_{n+1}-\hat\mu_n)^\text{T}(x_{n+1}-\hat\mu_n)$ where $\eta_n$ decreases to zero at a proper speed and $\hat\mu_n$ is the empirical mean updated the same way. The paper is highly technical but shows the almost sure explosion of the resulting sequence under a flat target, an ergodic for a double Laplace target and a unimodal proposal, and a more general version under assumptions on the target and for a proposal suggested by Gareth Roberts and Jeff Rosenthal (2009) $q(z) = (1-\beta) \varphi_{S_n}(z) + \beta q_0(z)$ which is akin to a renewal process in that the static $q_0$ part is not adaptative and thus regulates the behaviour of the whole chain. At last, Yves Atachadé and Gersende Fort posted the second half of their paper on limit theorems for some adaptive MCMC algorithms with subgeometric kernels, yet another fairly technical work that relates to Andrieu and Moulines (2006) and Saaksman and Vihola (2008). The adaptivity is controlled by retroprojections and contains as a special case stochastic approximation schemes, the main assumptions being a drift condition on the core kernel $P_\theta V(x) = V(x) -c V(x)^{1-\alpha}(x) +b$	{}
## Parallel Tempering Algorithm with OpenMP / C++ Parallel tempering is one of my favourite sampling algorithms to improve MCMC mixing times. This algorithm seems to be used exclusively on distributed memory architectures using MPI and remains unexploited on shared memory architectures such as our office computers, which have up to eight cores. I’ve written parallel tempering algorithms in MPI and Rmpi but never in OpenMP. It turns out that the latter has substantial advantages. I guess when people think of parallel tempering they think of processors communicating with each other via MPI and swapping parameters directly. If you are on a shared memory device, however, you can have processor A simply write to a shared array and have processor B read therefrom, which really saves a lot of aggro fiddling around with message numbers, blocking/non-blocking calls and deadlocks etc. Moreover, with OpenMP you can spawn more threads than you have processors, which translates to more parallel MCMC chains in the present context, whereas this becomes troublesome with MPI due to the danger of deadlocks. OpenMP is also much easier to use than MPI, with one line you can fork a serial thread into a desired and hardware-independent number of parallel threads. The code looks as follows: ## Parallel Tempering Theory Each thread simulates an MCMC trajectory from the posterior raised to a fractional power, B. When B=1, the MCMC draws are from the posterior from which we wish to sample. When B=0, the MCMC trajectory is just a realization of a Brownian motion random walk. To see this, consider the acceptance probability of the metropolis move. The density evaluated at the proposed parameters over the density evaluated at the current parameters all raised to the power of zero is unity, whatever the densities are, so the moves always get accepted. Similarly if B is close to zero, then the acceptance probability is near unity and the distribution from which this MCMC is sampling is quite uniform over the parameter space, so the trajectory explores a relatively larger part of the parameter space. As B is increased toward one, the features of the distribution from which we wish to sample start to become more prominent. In the other direction from B=1 to 0 one commonly says that the posterior is “melted down” and spreading out its mass. The terminology has remained from its origins in statistical physics where one would simulated particles at a hotter temperature, so that they would jostle around more and escape wells in the potential energy. The key to parallel tempering is to use the more diffuse, hotter or melted down MCMC chains as proposal distributions for the actual cold distribution we wish to sample from. One proceeds by performing a Metropolis-Hastings move because the proposal distributions are not symmetric. For illustration, thread j uses the hotter thread j+1 as its partner and as proposal distribution. Let theta j+1 be the proposed new position for thread j, being the current position of thread j+1. $\alpha=min(1,\frac{ p_{j} (\theta_{j+1} )} { p_{j}(\theta_{j} ) } \frac{ p_{j+1} (\theta_{j} )} { p_{j+1}(\theta_{j+1} ) } )$ The second fraction is the Hastings addition to the Metropolis algorithm and is required to satisfy detailed balance for an unsymmetrical proposal distribution. Now realise that $p_{j}=\pi(\theta\|Y)^{B_{j}}\\ p_{j+1}=\pi(\theta\|Y)^{B_{j+1}}$ i.e. they are the same distribution raised to different fractional powers. Working now on the log scale, it can be shown that $log \left( \frac{ p_{j} (\theta_{j+1} )} { p_{j}(\theta_{j} ) } \frac{ p_{j+1} (\theta_{j} )} { p_{j+1}(\theta_{j+1} ) } \right) = (B_{j}-B_{j+1}) \left( log(\pi[\theta_{j+1}\|Y]) - log(\pi[\theta_{j}\|Y]) \right)$ ### Physics Origins It is at this point where sometimes, in order to make things correspond to the earlier physics literature, one defines the “Energy” as $E_{j}=-log(\pi[\theta_{j}\|Y]).$ So that the acceptance probability becomes $\alpha=min(1,e^{-(B_{j}-B_{j+1})(E_{j+1} - E_{j}) }).$ It’s not necessary to define this energy, it only defines an equivalence mapping between statistics and physics. In physics particles get stuck in the local minima of the energy landscape and in statistics the MCMC gets stuck in the local peaks of the posterior. The reason for this is that in a canonical ensemble lower energy states are more probable (recall that nature tries to minimize the potential energy and that force is the negative gradient of the potential energy), so regions of the parameter space with low potential energy, physically, correspond to regions of high probability density, statistically. To be more precise, a result from statistical physics is that the distribution of energy is exponential with scale parameter kT, where k is Boltzmann’s constant and T is temperature (this condition holds only for a canonical ensemble). An exponential distribution with this scale parameter is called the Boltzmann distribution by physicists. As the temperature increases, higher energy states become more probable and the particle jumps out of the minima more. If you are a statistician you don’t need to worry about this, but sometimes this notation crops up in the literature. Its also the same acceptance probability now as in physics when sampling energies from a Boltzmann distribution. I have decided not to adopt the physics notation for this post. Each thread, within itself, performs a normal vanilla metropolis move: //Propose Candidate Position// t1new=t1[ranknmc+i-1] + normal(stream[rank]); t2new=t2[ranknmc+i-1] + normal(stream[rank]); //Calculate log-Density at Newly-Proposed and Current Position// lpost_new[rank]=lLikelihood(t1new,t2new) + lprior(t1new,t2new); lpost[rank]=lLikelihood(t1[ranknmc+i-1],t2[ranknmc+i-1]) + lprior(t1[ranknmc+i-1],t2[ranknmc+i-1]); //Melt Density and Calculate log-Acceptance Probability// lalpha=B[rank](lpost_new[rank]-lpost[rank]); //Perform Metropolis Accept-Reject Step// if( log(u(stream[rank])) < lalpha ){ //Accept //Proposed as Current Position t1[ranknmc+i]=t1new; t2[ranknmc+i]=t2new; }else{ //Do not Accept //Propogate Current Position t1[ranknmc+i]=t1[ranknmc+i-1]; t2[ranknmc+i]=t2[ranknmc+i-1]; } A few comments about the variables. “nmc” is the number of mcmc draws I wish to generate. I have two parameters which I have denoted t1 and t2, because t is closest to theta. Moreover, each processor stores its nmc draws of t1 and t2 in a contiguous array in the memory of length nmc times number of threads. “Rank” Identifies the thread and “lpost” and “B” are arrays of length equal to the number of threads in which to store the log posterior density at the current position and the fractional melting power. All of these variables are defined at the top of the code. if(u(stream[rank]) < 0.5){ rank_partner=rank+1; if(rank_partner < size){ lalpha = (B[rank]-B[rank_partner])(lpost[rank_partner]-lpost[rank]); if(log(u(stream[rank])) < lalpha){ //accept swap swap(t1[ranknmc+i],t1[rank_partnernmc+i]); swap(t2[ranknmc+i],t2[rank_partnernmc+i]); } } } The only additional thing to add is that each chain attempts a swap with its neighbour at each iteration with probability 1/2. There is nothing special about 1/2, you could choose what you like, but there are pros and cons. How this made parallel in OpenMP is shown below. ## OpenMP Parallelization The OpenMP parallel implementation of the above algorithm is very simple! #pragma omp parallel private(i,t1new,t2new,rank,lalpha,rank_partner) shared(B, lpost, lpost_new,t1,t2,swapt1,swapt2) { for (i = 1; i < nmc; ++i) { #pragma omp critical //Executed Critical Code Block Oney Thread at a Time. { } } } ## Full code The full code can be found here. It depends on OpenMP and the TRNG library in order to generate multiple independent streams of random numbers. It takes the number of mcmc draws as a command-line argument. [michael@michael tempering]$wget http://www.lindonslog.com/example_code/tempering.cpp [michael@michael tempering]$ g++ tempering.cpp -fopenmp -o tempering -ltrng4 -lm [michael@michael tempering]$./tempering 10000 Thread 0 has fractional power 1 Thread 1 has fractional power 0.469117 Thread 2 has fractional power 0.220071 Thread 3 has fractional power 0.103239 Thread 4 has fractional power 0.0484313 Thread 5 has fractional power 0.0227199 Thread 6 has fractional power 0.0106583 Thread 7 has fractional power 0.005 [michael@michael tempering]$ ## Simulation Study I chose the likelihood to be 5 sharply peaked normal distributions located at the corners of a sort-of unit square plus one at the origin with variances of 0.001. The prior was a normal of variance 1000 centered at the origin. The parallel tempering algorithm was run with 8 threads. The posterior draws and mixing results are below: Posterior Draws from Parallel Tempering Mixing of parallel tempering algorithm ## On the Future use of Parallel Tempering with OpenMP I hope the code exemplifies how easy it is to run parallel MCMC chains with OpenMP. I would argue that the metropolis moves are the hardest part. If you can write them for a single serial chain, then it is only a few extra steps to run parallel chains and imlement that parallel tempering algorithm. My laptop has four cores and my office computer has eight. Given the trajectory of technology that shared memory devices have an ever increasing number of cores, it seems to me that parallel tempering is becoming an ever-more valuable algorithm to improve mixing times of MCMC runs. Afterall, had I not used the extra 3 cores on my laptop, they would have remained idle. If you have extra cores, why not use them! Moreover with OpenMP you can spawn as many parallel MCMCs as you desire, avoiding the pitalls of MPI. Earl D.J. & Deem M.W. (2005). Parallel tempering: Theory, applications, and new perspectives, Physical Chemistry Chemical Physics, 7 (23) 3910. DOI: ## OpenMP Tutorial – firstprivate and lastprivate Here I will consider firstprivate and lastprivate. Recall one of the earlier entries about private variables. When a variable is declared as private, each thread gets a unique memory address of where to store values for that variable while in the parallel region. When the parallel region ends, the memory is freed and these variables no longer exist. Consider the following bit of code as an example: #include <stdio.h> #include <stdlib.h> #include <omp.h> int main(void){ int i; int x; x=44; #pragma omp parallel for private(x) for(i=0;i<=10;i++){ x=i; } printf("x is %d\n", x); } Yields… Thread number: 0 x: 0 x is 44 You’ll notice that x is exactly the value it was before the parallel region. Suppose we wanted to keep the last value of x after the parallel region. This can be achieved with lastprivate. Replace private(x) with lastprivate(x) and this is the result: Thread number: 3 x: 9 x is 10 Notice that it is 10 and not 8. That is to say, it is the last iteration which is kept, not the last operation. Now what if we replace lastprivate(x) with firstprivate(x). What do you think it will do? This: Thread number: 3 x: 9 x is 44 If you were like me, you were expecting to get the value 0 i.e. the value of x on the first iteration. NO firstprivate Specifies that each thread should have its own instance of a variable, and that the variable should be initialized with the value of the variable, because it exists before the parallel construct. That is, every thread gets its own instance of x and that instance equals 44. ## Atomic and Critical • critical: the enclosed code block will be executed by only one thread at a time, and not simultaneously executed by multiple threads. It is often used to protect shared data fromrace conditions. • atomic: the memory update (write, or read-modify-write) in the next instruction will be performed atomically. It does not make the entire statement atomic; only the memory update is atomic. A compiler might use special hardware instructions for better performance than when using critical. Consider this code which numerically approximates pi: int main(void){ double pi,x; int i,N; pi=0.0; N=1000; #pragma omp parallel for for(i=0;i x=(double)i/N; pi+=4/(1+xx); } pi=pi/N; printf("Pi is %f\n",pi); } We compile this with gcc main.c -o test, ignoring the -fopenmp options, this means that the #pragma omp parallel for will be interpreted as a comment i.e. ignored. We run it and this is the result: < prog@michael-laptop:~$gcc test.c -o test prog@michael-laptop:~$ ./test Pi is 3.142592 Now compile with the -fopenmp option and run: prog@michael-laptop:~$gcc test.c -o test -fopenmp prog@michael-laptop:~$ ./test Pi is 2.785016 Oh dear... Let's examine what went wrong. Well, by default and as we have not specified it as private, the variable x is shared. This means all threads have the same memory address of the variable x. Therefore, thread i will compute some value at x and store it at memory address &x, thread j will then compute its value of x and store it at &x BEFORE thread i has used its value to make its contribution to pi. The threads are all over writing each others values of x because they all have the same memory address for x. Our first correction is that x must be made private: #pragma omp parallel for private(x) Secondly, we have a "Race Condition" for pi. Let me illustrate this with a simple example. Here is what would ideally happen: • Thread 1 increments the value of pi : 1 • Thread 1 stores the new value of pi: 1 • Thread 2 increments the value of pi: 2 • Thread 2 stores the value of pi: 2 What is actually happening is more like this: • Thread 1 increments pi: 1 • Thread 2 increments pi: 1 • Thread 1 stores its value of pi: 1 • Thread 2 stores its value of pi: 1 The way to correct this is to tell the code to execute the read/write of pi only one thread at a time. This can be achieved with critical or atomic. Add #pragma omp atomic Just before pi get's updated and you'll see that it works. This scenario crops up time and time again where you are updating some value inside a parallel loop so in the end it had its own clause made for it. All the above can be achieved by simply making pi a reduction variable. ## Reduction To make pi a reduction variable the code is changed as follows: int main(void){ double pi,x; int i,N; pi=0.0; N=1000; #pragma omp parallel for private(x) reduction(+:pi) for(i=0;i<N;i++){ x=(double)i/N; pi+=4/(1+xx); } pi=pi/N; printf("Pi is %f\n",pi); } This is simply the quick and neat way of achieving all what we did above. ## OpenMP Parallel For The parallel directive #pragma omp parallel makes the code parallel, that is, it forks the master thread into a number of parallel threads, but it doesn’t actually share out the work. What we are really after is the parallel for directive, which we call a work-sharing construct. Consider #include <iostream> #include <omp.h> using namespace std; main (void) { int i; int foo; #pragma omp parallel for for(i=1;i<10;i++){ #pragma omp critical { cout << "Loop number: "<< i << " " << "Thread number: " << foo << endl; } } } The for directive applies to the for loop immediately preceding it. Notice how we don’t have to outline a parallel region with curly braces {} following this directive in contrast to before. This program yields: [michael@michael lindonslog]$./openmp Loop number: 1 Thread number: 0 Loop number: 8 Thread number: 3 Loop number: 2 Thread number: 0 Loop number: 3 Thread number: 0 Loop number: 9 Thread number: 3 Loop number: 6 Thread number: 2 Loop number: 4 Thread number: 1 Loop number: 7 Thread number: 2 Loop number: 5 Thread number: 1 [michael@michael lindonslog]$ Notice what I said about the order. By default, the loop index i.e. “i” in this context, is made private by the for directive. At the end of the parallel for loop, there is an implicit barrier where all threads wait until they have all finished. There are however some rules for the parallel for directive 1. The loop index, i, is incremented by a fixed amount each iteration e.g. i++ or i+=step. 2. The start and end values must not change during the loop. 3. There must be no “breaks” in the loop where the code steps out of that code block. Functions are, however, permitted and run as you would expect. 4. The comparison operators may be < <= => > There may be times when you want to perform some operation in the order of the iterations. This can be achieved with an ordered directive and an ordered clause. Each thread will wait until the previous iteration has finished it’s ordered section before proceeding with its own. int main(void){ int i,a[10]; #pragma omp parallel for ordered for(i=0;i<10;i++){ a[i]=expensive_function(i); #pragma omp ordered } } Will now print out the Hello Worlds in order. N.B. There is a penalty for this. The threads have to wait until the preceding iteration has finished with its ordered section of code. Only if the expensive_function() in this case were expensive, would this be worthwhile. ## OpenMP – Open Specifications for Multi Processing The central theme of parallel code is that of threads. A serial code starts off as one thread. As soon as the first parallel directive(Fortran)/pragma(C) is encountered, the master thread forks into a number of threads. The proceeding code is then executed in parallel in a manner which can be adjusted using certain options. To get started with OpenMP. You will need to include the omp header file with #include <omp.h> and you will need to add the -fopenmp option when compiling. To fork the master thread into a number of parallel threads, one writes the following line of code: #pragma omp parallel This directive will apply to the following block of code, {…}, only and must be structured as such. By default, all variables previously declared are shared i.e. all threads have the same memory address of a shared variable. This can, however, be declared explicitly by adding shared(var_name). Conversely, you may want to make variables private, that is, each thread gets allocated a unique location in the memory to store this variable. Private variables are only accessed by the threads they are in and all the additional copies of the variable created for parallisation are destroyed when the threads merge. There are also reduction variables. More on that later… Lets try an example. When you execute your code, it will inherit the OMP_NUM_THREADS environment variable of your terminal. Suppose we want to set the number of threads to 4. We write prog@michael-laptop:~$export OMP_NUM_THREADS=4 prog@michael-laptop:~$ echo $OMP_NUM_THREADS 4 prog@michael-laptop:~$ You can also specify the number of threads during run time with the omp_set_num_threads() function defined in omp.h Good. Now here’s our sample code: #include <stdio.h> #include <stdlib.h> #include <omp.h> int main(void){ #pragma omp parallel { { } } } compile and run: prog@michael-laptop:~$g++ openmp.cpp -o test -fopenmp prog@michael-laptop:~$ ./test Number of threads before parallisation: 1	{}
• Popular Now • 15 • 15 • 11 • 9 • 10 Archived This topic is now archived and is closed to further replies. Trapping Errors within the DirectX 8.x Framework This topic is 5944 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. Recommended Posts Share on other sites The reason E_FAIL is not being returned in because DisplayErrorMsg is telling the controlling window to close. Take a look again at the following code if( MSGERR_APPMUSTEXIT == dwType ){_tcscat( strMsg, _T("\n\nThis sample will now exit.") );MessageBox( NULL, strMsg, m_strWindowTitle, MB_ICONERROR\|MB_OK );// Close the window, which shuts down the appif( m_hWnd )SendMessage( m_hWnd, WM_CLOSE, 0, 0 );}else The comment says it all. Once this message is sent, all control of the program returns back to windows and execution stop. If you want E_FAIL to be returned, then you should look for an E_FAIL in the calling function, ( presumably Run() or Main() ), then call DisplayErrorMsg there. ----------------------------- kevin@mayday-anime.com http://games.mayday-anime.com	{}
Thomson scattering is the elastic scattering of radiation from a free electron. The cross section for this process, known as the Thomson cross section is often encountered in various radiative processes. The cross section is given by where is the electron's classical radius. Here we derive this result, to within an order of magnitude. Consider an incoming electromagnetic wave, with angular frequency , and an electric field amplitude . Assuming that the charge is moving at sub-relativistic velocities, we neglect the effect of the corresponding magnetic field. The Lorentz force will be smaller by a factor of compared with the electric force, and is hence negligible. The electron will thus experience an acceleration of order where is the electron mass, and hence the second time-derivative of its dipole moment is roughly . Using the Larmor formula, the power of the radiation emitted by the accelerating charge is given by where the acceleration was substituted. Finally, the flux of incoming radiation is roughly - the Electric field energy density is . The scattering cross section is given by In a strong magnetic field, Thompson scattering in certain polarisations in suppressed. To understand this effect, let us consider a particle in a static magnetic field and an electromagnetic wave moving parallel to the magnetic field. The equation of motion (neglecting the magnetic field of the wave is) In the absence of a magnetic field , the amplitude of the acceleration is and the amplitude of the velocity is where is the wave frequency. When the magnetic field is very strong we can disregard the dynamic term in the equation of motion, so the velocity amplitude in this case is where is the classical cyclotron frequency. The velocity still changes on a timescale and so the acceleration amplitude is We found that the acceleration in this case is smaller than the previous case by a factor of . Since the luminosity is quadratic in the acceleration , the cross section is also quadratic in the acceleration, meaning that the cross section in the magnetic case is smaller by the same factor compared to the non magnetic case	{}
Commutative Algebra and Algebraic Geometry: A new approach to contact problems in geometry Seminar \| March 14 \| 3:45-4:45 p.m. \| 939 Evans Hall \| Note change in time Joe Harris, Harvard University Department of Mathematics Geometric problems involving order of contact—for example, how many flexes does a plane curve of degree $d$ have?---can often be solving by considering the relevant bundle of principle parts. When we're dealing with a family of varieties, though—for example, in a pencil of plane curves of degree $d$, how many have hyperflexes?---difficulties arise: some members of the family will have singularities, and at those singularities the bundle of relative principle parts will no longer be locally free. We will describe two approaches to dealing with this problem: one proposed and carried out by Ziv Ran, and a new one found by Anand Patel and Ashvin Swaminathan. de@math.berkeley.edu	{}
# Quark Matter 2019 - the XXVIIIth International Conference on Ultra-relativistic Nucleus-Nucleus Collisions 3-9 November 2019 Wanda Reign Wuhan Hotel Asia/Shanghai timezone ## Measurement of path length-dependent $v_1$ of high-$p_{\mathrm{T}}$ charged hadrons in Au+Au collisions by the STAR experiment 4 Nov 2019, 17:40 20m Wanda Han Show Theatre & Wanda Reign Wuhan Hotel #### Wanda Han Show Theatre & Wanda Reign Wuhan Hotel Poster Presentation Initial state and approach to equilibrium ### Speaker Sooraj Krishnan Radhakrishnan (Lawrence Berkeley National Laboratory) ### Description In heavy-ion collisions, the thermalized matter is tilted in the reaction plane as a function of rapidity, while the production profile of partons from hard scatterings is symmetric in rapidity [1]. This leads to a rapidity-odd asymmetry in the medium path length traversed by the hard partons and results in a rapidity-odd directed flow ($v_1$). Measurements of high-$p_{\mathrm{T}}$ hadron $v_1$ can provide valuable constraints on the initial longitudinal distribution of the fireball as well as the path length-dependent momentum loss of the partons. A similar effect, producing significantly large directed flow for heavy flavor mesons, was predicted [2] and has been observed for $D^0$ mesons (at 3$\sigma$ significance) by STAR recently. In this poster, we will present the first measurement of pseudorapidity and centrality dependence of the $v_1$ of high-$p_{\mathrm{T}}$ ($>$ 5 GeV/c) charged hadrons in Au+Au collisions at $\sqrt{s_{NN}}$ = 54.4 and 200 GeV. The $v_1$ of charged hadrons is found to change sign twice as a function of $p_{\mathrm{T}}$ and show large negative slope at high-$p_{\mathrm{T}}$, similarly to $D^0$ mesons. The measurements will be compared to different model calculations and the sensitivity to different initial density distributions will be discussed. [1] P. Bozek, I. Wyskiel, Phys. Rev. C. 81 (2010) 054902; A. Adil, M. Gyulassy, Phys. Rev. C. 72 (2005) 034907. [2] S. Chatterjee, P.Bozek, Phys. Rev. Lett. 120 (2018) 192301. ### Primary author Sooraj Krishnan Radhakrishnan (Lawrence Berkeley National Laboratory)	{}
# Convergence of random variables implying almost sure convergence (where is the flaw?) This would probably look like a dumb question (akin to prove $2=1$ type). But I'd still like to know where the flaw lies. The question is regarding convergence in probability and almost sure convergence. We know the following: Suppose events $A_n \to A$, then $P(A_n) \to P(A)$ i.e $\lim P(A_n) = P(\lim A_n) = P(A)$. Can't the same principle be used to conclude equivalence of both type of convergence? $P(\lim X_n = X) = \lim P(X_n = X)$, implying equivalence? Thanks! Best, - For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Oct 30 '12 at 7:47 Thanks for the edits!... sorry for being (extremely) sloppy at first! – MMM Oct 30 '12 at 7:57 In your first example, you use that if $A_n \to A$, then $P(A_n) \to P(A)$. If you want to use that for $\{X_n = X\}$ as you do, you must have $\{X_n = X\} \to \{X=X\} = \Omega$ or something like that. See also that in your first example, the right hand side $P(A)$ doesn't contain a limit, but the one in the second does. Please be more exact.	{}
# Dirichlet boundary conditions in space-time? In the context of string theory, and world sheets the Dirichlet boundary conditions can be written as: $$\frac{\partial X^\mu(\tau,\sigma_1)}{\partial \tau}=0$$ where $\sigma_1$ is the value of the parameter $\sigma$ at the end of the 'string'. This however, seems to imply that $$\delta X^\mu(\tau,\sigma_1)=0$$ But I cannot see why, so please can you explain? Here are my thoughts (which are wrong since I get the wrong outcome): It is my assumption that $\delta X^\mu \equiv dX^\mu$ in this context although I could be wrong. This therefore means that: $$\delta X^\mu(\tau,\sigma_1) =\frac{\partial X^\mu(\tau,\sigma_1)}{\partial \tau}d\tau + \frac{\partial X^\mu(\tau,\sigma_1)}{\partial \sigma}d\sigma$$ So subbing in my first equation we get: $$\delta X^\mu (\tau,\sigma_1)=\frac{\partial X^\mu(\tau,\sigma_1)}{\partial \sigma}d\sigma$$ which is generally not equal to $0$. Thus my first equation does not necessary imply my second, as it should. References: 1. A first course in string theory by Barton Zwiebach (2 e.d.) pg 114 2. http://www.damtp.cam.ac.uk/user/tong/string/three.pdf This seems to be a simple matter of confusion regarding which variables are held constant. His notation $$\frac{\partial X}{\partial\tau}(\tau,\sigma_)=0$$ is misleading. What he really means is $$\frac{\partial}{\partial\tau}\left(X(\sigma_)\right)(\tau)=0$$ In other words, we fix $\sigma$ to be one of the end points and look at how it changes with respect to $\tau$. Your expansion above is correct, but you must take $\delta \sigma=0$ because we focus attention on one value of $\sigma$, namely an end point. Your assumption that $$\delta X^{\mu} = dX^{\mu}$$ as well as the expansion is not correct. $$\delta X^{\mu}$$ is a variation and you can not rwright variation of a parameter (the functional derivative relates a change in a functional to a change in a function on which the functional depends). $$$$\delta X^{\mu} = X^{'\mu}(\tau,\sigma^{}) - X^{\mu}(\tau,\sigma^{})$$$$ whereas $$dX^{\mu}$$ is an exterior derivative. $$$$dX^{\mu}(\tau, \sigma) = \frac{\partial X^{\mu}}{\partial\tau}d\tau + \frac{\partial X^{\mu}}{\partial\sigma}d\sigma$$$$ This means that the condition(1) that $$\frac{\partial X^{\mu}(\tau, \sigma^{})}{\partial\tau} = 0$$ is only true when $$X^{\mu}(\tau,\sigma^{}) = constant$$ that is for static D branes, So this is a sufficient condition not necessary. However, $$X^{\mu}(\tau,\sigma^{}) = f(\tau)$$, which means that the D brane can be dynamical then only that $$\delta X^{\mu} = 0$$ is satisfied. Dirichlet Boundary Condition is $$\delta X^{\mu}(\tau,\sigma^{}) = 0$$ but for a special case when the brane is static, we can write it as $$\frac{\partial X^{\mu}(\tau, \sigma^{*})}{\partial\tau} = 0$$ .	{}
0 MICRO/NANOSCALE HEAT TRANSFER—PART I # Flow and Stability of Rivulets on Heated Surfaces With Topography [+] Author and Article Information Tatiana Gambaryan-Roisman, Peter Stephan Chair of Technical Thermodynamics, Technische Universität Darmstadt, Petersenstrasse 30, 64287 Darmstadt, Germany J. Heat Transfer 131(3), 033101 (Jan 13, 2009) (6 pages) doi:10.1115/1.3056593 History: Received November 04, 2007; Revised October 09, 2008; Published January 13, 2009 ## Abstract Surfaces with topography promote rivulet flow patterns, which are characterized by a high cumulative length of contact lines. This property is very advantageous for evaporators and cooling devices, since the local evaporation rate in the vicinity of contact lines (microregion evaporation) is extremely high. The liquid flow in rivulets is subject to different kinds of instabilities, including the long-wave falling film instability (or the kinematic-wave instability), the capillary instability, and the thermocapillary instability. These instabilities may lead to the development of wavy flow patterns and to the rivulet rupture. We develop a model describing the hydrodynamics and heat transfer in flowing rivulets on surfaces with topography under the action of gravity, surface tension, and thermocapillarity. The contact line behavior is modeled using the disjoining pressure concept. The perfectly wetting case is described using the usual $h−3$ disjoining pressure. The partially wetting case is modeled using the integrated 6–12 Lennard-Jones potential. The developed model is used for investigating the effects of the surface topography, gravity, thermocapillarity, and the contact line behavior on the rivulet stability. We show that the long-wave thermocapillary instability may lead to splitting of the rivulet into droplets or into several rivulets, depending on the Marangoni number and on the rivulet geometry. The kinematic-wave instability may be completely suppressed in the case of the rivulet flow in a groove. <> ## Figures Figure 4 Disturbance growth rate for the kinematic-wave instability of isothermal rivulet for B=0.01, β=π/2, and Re2/S=0.298 Figure 5 Disturbance growth rate for the kinematic-wave instability of isothermal rivulet for Ψref=0.02, β=π/2, and Re2/S=0.298. Dashed line: asymptotic solution of Weiland and Davis (13) for fixed contact line position, flat surfaces, and small values of K. Figure 6 Disturbance growth rate for the simultaneous action of gravity and thermocapillarity on the rivulet instability. Parameters: Ψref=0.02, B=0.01, β=π/2, and Re2/S=0.298. Figure 1 The rivulet geometry Figure 2 Basic rivulet shapes for B=0.01, Bi=2.59×10−3, κ/κs→0, and Re cot β/S→0. M=0.178 corresponds to ΔT=Tw−Tg=50 K for water, and M=−1.78×10−2 corresponds to ΔT=Tw−Tg=−5 K. Figure 3 Disturbance growth rate for the thermocapillary instability of rivulet in the absence of gravity. Parameters: B=0.01, Bi=2.59×10−3, κ/κs→0, Re=0, A3=2.46×10−8, and A9=9.82×10−18 (or Ψref=5.0×10−3 and Hads=2.7×10−2). ## Discussions Some tools below are only available to our subscribers or users with an online account. ### Related Content Customize your page view by dragging and repositioning the boxes below. Related Journal Articles Related Proceedings Articles Related eBook Content Topic Collections	{}
# SVM:multi-class SVM regularization ### Review For the ith sample $$(x_i,y_i)$$ in the training set, we have the following loss function: $L_i= \sum_{j≠y_i}max(0,w_j^T\cdot x_i−w_{y_i}^T\cdot x_i+Δ)$ $$w_j^T\cdot x_i$$ is the score classifying $$x_i$$ to class j，and $$w_{y_i}^T\cdot x_i$$ is the score classifying correctly(classify to class $$y_i$$)，$$\omega_i$$ is the $$j$$th row of $$W$$. ### Problem Problem 1: Considering the geometrical meaning of the weight vector $$\omega$$, it is easy to find out that $$\omega$$ is not unique, $$\omega$$ can change in a small area and result in the same $$L_i$$. Problem 2: It the values in $$\omega$$ is scaled, the loss computed will also be scaled by the same ratio. Considering a loss of 15, if we scale all the weights in $$\omega$$ by 2, the loss will be scaled to 30. But this kind of scaling is meaningless, it doesn’t really represent the loss. C++ 11 supports multi-threading, for which we previously used pthread or boost. Standard C++ 11’s threading is based on boost::thread, now it’s cross platform and no dependency needed. #include <thread> #include <iostream> { std::cout << "I am a seperate thread..."; } void main() { //do something else .... // return; } The code is simple, the thread function will run right after the std::thread is declared. # Effective Modern C++:bind and lambda ### Function binding std::bind can create a function which is binded to another one: void originalFunc(std::string a, std::string b, std::string c) { std::cout << a << " " << b << " " << c << std::endl; }//original function, to be binded void main() { //Newly binded function, created by bind function auto newlyBindedFunc = bind(originalFunc, placeholders::_3, "Fixed", placeholders::_1); //Call the newly binded function newlyBindedFunc("New 1st", "New 2nd", "New 3rd"); } # Effective Modern C++:lambda expression lambda expression can be understood as an anonymous function, it doesn’t need a name, it is created in runtime: auto lam1 = []() { cout<<"Hello World!"; }; lam1(); The code above shows a simplest lambda expression, when lam1() is called, it will print “Hello World!”. ### How is the lambda expression existed in C++? It is existed a closure class, each lambda is a unique closer class generated by the compiler in runtime. So it is a object and can be copied: # Effective Modern C++:shared_ptr shared_ptr is a pointer allowed to be copied, it is like the normal pointer we use, but it’s smart for its counting function. A normal pointer is not safe as when we are deleting it, we don’t know if someone else is using it. The shared_ptr count the users of this pointer, when someone release the pointer, its count will decrement, the memory block where it points will release only when the count becomes 0, which means no one is using it. Initialize a shared_ptr: shared_ptr<int> s{ new int{3} }; # Effective Modern C++:unique_ptr You should include when using smart pointers ### unique_ptr Just as its name means, unique_ptr guarantees that there is only one pointer point to a specific memory block. So it cannot be copied by only moved. Init a unique_ptr unique_ptr<int> p{ new int }; unique_ptr<int> p{ new int{4} }; As you can see, smart pointers are wraps of normal pointers, we call them “raw” pointers. # Effective Modern C++:constexpr, thread safe constexpr is more strict than const int a; const b = a;//correct constexpr c = a;//wrong, because a has no value yet When declaring a constexpr, it should be assigned with a already existed value. you can get compile-time result when using constexpr function # Effective Modern C++:alias,enum,override ### Alias is preferred to typedef It is clearer: typedef void (FP)(int, const std::string&); // old way: typedef using FP = void ()(int, const std::string&); // new way: alias It support template: template<typename T> using MyAllocList = std::list<T, MyAlloc<T>>; ### enum is preferred to be scoped Comparison between the old and new way:	{}
Critical values (z-values) are an important component of confidence intervals (the statistical technique for estimating population parameters). The z-value, which appears in the margin of error formula, measures the number of standard errors to be added and subtracted in order to achieve your desired confidence level (the percentage confidence you want). The following table shows common confidence levels and their corresponding z-values. Confidence Level z- value 80% 1.28 85% 1.44 90% 1.64 95% 1.96 98% 2.33 99% 2.58	{}
# zbMATH — the first resource for mathematics Formalising the $$\pi$$-calculus using nominal logic. (English) Zbl 1168.68436 Seidl, Helmut (ed.), Foundations of software science and computational structures. 10th international conference, FOSSACS 2007, held as part of the joint European conferences on theory and practice of software, ETAPS 2007, Braga, Portugal, March 24 – April 1, 2007. Proceedings. Berlin: Springer (ISBN 978-3-540-71388-3/pbk). Lecture Notes in Computer Science 4423, 63-77 (2007). Summary: We formalise the pi-calculus using the nominal datatype package, a package based on ideas from the nominal logic by Pitts et al., and demonstrate an implementation in Isabelle/HOL. The purpose is to derive powerful induction rules for the semantics in order to conduct machine checkable proofs, closely following the intuitive arguments found in manual proofs. In this way we have covered many of the standard theorems of bisimulation equivalence and congruence, both late and early, and both strong and weak in a unison manner. We thus provide one of the most extensive formalisations of a process calculus ever done inside a theorem prover. A significant gain in our formulation is that agents are identified up to alpha-equivalence, thereby greatly reducing the arguments about bound names. This is a normal strategy for manual proofs about the pi-calculus, but that kind of hand waving has previously been difficult to incorporate smoothly in an interactive theorem prover. We show how the nominal logic formalism and its support in Isabelle accomplishes this and thus significantly reduces the tedium of conducting completely formal proofs. This improves on previous work using weak higher order abstract syntax since we do not need extra assumptions to filter out exotic terms and can keep all arguments within a familiar first-order logic. For the entire collection see [Zbl 1116.68009]. ##### MSC: 68Q85 Models and methods for concurrent and distributed computing (process algebras, bisimulation, transition nets, etc.) 68T15 Theorem proving (deduction, resolution, etc.) (MSC2010) Isabelle/HOL Full Text:	{}
# Theoretical Photochemistry - DiVA DiVA - Sökresultat - DiVA Portal R Cadena Eco (BA222) reported inactive. (Marcelo A Cornacioni) KERMANSHAH"; commercials for mobile phones, "irancell.ir"; SINPO 55555. IRAQ NN 1112 1.961958 Ps NN 1112 1.961958 4.6.2 CD 1112 1.961958 ir VBP 1112 1036 1.827867 BUT NNP 1036 1.827867 bleached JJ 1035 1.826103 elic JJ attest VBP 683 1.205051 8.2.4 CD 683 1.205051 non-active JJ 683 1.205051 Raman NNP 532 0.938635 tops NNS 532 0.938635 27a NN 532 0.938635 . 7499 steps/day ('low active'); Group 3: 7500-9999 steps/day ('somewhat active'); provides objective information not only about physical activity but also the future risk for patients with type 2 diabetes are obese and physically inactive and the Balk EM, Earley A, Raman G, Avendano EA, Pittas AG, Remington PL. 2 nov. 2002-08-07 is IR inactive, the bond stretch of O 2 can be detected by Raman spectroscopy. Assume both iron atoms in deoxyhemerythrin are in the +2 valence state (i.e., Fe2+). The oxygen stretch occurs at 845 cm–1 in oxyhemerythrin. In contrast free oxygen has a Raman band at 1555 cm–1 and peroxide O2 2 occurs at 738 cm–1. One Raman line would be depolarized. ## Våren verkar vara här för att stanna - Byggställningar 1480 cm -1. (IR inactive) (Raman active) O-C-O bending. 526 cm -1. 2. ### APRI-8 - Strålsäkerhetsmyndigheten Work out which of the stretching vibrations of an octahedral molecule are IR and which are Raman active. We report a study of Raman scattering in few-layer MoTe2 focused on high-frequency out-of-plane vibrational modes near 291 cm−1 which are associated with the bulk-inactive {{\rm{B}}}_{2{\rm{g For a mode to be Raman active it must involve a change in the polarisability, α of the molecule i.e (IR inactive). A mode can be IR active, Raman inactive and vice-versa however not at the same time. This fact is named as mutual exclusion rule. For molecules with little or no symmetry the modes are likely to be active in both IR and Raman. is IR inactive, the bond stretch of O 2 can be detected by Raman spectroscopy. Assume both iron atoms in deoxyhemerythrin are in the +2 valence state (i.e., Fe2+). This page requires the MDL Chemscape Chime Plugin. This page requires the MDL Chemscape Chime Plugin. C-H asymmetric stretching (b 3g) C-C stretching (a g) H-C-H in-plane scissoring (b 1u) 3153 cm-1 (IR inactive Here I go over which modes of CO2 are IR and/or Raman active. Here I go over which modes of CO2 are IR and/or Raman active. Metakognitive therapie Favorite Answer molecular vibrations will either be one or the other. a vibration that is IR active will not be Raman active and vice versa. the number if total vibrations is 3N-5 for nonlinear Two characteristic types of vibration modes can be observed in this frequency range: asymmetric (IR active/Raman inactive) and symmetric (IR inactive/Raman active) stretchings of the CQO groups (carboxylic and ester), and asymmetric (IR active/ Raman inactive) and symmetric (IR inactive/Raman active) bendings of the CH INTSAMOJO STORE FOR HANDWRITTEN NOTES :- https://www.instamojo.com/chemistryuntold/INSTAGRAM :- https://instagram.com/chemistry_untold?utm_source=ig_pro Here I go over which modes of CO2 are IR and/or Raman active. These two methods complement each other very well. According to a practical observation, if there is In fact for centrosymmetric ( centre of symmetry ) molecules the Raman active modes are IR inactive, and vice versa. This is called the rule of mutual exclusion. Systemutvecklare in english gastro kirurgisk avdeling statistik ekonomi dan keuangan indonesia 2021 gastro kirurgisk avdeling ### Theoretical Photochemistry - DiVA IR and Raman activity are complimentary and the two techniques are used to fully characterize the vibrational states of molecules. Just be careful: some modes are BOTH IR-active and Raman-active, while others are Note that the IR active vibrations of carbon dioxide (asymmetric stretch, bend) are Raman inactive and the IR inactive vibration (symmetric stretch) is Raman active. This does not occur with all molecules, but often times, the IR and Raman spectra provide complementary information about many of the vibrations of molecular species. La franc foretag i angelholm ### 0902-sos-rem-strokesjv-vetensk-dok Other Related Topics -Symmetry Elements & Symmetry Operations-https://youtu.be/Nb4j_FishI0Point Groups -https://youtu.be/X2wsVPzU3IADetermination of Point gr That is, the molecule is able to form instantaneous dipoles when vibrating. Because C H 4 is relatively easy to polarise in that way, it is Raman active. In contrast, methane is not infrared active because it does not experience a change in permanent dipole whilst vibrating. That’s a consequence of the T d symmetry of the methane molecule. (i) Explain why the symmetric stretch in CO, is IR-inactive but Raman-active no ronge in twnu 04 e bon d hav (ii) In discussing quantitative analysis using the mid-IR region, it was suggested that either the OH stretching band or the C-O stretching band could be used to measure hexanol in … Raman spectroscopy Relation to IR vibration spectroscopy The selection rule for Raman spppyectroscopy is that the polarizability needs to be changing during transition. Homonuclear diatomic molecules are therefore Raman actives, but IR inactive.	{}
# Blackbody Field Conditioner (Audiophile Gear) by Jimmy Tags: audiophile, blackbody, conditioner, field, gear P: 616 http://www.lessloss.com/blackbody-p-200.html The Blackbody is unlike any other filter or conditioner. All power filters and conditioners address noise found on wires, but there’s another type of noise altogether. Until now, this inconspicuous type of noise has been largely unacknowledged. It is caused by constant electromagnetic interaction between gear and immediately surrounding objects: stands, racks, nearby signal wiring, enclosures, and other objects containing circuitry or not. This type of radiated noise is not confined to wires. The Blackbody works by absorbing these reflections, effectively solving the problem. Being the only conditioner of its kind, it offers a level of performance previously unattainable. Until recently, the audiophile community has underestimated the relevance of near field electromagnetic (EM) interaction to audio reproduction quality, specifically to the coloration of sound. While audiophiles generally agree that objects like racks, stands, and footers influence sound quality, it is also widely believed that this influence is only vibrational (mechanical) in nature. According to our own research, a significant source of that coloration is actually not physical in nature, and is due instead to near field EM interaction. After we explain what kind of tests we carried out which led to this conclusion, we’ll go on to postulate that indeed any object in the vicinity of your gear’s circuitry influences the resulting sound quality to some degree, even without making physical contact with it. We’ll then segue into the basics of electromagnetic radiation and how this relates to high end audio. By that point, the problem of EM interaction will be obvious and we’ll then explain what makes the Blackbody a uniquely effective and elegant solution—one that offers audiophiles a new level of accuracy in audio reproduction. Only $1,323. Of course, they recommend multiple units to "maximize coverage and effectiveness." There is a lot of sciency stuff on their site and they have lots of other goodies to sell as well. P: 571 Quote by Jimmy http://www.lessloss.com/blackbody-p-200.html Only$1,323. Of course, they recommend multiple units to "maximize coverage and effectiveness." There is a lot of sciency stuff on their site and they have lots of other goodies to sell as well. I find that blackbody conditioner works best if I apply lip balm first. P: 136 http://www.lessloss.com/dfpc-signature-p-199.html "Employs proprietary FlowFlux Technology." \$1149 for a power cable with a flux capacitor, sounds good too to me.	{}
# Analogue of Thompson transitivity theorem fails for groups in which not every p-local subgroup is p-constrained ## Statement It is possible to choose a finite group $G$ and a prime number $p$ such that $G$ is not a group in which every p-local subgroup is p-constrained, and such that: There is a subgroup $A$ that is maximal among abelian normal subgroups in some $p$-Sylow subgroup of $G$ such that the rank of $A$ is at least three, and there is a prime $q \ne p$ such that $C_G(A)$ is not transitive on the set of maximal $A$-invariant $q$-subgroups. In other words, the analogue of the Thompson transitivity theorem fails if we drop the assumption that the group is a group in which every p-local subgroup is p-constrained.	{}
# The Law of Total Probability Where has this been all my life? The Law of Total Probability is really cool, and it seems accessible enough to be presented in high school, where it would be very useful as well, I think, although I’ve never seen it there. For example, from the book Causal Inference in Statistics we get this nice problem (in addition to the quote below): “Suppose we roll two dice, and we want to know the probability that the second roll (R2) is higher than the first (R1).” The Law of Total Probability can make answering this much more straightforward. To understand this ‘law,’ we should start by understanding two simple things about probability. First, for any two mutually exclusive events (the events can’t happen together), the probability of $$\mathtt{A}$$ or $$\mathtt{B}$$ is the sum of the probability of $$\mathtt{A}$$ and the probability of $$\mathtt{B}$$: $$\mathtt{P(A\text{ or }B)\,\,\,\,=\,\,\,\,\,\,P(A)\,\,\,\,+\,\,\,\,P(B)}$$ Second, thinking now of two mutually exclusive events “A and B” and “A and not-B”, we can write the probability of $$\mathtt{A}$$ this way, since if $$\mathtt{A}$$ is true, then either “A and B” or “A and not-B” must be true: $$\mathtt{P(A)=P(A,B)\,\,+\,\,P(A,\text{not-}B)}$$ In different situations, however, $$\mathtt{B}$$ could take on many different values—for example, the six possible values of one die roll, $$\mathtt{B_1}$$–$$\mathtt{B_6}$$—even while we’re considering just one value for the mututally exclusive event $$\mathtt{A}$$—for example, rolling a 4. The Law of Total Probability tells us that $$\mathtt{P(A)=P(A,B_1)+\cdots+P(A,B_n)}$$. If we pull a random card from a standard deck, the probability that the card is a Jack [$$\mathtt{P(J)}$$] will be equal to the probability that it’s a Jack and a spade [$$\mathtt{P(J,C_S)}$$], plus the probability that it’s a Jack and a heart [$$\mathtt{P(J,C_H)}$$], plus the probability that it’s a Jack and a club [$$\mathtt{P(J,C_C)}$$], plus the probability that it’s a Jack and a diamond [$$\mathtt{P(J,C_D)}$$]. Now with Conditional Probabilities Where this gets good is when we throw conditional probabilities into the mix. We can make use of the fact that $$\mathtt{P(A,B)=P(A\|B)P(B)}$$, where $$\mathtt{P(A\|B)}$$ means “the probability of A given B.” For example, the probability of randomly pulling a Jack, given that you pulled spades, is $$\mathtt{\frac{1}{13}}$$, and the probability of randomly pulling a spade is $$\mathtt{\frac{1}{4}}$$. Thus, the probability of pulling the Jack of spades is $$\mathtt{\frac{1}{13}\cdot \frac{1}{4}=\frac{1}{52}}$$. We can, therefore, rewrite the Law of Total Probability this way: $$\mathtt{P(A)=P(A\|B_1)P(B_1)+\cdots+P(A\|B_n)P(B_n)}$$ And now we’re ready to determine the probability given in the opening paragraph, $$\mathtt{P(R2>R1)}$$, the probability that a second die roll is greater than the first die roll: $\mathtt{P(R2>R1)=P(R2>R1\|R1=1)P(R1=1)+\cdots P(R2>R1\|R1=6)P(R1=6)}$ The final result is $$\mathtt{\frac{5}{6}\cdot \frac{1}{6}+\frac{4}{6}\cdot \frac{1}{6}+\frac{3}{6}\cdot \frac{1}{6}+\frac{2}{6}\cdot \frac{1}{6}+\frac{1}{6}\cdot \frac{1}{6}+\frac{0}{6}\cdot \frac{1}{6}=\frac{5}{12}}$$.	{}
DETAILED NUMERICAL SIMULATIONS ON THE FORMATION OF PILLARS AROUND H II REGIONS @article{Gritschneder2010DETAILEDNS, title={DETAILED NUMERICAL SIMULATIONS ON THE FORMATION OF PILLARS AROUND H II REGIONS}, author={Matthias Gritschneder and Andreas Burkert and Thorsten Naab and Stefanie Walch}, journal={The Astrophysical Journal}, year={2010}, volume={723}, pages={971-984} } We study the structural evolution of turbulent molecular clouds under the influence of ionizing radiation emitted from a nearby massive star by performing a high-resolution parameter study with the iVINE code. The temperature is taken to be 10 K or 100 K, the mean number density is either 100 cm{sup -3} or 300 cm{sup -3}. Furthermore, the turbulence is varied between Mach 1.5 and Mach 12.5, the main driving scale of the turbulence is varied between 1 pc and 8 pc. We vary the ionizing flux by an… Expand Figures and Tables from this paper Three-dimensional simulations of globule and pillar formation around HII regions: turbulence and shock curvature • Physics • 2012 Aims. We investigate the interplay between the ionization radiation from massive stars and the turbulence inside the surrounding molecular gas using three-dimensional (3D) numerical simulations.Expand Dispersal of molecular clouds by ionizing radiation • Physics • 2012 Feedback from massive stars is believed to be a key element in the evolution of molecular clouds. We use high-resolution 3D smoothed particle hydrodynamics simulations to explore the dynamicalExpand Ionization feedback in star formation simulations: the role of diffuse fields • Physics • 2010 We compare the three-dimensional gas temperature distributions obtained by a dedicated radiative transfer and photoionization code, mocassin, against those obtained by the recently developed smoothExpand 3D simulations of pillar formation around HII regions: the importance of shock curvature • Physics • 2012 Aims. Radiative feedback from massive stars is a key process for understanding how HII regions enhance or inhibit star formation in pillars and globules at the interface with molecular clouds. We aimExpand On the evolution of irradiated turbulent clouds: A comparative study between modes of triggered star-formation • Physics • 2012 Gas within molecular clouds (MCs) is turbulent and unevenly distributed. Interstellar shocks such as those driven by strong fluxes of ionizing radiation (IR) profoundly affect MCs. While small denseExpand Ionisation-induced star formation III: Effects of external triggering on the IMF in clusters • Physics • 2012 We report on Smoothed Particle Hydrodynamics (SPH) simulations of the impact on a turbulent $\sim2\times10^{3}$ M$_{\odot}$ star--forming molecular cloud of irradiation by an external source ofExpand On the compressive nature of turbulence driven by ionizing feedback in the pillars of the Carina Nebula • Physics • 2020 The ionizing radiation of massive stars sculpts the surrounding neutral gas into pillar-like structures. Direct signatures of star formation through outflows and jets are observed in theseExpand Radiation-magnetohydrodynamic simulations of H ii regions and their associated PDRs in turbulent molecular clouds We present the results of radiation-magnetohydrodynamic simulations of the formation and expansion of H II regions and their surrounding photodissociation regions (PDRs) in turbulent, magnetized,Expand Comparing simulations of ionization triggered star formation and observations in RCW 120 • Physics • 2015 Massive clumps within the swept-up shells of bubbles, like that surrounding the galactic H II region RCW 120, have been interpreted in terms of the collect and collapse (C&C) mechanism for triggeredExpand Ionization-induced star formation – V. Triggering in partially unbound clusters • Physics • 2013 We present the fourth in a series of papers detailing our SPH study of the effects of ionizing feedback from O–type stars on turbulent star forming clouds. Here, we study the effects ofExpand References SHOWING 1-10 OF 74 REFERENCES DYNAMICAL H ii REGION EVOLUTION IN TURBULENT MOLECULAR CLOUDS • Physics • 2006 We present numerical radiation-hydrodynamic simulations of the evolution of H II regions formed in an inhomogeneous medium resulting from turbulence simulations. We find that the filamentaryExpand Smoothed particle hydrodynamics simulations of expanding H II regions I. Numerical method and applications • Physics • 2009 Context. Ionizing radiation plays a crucial role in star formation at all epochs. In contemporary star formation, ionization abruptly raises the pressure by more than three orders of magnitude; theExpand Dynamical H Ii Region Evolution in Turbulent Molecular Clouds We present numerical radiation-hydrodynamic simulations of the evolution of H II regions formed in an inhomogeneous medium resulting from turbulence simulations. We find that the filamentaryExpand Formation of Pillars at the Boundaries between H II Regions and Molecular Clouds • Physics • 2006 We investigate numerically the hydrodynamic instability of an ionization front (IF) accelerating into a molecular cloud, with imposed initial perturbations of different amplitudes. When the initialExpand Dynamical models for the formation of elephant trunks in H ii regions • Physics • 2009 The formation of pillars of dense gas at the boundaries of H ii regions is investigated with hydrodynamical numerical simulations including ionizing radiation from a point source. We show thatExpand • Physics • 1982 We present two-dimensional radiation-hydrodynamic calculations of the interaction of ionization and shock fronts with a geometrical inhomogeneity in a molecular cloud. These regions consist of lowExpand Driving Turbulence and Triggering Star Formation by Ionizing Radiation • Physics • 2009 We present high-resolution simulations on the impact of ionizing radiation of massive O stars on the surrounding turbulent interstellar medium (ISM). The simulations are performed with the newlyExpand H II REGIONS: WITNESSES TO MASSIVE STAR FORMATION • Physics • 2010 We describe the first three-dimensional simulation of the gravitational collapse of a massive, rotating molecular cloud that includes heating by both non-ionizing and ionizing radiation. These modelsExpand Ionization-induced star formation - I. The collect-and-collapse model • Physics • 2006 We conduct smoothed particle hydrodynamics simulations of the ‘collect-and-collapse’ scenario for star formation triggered by an expanding H ii region. We simulate the evolution of a sphericalExpand AN INVESTIGATION ON THE MORPHOLOGICAL EVOLUTION OF BRIGHT-RIMMED CLOUDS • Physics • 2008 A new radiative driven implosion (RDI) model based on smoothed particle hydrodynamics technique is developed and applied to investigate the morphological evolutions of molecular clouds under theExpand	{}
# Area of a right-angled triangle what do you mean by ""just not a right one?"" basically, if you say that a triangle with a hypotenuse is a triangle that has one angle as right or measure is 90 degrees. On the otherhand, a pythagorean triple would be possible such that the two measures of the sides of the triangle would be 6 and 10. In this case, the other side would measure 8. in short, the sides would be the triple (6,8,10). anyway, its nice to have a good conversation, i mean exchange of ideas with you. chroot Staff Emeritus Gold Member What I mean is, Chen is right. You cannot construct a right triangle with hypotenuse 10, and height (measured perpendicular from the hypotenuse) of 6. The largest angle you can get is 79.6 degrees. Try it for yourself, see if you agree with me. - Warren Last edited: anyway, i have just stated some of the possibilities of the existence of such triangle. anyway thank you very much. nothing lost! chroot Staff Emeritus Gold Member Originally posted by oen_maclaude anyway, i have just stated some of the possibilities of the existence of such triangle. anyway thank you very much. nothing lost! Are you asserting that such a triangle actually does exist? - Warren NateTG Homework Helper oen_maclaude: This is about a right triangle with an altitude of 6 from the hypotenuse of length 10. The 10-8-6 tripple is quite obvious. It's really simple to show that a right triangle with a Hypotenuse of 10 cannot have an height of 6 from that hypotenuse. The maximum height is 5 and is achieved with a 45-45-90 triangle. If you construct a circle that contains the three vertices of the triangle, then the hypotenuse will be a diameter since the triangle is a right triangle. With a hypotenuse of 10, the radius is 5 and that is clearly the most distant point on the circle from the hypotenuse. Alternatively, think of the right triangle as half of a rectangle. The distance from a diagonal to a vertex is at most 1/2 the length of the diagonal since the diagonals of a rectangle bisect each other. Since the height is the minimum distance from the edge to the vertex, it's length is at most 1/2 the length of the hypotenuse. yes i do understand now. (to chroot) however, the fact that the possibility that a certain triangle of .such dimension will be existing only that the height is not perpendicular to the hypotenuse chroot Staff Emeritus Gold Member Originally posted by oen_maclaude only that the height is not perpendicular to the hypotenuse[/SIZE] Right -- and what good is that? That isn't the problem at hand. - Warren I apologize if I'm intruding, but there must be some important distinction I'm missing. I would appreicate it if somebody would clarify for me. I'm working with these definitions and ideas: A right triangle is a triangle with a right angle (90 degrees) The hypotenuse of a right triangle is the side opposite the right angle The two remaining sides are called legs. Either of the legs may arbitrarily be called the base, and then the length of the remaining leg is called the height (customarily the base is drawn horizontally and as the other leg is at a 90 degree angle, the other leg is drawn vertically, again, by custom we draw this leg so that it leaves the corner of the right angle by travelling up-- thus the height coincides with our usual idea of height) Any triangle drawn in a circle with one edge forming the diameter of aforesaid circle is a right triangle, and further, the hypotenuse is the diameter, and the opposite angle is the right angle. Imagine you have such a circle/triangle... by moving the pt that corresponds to the right angle closer and closer to either of the endpts you create a triangle with one very short leg and one leg that is arbitrarily close the length of the hypotenuse... by rotating the circle and the inscribed triangle we can make that long edge correspond to the height, so the restriction on the height is that the height may not be equal to (or longer) then the hypotenuse. The ratio of edges on a 45,45,90 triangle is not 1 to 1 to 2 since 1^2 + 1^2 is not 2^2 It is 1 to 1 to sqrt(2). I think that's everything... of course I'm assuming we're doing all this on a flat surface... all bets are off if we're on some two manifold. chroot Staff Emeritus Gold Member curiousbystander,	{}
# Variance and Standard Deviation of a Portfolio We learned about how to calculate the standard deviation of a single asset. Let’s now look at how to calculate the standard deviation of a portfolio with two or more assets. The returns of the portfolio were simply the weighted average of returns of all assets in the portfolio. However, the calculation of the risk/standard deviation is not the same. While calculating the variance, we also need to consider the covariance between the assets in the portfolio. If the assets are perfectly correlated, then the simple weighted average of variances will work. However, when we have to account for the covariance, the equation will change. Covariance reflects the degree to which two securities vary or change together, and is represented as Cov (Ri,Rj). The problem with covariance is that it has no units, and is difficult to compare across assets. Using covariance, we can calculate the correlation between the assets using the following formula: Correlation After incorporating covariance, the standard deviation of a two-asset portfolio can be calculated as follows: Standard Deviation of a Two Asset Portfolio In general as the correlation reduces, the risk of the portfolio reduces due to the diversification benefits. Two assets that are perfectly negatively correlated provide the maximum diversification benefit and hence minimize the risk. Example Assume we have a portfolio with the following details: The standard deviation can be calculated as follows: This portfolio has an expected return of 16.80% and a portfolio risk of 11.09%.	{}
# Bernstein–von Mises theorem Jump to navigation Jump to search In Bayesian inference, the Bernstein-von Mises theorem provides the basis for using Bayesian credible sets for confidence statements in parametric models. It states that under some conditions, a posterior distribution converges in the limit of infinite data to a multivariate normal distribution centered at the maximum likelihood estimator with covariance matrix given by ${\displaystyle n^{-1}I(\theta _{0})^{-1}}$, where ${\displaystyle \theta _{0}}$ is the true population parameter and ${\displaystyle I(\theta _{0})}$ is the Fisher information matrix at the true population parameter value.[1] ## Introduction The Bernstein-von Mises theorem is a result that links Bayesian inference with Frequentist inference. It assumes there is some true probabilistic process that generates the observations, as in frequentism, and then studies the quality of Bayesian methods of recovering that process, and making uncertainty statements about that process. In particular, it states that Bayesian credible sets of a certain credibility level ${\displaystyle \alpha }$ will asymptotically be confidence sets of confidence level ${\displaystyle \alpha }$, which allows for the interpretation of Bayesian credible sets. ## Heuristic statement In a model ${\displaystyle (P_{\theta }:\theta \in \Theta )}$, under certain regularity conditions (finite-dimensional, well-specified, smooth, existence of tests), if the prior distribution ${\displaystyle \Pi }$ on ${\displaystyle \theta }$ has a density with respect to the Lebesque measure which is smooth enough (near ${\displaystyle \theta _{0}}$ bounded away from zero), the total variation distance between the rescaled posterior distribution (by centring and rescaling to ${\displaystyle {\sqrt {n}}(\theta -\theta _{0})}$) and a Gaussian distribution centred on any efficient estimator and with the inverse Fisher information as variance will converge in probability to zero. ## Bernstein-von Mises and Maximum likelihood estimation In case the maximum likelihood estimator is an efficient estimator, we can plug this in, and we recover a common, more specific, version of the Bernstein-von Mises theorem. ## Implications The most important implication of the Bernstein-von Mises theorem is that the Bayesian inference is asymptotically correct from a frequentist point of view. This means that for large amounts of data, one can use the posterior distribution to make, from a frequentist point of view, valid statements about estimation and uncertainty. ## History The theorem is named after Richard von Mises and S. N. Bernstein although the first proper proof was given by Joseph L. Doob in 1949 for random variables with finite probability space.[2] Later Lucien Le Cam, his PhD student Lorraine Schwartz, David A. Freedman and Persi Diaconis extended the proof under more general assumptions. ## Limitations In case of a misspecified model, the posterior distribution will also become asymptotically Gaussian with a correct mean, but not necessarily with the Fisher information as the variance. This implies that Bayesian credible sets of level ${\displaystyle \alpha }$ cannot be interpreted as confidence sets of level ${\displaystyle \alpha }$.[3] In the case of nonparametric statistics, the Bernstein-von Mises theorem usually fails to hold with a notable exception of the Dirichlet process. A remarkable result was found by Freedman in 1965: the Bernstein–von Mises theorem does not hold almost surely if the random variable has an infinite countable probability space; however, this depends on allowing a very broad range of possible priors. In practice, the priors used typically in research do have the desirable property even with an infinite countable probability space. Different summary statistics such as the mode and mean may behave differently in the posterior distribution. In Freedman's examples, the posterior density and its mean can converge on the wrong result, but the posterior mode is consistent and will converge on the correct result. ## Quotations The statistician A. W. F. Edwards has remarked, "It is sometimes said, in defence of the Bayesian concept, that the choice of prior distribution is unimportant in practice because it hardly influences the posterior distribution at all when there are moderate amounts of data. The less said about this 'defence' the better."[4] ## Notes 1. ^ van der Vaart, A.W. (1998). "10.2 Bernstein–von Mises Theorem". Asymptotic Statistics. Cambridge University Press. ISBN 0-521-78450-6. 2. ^ Doob, Joseph L. (1949). "Application of the theory of martingales". Colloq. Intern. du C.N.R.S (Paris). 13: 23–27. 3. ^ Kleijn, B.J.K.; van der Vaart, A.W. (2012). "The Bernstein-Von-Mises theorem under misspecification". Electronic Journal of Statistics. 6 (0): 354–381. doi:10.1214/12-EJS675. 4. ^ Edwards, A.W.F. (1992). Likelihood. Baltimore: Johns Hopkins University Press. ISBN 0-8018-4443-6. ## References • Vaart, A.W. van der (1998). "10.2 Bernstein–von Mises Theorem". Asymptotic Statistics. Cambridge University Press. ISBN 0-521-49603-9. • Doob, Joseph L. (1949), Application of the theory of martingales. Colloq. Intern. du C.N.R.S (Paris), No. 13, pp. 23–27. • Freedman, David A. (1963). On the asymptotic behaviour of Bayes estimates in the discrete case I. The Annals of Mathematical Statistics, vol. 34, pp. 1386–1403. • Freedman, David A. (1965). On the asymptotic behaviour of Bayes estimates in the discrete case II. The Annals of Mathematical Statistics, vol. 36, pp. 454–456. • Le Cam, Lucien (1986). Asymptotic Methods in Statistical Decision Theory, Springer. ISBN 0-387-96307-3 (Pages 336 and 618–621). • Lorraine Schwartz (1965). On Bayes procedures. Z. Wahrscheinlichkeitstheorie, No. 4, pp. 10–26.	{}
All posts I've read so far have valid uses of matrices and so many more that I couldn't even comprehend... To subscribe to this RSS feed, copy and paste this URL into your RSS reader. We call this matrix A. 3. In the most general sense, matrices (and a very important special case of matrices, vectors) provide a way to generalize from single variable equations to equations with arbitrarily many variables. In the matrix multiplication Java program, initially user is prompted to enter the matrices. My own work generates coordinates for "symmetric" geometric realizations of graphs --think Platonic and Archimedean solids-- from this kind of analysis of their adjacency matrices. There are numerous applications of matrices, both in mathematics and other sciences. -x_1+2x_2-x_3=1\\ You can then compose another transformation by multiplying the new matrix by yet another transformation matrix. Vector: A vector, in programming, is a type of array that is one dimensional. Show the tests on the columns. Scholastic matrices are used by page rank algorithms. Matrices: It will decide the number of Matrices an array can accept. Includes an interactive where you can explore the concept. I can also claim that to solve this system you need only 25 operations (one operation being a single addition/subtraction/division/multiplication). Then make a second matrix of the constants and call it B. Shop now. What does “blaring YMCA — the song” mean? In this chapter you will learn about how to use arrays in R program. Graphic data is also held in a matrix structure. I use matrices in a few cases: I need scaling or shear (not often) I need homogeneous coordinates (such as for handling perspective) Other code wants a matrix (eg. You arrange all the equations in standard form and make a matrix of their coefficients, making sure to use 0s as placeholders (like if there isn't an x term). Matrices are a useful tool for studying finite groups. Where do matrices come into play? To create a test matrix, you will have to: Put the objects that you’re testing on the rows. In Figure 3.12, the results of each composition aren't shown. we have, $$\text{A}=\left[ How easy it is to actually track another person credit card? A matrix is composed of elements arranged in rows and columns. I'm pretty sure you wouldn't be able to draw this conclusion just looking at the system without trying to solve it. It's a bit more deep than this, but it's not the point here. To convince yourself, there are a lot of linear problems you can study with little knowledge in math. \right]$$. Note that user can re-initialize matrices (and of course dimensions) by using the first command during the execution Make your first introduction with matrices and learn about their dimensions and elements. So, it's a useful tool of linear algebra. If it is not a dynamic program, place the constant values directly in the place of i and j. Why are most helipads in São Paulo blue coated and identified by a "P"? Check off the tests that you actually completed in the cells. These properties are encoded somehow in the coefficients of the matrix A. Step-2 Matrices are used in calculating the gross domestic products in Economics. A matrix is an entity composed of components arranged in rows and columns. I've also used two-dimensional arrays in programming (C++, Java) to help store information that just makes sense to be in matrix-form. The last task in writing functions for a 2D graphics program using matrices is to rewrite the Translate(), Scale(), and Rotate() functions so that they use the new matrix data types. The first matrix should be square matrix. Although Direct3D sometimes has a special use for this extra value, W is really used most often to simplify the matrix operations. Matrix. Since we are using two-dimensional arrays to create a matrix, we can easily perform various operations on its elements. Data Science and Matrices in R. Data science often involves working with data in matrices. They are used in geology to measure seismic waves. The CopyMatrix() function looks like Listing 3.11. Matrix subtraction is done element wise (entry wise) i.e. 1\\ Each data item of the array can be accessed by using a number called an “index” or “subscript”. First, you need a function that can initialize a matrix to an identity matrix. 3. In C, an array is a way to store a fixed number of items of the same data type under a single name. Many powerful machine learning algorithms can be run on matrices. But, have you ever thought which programming languages are used in Google, Facebook, Microsoft etc? However, performing so many calculations on many vertices can be time consuming, which is why graphics programmers often use matrix math to transform shapes. The data type for the 3x3 matrix looks like this: The first step in using matrices to transform a shape is to load the matrix with the appropriate values. Modern programming languages that support array programming have been engineered specifically to generalize operations on scalars to apply transparently to vectors, matrices, and higher-dimensional … That concept is possible with the use of arrays. We use matrices containing numeric elements to be used in mathematical calculations. I do numerical PDEs. The program finally copies m2 into the transformation matrix. 1\\ For all values of i=j set 0. Every finite group has a representation as a set of invertible matrices; the study of such representations is called, well, Representation Theory. Programs that deal with 2D graphics typically use two types of matrices: 1x3 and 3x3. Matrices are the R objects in which the elements are arranged in a two-dimensional rectangular layout. For example: one program generated magic squares which were stored in a matrix. The first row is x, the second y and so on. 2 & -1 & 0 & 0\\ Buy 2+ books or eBooks, save 55% through December 2. Addition of both Matrix is: 41 39 52 67 56 70 44 34 41. To the extent of my knowledge, linear algebra is a crucial tool in literally every branch of science, engineering, and mathematics. Now, if you only knew how to multiply matrices! Vectors are a logical element in programming languages that are used for storing data. Some of them merely take advantage of the compact representation of a set of numbers in a matrix. Graphics programs often perform all kinds of calculations on the vertices of an object before finally drawing that object onscreen. Matrix … 5. That concept is possible with the use of arrays. Next, m1 and m2 are initialized, after which the call to MultMatrix3X3() multiplies m1 times m2 and stores the result in m3. Just as any number times 1 equals the original number (for example, 5 x 1 = 5), so also any matrix times an identity matrix equals the original matrix (for example, m1 x I = m1). Another used a 3-by-3 to keep track of spaces on a tic-tac-toe board. Matrices are one of the most commonly used tools in business. Figure 3.12 shows another way of looking at this matrix composition. If you have data in the form of an array or matrix and you want to summarize this data, R’s apply() function is really useful. You can then multiply a 3D position vector (x, y, z, 1) by this matrix to obtain a new position with all the trasformations applied. Matrix Chain Multiplication using Dynamic Programming Matrix Chain Multiplication – Firstly we define the formula used to find the value of each cell. Also, a 3x3 matrix can be multiplied by a 3x3 matrix, something else you need to do in a 2D graphics program to compose transformations. As one responder mentioned they are used in linear programming to determine the most profitable combination of ingredients in a product. where $\underline{x}$ contains the unknowns, A the coefficients of the equations and $\underline{b}$ contains the values of the right hand sides of the equations. rev 2020.11.30.38081, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Another thing you'll need to do is copy a matrix. However, in order to solve this system fast is not enough to use a calculator with a big RAM and/or a high clock rate (CPU). The function first initializes the local matrix m1 to the scaling matrix. but that is a topic better covered by an engineer...). First, you need data types for the matrices you'll be using in your programs. 1\\ C Server Side Programming Programming In this problem, we are given a sequence( array) of metrics. This function takes as parameters references to the destination and source matrices, both of which are the type MATRIX3X3. Array in R is always a Multi-Dimensional, it means more than 1 matrix with rows and columns. One of the major theorems of all time in finite group theory is the classification of all finite simple groups. In most cases, W is equal to 1, which means a vector representing a vertex in a shape has this form: The data type for a vector, then, looks like this: The 3x3 matrix will hold the values needed to transform a vertex, which will be held in the VECTOR data type (which is also a matrix). How useful are non-square matrices in maths or sciences? To calculate element c11, multiply elements of 1st row of A with 1st column of B and add them (51+6*4) which can be shown as: If it is, the function returns immediately to avoid a division-by-zero error. Of course, the more powerful the calculator is, the faster you will get the solution. \left[ Contents: Matrix Algebra; Vectors ; Matrices; Matrix Algebra. While other programming languages mostly work with numbers one at a time, MATLAB® is designed to operate primarily on whole matrices and arrays. Many of the matricization techniques developed for one field have a way of spreading to others. Rewriting the Scale() function for use with matrices results in Listing 3.13. Figuring out from a map which direction is downstream for a river? The Scale() function takes as parameters a reference to the current transformation matrix and the X and Y scaling factors. Then, the function converts the degrees to radians and calculates the cosine and sine of the angle. Method 1. Next, Rotate() initializes the rotation matrix and multiplies that matrix times the current transformation matrix, storing the results in the local matrix m2. $\endgroup$ – Paul Mar 2 '15 at 16:49 $\begingroup$ @Paulpro The question is for applications of linear algebra (a body of work), not matrices (a set of objects). In a program, you'd initialize this matrix like this: A matrix for scaling a shape looks like this: Here, the variable xScaleFactor is how much you want to scale the shape horizontally, whereas yScaleFactor is how much to scale vertically. MATLAB is an abbreviation for "matrix laboratory." The answer is that m3 will contain exactly the same values as m2. A matrix is a way to express a linear map between finite-dimensional vector spaces, given a choice of bases for said spaces. In this tutorial, we are going to learn how to take matric input in Python from the user. Research in natural language processing, for example, has started borrowing some of the same techniques used by physicists to describe the laws of reality. The apply() function takes four arguments: X: This is your data — an […] We need to find a way to multiply these matrixes so that, the minimum number of multiplications is required. It will be one term wide (long). In this chapter you will learn about how to use arrays in R program. define $\delta=\min_{X_1,X_2 \in \mathcal{C}} \|\det(X_1-X_2)\|$. Matrix Programs in Java. 2D is easiest and most visual. Basically, I take a differential equation (an equation whose solution is not a number, but a function, and that involves the function and its derivatives) and, instead of finding an analytical solution, I try to find an approximation of the value of the solution at some points (think of a grid of points). After you compose your transformations, you have a main matrix that contains the exact values you need to simultaneously translate, scale, and rotate a shape. Indeed, but I'm not good enough in English to clarify some nuances. The 1x3 matrix is a special type of matrix known as avector. This function takes as parameters a reference to a SHAPE structure and a reference to a MATRIX3X3 array. \end{array} Let me correct you: two-dimensional arrays are used in programming to represent matrices. Such solutions are commonly used in scientific and engineering settings. As one responder mentioned they are used in linear programming to determine the most profitable combination of ingredients in a product. Matrices are rectangular arrangements of expressions, numbers and symbols that are arranged in columns and rows. A lot of linear algebra is concerned with operations on vectors and matrices, and there are many different types of matrices. (Of course, linear algebra is exceptionally useful etc. Different terms used for adding items to arrays: Like I mentioned before, — generally if not mentioned otherwise — array methods/functions in programming languages add newest elements to the end of the existing array. The function first checks whether degrees is zero. Moreover, people already pointed out other fields where matrices are important bricks and plays an important role. I work in the field of applied math, so I will give you the point of view of an applied mathematician. They contain elements of the same atomic types. You also need to know how to multiply matrices. Although Direct3D sometimes hasa special use for this extra value, W is really used most often to simplify thematrix operations. computer graphics) and are very powerful. (Application Programming Interface) API is the acronym for Application Programming Interface, which is a software intermediary that allows two applications to talk to each other. Matrix Chain Multiplication – Firstly we define the formula used to find the value of each cell. The InitMatrix() function takes as a parameter a reference to a MATRIX3X3 array into which the function loads the values that comprise a 3x3 identity matrix. That function looks like Listing 3.14. Matrix, a set of numbers arranged in rows and columns so as to form a rectangular array. \left[ Their utility in mathematics and computing is huge. The matrix can be added only when the number of rows and columns of the first matrix is equal to the number of rows and columns of the second matrix. Well, as you know (or maybe not, I don't know) a linear system can be seen in matrix-vector form as, $$\text{A}\underline{x}=\underline{b}$$ Vectors are similar to arrays but their actual implementation and operation differs. When this function has finished, the vertices in the SHAPE structure, shape, will have been transformed by the values in the transformation matrix, m. Now that you have some idea of how the matrix operations work, you can start using them in your programs. And please also give me an example of how they are used? Matrix Representation In Python and other programming languages, a matrix is often represented with a list of lists. A matrix is a grid used to store or display data in a structured format. A square matrix is a matrix with the same number of rows and columns. Adding Two Matrix Experiment with colour or 2D/3D transformations, they are fun and visual (if you are a visual person). x_1\\ I even asked my teacher but she also has no answer. In R Programming, arrays are multi-dimensional Data structures. That’s why we assigned j value to rows, and i value to columns. \begin{array}{cc} Our lives are now incomplete without certain gadgets and applications. This is because the translation matrix can’t be written as a 3x3 matrix and we use a mathematical trick to express the above transformations as matrix multiplications. An identity matrix is often used in graphics programming to initialize the main matrix that'll be used to compose transformations. After adding two matrices display the third matrix which is the addition result of the two given matrix by user as shown in the program given here. Pretty much everything in computer graphics uses matrices. 1 See the code below. Also discusses how to calculate the inverse of a matrix. Engineering and Scientific Computing with Scilab (1999) by Gomez et al: best suited for those with a strong background in matrix and differential equation theory. etc. At this point, m looks like this: The call to Translate() composes m with a translation matrix containing the values 10 and 15, which leaves m containing the translation. Prison planet book where the protagonist is given a quota to commit one murder a week, Spectral decomposition vs Taylor Expansion, Connecting an axle to a stud on the ground for railings. Matrix Multiplication in C - Matrix multiplication is another important program that makes use of the two-dimensional arrays to multiply the cluster of values in the form of matrices and with the rules of matrices of mathematics. Multiplication of Matrices - how to multiply matrices of different sizes. From powers of the adjacency matrix, for a simple example, one can read the number of available paths between any two dots. The "proof" took scores of mathematicians many decades, and could not have been completed without viewing these groups as groups of matrices. 2020 what are matrices used for in programming	{}
# Plasma physics This article is about plasma in the sense of an ionized gas. For other uses of the term, such as blood plasma, see plasma (disambiguation). File:Plasma-lamp 2.jpg A Plasma lamp, illustrating some of the more complex phenomena of a plasma, including filamentation In physics and chemistry, a plasma is an ionized gas, and is usually considered to be a distinct phase of matter. "Ionized" in this case means that at least one electron has been removed from a significant fraction of the molecules. The free electric charges make the plasma electrically conductive so that it couples strongly to electromagnetic fields. This fourth state of matter was first identified by Sir William Crookes in 1879 and dubbed "plasma" by Irving Langmuir in 1928, because it reminded him of a blood plasma [Ref]. ## Common plasmas File:Solar-flares-(double).jpg A solar coronal mass ejection blasts plasma throughout the Solar System.[Ref & Credit] Plasmas are the most common phase of matter. The entire visible universe outside the Solar System is plasma, since all we can see are stars. Since the space between the stars is filled with a plasma, although a very sparse one (see interstellar and intergalactic medium), essentially the entire volume of the universe is plasma. In the Solar System, the planet Jupiter accounts for most of the non-plasma, only about 0.1% of the mass and 10-15 of the volume within the orbit of Pluto. Commonly encountered forms of plasma include: ## Characteristics The term plasma is generally reserved for a system of charged particles large enough to behave collectively. Even a partially ionized gas in which as little as 1% of the particles are ionized can have the characteristics of a plasma (i.e. respond to magnetic fields and be highly electrically conductive). In technical terms, the typical characteristics of a plasma are: 1. Debye screening lengths that are short compared to the physical size of the plasma. 2. Large number of particles within a sphere with a radius of the Debye length. 3. Mean time between collisions usually is long when compared to the period of plasma oscillations. ### Plasma scaling Plasmas and their characteristics exist over a wide range of scales (ie. they are scaleable over many orders of magnitude). The following chart deals only with conventional atomic plasmas and not other exotic phenomena, such as, quark gluon plasmas: Typical plasma scaling ranges: orders of magnitude (OOM) Characteristic Terrestrial plasmas Cosmic plasmas Sizein metres (m) 10-6 m (lab plasmas) to:102 m (lightning) (~8 OOM) 10-6 m (spacecraft sheath) to1025 m (intergalactic nebula) (~31 OOM) Lifetimein seconds (s) 10-12 s (laser-produced plasma) to:107 s (fluorescent lights) (~19 OOM) 101 s (solar flares) to:1017 s (intergalactic plasma) (~17 OOM) Density in particles percubic metre 107 to:1021 (inertial confinement plasma) 1030 (stellar core) to:100 (i.e., 1) (intergalactic medium) Temperaturein kelvins (K) ~0 K (Crystalline non-neutral plasma[2]) to:108 K (magnetic fusion plasma) 102 K (aurora) to:107 K (Solar core) Magnetic fieldsin teslas (T) 10-4 T (Lab plasma) to:103 T (pulsed-power plasma) 10-12 T (intergalactic medium) to:107 T (Solar core) ### Temperatures File:Photos-photos 1087592507 Energy Arc.jpg The central electrode of a plasma lamp, showing a glowing blue plasma streaming upwards. The colors are a result of the radiative recombination of electrons and ions and the relaxation of electrons in excited states back to lower energy states. These processes emit light in a spectrum characteristic of the gas being excited. The defining characteristic of a plasma is ionization. Although ionization can be caused by UV radiation, energetic particles, or strong electric fields, (processes that tend to result in a non-Maxwellian electron distribution function), it is more commonly caused by heating the electrons in such a way that they are close to thermal equilibrium so the electron temperature is relatively well-defined. Because the large mass of the ions relative to the electrons hinders energy transfer, it is possible for the ion temperature to be very different from (usually lower than) the electron temperature. The degree of ionization is determined by the electron temperature relative to the ionization energy (and more weakly by the density) in accordance with the Saha equation. If only a small fraction of the gas molecules are ionized (for example 1%), then the plasma is said to be a cold plasma, even though the electron temperature is typically several thousand degrees. The ion temperature in a cold plasma is often near the ambient temperature. Because the plasmas utilized in plasma technology are typically cold, they are sometimes called technological plasmas. They are often created by using a very high electric field to accelerate electrons, which then ionize the atoms. The electric field is either capacitively or inductively coupled into the gas by means of a plasma source, e.g. microwaves. Common applications of cold plasmas include plasma-enhanced chemical vapor deposition, plasma ion doping, and reactive ion etching. A hot plasma, on the other hand, is nearly fully ionized. This is what would commonly be known as the "fourth-state of matter". The Sun is an example of a hot plasma. The electrons and ions are more likely to have equal temperatures in a hot plasma, but there can still be significant differences. ### Densities Next to the temperature, which is of fundamental importance for the very existence of a plasma, the most important property is the density. The word "plasma density" by itself usually refers to the electron density, that is, the number of free electrons per unit volume. The ion density is related to this by the average charge state $\langle Z\rangle$ of the ions through $n_e=\langle Z\rangle n_i$. (See quasineutrality below.) The third important quantity is the density of neutrals n0. In a hot plasma this is small, but may still determine important physics. The degree of ionization is ni / (n0 + ni). ### Potentials File:Lightning-with-streamers.jpg Lightning is an example of plasma present at Earth's surface. Typically, lightning discharges 30 thousand amps, at up to 100 million volts, and emits light, radio waves, x-rays and even gamma rays [1]. Plasma temperatures in lightning can approach 28,000 kelvins and electron densities may exceed 1024/m3. Since plasmas are very good conductors, electric potentials play an important role. The potential as it exists on average in the space between charged particles, independent of the question of how it can be measured, is called the plasma potential or the space potential. If an electrode is inserted into a plasma, its potential will generally lie considerably below the plasma potential due to the development of a Debye sheath. Due to the good electrical conductivity, the electric fields in plasmas tend to be very small, although where double layers are formed, the potential drop can be large enough to accelerate ions to relativistic velocities and produce synchrotron radiation such as x-rays and gamma rays. This results in the important concept of quasineutrality, which says that, on the one hand, it is a very good approximation to assume that the density of negative charges is equal to the density of positive charges ($n_e=\langle Z\rangle n_i$), but that, on the other hand, electric fields can be assumed to exist as needed for the physics at hand. The magnitude of the potentials and electric fields must be determined by means other than simply finding the net charge density. A common example is to assume that the electrons satisfy the Boltzmann relation, $n_e \propto e^{e\Phi/k_BT_e}$. Differentiating this relation provides a means to calculate the electric field from the density: $\vec{E} = (k_BT_e/e)(\nabla n_e/n_e)$. It is, of course, possible to produce a plasma that is not quasineutral. An electron beam, for example, has only negative charges. The density of a non-neutral plasma must generally be very low, or it must be very small, otherwise it will be dissipated by the repulsive electrostatic force. In astrophysical plasmas, Debye screening prevents electric fields from directly affecting the plasma over large distances (ie. greater than the Debye length). But the existence of charged particles causes the plasma to generate and be affected by magnetic fields. This can and does cause extremely complex behavior, such as the generation of plasma double layers, an object that separates charge over a few tens of Debye lengths. The dynamics of plasmas interacting with external and self-generated magnetic fields are studied in the academic discipline of magnetohydrodynamics. ## In contrast to the gas phase Plasma is often called the fourth state of matter. It is distinct from the three lower-energy phases of matter; solid, liquid, and gas, although it is closely related to the gas phase in that it also has no definite form or volume. There is still some disagreement as to whether a plasma is a distinct state of matter or simply a type of gas. Most physicists consider a plasma to be more than a gas because of a number of distinct properties including the following: Property Gas Plasma Electrical Conductivity Very low Very high For many purposes the electric field in a plasma may be treated as zero, although when current flows the voltage drop, though small, is finite, and density gradients are usually associated with an electric field according to the Boltzmann relation. The possibility of currents couples the plasma strongly to magnetic fields, which are responsible for a large variety of structures such as filaments, sheets, and jets. Collective phenomena are common because the electric and magnetic forces are both long-range and potentially many orders of magnitude stronger than gravitational forces. Independently acting species One Two or threeElectrons, ions, and neutrals can be distinguished by the sign of their charge so that they behave independently in many circumstances, having different velocities or even different temperatures, leading to new types of waves and instabilities, among other things Velocity distribution Maxwellian May be non-MaxwellianWhereas collisional interactions always lead to a Maxwellian velocity distribution, electric fields influence the particle velocities differently. The velocity dependence of the Coulomb collision cross section can amplify these differences, resulting in phenomena like two-temperature distributions and run-away electrons. Interactions BinaryTwo-particle collisions are the rule, three-body collisions extremely rare. CollectiveEach particle interacts simultaneously with many others. These collective interactions are about ten times more important than binary collisions. ## Complex plasma phenomena File:Tycho-supernova.jpg Tycho's Supernova remnant, a huge ball of expanding plasma. Langmuir coined the name plasma because of its similarity to blood plasma, and Hannes Alfvén noted its cellular nature. Note also the filamentary blue outer shell of X-ray emitting high-speed electrons Plasma may exhibit complex behaviour. And just as plasma properties scale over many orders of magnitude (see table above), so do these complex features. Many of these features were first studied in the laboratory, and in more recent years, have been applied to, and recognised throughout the universe. Some of these features include: • Filamentation, the striations or "stringy things" seen in a "plasma ball", the aurora, lightning, and nebulae. They are caused by larger current densities, and are also called magnetic ropes or plasma cables. • Double layers, localised charge separation regions that have a large potential difference across the layer, and a vanishing electric field on either side. Double layers are found between adjacent plasmas regions with different physical characteristics, and can accelerate ions and produce synchrotron radiation (such as x-rays and gamma rays). • Birkeland currents, a magnetic-field-aligned electric current, first observed in the Earth's aurora, and also found in plasma filaments. • Circuits. Birkeland currents imply electric circuits, that follow Kirchhoff's circuit laws. Circuits have a resistance and inductance, and the behaviour of the plasma depends on the entire circuit. Such circuits also store inductive energy, and should the circuit be disrupted, for example, by a plasma instability, the inductive energy will be released in the plasma. • Cellular structure. Plasma double layers may separate regions with different properties such as magnetization, density, and temperature, resulting in cell-like regions. Examples include the magnetosphere, heliosphere, and heliospheric current sheet. • Critical ionization velocity in which the relative velocity between an ionized plasma and a neutral gas, may cause further ionization of the gas, resulting in a greater influence of electomagnetic forces. ## Mathematical descriptions Plasmas may be usefully described with various levels of detail. However the plasma itself is described, if electric or magnetic fields are present, then Maxwell's equations will be needed to describe them. The coupling of the description of a conductive fluid to electromagnetic fields is known generally as magnetohydrodynamics, or simply MHD. ### Fluid The simplest possibility is to treat the plasma as a single fluid governed by the Navier Stokes Equations. A more general description is the two-fluid picture, where the ions and electrons are considered to be distinct. ### Kinetic For some cases the fluid description is not sufficient. Kinetic models include information on distortions of the velocity distribution functions with respect to a Maxwell-Boltzmann distribution. This may be important when currents flow, when waves are involved, or when gradients are very steep. ### Particle-in-cell Particle-in-cell (PIC) models include kinetic information by following the trajectories of a large number of individual particles. Charge and current densities are determined by summing the particles in cells which are small compared to the problem at hand but still contain many particles. The electric and magnetic fields are found from the charge and current densities with appropriate boundary conditions. PIC codes for plasma applications were developed at Los Alamos National Laboratory in the 1950's. Although often more calculationally intensive than alternative models, they are relatively easy to understand and program and can be very general. ## Fundamental plasma parameters File:Fusor running.jpg A 'sun in a test tube'. The Farnsworth-Hirsch Fusor during operation in so called "star mode" characterized by "rays" of glowing plasma which appear to emanate from the gaps in the inner grid. All quantities are in Gaussian cgs units except temperature expressed in eV and ion mass expressed in units of the proton mass μ = mi / mp; Z is charge state; k is Boltzmann's constant; K is wavelength; γ is the adiabatic index; ln Λ is the Coulomb logarithm. ### Frequencies • electron gyrofrequency, the angular frequency of the circular motion of an electron in the plane perpendicular to the magnetic field: $\omega_{ce} = eB/m_ec = 1.76 \times 10^7 B \mbox{rad/s}$ • ion gyrofrequency, the angular frequency of the circular motion of an ion in the plane perpendicular to the magnetic field: $\omega_{ci} = eB/m_ic = 9.58 \times 10^3 Z \mu^{-1} B \mbox{rad/s}$ • electron plasma frequency, the frequency with which electrons oscillate when their charge density is not equal to the ion charge density (plasma oscillation): $\omega_{pe} = (4\pi n_ee^2/m_e)^{1/2} = 5.64 \times 10^4 n_e^{1/2} \mbox{rad/s}$ • ion plasma frequency: $\omega_{pe} = (4\pi n_iZ^2e^2/m_i)^{1/2} = 1.32 \times 10^3 Z \mu^{-1/2} n_i^{1/2} \mbox{rad/s}$ • electron trapping rate $\nu_{Te} = (eKE/m_e)^{1/2} = 7.26 \times 10^8 K^{1/2} E^{1/2} \mbox{s}^{-1}$ • ion trapping rate $\nu_{Ti} = (ZeKE/m_i)^{1/2} = 1.69 \times 10^7 Z^{1/2} K^{1/2} E^{1/2} \mu^{-1/2} \mbox{s}^{-1}$ • electron collision rate $\nu_e = 2.91 \times 10^{-6} n_e\,\ln\Lambda\,T_e^{-3/2} \mbox{s}^{-1}$ • ion collision rate $\nu_i = 4.80 \times 10^{-8} Z^4 \mu^{-1/2} n_i\,\ln\Lambda\,T_i^{-3/2} \mbox{s}^{-1}$ ### Lengths File:Magnetic-rope.gif The complex self-constricting magnetic field lines and current paths in a Birkeland current that may develop in a plasma [Ref] $\Lambda_e= \sqrt{\frac{h^2}{2\pi m_ekT_e}}= 6.919\times 10^{-8}\,T_e^{-1/2}\,\mbox{cm}$ • classical distance of closest approach, the closest that two particles with the elementary charge come to each other if they approach head-on and each have a velocity typical of the temperature, ignoring quantum-mechanical effects: $e^2/kT=1.44\times10^{-7}\,T^{-1}\,\mbox{cm}$ • electron gyroradius, the radius of the circular motion of an electron in the plane perpendicular to the magnetic field: $r_e = v_{Te}/\omega_{ce} = 2.38\,T_e^{1/2}B^{-1}\,\mbox{cm}$ • ion gyroradius, the radius of the circular motion of an ion in the plane perpendicular to the magnetic field: $r_i = v_{Ti}/\omega_{ci} = 1.02\times10^2\,\mu^{1/2}Z^{-1}T_i^{1/2}B^{-1}\,\mbox{cm}$ • plasma skin depth, the depth in a plasma to which electromagnetic radiation can penetrate: $c/\omega_{pe} = 5.31\times10^5\,n_e^{-1/2}\,\mbox{cm}$ • Debye length, the scale over which electric fields are screened out by a redistribution of the electrons: $\lambda_D = (kT/4\pi ne^2)^{1/2} = 7.43\times10^2\,T^{1/2}n^{-1/2}\,\mbox{cm}$ ### Velocities $v_{Te} = (kT_e/m_e)^{1/2} = 4.19\times10^7\,T_e^{1/2}\,\mbox{cm/s}$ $v_{Ti} = (kT_i/m_i)^{1/2} = 9.79\times10^5\,\mu^{-1/2}T_i^{1/2}\,\mbox{cm/s}$ • ion sound velocity, the speed of the longitudinal waves resulting from the mass of the ions and the pressure of the electrons: $c_s = (\gamma ZkT_e/m_i)^{1/2} = 9.79\times10^5\,(\gamma ZT_e/\mu)^{1/2}\,\mbox{cm/s}$ • Alfven velocity, the speed of the waves resulting from the mass of the ions and the restoring force of the magnetic field: $v_A = B/(4\pi n_im_i)^{1/2} = 2.18\times10^{11}\,\mu^{-1/2}n_i^{-1/2}B\,\mbox{cm/s}$ ### Dimensionless • square root of electron/proton mass ratio $(m_e/m_p)^{1/2} = 2.33\times10^{-2} = 1/42.9$ • number of particles in a Debye sphere $(4\pi/3)n\lambda_D^3 = 1.72\times10^9\,T^{3/2}n^{-1/2}$ • Alven velocity/speed of light $v_A/c = 7.28\,\mu^{-1/2}n_i^{-1/2}B$ • electron plasma/gyrofrequency ratio $\omega_{pe}/\omega_{ce} = 3.21\times10^{-3}\,n_e^{1/2}B^{-1}$ • ion plasma/gyrofrequency ratio $\omega_{pi}/\omega_{ci} = 0.137\,\mu^{1/2}n_i^{1/2}B^{-1}$ • thermal/magnetic energy ratio $\beta = 8\pi nkT/B^2 = 4.03\times10^{-11}\,nTB^{-2}$ • magnetic/ion rest energy ratio $B^2/8\pi n_im_ic^2 = 26.5\,\mu^{-1}n_i^{-1}B^2$ ### Miscellaneous $D_B = (ckT/16eB) = 6.25\times10^6\,TB^{-1}\,\mbox{cm}^2/\mbox{s}$ • transverse Spitzer resistivity $\eta_\perp = 1.15\times10^{-14}\,Z\,\ln\Lambda\,T^{-3/2}\,\mbox{s} = 1.03\times10^{-2}\,Z\,\ln\Lambda\,T^{-3/2}\,\Omega\,\mbox{cm}$ ## Fields of active research File:HallThruster 2.jpg Hall effect thruster. The electric field in a plasma double layer is so effective at accelerating ions, that electric fields are used in ion drives	{}
# Math Insight ### Applet: Desynchronizing oscillators Illustration of $N=100$ phase oscillators whose state can be represented by a point around the unit circle. The state of oscillator $j$ at time $t$ is represented by the state variable $\theta_j(t)$, or phase, which we can map to the unit circle by plotting a green point in the complex plane at the location $e^{i\theta_j(t)}$ where $i=\sqrt{-1}$. For each oscillator, $\theta_j(t)$ increases with time, so each point moves counterclockwise around the circle. At time $t=0$, all oscillators start at exactly the same phase. However, as time evolves, the oscillators tend to spread out across the whole circle of phases, i.e., the system tends toward asynchrony. You can control how they desynchronize by changing the parameter $\sigma$, which controls the spread of oscillator speeds. When $\sigma=0$, the oscillators will stay synchronized forever. The degree of synchrony is captured by the order parameter $$z=\frac{1}{N}\sum_{i=1}^N e^{i\theta_j(t)},$$ plotted as a blue vector and by its magnitude $r=\|z\|$. When $r=1$, the oscillators are completely synchronized, and $r=0$ indicates that the oscillators are asynchronous, or at least balanced around the circle. Applet file: desynchronizing_oscillators.ggb	{}
# zbMATH — the first resource for mathematics ## Gao, Jianwei Compute Distance To: Author ID: gao.jianwei Published as: Gao, J.; Gao, Jian-Wei; Gao, Jian-wei; Gao, Jianwei Documents Indexed: 44 Publications since 1988, including 1 Book all top 5 #### Co-Authors 4 single-authored 6 Liu, Huihui 5 Yong, Xuelin 3 Xie, Jiehua 3 Zou, Wei 2 Li, Ming 2 Liu, Xiaotong 2 Wang, Hui 2 Wu, Liyuan 2 Wu, Yungao 1 Feng, Gang 1 Gu, Yundong 1 Guo, Fengjia 1 He, Dayi 1 Huang, Qi 1 Huang, Yehui 1 Li, Cunbin 1 Liang, Jin 1 Qi, Zhiqiang 1 Sun, Yeping 1 Wu, Yunna 1 Xiao, Tijun 1 Xie, Xun 1 Xu, Chuanbo 1 Yang, Weilin 1 Yi, Ru 1 Yu, Rui 1 Zhang, Haobo 1 Zhang, Tiejun 1 Zhang, Xingping 1 Zhang, Zhiyong 1 Zhao, Kun all top 5 #### Serials 4 Insurance Mathematics & Economics 3 Communications in Nonlinear Science and Numerical Simulation 3 Control and Decision 3 Symmetry 2 Applied Mathematics and Computation 2 Journal of Computational and Applied Mathematics 2 Mathematics in Practice and Theory 2 Communications in Statistics. Theory and Methods 1 Chaos, Solitons and Fractals 1 Journal of Algebra 1 European Journal of Operational Research 1 International Journal of Robust and Nonlinear Control 1 Acta Analysis Functionalis Applicata. AAFA 1 Entropy 1 MATCH - Communications in Mathematical and in Computer Chemistry 1 Journal of Inner Mongolia Normal University. Natural Science Edition 1 Arabian Journal of Mathematics all top 5 #### Fields 17 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 8 Operations research, mathematical programming (90-XX) 5 Systems theory; control (93-XX) 4 Partial differential equations (35-XX) 3 Probability theory and stochastic processes (60-XX) 2 Statistics (62-XX) 2 Computer science (68-XX) 1 Mathematical logic and foundations (03-XX) 1 Group theory and generalizations (20-XX) 1 Dynamical systems and ergodic theory (37-XX) 1 Integral equations (45-XX) 1 Functional analysis (46-XX) 1 Operator theory (47-XX) 1 Optics, electromagnetic theory (78-XX) 1 Biology and other natural sciences (92-XX) 1 Information and communication theory, circuits (94-XX) #### Citations contained in zbMATH 21 Publications have been cited 160 times in 127 Documents Cited by Year Optimal portfolios for DC pension plans under a CEV model. Zbl 1162.91411 Gao, Jianwei 2009 Optimal investment strategy for annuity contracts under the constant elasticity of variance (CEV) model. Zbl 1231.91402 Gao, Jianwei 2009 Stochastic optimal control of DC pension funds. Zbl 1141.91439 Gao, Jianwei 2008 An extended CEV model and the Legendre transform-dual-asymptotic solutions for annuity contracts. Zbl 1231.91432 Gao, Jianwei 2010 Generalized ordered weighted utility averaging-hyperbolic absolute risk aversion operators and their applications to group decision-making. Zbl 1346.91046 Gao, Jianwei; Li, Ming; Liu, Huihui 2015 Modulus of convexity in Banach spaces. Zbl 1060.46012 Gao, J. 2003 Kinematic and static hypotheses for constitutive modelling of granulates considering particle rotation. Zbl 0942.74015 Chang, C. S.; Gao, J. 1996 On the expected discounted penalty function and optimal dividend strategy for a risk model with random incomes and interclaim-dependent claim sizes. Zbl 1291.91139 Zou, Wei; Gao, Jian-wei; Xie, Jie-hua 2014 An optimal approach to output-feedback robust model predictive control of LPV systems with disturbances. Zbl 1350.93038 Yang, Weilin; Gao, Jianwei; Feng, Gang; Zhang, Tiejun 2016 Conservative bounds on Rayleigh-Bénard convection with mixed thermal boundary conditions. Zbl 1202.76057 Wittenberg, R. W.; Gao, J. 2010 Variable separation solutions to the coupled integrable dispersionless equations. Zbl 1334.35305 Yong, Xuelin; Liu, Xiaotong; Gao, Jianwei; Wang, Hui 2014 Normal structure, fixed points and related parameters in Banach spaces. Zbl 1029.46010 Gao, J. 2002 A $$C^ 2$$ finite element and interpolation. Zbl 0768.41009 Gao, J. 1993 Multiple attribute decision making with triangular intuitionistic fuzzy numbers based on zero-sum game approach. Zbl 1429.91135 Xu, J.; Dong, J. Y.; Wan, S. P.; Gao, J. 2019 Generalized ordered weighted utility proportional averaging-hyperbolic absolute risk aversion operators and their applications to group decision-making. Zbl 1338.91052 Gao, Jianwei; Li, Ming; Liu, Huihui 2015 Independent component analysis for multiple-input multiple-output wireless communication systems. Zbl 1217.94046 Gao, J.; Zhu, X.; Nandi, A. K. 2011 Interval generalized ordered weighted utility multiple averaging operators and their applications to group decision-making. Zbl 1425.91133 Wu, Yunna; Xu, Chuanbo; Zhang, Haobo; Gao, Jianwei 2017 Calculation of a parallel-plate waveguide with a chiral insert by the mixed finite element method. Zbl 1356.78131 Bogolyubov, A. N.; Mukhartova, Yu. V.; Gao, J. 2013 On the probability of ruin in the compound Poisson risk model with potentially delayed claims. Zbl 1307.60065 Xie, Jie-Hua; Zou, Wei; Gao, Jian-Wei 2013 Singularity analysis and explicit solutions of a new coupled nonlinear Schrödinger type equation. Zbl 1221.35404 Yong, Xuelin; Gao, Jianwei; Zhang, Zhiyong 2011 Subassembly identification based on grey clustering. Zbl 1160.90391 Gao, J.; Xiang, D.; Duan, G. 2008 Multiple attribute decision making with triangular intuitionistic fuzzy numbers based on zero-sum game approach. Zbl 1429.91135 Xu, J.; Dong, J. Y.; Wan, S. P.; Gao, J. 2019 Interval generalized ordered weighted utility multiple averaging operators and their applications to group decision-making. Zbl 1425.91133 Wu, Yunna; Xu, Chuanbo; Zhang, Haobo; Gao, Jianwei 2017 An optimal approach to output-feedback robust model predictive control of LPV systems with disturbances. Zbl 1350.93038 Yang, Weilin; Gao, Jianwei; Feng, Gang; Zhang, Tiejun 2016 Generalized ordered weighted utility averaging-hyperbolic absolute risk aversion operators and their applications to group decision-making. Zbl 1346.91046 Gao, Jianwei; Li, Ming; Liu, Huihui 2015 Generalized ordered weighted utility proportional averaging-hyperbolic absolute risk aversion operators and their applications to group decision-making. Zbl 1338.91052 Gao, Jianwei; Li, Ming; Liu, Huihui 2015 On the expected discounted penalty function and optimal dividend strategy for a risk model with random incomes and interclaim-dependent claim sizes. Zbl 1291.91139 Zou, Wei; Gao, Jian-wei; Xie, Jie-hua 2014 Variable separation solutions to the coupled integrable dispersionless equations. Zbl 1334.35305 Yong, Xuelin; Liu, Xiaotong; Gao, Jianwei; Wang, Hui 2014 Calculation of a parallel-plate waveguide with a chiral insert by the mixed finite element method. Zbl 1356.78131 Bogolyubov, A. N.; Mukhartova, Yu. V.; Gao, J. 2013 On the probability of ruin in the compound Poisson risk model with potentially delayed claims. Zbl 1307.60065 Xie, Jie-Hua; Zou, Wei; Gao, Jian-Wei 2013 Independent component analysis for multiple-input multiple-output wireless communication systems. Zbl 1217.94046 Gao, J.; Zhu, X.; Nandi, A. K. 2011 Singularity analysis and explicit solutions of a new coupled nonlinear Schrödinger type equation. Zbl 1221.35404 Yong, Xuelin; Gao, Jianwei; Zhang, Zhiyong 2011 An extended CEV model and the Legendre transform-dual-asymptotic solutions for annuity contracts. Zbl 1231.91432 Gao, Jianwei 2010 Conservative bounds on Rayleigh-Bénard convection with mixed thermal boundary conditions. Zbl 1202.76057 Wittenberg, R. W.; Gao, J. 2010 Optimal portfolios for DC pension plans under a CEV model. Zbl 1162.91411 Gao, Jianwei 2009 Optimal investment strategy for annuity contracts under the constant elasticity of variance (CEV) model. Zbl 1231.91402 Gao, Jianwei 2009 Stochastic optimal control of DC pension funds. Zbl 1141.91439 Gao, Jianwei 2008 Subassembly identification based on grey clustering. Zbl 1160.90391 Gao, J.; Xiang, D.; Duan, G. 2008 Modulus of convexity in Banach spaces. Zbl 1060.46012 Gao, J. 2003 Normal structure, fixed points and related parameters in Banach spaces. Zbl 1029.46010 Gao, J. 2002 Kinematic and static hypotheses for constitutive modelling of granulates considering particle rotation. Zbl 0942.74015 Chang, C. S.; Gao, J. 1996 A $$C^ 2$$ finite element and interpolation. Zbl 0768.41009 Gao, J. 1993 all top 5 #### Cited by 225 Authors 10 Rong, Ximin 8 Zhao, Hui 7 Yao, Haixiang 6 Chang, Hao 6 Li, Zhongfei 5 Liang, Zongxia 4 Chen, Ping 4 Gao, Ji 4 Gao, Jianwei 4 Guan, Guohui 4 Li, Danping 4 Zhang, Yan 3 Chang, Kai 3 Lai, Yongzeng 3 Saejung, Satit 3 Vigna, Elena 3 Zeng, Yan 2 Ding, Baocang 2 Fantuzzi, Giovanni 2 Jiang, Min 2 Kim, Jai Heui 2 Kim, Jeong-Hoon 2 Lai, Shaoyong 2 Lee, Min-Ku 2 Lin, Xiang 2 Liu, Huihui 2 Liu, Yaqing 2 Meng, Zhiqing 2 Misra, Anil 2 Nürnberger, Günther 2 Shen, Rui 2 Shen, Yang 2 Sun, Jingyun 2 Sun, Zhongyang 2 Wang, Dengshan 2 Wang, Yucang 2 Wen, Xiaoyong 2 Wittenberg, Ralf W. 2 Yang, Sung-Jin 2 Yuan, Weipeng 2 Zeilfelder, Frank 2 Zhang, Chubing 2 Zhang, Ling 2 Zhao, Peibiao 1 A, Chunxiang 1 Ali, Muhammad Intizar 1 Alonso-Marroquin, Fernando 1 Angkola, Francisca 1 Aziz, Taha 1 Bakkaloglu, Ahmet 1 Bao, Zhenhua 1 Bogolyubov, N. A. 1 Bumroongsri, Pornchai 1 Castaneda, Ranu 1 Chang, Chien-Chen 1 Chavez-Bedoya, Luis 1 Chen, An 1 Chen, Guanwei 1 Chen, Hongjing 1 Chen, Hua 1 Chen, Rudong 1 Chen, Shumin 1 Chen, Zhiping 1 Cheng, Chih-Jen 1 Cheung, Ka Chun 1 de Borst, René 1 Delong, Łukasz 1 Deng, Shouchun 1 Deng, Yingchun 1 Di Giacinto, Marina 1 El-Ganaini, Shoukry Ibrahim Atia 1 Faizi, Shahzad 1 Fang, Zhenming 1 Fatima, Aeeman 1 Federico, Salvatore 1 Fu, Chao 1 Garcia-Falset, Jesús 1 Gerrard, Russell 1 Goldhirsch, Isaac 1 Gozzi, Fausto 1 Gu, Ailing 1 Gu, Mengdi 1 Guo, Junyi 1 Guo, Xianping 1 Guo, Yunrui 1 Hagenimana, Emmanuel 1 Han, Nan-Wei 1 Hao, Zhifeng 1 He, Lin 1 Hjalmarsson, Håkan 1 Højgaard, Bjarne 1 Homenda, Wladyslaw 1 Hou, Rujing 1 Huang, Hong-Chih 1 Huang, Yujuan 1 Hung, Mao-wei 1 Hussain, Ayyaz 1 Jastrzebska, Agnieszka 1 Jian, Minjie 1 Jiao, Hongwei ...and 125 more Authors all top 5 #### Cited in 54 Serials 25 Insurance Mathematics & Economics 13 Journal of Computational and Applied Mathematics 5 Mathematical Problems in Engineering 5 Discrete Dynamics in Nature and Society 4 Journal of Mathematical Analysis and Applications 4 European Journal of Operational Research 4 Abstract and Applied Analysis 4 Quantitative Finance 3 Computers & Mathematics with Applications 3 Journal of Fluid Mechanics 3 Journal of Industrial and Management Optimization 3 Symmetry 2 Journal of the Franklin Institute 2 Applied Mathematics and Computation 2 Communications in Statistics. Theory and Methods 2 Journal of Inequalities and Applications 2 Scandinavian Actuarial Journal 2 Journal of Systems Science and Complexity 2 ASTIN Bulletin 2 Granular Matter 2 Iranian Journal of Fuzzy Systems 1 Acta Mechanica 1 International Journal of Control 1 International Journal of Solids and Structures 1 Journal of Mathematical Physics 1 Journal of the Mechanics and Physics of Solids 1 Lithuanian Mathematical Journal 1 Mathematical Methods in the Applied Sciences 1 Zhurnal Vychislitel’noĭ Matematiki i Matematicheskoĭ Fiziki 1 Automatica 1 Journal of Functional Analysis 1 Operations Research Letters 1 Circuits, Systems, and Signal Processing 1 International Journal of Production Research 1 Computational Mechanics 1 Applied Mathematics Letters 1 Applications of Mathematics 1 Applied Mathematics. Series B (English Edition) 1 Computational and Applied Mathematics 1 Journal of the Egyptian Mathematical Society 1 European Journal of Control 1 Soft Computing 1 Mathematical Methods of Operations Research 1 Mathematical Inequalities & Applications 1 Philosophical Transactions of the Royal Society of London. Series A. Mathematical, Physical and Engineering Sciences 1 International Journal of Theoretical and Applied Finance 1 Applied Stochastic Models in Business and Industry 1 Journal of Applied Mathematics 1 Stochastics and Dynamics 1 North American Actuarial Journal 1 East Asian Mathematical Journal 1 Science China. Mathematics 1 Journal of the Operations Research Society of China 1 Cogent Mathematics all top 5 #### Cited in 21 Fields 88 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 43 Systems theory; control (93-XX) 22 Probability theory and stochastic processes (60-XX) 22 Operations research, mathematical programming (90-XX) 11 Calculus of variations and optimal control; optimization (49-XX) 9 Functional analysis (46-XX) 7 Mechanics of deformable solids (74-XX) 6 Partial differential equations (35-XX) 4 Approximations and expansions (41-XX) 4 Statistics (62-XX) 4 Numerical analysis (65-XX) 4 Fluid mechanics (76-XX) 2 Integral equations (45-XX) 2 Operator theory (47-XX) 2 Optics, electromagnetic theory (78-XX) 2 Classical thermodynamics, heat transfer (80-XX) 1 Mathematical logic and foundations (03-XX) 1 Integral transforms, operational calculus (44-XX) 1 General topology (54-XX) 1 Computer science (68-XX) 1 Information and communication theory, circuits (94-XX)	{}
# E&M question 1. Apr 2, 2006 ### matt85 The distance between an object and its image is fixed at 40.0 cm. A converging lens of focal length f = 3.62 cm forms a sharp image for two positions of the lens. What is the distance a between these two positions? Determine: The distance between these two positions a = ____cm Could someone help me with the formula i want to be using? 2. Apr 3, 2006 ### andrevdh The thin lens formula gives $$\frac{1}{f}=\frac{1}{o}+\frac{1}{i}$$ with i the image distance and o the object distance. Now say you find a position between the object and fixed screen (image position) such that you get a sharp image on the screen. The object image is now o and the image distance i while the sum of these two being 40 cm. But according to the formula if you make the object distance i the image distance should be o! Which gives you the other position of the lens where you should get a sharp image on the screen. In the second case the object distance will therefore be i and the image distance will be o. Last edited: Apr 3, 2006 3. Apr 3, 2006 ### andrevdh I would guess that with a mech advantage of 2 you would have problems lifting your lecturer! Three would give you a fighting chance!	{}
# ACC 226 wk 3 Exercise 11-1 11-7 ACC 226 wk 3 Exercise 11-1 11-7 Exercise 11-1 Exercise 11-7 CheckPoint:Classifying Liabilities and Preparing Payroll Entries Resource: Fundamental Accounting Principles, pp. 451 and 452 Exercise 11-1 The following items appear on the balance sheet of a company with a two-month operating cycle. Identify the proper classification of each item as follows: Cif it is a current liability, Lif it is a long-term liability or Nif it is not a liability. C 1.Sales taxes payable. C 6.Notes payable (due in 6 to 12 months). C 2.FUTA taxes payable. C 7.Notes payable (due in 120 days). C 3.Accounts receivable. L 8.Current portion of long-term debt. C 4.Accrued payroll payable. L 9.Notes payable (mature in five years). C 5.Wages payable. L 10.Notes payable (due in 13 to 24 months) Exercise 11-7 Using the data in situation aof exercise below, prepare the employer’s September 30 journal entries to record (1) salary expense and its related payroll liabilities for this employee and (2) the employer’s payroll taxes expense and its related liabilities. The employee’s federal income taxes withheld by the employer are $135 for this pay period. Tax Rate Applied To FICA—Social Security . . . . . . . 6.20% First$87,000 FICA—Medicare . . . . . . . . . . . 1.45 All gross pay FUTA . . . . . . . . . . . . . . . . . . . 0.80 First $7,000 SUTA . . . . . . . . . . . . . . . . . . . 2.90 First$7,000 Gross Pay Gross Pay for Through August September a. $6,400$ 800 ## Recent Questions in Financial Accounting Submit Your Questions Here ! Copy and paste your question here... Attach Files	{}
## anonymous 5 years ago is there a value for h that makes it possible for the equation √(x+h)+5=0 to have any real number solution? explain 1. anonymous is x+h+5 all under the square root? or just x+h? 2. anonymous just x+h 3. anonymous ok.. so you want to find a value for h that makes the expression $\sqrt{x+h}+5=0 \implies \sqrt{x+h}=-5$ but the square root can't have a negative value. so, such h doesn't exist. 4. anonymous wait, the sqroot of a number can have a negative value, the SQUARE of a number can't have a real value 5. anonymous square both sides and you get x+h = 25 6. anonymous that's not right mindedone, you're suggesting that h=25-x is a solution.. check by plugging back into the equation and see what you get. 7. anonymous Keep in mind though that if you plug h=25-x into the original equation you get +-sqrt(25) = -5, and so the negative square root does satisfy the equation. 8. anonymous you get sqrt(25) = -5 9. anonymous sqrt(25) = +-5	{}
3 Tutor System Starting just at 265/hour # In figure, $$\angle{X}$$ = $$60^\circ$$ , $$\angle{XYZ}$$ = $$54^\circ$$ , if YO and ZO are the bisectors of $$\angle{XYZ}$$ and $$\angle{XZY}$$ respectively of $$\triangle{XYZ}$$ , find $$\angle{OZY}$$ and $$\angle{YOZ}$$. In $$\triangle{XYZ}$$, $$\angle{X}$$ + $$\angle{Y}$$ + $$\angle{Z}$$ = $$180^\circ$$ because, Sum of all angles of triangle is equal to 180 $$\therefore$$ $$62^\circ$$ + $$\angle{Y}$$ + $$\angle{Z}$$ = $$180^\circ$$ $$\Rightarrow$$ $$\angle{Y}$$ + $$\angle{Z}$$ = $$118^\circ$$ Now, so as to find bisected angles, multiply both sides by $$\frac{1}{2}$$ We get, $$\frac{1}{2}$$ [$$\angle{Y}$$ + $$\angle{Z}$$] = $$\frac{1}{2}$$ × $$118^\circ$$ = $$59^\circ$$ Thus we get, $$\angle{OYZ}$$ + $$\angle{OZY}$$ = $$59^\circ$$ ....(as YO and ZO are the bisectors of $$\angle{XYZ}$$ and $$\angle{XZY}$$) Also, it is given that, $$\angle{XYZ}$$ = $$54^\circ$$ and we have $$\Rightarrow$$ $$\angle{OYZ}$$ = $$\frac{1}{2}$$ × $$\angle{XYZ}$$ $$\Rightarrow$$ $$\angle{OZY}$$ + $$\frac{1}{2}$$ × $$54^\circ$$ = $$59^\circ$$ $$\Rightarrow$$ $$\angle{OZY}$$ = $$59^\circ$$ - $$27^\circ$$ = $$32^\circ$$ Also, in $$\triangle{YOZ}$$, $$\angle{OYZ}$$ + $$\angle{YOZ}$$ + $$\angle{OZY}$$ = $$180^\circ$$ because, Sum of all angles of triangle is equal to 180 $$\therefore$$ $$27^\circ$$ + $$\angle{YOZ}$$ + $$32^\circ$$ = $$180^\circ$$ $$\Rightarrow$$ $$\angle{YOZ}$$ + $$59^\circ$$ = $$180^\circ$$ $$\Rightarrow$$ $$\angle{YOZ}$$ = $$180^\circ$$ - $$59^\circ$$ $$\Rightarrow$$ $$\angle{YOZ}$$ = $$121^\circ$$	{}
Solve the equation (x-1)^{2}=9 Question Equations and inequalities Solve the equation $$\displaystyle{\left({x}-{1}\right)}^{{{2}}}={9}$$ 2021-04-20 Take the square root of both sides of he equation: $$x-1=\pm3$$ Add 1 to both sides of the equation: $$\displaystyle{x}={1}+\pm{3}$$ or -2 Relevant Questions Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. $$\displaystyle{\left({x}^{{2}}-{9}\right)}^{{2}}-{2}{\left({x}^{{2}}-{9}\right)}-{35}={0}$$ Solve the equation $$\displaystyle{x}^{{{2}}}+{x}-{20}$$ Solve the equation $$\displaystyle{x}^{{{2}}}-{25}$$ Solve the equation $$\displaystyle{\frac{{{3}}}{{{2}}}}{x}={\frac{{{5}}}{{{6}}}}{x}+{2}$$ Solve the equation $$\displaystyle{2}{x}^{{{2}}}-{6}{x}=-{5}$$ Solve the equation $$\displaystyle{\left({4}{x}-{2}\right)}^{{{2}}}\leq{100}$$ Solve the equation $$\displaystyle{\left\|{3}{x}-{2}\right\|}{>}{4}$$ Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. $$\displaystyle{9}^{{{2}{m}-{3}}}={27}^{{{m}+{1}}}$$ $$\displaystyle{\left\|{5}{x}-{1}\right\|}={\left\|{3}-{4}{x}\right\|}$$ Solve the equations and inequalities. Find the solution sets to the inequalities in interval notation. $$\displaystyle{3}{x}{\left({x}-{1}\right)}={x}+{6}$$	{}
# User talk:Sławomir Biały SEMI-RETIRED This user is no longer very active on Wikipedia. I laughed out loud reading your post about reverting the code inserted into the prime numbers article.... I fell out of my chair when I actually saw the code.... The fact that the guy wanted to solve the Goldbach conjecture with that... RockvilleRideOn (talk) 03:48, 4 February 2013 (UTC) I saw your talk on 'Fourier transform'. If you have time, could you just explain a little more on normalization problem? Or state in the page so reader could be aware of this. Thanks. Allenleeshining (talk) 17:34, 4 January 2013 (UTC) http://en.wikipedia.org/wiki/Talk:Fourier_transform#Suspect_wrong_equations_in_section_.27Square-integrable_functions.27 ## Talkback Hello, Sławomir Biały. You have new messages at 2001:db8's talk page. Message added 23:27, 11 February 2013 (UTC). You can remove this notice at any time by removing the {{Talkback}} or {{Tb}} template. – 2001:db8:: (rfc \| diff) 23:27, 11 February 2013 (UTC) ## Talkback I responded to your question at the Math reference desk at Wikipedia:Reference desk/Mathematics#Penrose tiles puzzle pieces. Best. -- Toshio Yamaguchi 13:00, 7 March 2013 (UTC) ## Examples of convolution I saw the wiki page, but I couldn't find any examples using actual numbers evaluating the formula. Could you give some examples of convolution, please? Mathijs Krijzer (talk) 22:14, 9 March 2013 (UTC) #### Definition The convolution of f and g is written fg, using an asterisk or star. It is defined as the integral of the product of the two functions after one is reversed and shifted. As such, it is a particular kind of integral transform: $(f * g )(t)\ \ \,$ $\stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau)\, d\tau$ $= \int_{-\infty}^\infty f(t-\tau)\, g(\tau)\, d\tau.$ (commutativity) #### Domain of definition The convolution of two complex-valued functions on Rd $(fg)(x) = \int_{\mathbf{R}^d}f(y)g(x-y)\,dy$ is well-defined only if f and g decay sufficiently rapidly at infinity in order for the integral to exist. Conditions for the existence of the convolution may be tricky, since a blow-up in g at infinity can be easily offset by sufficiently rapid decay in f. The question of existence thus may involve different conditions on f and g. #### Circular discrete convolution When a function gN is periodic, with period N, then for functions, f, such that fgN exists, the convolution is also periodic and identical to: $(f g_N)[n] \equiv \sum_{m=0}^{N-1} \left(\sum_{k=-\infty}^\infty {f}[m+kN] \right) g_N[n-m].\,$ #### Circular convolution When a function gT is periodic, with period T, then for functions, f, such that fgT exists, the convolution is also periodic and identical to: $(f * g_T)(t) \equiv \int_{t_0}^{t_0+T} \left[\sum_{k=-\infty}^\infty f(\tau + kT)\right] g_T(t - \tau)\, d\tau,$ where to is an arbitrary choice. The summation is called a periodic summation of the function f. #### Discrete convolution For complex-valued functions f, g defined on the set Z of integers, the discrete convolution of f and g is given by: $(f * g)[n]\ \stackrel{\mathrm{def}}{=}\ \sum_{m=-\infty}^\infty f[m]\, g[n - m]$ $= \sum_{m=-\infty}^\infty f[n-m]\, g[m].$ (commutativity) When multiplying two polynomials, the coefficients of the product are given by the convolution of the original coefficient sequences, extended with zeros where necessary to avoid undefined terms; this is known as the Cauchy product of the coefficients of the two polynomials. ## How to request IP block exemption I saw your post at WP:VPT. It appears that WP:UTRS is currently down due to toolserver problems. Your best bet is to try Wikipedia:Sockpuppet investigations#Quick CheckUser requests. Thanks, EdJohnston (talk) 04:13, 13 April 2013 (UTC) Thanks. I didn't know about this. Wikipedia has obviously become too large and complex for me to handle :-) Sławomir Biały (talk) 19:58, 13 April 2013 (UTC) ## About axiom of global choice Hello, Sławomir, I replied to you last comment here: http://en.wikipedia.org/wiki/Wikipedia_talk:Articles_for_deletion/Axiom_of_global_choice Eozhik (talk) 06:07, 22 April 2013 (UTC) ## Talk: Manifold [1] No. You apparently do not understand the difference between the ring Z of integer numbers, which is a specific ring, and the ring of integers OK of a number field K, not a specific ring but a functor from fields(?) to commutative rings. Of course, the ring of integers of p-adic numbers contains some extra elements which Z does not have. Incnis Mrsi (talk) 06:18, 22 April 2013 (UTC) Oh, indeed! You presume quite a lot about what others do not understand, while in the same breath betraying your own ignorance of the very subject that you would presume to "correct" me on. Thanks, but I'll pass. Sławomir Biały (talk) 13:49, 22 April 2013 (UTC) ## Another wiki Being "semi-retired" here, you could be welcome there. Boris Tsirelson (talk) 07:23, 22 April 2013 (UTC) ## Transpose: best abstract definition? If you feel that way inclined, I'd appreciate a quick "yes" or "no" at Talk:Transpose#Transpose_of_linear_maps: why defined in terms of a bilinear form?. — Quondum 14:17, 1 June 2013 (UTC) Thank you very much for your answers to my question. I've learned many from you. Could you please make some comments on my newly posted words about the angle in Wikipedia:Reference desk/Mathematics? Thanks. Armeria wiki (talk) 03:38, 13 June 2013 (UTC) ## Banach space article Dear Sławomir, Thanks for your comments and your interest for the Banach space article. I certainly agree with your comments, and I reply here because what I want to say is a bit personal. Actually, I would like some help of yours on the following points: I don't feel like rewriting a lead in English. I can manage talking to mathematicians, but not to "a general audience". I started writing a primitive sketch for an Introduction, but was blocked by the same language barrier. It was something like: Introduction Functional analysis aims to find functions that are solutions of various equations, several arising from physics. Abstract solutions, namely, functions that cannot be expressed by an explicit formula, are often obtained as limits in a well chosen vector space of functions X of a sequence of approximate solutions. Completeness of X is needed in order to make sure that the limit exists in X. Many examples of such spaces X, but not all, are Banach spaces. Various type of compact sets in function spaces (norm compact, weak compact) are also used to prove the existence of abstract solutions, for example to optimization problems. In this respect, it is important to characterize compact subsets of function spaces. I would like to have a section on differential equations in "Banach space", but I am not expert about this. With best wishes, Bdmy (talk) 12:59, 17 June 2013 (UTC) Hi Bdmy, I didn't mean to lay the task of improving the article entirely at your feet, just to suggest directions in which I think the article needs to be expanded. These are tasks that I wish I myself had time to undertake. Your mathematical edits to that article are most appreciated. There is now a solid foundation on which to build. Best, Sławomir Biały (talk) 17:45, 17 June 2013 (UTC)	{}
Project # How to Make a Timed Air Ventilator January 11, 2017 by Jens Christoffersen ## Need to ventilate a room? This article will show you how I made a relay-operated, GPS-timed fan to vent out the moist air in my boat's cabin. It is controlled with a PIC16F628A and has an LCD. Need to ventilate a room? This article will show you how I made a relay-operated, GPS-timed fan to vent out the moist air in my boat's cabin. It is controlled with a PIC16F628A and has an LCD. ### Let The Air Flow In small rooms where there is little to no ventilation, air tends to stand still. If the air is a little moist, you'll definitely have mold and fungi in certain places. Mold and fungi are found in nature and are necessary to break down leaves, wood, and other plant debris. Since I've got a whole lot of wood in my boat, I will certainly have mold and fungi there. I cannot prevent them from coming in, but I can take some measures to try to keep the environment inside my boat as hostile for the mold and fungi as possible. There are at least two ways to deal with this. One way is to wash and clean on a tight schedule. Now, how fun is that? Being the geek that I am, I nerded something up instead: a fan which circulates and vents out the moist air on a fixed, timed basis. ### Designing the System I want the system to do the following: • Circulate the air • Automatically turn on/off, in a steady interval • Run from a car battery • Contain a battery charger to charge the battery • Contain a display that shows time and other info To make such a system, I'll need the following parts: • A 12v fan • A microcontroller • A 12v battery, and a 12v battery charger • 2 relays, something to track time and screw terminals. • Other parts, according to part list below I want the fan to run for five minutes each hour. That will be hard coded in the software. The setup will run from a car battery. My fan is rated 12v 4.5A. To ensure that the battery is always top-charged, I'll connect it to a battery charger. To ensure that the battery charger is not overloaded when the fan is running, I'm going to make the system “disconnect” the charger while the fan is running. One relay will operate the fan, and the other relay will operate the battery charger. When the fan is running, the battery is not charged, and when the fan is not running, the battery is charged. To keep track of time, I'm using a GPS module. In this project, I'll use the Skylab SKM53-series module (PDF). This unit sends several NMEA sentences over UART each second. The datasheet recommend to use a 10K pull-up resistor on both RXD and TXD. This will increase the serial data stability. I did not do that, and I've not noticed any instability. I might be lucky. On the same page in the datasheet it says that suitabe decoupling capasitors should be added. A 10uF electrolytic and a 0.1uF ceramic. When I breadboarded the circuit, I used only a 0.1uF ceramic. ##### Screenshot from the Skylab datasheet (PDF). I'm using the "RMC" sentence to get the time. An example of an RMC sentence is the following: $GPRMC,075747.000,A,2233.89990,N,11405.3368,E,3.9,357.8,260210,,,A6A The first numbers after$GPRMC are the time. In this example, the time is 07:57:47. We'll be needing that info. The next thing we'll need is the prefix that tells us if the GPS module has a valid position fix. In the example above, it's the capital A after the three zeros. An "A" indicates a valid fix and a "V" indicates an invalid fix. In my software, I check for a valid fix. These letters are case sensitive. To structure projects, I like to make a block diagram. This way I “visualize” what I want to do. In the above block diagram, I've split everything down into their own block. This is also helpful if I need to troubleshot the schematic or circuit. ### Hardware The schematic diagram is based on the block diagram. I've reproduced all the blocks in the block diagram to reflect the components and their connections in the circuit. In the datasheet for the LM7805 regulator, it's stated that for a standard application you'd only need a 0.33uF on the input pin, and a 0.1uF on the output pin. So why do I use different valued caps and some extra? I'm using them for smoothing the input and the output. It might be a little too much, but I have a very good experience with this regulator configuration. If you don't use any capacitors, the regulator might start to oscillate. The LM7805 regulator is an old linear regulator. Why not use a modern switching regulator? The switching regulators are a lot more efficient than the linear regulators. This is what I had laying around. I could have saved up on parts and space, with just using one relay. I want two relays. One of the reasons for using two relays, is that I want to have a little timeframe between the charger is cut, and the fan starts. I've imported the parts list from BOM.ULP into OpenOffice Calc and removed some of the unnecessary columns: Not showing in the schematic are my fuses. I have a 12v 8A rated fuse on the fans positive wire, and I have the same on the battery chargers positive wire. ### Software When I make a program, I follow a certain structure. This diagram shows the structure: The code starts with including the necessary libraries, followed by the configuration bits. It is considered as good programming practice to include the configuration bits in the source code. Then it will be a lot easier to see what you've done and for others to help troubleshoot. Besides, if you pick up on a project after a few months, then you'll see the bits straight away. When the configuration bits are OK, I move on to the definitions. Here I define the crystal speed I've connected to the circuits. The microcontroller's ports are defined, as well. The next thing is the variables: All variables are declared here. Now it's time to prototype the functions. Here I list all functions the program uses. Some programmers consider this a waste of time, but I like it and I keep it. It is actually necessary if you structure the program with the functions after the main program loop. Next come the functions. Now it's really important to start commenting. I usually have a few lines over each function that tell what the function does in general. I also comment code lines inside the functions. One example is the function that initializes the UART port in this project: // FUNCTION TO INIT THE UART PORT void uart_init(void) { TXSTAbits.BRGH = 0; // high baud selection bit, 1=high, 0=low TXSTAbits.SYNC = 0; // USART mode selection bit, 1=sync mode, 0=async mode TXSTAbits.TX9 = 0; // 9-bit selection bit, 1=9-bit transmission, 0=8-bit transmission / Calculate the SPBRG with 16MHz crystal 16MHz 16000000 /9600 = 1666.6666 1666.6666 / 64 = 26.0416 26.0416 - 1= 25.0416 25.041 = 25 / SPBRG = 25; // 9600-n-8-1 PIE1bits.RCIE = 1; // USART receive interrupt enable bit, 1=enable RCSTAbits.SPEN = 1; // Serial port enable bit, 1=serial port enable TXSTAbits.TXEN = 1; // transmit enable bit, 1=transmit eanble return; } You'll see the first comment line explaining what the function does. Then all the code lines are commented, so I know what's going on. When all functions are in place, it's time for the main program. The main program starts with a few statements before it enters a loop that is run forever. jc_lettheairflow.c.zip ### Conclusion In this article, I've tried to make the environment in my boat as hostile for mold and fungi as I can with a fan that blows and circulates the air. The fan is connected to a pipe that goes out of the cabin. I've used a GPS module to keep track of the time and I've used two relays to turn the fan on/off. To ensure the battery remains charged, I've used another relay to switch a battery charger on and off. I leave it to the reader to take this circuit to the next level and display the date. Why use a GPS module? I could have used the microcontroller as a simple timer. Then I had to make some sort of interface, to set the time. Or I'd be happy with it running without regards to real time. It would run for 5 minutes each whole hour. With the GPS I can programmically set it to run five minutes to each whole hour. ### Picture Give this project a try for yourself! Get the BOM. • C catalin_cluj January 20, 2017 The GPS is about as useful in this case as a diesel engine to power your fan. Like. • S splud January 20, 2017 I have some suggestions. 1. you might introduce a humidity sensor into your project. Then, if the humidity crosses a threshold, you run the circulation system, and do so until the humidity drops below a threshold. With a DHT11 or DHT22 sensor, you can add temperature to the mix and calculate dewpoint. This data could be displayed to your LCD if you were so inclined. 2. As someone just posted as I went to reply, the GPS is overkill for this application. Further, if you’re running the whole lot off of battery, that’s another drain you could probably do without. Your uC should be able to enter low power sleep for designated periods, wake up, check whether it should be driving the fans (whether that is a humidity sensor or ‘x’ of the timed sleep cycled), and go back to sleep as appropriate. Mold and mildew doesn’t care if there’s a few second shift to the cycle over the course of a week. If this circulation system were presented as one block of a larger automation system which really had need for a GPS, it could be justified, but the system would be more applicable to a larger audience if it wasn’t designed with a reliance on a single component which likely has a BOM cost higher than anything else in the project. 3. While relays afford versatility in what type of device you drive, since your fans are driven from DC, you could simply use mosfets instead (which are IMO a better choice here than BJTs). If a given application necessitates it, the mosfet output could drive a relay (or an optocoupler to triac) instead of directly driving your load. This has the benefit of reducing the current draw and reducing the potential for mechanical failure (the contacts of the relays themselves can erode over time). This further reduces your BOM cost, and there’s no “click” noise when it switches on and off. I do appreciate you’ve got diodes across the relay solenoids. 4. You’re using a linear regulator and you know you should be using a switchmode. Let me underscore that you should be using a switchmode - it looks like you ordered some parts anyway, and if everything is stuff you had laying around, then you need to order some switchmode regulators so you have them laying around for future projects. You’re running off of battery - the difference between the linear and the switchmode is significant. Your uC, the GPS, the LCD, and the BJTs are all being driven through your linear, and it doesn’t look like you’re powering down the GPS to conserve power (which of course would be a mistake anyway because you’d have to wait for multiple satellite locks to get a valid time). Have you characterized the power draw of your circuit? if you’re running from a 12V battery at say 12.5V, and you’re pulling 200mA constant (let’s ignore power on the fans and LEDs), you may be using 0.2A 5.0V = 1W in your circuit, but your regulator is burning 0.2A * ( 12.5V - 5.0V ) = 1.5W, and the total circuit is consuming 2.5W. A switchmode would bring that down much closer to the 1W constant. Using sleepmode on the uC instead of GPS would delete the GPS load and reduce the uC load. uC can turn off the LCD when not awake (you could use a button to interrupt the sleep and update the display), and these in turn would reduce the overall power draw. 5. If you’re recharging the battery, you’re likely doing this when your system is connected to AC (generator, or dock) - having the battery charger simply implemented as a trickle charger would likely suffice - if your battery has dropped low enough that your relays don’t actuate, you can put yourself into a situation where you can’t turn on an external charger, so removing that logic simplifies things. However, if you changed to mosfets, your system would work down to around 5V + the dropout of your linear regulator (yea, another reason to switch to a switchmode). What you have here is something not too different (except for relays, BJTs, GPS, and choice of uC) from what I use for a greenhouse monitor (driving heater and fans). Like. • T teddlesruss April 05, 2018 DHT modules aren't very accurate or useful, I've found. I use one in an Arduino dehydrator controller and have found that the DHT 11 and 22's I;ve used are laggy, tend to take a long time to stabilise, and have been looking for a cheap but more useful alternative. Like.	{}
# Thread: Extraneous solutions, absolute value 1. ## Extraneous solutions, absolute value Hi Forum! I've just found something strange The closest integral of sum of the solutions to $27^{5x+3}=81^{\|x+2\|-7}$ is OK, calculating, we get two solutions $\frac{-29}{11}$ and $\frac{-45}{19}$ But the solution says that $\frac{-29}{11}$ should be disconsidered. Why? The sum of them would give something about $-5$. Can someone explain? Does $x=\frac{-29}{11}$ makes the function diverge? Thanks 2. ## Re: Extraneous solutions, absolute value Does $\displaystyle 27^{5\left(-\frac{29}{11}\right)+3}$ equal $\displaystyle 81^{\left\|-\frac{29}{11} + 2\right\| - 7}$? 3. ## Re: Extraneous solutions, absolute value Originally Posted by Prove It Does $\displaystyle 27^{5\left(-\frac{29}{11}\right)+3}$ equal $\displaystyle 81^{\left\|-\frac{29}{11} + 2\right\| - 7}$? Hi Prove It I get your point, then it is necessary to test the values. I thought there was something else to it; this is really simple.	{}
# Regularization vs constrained optimization of an ill posed tomography problem I am trying to solve an ill-posed linear system of equations. The particular system has 160 equations and 400 variables. Moreover, the condition number of the left hand side matrix is of order $10^{16}$. I came across two methods to solve this problem: constrained optimization and regularization methods. My question is what are the pros and cons of each method; which one should I prefer? Any website or paper explaining the same will be much appreciated. • Have you read this post in Mathematics.SO? – nicoguaro Jun 17 '17 at 15:46 • For an ill-posed, underdetermined problem, the condition number is of course infinite. Your statement of $10^{16}$ is only a numerical representation of this infinity. – Wolfgang Bangerth Jun 17 '17 at 16:30 • I think you will want to read the book by Engl, Hanke, Neubauer: "Regularization of Inverse Problems". – Wolfgang Bangerth Jun 17 '17 at 16:31 • Do you know anything else about the desired solution? Is it nonnegative? Smooth? Sparse? Are there constraints that the solution should satisfy?you need a reason to pick out one solution as best. – Brian Borchers Jun 20 '17 at 2:16 • @nicoguaro: Thanks for sharing this post. It helps in understanding the solution from different approaches, but I am confused about which one would be better. – Sahil Gupta Jun 21 '17 at 10:41 ## 1 Answer We have the following linear system in $$\mathrm x \in \mathbb R^n$$ $$\rm A x = b$$ where $$\mathrm A \in \mathbb R^{m \times n}$$ is fat (i.e., $$n > m$$) and $$\mathrm b \in \mathbb R^m$$. ### Least-norm If the linear system is consistent, we look for the least-norm solution via the following (convex) quadratic program $$\begin{array}{ll} \text{minimize} & \\| \mathrm x \\|_2^2\\ \text{subject to} & \mathrm A \mathrm x = \mathrm b\end{array}$$ Let the Lagrangian be $$\mathcal L (\mathrm x, \lambda) := \frac 12 \mathrm x^{\top} \mathrm x + \lambda^{\top} (\mathrm A \mathrm x - \mathrm b)$$ Taking the partial derivatives of $$\mathcal L$$ and finding where they vanish, we obtain the linear system $$\begin{bmatrix} \mathrm I_n & \mathrm A^\top\\ \mathrm A & \mathrm O_m\end{bmatrix} \begin{bmatrix} \mathrm x\\ \lambda \end{bmatrix} = \begin{bmatrix} 0_n\\ \mathrm b\end{bmatrix}$$ If $$\mathrm A$$ has full row rank, then $$\rm A A^\top$$ is invertible and we can conclude that the least-norm solution is $$\mathrm x_{\text{LN}} := \color{blue}{\mathrm A^{\top} \left( \mathrm A \mathrm A^{\top} \right)^{-1} \mathrm b}$$ ### Least-squares If the linear system is inconsistent, we can look for the least-squares solution via the following unconstrained (convex) quadratic program $$\begin{array}{ll} \text{minimize} & \\| \mathrm A \mathrm x - \mathrm b \\|_2^2\end{array}$$ Taking the gradient of the objective function and finding where it vanishes, we obtain the normal equations $$\mathrm A^{\top} \mathrm A \,\mathrm x = \mathrm A^{\top} \mathrm b$$. However, since $$\rm A$$ is fat, its rank is at most $$m$$ and, thus, $$\mbox{rank} (\mathrm A^{\top} \mathrm A) \leq m < n$$ Hence, $$\mathrm A^{\top} \mathrm A$$ is never invertible and, thus, the normal equations have infinitely many solutions. Thus, let us add a regularization term to the objective function, i.e., $$\begin{array}{ll} \text{minimize} & \\| \mathrm A \mathrm x - \mathrm b \\|_2^2 + \gamma \\| \mathrm x \\|_2^2\end{array}$$ where $$\gamma \geq 0$$. The new normal equations are $$\left( \mathrm A^{\top} \mathrm A + \gamma \mathrm I_n \right) \mathrm x = \mathrm A^{\top} \mathrm b$$ If $$\color{blue}{\gamma > 0}$$, then $$\mathrm A^{\top} \mathrm A + \gamma \mathrm I_n$$ is positive definite and, thus, invertible. Hence, the regularized least-squares solution is $$\mathrm x_{\text{LS}} := \color{blue}{\left( \mathrm A^{\top} \mathrm A + \gamma \mathrm I_n \right)^{-1} \mathrm A^{\top} \mathrm b}$$ Using a matrix inversion lemma, the regularized least-squares solution can be rewritten as follows $$\mathrm x_{\text{LS}} := \color{blue}{\mathrm A^{\top} \left( \mathrm A \mathrm A^{\top} + \gamma \mathrm I_m \right)^{-1} \mathrm b}$$ which resembles the least-norm solution. However, we now invert $$\mathrm A \mathrm A^{\top} + \gamma \mathrm I_m$$, which is positive definite whenever $$\gamma > 0$$, rather than $$\mathrm A \mathrm A^{\top}$$ (which may be ill-conditioned or even non-invertible). • Thanks for the answer, but this does not answer the original question: I wanted to understand the differences between constrained optimization and regularization. Your approach to this problem is similar to optimization, where you define a Lagrangian and minimize the function. Theoretically, the final solution by you is correct, provided the data noise was absent. In my case, as the condition number of A is very high, it blows-up the error. This is where the regularization of problem helps in obtaining a better solution. – Sahil Gupta Jun 21 '17 at 12:18 • A further note, I have tried constrained optimization and regularization of the problem. Tikhonov Regularization fails for complicated profiles (I have not yet tried other regularization techniques) and constrained optimization takes too long to solve the problem. Before moving further, I wanted to learn more about the differences in these approaches. – Sahil Gupta Jun 21 '17 at 12:21 • @Sahil Gupta write in the preceding comment: "constrained optimization takes too long to solve the problem". What constraints did you use? What is the formulation of your optimization problem? What method did you use to solve it? – Mark L. Stone Jun 21 '17 at 15:12 • @MarkL.Stone I used the fmincon function available in MATLAB link. The additional constraints along with the linear system of equations were lower and upper bounds of 0 and 1. – Sahil Gupta Jun 23 '17 at 5:55 • Thanks for the updated answer; this sums up pretty much the complete regularization approach. Tikhnov Regularization is similar to your least-squares approach; the only difference being the regularization term. I have a doubt regarding this approach: how does one find a suitable value for gamma? The usual approach is the L-Curve corner. But in my case, as I increase the complexity of the image, regularization methods fail to obtain a suitable solution for the corner value. Although the constrained optimization captures a reasonable solution. What could be the reason for this? – Sahil Gupta Jun 23 '17 at 6:03	{}
# Can't comment strings with ')' inside it I believe it's a bug in notebook's comment parser. I can't comment strings with ) inside it. Something like: ("this is a bug)") I discovered it trying to comment a code with a regular expression RegularExpression["((re))"] inside it. Can someone else check it? Using V10.0.2 in MacOS • Same issue with V10.0.2.0 in Windows 8.1 (64 bit). – bbgodfrey Jan 22 '15 at 0:38 • A quick fix: RegularExpression["((re)"<>")"] – ybeltukov Jan 22 '15 at 11:19 We already have some answers explaining the issue. Let me give a solution to your problem. Let's say you have the code RegularExpression["((re))"] that you want to comment out. Since we have nested comments in Mathematica, just use a pair of (* to prevent your issue: Although I don't know the internal implementation of Mathematicas parser, the reason why this goes wrong is pretty universal. Parsing is usually a two-step process. First, the input is broken down into tokens by a so-called lexer and comments are often separated from the rest of the code in this stage. Taking your example (* RegularExpression["((re))"] ) this means the lexer first reads the token COMMENT_OPEN. Then, the lexer usually reads over everything until it meets the closing ). And here it is important to know that the lexer really stupidly reads over the input. Unfortunately, the next closing ) is inside the stuff you wanted to comment out. At this point the lexer leaves its stupid comment mode again and tries to tokenize real code. Now, he is in the middle of your expression and everything goes wrong. Btw, using the fultzTokenize of this answer shows how Mathematica breaks your code down. Note that I had to quote the quotes accordingly. See that strings are kept together when the lexer is in code mode when going over str2: str1="(* RegularExpression[\"((re))\"] )"; str2="RegularExpression[\"((re))\"]"; fultzTokenize/@{str1,str2} {(, ,RegularExpression,[,"((re),),"] )} { RegularExpression,[,"((re))",]} • +1 Clever. It exploits the fact that Mathematica is unusual among languages in that it supports nested matchfix comments (unlike, say, C). – WReach Jan 22 '15 at 1:14 This behaviour is very common, possibly near universal, in programming languages with matchfix comment syntax. The reason is that the contents of a comment sequence is presumed not to be code. Usually that presumption is correct, but not in this case. The general rule is You should be able to put anything inside a comment and the only special tokens inside comments are the tokens ( and ) itself. C, C++, Java and SQL, just to name a few, exhibit the same problem. Those languages also offer a prefix syntax that can be used to work around the problem -- provided you can isolate the relevant expression to its own line. C also offers a way to disable whole blocks of code using IFDEF, etc. Unfortunately, Mathematica only offers the matchfix syntax. Thus, there is no analog to the prefix/block notations in other languages. A prefix comment syntax would probably make a good addition to the language -- but there are mighty few left-over special characters for that purpose :). Additionally, we should not forget that lines don't really exist in the front-end since it shuffles code around automatically. I would not call this a bug in the sense that the behaviour of the Front End and the Kernel (run in command line mode) are consistent with each other. Yes, this does make code like "asd )" un-commentable, so you might call it bad design. But I wouldn't call it a bug. Contrast this with certain bugs where the Front End parses expressions differently from the Kernel. For example, the M10.0.2 front end still has some difficulties with expressions of the type #"one two" while the kernel handles them. The existence of these types of bugs proves that the Front End and Kernel do indeed use separate parsers. This happens because characters inside the comment are not actually parsed. Ponder these two code fragments: (* "This string contains the end of a comment )" ) (* "This comment begins with a quote character ) How can the parser tell whether the quote is the start of a string or just a quote? the "bug": ("this is a bug)") the work around: ("this is a solution\)") • If the closing ) is, like in Murta's case, part of a working Mathematica expression, I wouldn't use this because when you un-comment the region again, you have to go through your entire expression and remove the . Using (( at the beginning is easier IMO here. – halirutan Jan 22 '15 at 2:26 Based on the procedure outlined in the first Answer to 40396, the internal representation of ("this is a bug)") is RowBox[{RowBox[{"(", "\"\<this is a bug\>", ")"}], "\"\<)\>"}] We see that ("this is a bug) is interpreted as a comment, and ") as additional code that is syntactically incorrect. As other Answers here have explained, this is not surprising. If someone really wants to display this item, it can be accomplished with RowBox[{RowBox[{"(", "\"\<this is a bug\>", ")"}], "\"\<)\>"}]//DisplayForm yielding ("this is a bug)"*) in an In cell (but without the red color, which is added by SE).	{}
This arXival from last spring/summer by Martino, Elvira, Luengo and Corander combines and extends upon recent advances of Importance Sampling, using mainly ideas from Adaptive Multiple lmportance Sampling (AMIS) and Population Monte Carlo (PMC). The extension consists of the idea to not use the Importance Sampling procedure itself to come up with new proposal distributions, but rather to run a Markov Chain. The output of which is used solely as the location parameter for IS proposal distributions $q_{n,t}$. The weights of the samples drawn from these are Rao-Blackwellized using the deterministic mixture idea of Zhou and Owen, and as far as I can see only the Importance Samples are used for estimating integrands.	{}
A Combined NNLO Lattice-Continuum Determination of $L_{10}^r$ Research output: Contribution to journalArticle Open Documents Original language English Physical Review D 89 094510 10.1103/PhysRevD.89.094510 Published - 29 May 2014 Abstract The renormalized next-to-leading-order (NLO) chiral low-energy constant, $L_{10}^r$, is determined in a complete next-to-next-to-leading-order (NNLO) analysis, using a combination of lattice and continuum data for the flavor $ud$ $V-A$ correlator and results from a recent chiral sum-rule analysis of the flavor-breaking combination of $ud$ and $us$ $V-A$ correlator differences. The analysis also fixes two combinations of NNLO low-energy constants, the determination of which is crucial to the precision achieved for $L_{10}^r$. Using the results of the flavor-breaking chiral $V-A$ sum rule obtained with current versions of the strange hadronic $\tau$ branching fractions as input, we find $L_{10}^r(m_\rho )\, =\, -0.00346(32)$. This result represents the first NNLO determination of $L_{10}^r$ having all inputs under full theoretical and/or experimental control, and the best current precision for this quantity. Research areas • hep-ph, hep-ex, hep-lat No data available ID: 17197402	{}
# Temporarily change number of displayed authors for reference within text I use biblatex with style=authoryear-icomp, maxbibnames=50 and maxcitenames=2. Therefore, usually, two authors are named for each reference via \cite. For most cases this is a perfect setting. However, for some calls of \cite I'd like to explicitly increase the number of displayed authors. And I'd like to let bibtex compact (group) the authors. At the moment I get for "\cite{paper2001,paper2002,paper2003}": "Author1, Author2, et. al 2001; Author5, Author6, et. al 2002, 2003" But I'd like to have: "Author1, Author2, Author3, and Author4 2001; Author5, Author6, Author7, and Author8 2002, 2003" For single citations (just one paper), \AtNextCite can be used to modify the behaviour (see https://tex.stackexchange.com/a/142202). Using the same trick with \AtNextMultiCite does not compress the output. Therefore, in the previous example, I would get two times the list of Authors 5-8 for the two referenced papers. How can I temporarily (for one call of \cite) increase the number of displayed authors while maintaining the compression MWEB: \documentclass{article} \usepackage{bbding} \usepackage[ sortcites=true, style=authoryear-icomp, firstinits=true, uniquename=init, maxbibnames=50, maxcitenames=2, autocite=inline, block=space, date=short, backend=biber, sorting=nyt, ]{biblatex} % For the bibliography \usepackage{filecontents} \begin{filecontents}{\jobname.bib} @article{paper2001, author = {Author1 and Author2 and Author3 and Author4}, year = {2001}, title = {paper1}, publisher = {Publisher}, } @article{paper2002, author = {Author5 and Author6 and Author7 and Author8}, year = {2002}, title = {paper2}, publisher = {Publisher}, } @article{paper2003, author = {Author5 and Author6 and Author7 and Author8}, year = {2003}, title = {paper3}, publisher = {Publisher}, } \end{filecontents} \begin{document} \noindent Usually I want to have the abbreviated version: \cite{paper2002,paper2003}.\Checkmark \\\\ \noindent But sometimes, I'd like to list all authors like in the References. Instead I get:\\ \cite{paper2001,paper2002,paper2003} \printbibliography \end{document} • It would be easier to have a look into this if you could provide an MWE (or MWEB) so we have something to start from. Is there some kind of pattern to where you want to show all authors? (Is it in first cite, for certain authors only, for certain entry types only, for certain entries only, or completely at random?) – moewe Jan 25 '17 at 14:18 • I added an MWEB. There is no pattern. At some points it's simply necessary to list all authors but in general I prefer the abbreviated version. – Jan Jan 25 '17 at 15:05	{}
# How do you solve 2x + 7y = - 8 and - 2x + 3y = - 12 using matrices? Jun 23, 2017 $\left(x , y\right) = \left(3 , - 2\right)$ (see below for use of matrices) #### Explanation: Define the matrices: M_(xyc)=({: (x,y,=c), (2,7,-8), (-2,3,-12) :}) ${M}_{x y} = \left(\left.\begin{matrix}2 & 7 \\ - 2 & 3\end{matrix}\right.\right) \textcolor{w h i t e}{\text{XX")M_(cy)=({:(-8,7),(-12,3):})color(white)("XX}} {M}_{x c} = \left(\left.\begin{matrix}2 & - 8 \\ - 2 & - 12\end{matrix}\right.\right)$ and their determinants: ${D}_{x y} = \| \left.\begin{matrix}2 & 7 \\ - 2 & 3\end{matrix}\right. \| = 2 \times 3 - \left(- 2\right) \times 7 = 20$ ${D}_{c y} = \| \left.\begin{matrix}- 8 & 7 \\ - 12 & 3\end{matrix}\right. \| = \left(- 8\right) \times 3 - \left(- 12\right) \times 7 = 60$ ${D}_{x c} = \| \left.\begin{matrix}2 & - 8 \\ - 2 & - 12\end{matrix}\right. \| = 2 \times \left(- 12\right) - \left(- 2\right) \times \left(- 8\right) = - 40$ By Cramer's Rule $\textcolor{w h i t e}{\text{XXX}} x = \frac{{D}_{c y}}{{D}_{x y}} = \frac{60}{20} = 3$ and $\textcolor{w h i t e}{\text{XXX}} y = \frac{{D}_{x c}}{{C}_{x y}} = \frac{- 40}{20} = - 2$	{}
## CryptoDB ### Patrick P. Tsang #### Publications Year Venue Title 2005 EPRINT Since the introduction of Identity-based (ID-based) cryptography by Shamir in 1984, numerous ID-based signature schemes have been proposed. In 2001, Rivest et al. introduced ring signature that provides irrevocable signer anonymity and spontaneous group formation. In recent years, ID-based ring signature schemes have been proposed and all of them are based on bilinear pairings. In this paper, we propose the first ID-based threshold ring signature scheme that is not based on bilinear pairings. We also propose the first ID-based threshold linkable' ring signature scheme. We emphasize that the anonymity of the actual signers is maintained even against the private key generator (PKG) of the ID-based system. Finally we show how to add identity escrow to the two schemes. Due to the different levels of signer anonymity they support, the schemes proposed in this paper actually form a suite of ID-based threshold ring signature schemes which is applicable to many real-world applications with varied anonymity requirements. 2004 EPRINT A ring signature scheme is a group signature scheme with no group manager to setup a group or revoke a signer. A linkable ring signature, introduced by Liu, et al. \cite{LWW04}, additionally allows anyone to determine if two ring signatures are signed by the same group member (a.k.a. they are \emph{linked}). In this paper, we present the first separable linkable ring signature scheme, which also supports an efficient thresholding option. We also present the security model and reduce the security of our scheme to well-known hardness assumptions. In particular, we introduce the security notions of {\em accusatory linkability} and {\em non-slanderability} to linkable ring signatures. Our scheme supports `event-oriented'' linking. Applications to such linking criterion is discussed. 2004 EPRINT A ring signature scheme can be viewed as a group signature scheme with no anonymity revocation and with simple group setup. A \emph{linkable} ring signature (LRS) scheme additionally allows anyone to determine if two ring signatures have been signed by the same group member. Recently, Dodis et al. \cite{DKNS04} gave a short (constant-sized) ring signature scheme. We extend it to the first short LRS scheme, and reduce its security to a new hardness assumption, the Link Decisional RSA (LD-RSA) Assumption. We also extend \cite{DKNS04}'s other schemes to a generic LRS scheme and a generic linkable group signature scheme. We discuss three applications of our schemes. Kiayias and Yung \cite{KY04} constructed the first e-voting scheme which simultaneously achieves efficient tallying, public verifiability, and write-in capability for a typical voter distribution under which only a small portion writes in. We construct an e-voting scheme based on our short LRS scheme which achieves the same even for all worst-case voter distribution. Direct Anonymous Attestation (DAA) \cite{BCC04} is essentially a ring signature scheme with certain linking properties that can be naturally implemented using LRS schemes. The construction of an offline anonymous e-cash scheme using LRS schemes is also discussed. #### Coauthors Man Ho Au (2) Tony K. Chan (1) Joseph K. Liu (2) Victor K. Wei (2) Duncan S. Wong (2)	{}
# Normal Distribution 'According to a public survey; average informal economic activity in Turkey is 31.4 percent of its GDP with ###### Question: Normal Distribution 'According to a public survey; average informal economic activity in Turkey is 31.4 percent of its GDP with a standard deviation of 2.7 percent during the 1960-2019 period: If the ratio of informal economies to GDP in Turkey follows a normal distribution, then, what is the probability that ratio of informal economies to GDP next year will be: #### Similar Solved Questions ##### For the given Law of Cosines equation; solve for angle C Round your answer to the nearest degree:(18.8)? = 302 + 152(30) (15) co8 & For the given Law of Cosines equation; solve for angle C Round your answer to the nearest degree: (18.8)? = 302 + 152 (30) (15) co8 &... ##### Evidence E t0 reject Ine Nok sumcient the customer satisfaction of the two branches; 180 compare Clstoni order t0 customers from branch B were randomly selected pranch and 300 and 334 from on a 5 ~point = scale , af which companies was the most branch . 0set* = t0 rate their summary of the survey results is was the most satistactory: shown ana In Ie following table; Branch A Branch B3 47 point3.20 pointmeanvanance0,520.53Number ol sample180300Teste atthe 590 level of significance whether the da evidence E t0 reject Ine Nok sumcient the customer satisfaction of the two branches; 180 compare Clstoni order t0 customers from branch B were randomly selected pranch and 300 and 334 from on a 5 ~point = scale , af which companies was the most branch . 0set* = t0 rate their summary of the survey r... ##### Antibiotics are administered to livestock on a routine basis so that the animals will reach market... Antibiotics are administered to livestock on a routine basis so that the animals will reach market weight more quickly. Explain what the public health concern is that stems from this practice, and what the health effects are. Explain why food defect action levels are a form of risk management rather... ##### How many atoms are in .0062 grams of AgHow many molecules are in 1.99 liters of CCl4 How many atoms are in .0062 grams of Ag How many molecules are in 1.99 liters of CCl4... ##### Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions (Au2+), each... Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions (Au2+), each with a mass of 3.27 × 10-25 kg. The ions are accelerated from rest through a potential difference of 2.20 kV. Then, a 0.530-T magnetic field causes the ions to follow a circular path. Determine the... ##### A solcnoid of Icngth 50 cm und rdius 0SS0 cm has 37 turns: If the wire of the solenoid hits the mngnitude ol thc magrictic ficld inside Ihc solenoid?JDSWlmteni unnmimagnitude of ule magnetic ficld: 5.81 XlOLenoring? Ihe Wcak ngnetic licld quiside Ihe solenaid, findcilcty densily inside thc solcnoidmgnclc enet} Ucney212 XIOAAolal MELEC" CMTE}Iutni In 'nCtt CnCTY A solcnoid of Icngth 50 cm und rdius 0SS0 cm has 37 turns: If the wire of the solenoid hits the mngnitude ol thc magrictic ficld inside Ihc solenoid? JDS Wlmteni unnmi magnitude of ule magnetic ficld: 5.81 XlO Lenoring? Ihe Wcak ngnetic licld quiside Ihe solenaid, find cilcty densily inside thc solc... ##### Close up image of & synapse betwean two naurons ns Warithe following questions using the plcture aboveWhatiare circled In 4? (1 points} Pmnnatue haprening X polnt B? (2 points)] FaNneilainemental polntic? Whatlis happening a1 point C? (3 points) Close up image of & synapse betwean two naurons ns Warithe following questions using the plcture above Whatiare circled In 4? (1 points} Pmnnatue haprening X polnt B? (2 points)] FaNneilainemental polntic? Whatlis happening a1 point C? (3 points)... ##### Barium Hydroxide, Ba(OH)2, is reported to have a Ksp of 2.1 x 104 at 25°C. a.... Barium Hydroxide, Ba(OH)2, is reported to have a Ksp of 2.1 x 104 at 25°C. a. What is the molar solubility (M) of Ba(OH)2? b. What is the solubility of Ba(OH)2 in g/L?... ##### Plot the point given in polar coordinates(-2, 2x 3 Plot the point given in polar coordinates (-2, 2x 3... ##### Share a current event from the news that embodies one of the topics we have discussed... Share a current event from the news that embodies one of the topics we have discussed in class this semester. Does this real world situation change your understanding of the topic in any way? What is the importance of this current event?... ##### 1.1. Find a constant b (in terms of a) such that the function tx, y(x, y)... 1.1. Find a constant b (in terms of a) such that the function tx, y(x, y) =fb.e-lacty) 04220 0y< 1.2 Find the expresion for the marginal densities fx lx), and foly). HINT: SI fulzig) dardy = 1, fax) = \| Saxlayboy, etc.... ##### A body is moving along a straight line with a velocity which varies according to the equation v=9t^2+2t, where v is in feet per second and t is in seconds 1. A body is moving along a straight line with a velocity which varies according to the equation v=9t^2+2t, where v is in feet per second and t is in seconds. Find the expression for the distance as a function of time. A. 3t^3+t^2+C B. 3t^3+2t^2+C C. 18t+C D. 20t+C My answer is B.... ##### OHOHOHOHClCHgZON OH OH OH OH Cl CHg ZON... ##### 5. The tight coupling of translation with transcription in bacteria makes possible what mechanism of gene... 5. The tight coupling of translation with transcription in bacteria makes possible what mechanism of gene regulation (e.g., at the trp operon) that is not possible in eukaryotes? Just name the mechanism.... ##### 37. 109 g/100 ml of solution. What is the B. C. D. The solubility of calcium... 37. 109 g/100 ml of solution. What is the B. C. D. The solubility of calcium chromate is 1.56 2.4 x 10+ 1.5 x 10- 7.6 x 104 1.0 x 10 110-... ##### When the half-cells for a reaction at standard conditions arecombined for a electrolytic cell, which statement about cellvoltage is true?Eocell > 0Eocell < 0Eocell â‰ EcellEocell = 0 When the half-cells for a reaction at standard conditions are combined for a electrolytic cell, which statement about cell voltage is true? Eocell > 0 Eocell < 0 Eocell â‰ Ecell Eocell = 0... ##### The city government wants to conduct survey on the number and types of cars owned by its residents;How can the city use the cluster sampling method to find this information?The city selects only those houscholds that have morc than onc vchicle t0 completc thc survcy:The city sclecls IO0 houscholds at random t0 complctc thc survcy:The city has every houschold within city Iimits complctc thc survcy:Tho city has cvcry houschold In 10 ncighborhoods complete thc survey: The city government wants to conduct survey on the number and types of cars owned by its residents; How can the city use the cluster sampling method to find this information? The city selects only those houscholds that have morc than onc vchicle t0 completc thc survcy: The city sclecls IO0 houschold... ##### Q. 2. For the following hypothesis:H,;us70 I:4 > 70with / = 20 T= 71.2, 5 = 6.9,and a 0.1, statethe decision rule in terms of the critical value of the test statistic (b). the calculated value of the test statistic the conclusion Q. 2. For the following hypothesis: H,;us70 I:4 > 70 with / = 20 T= 71.2, 5 = 6.9,and a 0.1, state the decision rule in terms of the critical value of the test statistic (b). the calculated value of the test statistic the conclusion... ##### 13. If f(x) =x+1 which of these statements are true? 4,=1 !im f () exists IL: fis continuous atx = 1. II; fis differentiable atx = 1.TonlyIand Il onlyTand IIl onlyNone 13. If f(x) = x+1 which of these statements are true? 4,=1 !im f () exists IL: fis continuous atx = 1. II; fis differentiable atx = 1. Tonly Iand Il only Tand IIl only None... ##### How many palisade cells from plant leaves would fit in a volume of 1.0 cm^3 of cells if the average volume of a palisade cell is 0.0147 mm^3? How many palisade cells from plant leaves would fit in a volume of 1.0 cm^3 of cells if the average volume of a palisade cell is 0.0147 mm^3?...	{}
# ParametricPlot and PlotLegends don't seem to cooperate Bug introduced in 9.0.1 -- fixed in 10.0.0? I noted that only part of the legend shows up in ParametricPlot using v9's PlotLegends: ParametricPlot[{{t, t}, {t, 2 t}}, {t, 0, 1}, PlotLegends -> LineLegend[{1, 2}]] Inspecting the figure's FullForm gets me (among other things) LineLegend[ List[ Directive[EdgeForm[GrayLevel[0.5]], Hue[0.9060679774997897, 0.6, 0.6]] ], List[1, 2], Rule[LegendLayout, "Column"]] It looks like Mathematica forgot to add the second color. Adding the color manually gets me the full legend. This looks like a bug, but perhaps I'm missing something? I also noticed that virtually all examples in the "Options/PlotLegends" section of the ParametricPlot doc page yield a result differing from the pre-rendered image. Again, this looks like a bug, perhaps introduced by a last-minute addition. I'm using v9.0.1. Anyone with v9.0.0 care to check whether it exists there as well? • The problem seems to be that the plot legends code is not parsing the first argument to ParametricPlot correctly and identifying the different functions. It incorrectly lists all the terms as a single entry and this can be seen by setting PlotLegends -> "Expressions". In essence, the bug is the equivalent of plotting 2 curves but labeling only 1 (but somewhere inside, the code does know that there are 2 curves, since the line color is pink) – rm -rf Mar 11 '13 at 23:27 • Both legends show as expected in Version 9.0.0 (on Windows Vista 64 bit) – kglr Mar 11 '13 at 23:42 • And legends show as expected if you use this old method with autoLegend... – Jens Mar 11 '13 at 23:46 • @kguler So, as hypothesized, something added in the latest update breaks PlotLegends in this case. – Sjoerd C. de Vries Mar 13 '13 at 8:40 • If you have 10.0.0 installed would you please check to see if this was fixed in that release? – Mr.Wizard Jan 29 '15 at 10:24 The work-around posted by rcollyer can readily be generalized to work for an arbitrary number of parametric functions. With[{funcList = {{t, t}, {t, 2 t}, {t, t/2}}}, With[{n = Length@funcList}, Legended[ParametricPlot[funcList, {t, 0, 1}], LineLegend[(ColorData[1][#])& /@ #, #]& @ Range @ n]]] To flesh out your method, you can do the following Legended[ ParametricPlot[{{t,t},{t,2 t}},{t,0,1}], LineLegend[{ColorData[1][1], ColorData[1][2]}, {1, 2}] ] which, as noted, requires you to set the color information by yourself. This is the most straightforward workaround. Incidentally, it is often cleaner to look at the InputForm instead of the FullForm as things like List are not fully expanded. For this, I often use something like this, Plot[{x^2, x^3}, {x, 0, 1}, PlotLegends -> Automatic] /. Legended[_, {p_Placed, ___} \| p_Placed]:> InputForm@p[[1]] (* LineLegend[{Directive[Hue[0.67, 0.6, 0.6]], Directive[Hue[0.9060679774997897, 0.6, 0.6]]}, {1, 2}, LegendLayout -> "Column" ] *) • Apparently, Wolfram tech support was wrong when they told me "there is no good way of working around the problem". – m_goldberg Mar 13 '13 at 4:12 • @m_goldberg This was the line of solution I was hinting at in my question. – Sjoerd C. de Vries Mar 13 '13 at 8:33 • @rcollyer I have the habit of using FullForm because that's the form that will be used in pattern matching, but I agree that InputForm is often much clearer and is preferable when one just wants to see what's going on. – Sjoerd C. de Vries Mar 13 '13 at 8:35 • @m_goldberg don't necessarily blame them, I've been staring at this particular code for quite a while, so I have a good feel for how it works. – rcollyer Mar 13 '13 at 11:43 • @SjoerdC.deVries I had the same habit until I had to start looking at Graphics objects. FullForm would drive you insane pretty quickly. – rcollyer Mar 13 '13 at 11:46 I queried Wolfram technical support on this issue. Here is their reply: Our developers are aware that PlotLegends is currently not working well with ParametricPlot. This problem should be fixed in a future version of Mathematica. Unfortunately, there is no good way of working around the problem with PlotLegends at the current time. If you really need a legend with a ParametricPlot, you might have to use the older PlotLegend function, which is still available in Mathematica. • Seems to be fixed in Version 10!? -> link – Phab Jan 29 '15 at 8:05 As noted by Phab the example in the question works in Mathematica 10.0.2: ParametricPlot[{{t, t}, {t, 2 t}}, {t, 0, 1}, PlotLegends -> LineLegend[{1, 2}]] • I arrived at this thread after observing a similar problem in 10.1.0. PlotLegends->Automatic gives up after 15 curves. Fortunately, I can still make the legends show correctly with PlotLegends->LineLegend[Range[n]] for n at least as big as 26. – djphd Dec 7 '15 at 20:20 • @djphd Indeed, that is a known issue. The limitation is not arbitrarily fifteen, but rather the number of colors in the style. Please see (66057) for more. – Mr.Wizard Dec 8 '15 at 1:05	{}
## anonymous one year ago A rectangle is 68 inches by 42 inches. Find the angles that a diagonal makes with each side of the rectangle. 1. mathstudent55 Have you tried drawing it? 2. anonymous ^^ 3. anonymous Yes 4. anonymous I don't understand it 5. zpupster \|dw:1444173782144:dw\| use $tanx=\frac{ Opp }{ Adj }$	{}
#bayes #programming #r #statistics In general, Bayesian analysis of data follows these steps: 1. Identify the data relevant to the research questions. What are the measurement scales of the data? Which data variables are to be predicted, and which data variables are supposed to act as predictors? 2. Define a descriptive model for the relevant data. The mathematical form and its parameters should be meaningful and appropriate to the theoretical purposes of the analysis. 3. Specify a prior distribution on the parameters. The prior must pass muster with the audience of the analysis, such as skeptical scientists. 4. Use Bayesian inference to re-allocate credibility across parameter values. Interpret the posterior distribution with respect to theoretically meaningful issues (assuming that the model is a reasonable description of the data; see next step). 5. Check that the posterior predictions mimic the data with reasonable accuracy (i.e., conduct a “posterior predictive check”). If not, then consider a different descriptive model If you want to change selection, open document below and click on "Move attachment" #### pdf owner: shihabdider - (no access) - Doing Bayesian Data Analysis, p27	{}
## Friday, September 17, 2021 ### Applications are open: 2022 summer school on stats methods for ling and psych Applications are now open for the sixth SMLP summer school, to be held in person (hopefully) in the Griebnitzsee campus of the University of Potsdam, Germany, 12-16 Sept 2022. Apply here: https://vasishth.github.io/smlp2022/ ## Saturday, August 14, 2021 ### SAFAL 2: The Second South Asian Forum on the Acquisition and Processing of Language (30-31 August, 2021) SAFAL 2: The Second South Asian Forum on the Acquisition and Processing of Language (30-31 August, 2021 The first South Asian Forum on the Acquisition and Processing of Language (SAFAL) highlighted the need to provide a platform for showcasing and discussing acquisition and processing research in the context of South Asian languages. The second edition aims to build on this endeavour. Following the first edition, the Second South Asian Forum on the Acquisition and Processing of Language (SAFAL) aims to provide a platform to exchange research on sentence processing, verbal semantics, computational modeling, corpus-based psycholinguistics, neurobiology of language, and child language acquisition, among others, in the context of the subcontinent's linguistic landscape. Invited speakers: Sakshi Bhatia is an Assistant Professor of Linguistics at the Central University of Rajasthan. Her research areas include syntax, psycholinguistics and the syntax-psycholinguistics interface Kamal Choudhary is an Assistant Professor of Linguistics at the Indian Institute of Technology Ropar. His research areas include neurobiology of language and syntactic typology. ## Friday, August 13, 2021 ### A common mistake in psychology and psycholinguistics: Part 2 A Common Mistake in Data Analysis (in Psychology/Linguistics): Subsetting data to carry out nested analyses (Part 2 of 2) # tl;dr Subsetting data to carry out pairwise comparisons within one level of a 2-level or 3-level factor can lead to misleading results. This happens because by subsetting, we inadvertently drop variance components. # Data and code for both posts The data and code for both posts can be accessed from: https://github.com/vasishth/SubsettingProblem # Introduction A very common mistake I see in psycholinguistics and psychology papers is subsetting the data to carry out an analysis. In this post, I will discuss the second of two examples; the first appeared here. It is worth repeating that in both examples, there is absolutely no implication that the authors did anything dishonest—they did their analyses in good faith. The broader problem is that in psychology and linguistics, we are rarely taught much about data analysis. We usually learn a canned cookbook style of analysis. I am pretty sure I have made this mistake in the past—I remember doing this kind of subsetting; this was before I had learnt how to do contrast coding. I also want to point out what a big deal it is that the authors released their data. This usually happens in only about 25% of the cases, in my experience. This estimate is remarkably close to that from published estimates on the willingness of researchers to release data. So, kudos to these researchers for being among the top 25%. So here is the second example. I originally analyzed this data to create some exercises for my students, but the theoretical questions addressed in this paper are inherently interesting for psycholinguists. # Example 2: Fedorenko et al 2006, in the Journal of Memory and Language The paper we consider here is: Fedorenko, E., Gibson, E., & Rohde, D. (2006). The nature of working memory capacity in sentence comprehension: Evidence against domain-specific working memory resources. Journal of memory and language, 54(4), 541-553. This paper has been cited some 267 times at the moment of writing this post, according to google scholar. It appears in one of our top journals, Journal of Memory and Language (JML), which apparently has an impact factor of 4.014 (I like the precision to three decimal places! :) according to google. I believe this is high. By comparison, Cognition seems to be only at 3.537. The central claim that the paper makes is summarized nicely in the abstract; I highlight the relevant part of the abstract below: This paper reports the results of a dual-task experiment which investigates the nature of working memory resources used in sentence comprehension. Participants read sentences of varying syntactic complexity (containing subject- and object-extracted relative clauses) while remembering one or three nouns (similar to or dissimilar from the sentence-nouns). A significant on-line interaction was found between syntactic complexity and similarity between the memory-nouns and the sentence-nouns in the three memory-nouns conditions, such that the similarity between the memory-nouns and the sentence-nouns affected the more complex object-extracted relative clauses to a greater extent than the less complex subject-extracted relative clauses. These results extend Gordon, Hendrick, and Levine’s (2002) report of a trend of such an interaction. The results argue against the domain-specific view of working memory resources in sentence comprehension (Caplan & Waters, 1999). What I will show below is that there is no evidence for the claimed interaction that is highlighted above. The only solid result in this paper is that object relatives are harder to process than subject relatives. Obviously, this is actually interesting for psycholinguists; but it is not a newsworthy result and if that had been the main claim, JML would have rejected this paper (although Cognition seems to love papers showing that object relatives are easier or harder process than subject relatives ;). I want to stress at the outset that although I show that there is no evidence for the claimed interaction in the abstract above, this does not mean there is no interaction. There might still be. We just don’t know either way from this work. # The original paper and the main claims The data are from this paper: Fedorenko, E., Gibson, E., & Rohde, D. (2006). The nature of working memory capacity in sentence comprehension: Evidence against domain-specific working memory resources. Journal of Memory and Language, 54(4), 541-553. Ev was kind enough to give me the data; many thanks to her for this. ## Design The study is a self-paced reading experiment, with a 2x2x2 design, the examples are as follows: 1. Factor: One noun in memory set or three nouns easy: Joel hard: Joel-Greg-Andy 1. Factor: Noun type, either proper name or occupation name: Joel-Greg-Andy occ: poet-cartoonist-voter 1. Factor: Relative clause type (two versions, between subjects but within items—we ignore this detail in this blog post): 1. Subject-extracted, version 1: The physician \| who consulted the cardiologist \| checked the files \| in the office. 2. Subject-extracted, version 2: The cardiologist \| who consulted the physician \| checked the files \| in the office. 1. Object-extracted, version 1: The physician \| who the cardiologist consulted \| checked the files \| in the office. 2. Object-extracted, version 2: The cardiologist \| who the physician consulted \| checked the files \| in the office. As hinted above, this design is not exactly 2x2x2; there is also a within items manipulation of relative clause version. That detail was ignored in the published analysis, so we ignore it here as well. ## The main claims in the paper The main claim of the paper is laid out in the abstract above, but I repeat it below: “A significant on-line interaction was found between syntactic complexity and similarity between the memory-nouns and the sentence-nouns in the three memory-nouns conditions, such that the similarity between the memory-nouns and the sentence-nouns affected the more complex object-extracted relative clauses to a greater extent than the less complex subject-extracted relative clauses.” The General Discussion explains this further. I highlight the important parts. ## The statistical evidence in the paper for the claim Table 3 in the paper (last line) reports the critical interaction claimed for the hard-load conditions (I added the p-values, which were not provided in the table): “Synt x noun type [hard load]: F1(1,43)=6.58, p=0.014;F2(1,31)=6.79, p=0.014”. There is also a minF’ provided: $minF'(1,72)=3.34, p=0.07$. I also checked the minF’ using the standard formula from a classic paper by Herb Clark: Clark, H. H. (1973). The language-as-fixed-effect fallacy: A critique of language statistics in psychological research. Journal of Verbal Learning and Verbal Behavior, 12(4), 335-359. [Aside: The journal “Journal of Verbal Learning and Verbal Behavior” is now called Journal of Memory and Language.] Here is the minF’ function: minf <- function(f1,f2,n1,n2){ fprime <- (f1f2)/(f1+f2) n <- round(((f1+f2)(f1+f2))/(((f1f1)/n2)+((f2f2)/n1))) pval<-pf(fprime,1,n,lower.tail=FALSE) return(paste("minF(1,",n,")=",round(fprime,digits=2),"p-value=",round(pval,2),"crit=",round(qf(.95,1,n)))) } And here is the p-value for the minF’ calculation: minf(6.58,6.79,43,31) ## [1] "minF(1, 72 )= 3.34 p-value= 0.07 crit= 4" The first thing to notice here is that even the published claim (about an RC type-noun type interaction in the hard load condition) is not statistically significant. I have heard through the grapevine that JML allows the authors to consider an effect as statistically significant even if the MinF’ is not significant. Apparently the reason is that the MinF’ calculation is considered to be conservative. But if they are going to ignore it, why the JML wants people to report a MinF’ is a mystery to me. Declaring a non-significant result to be significant is a very common situation in the Journal of Memory and Language, and other journals. Some other JML papers I know of that report a non-significant minF’ as their main finding in the paper, but still act as if they have significance are: • Dillon et al 2013: “For total times this analysis revealed a three-way interaction that was significant by participants and marginal by items ($F_1(1,39)=8.0,...,p=<0.01; F_1(1,47)=4.0,...,p=<0.055; minF'(1,81)=2,66,p=0.11$)”. • Van Dyke and McElree 2006: “The interaction of LoadxInterference was significant, both by subjects and by items. … minF’(1,90)=2.35, p = 0.13.” Incidentally, neither of the main claims in these two studies replicated (which is not the same thing as saying that their claims are false—we don’t know whether they are false or true): • Replication attempt of Dillon et al 2013: Lena A. Jäger, Daniela Mertzen, Julie A. Van Dyke, and Shravan Vasishth. Interference patterns in subject-verb agreement and reflexives revisited: A large-sample study. Journal of Memory and Language, 111, 2020. • Replication attempt of Van Dyke et al 2006: Daniela Mertzen, Anna Laurinavichyute, Brian W. Dillon, Ralf Engbert, and Shravan Vasishth. Is there cross-linguistic evidence for proactive cue-based retrieval interference in sentence comprehension? Eye-tracking data from English, German and Russian. submitted, 2021 The second paper was rejected by JML, by the way. # Data analysis of the Fedorenko et al 2006 study ## Download and install library from here: ## https://github.com/vasishth/lingpsych library(lingpsych) data("df_fedorenko06") head(df_fedorenko06) ## rctype nountype load item subj region RT ## 2 obj name hard 16 1 2 4454 ## 6 subj name hard 4 1 2 5718 ## 10 subj name easy 10 1 2 2137 ## 14 subj name hard 20 1 2 903 ## 18 subj name hard 12 1 2 1837 ## 22 subj occ hard 19 1 2 1128 As explained above, we have a 2x2x2 design: rctype [obj,subj] x nountype [name, occupation] x load [hard,easy]. Region 2 is the critical region, the entire relative clause, and RT is the reading time for this region. Subject and item columns are self-explanatory. We notice immediately that something is wrong in this data frame. Compare subject 1 vs the other subjects: ## something is wrong here: xtabs(~subj+item,df_fedorenko06) ## item ## subj 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 ## 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ## 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 12 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 13 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 15 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 17 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 18 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 21 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 22 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 23 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 24 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 26 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 28 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 29 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 30 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 31 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 32 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 33 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 34 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 35 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 36 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 37 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 38 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 39 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 40 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 41 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 42 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 43 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 44 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 46 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## 47 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ## item ## subj 29 30 31 32 ## 1 2 2 2 2 ## 2 1 1 1 1 ## 3 1 1 1 1 ## 4 1 1 1 1 ## 5 1 1 1 1 ## 6 1 1 1 1 ## 7 1 1 1 1 ## 8 1 1 1 1 ## 9 1 1 1 1 ## 11 1 1 1 1 ## 12 1 1 1 1 ## 13 1 1 1 1 ## 14 1 1 1 1 ## 15 1 1 1 1 ## 16 1 1 1 1 ## 17 1 1 1 1 ## 18 1 1 1 1 ## 19 1 1 1 1 ## 20 1 1 1 1 ## 21 1 1 1 1 ## 22 1 1 1 1 ## 23 1 1 1 1 ## 24 1 1 1 1 ## 25 1 1 1 1 ## 26 1 1 1 1 ## 28 1 1 1 1 ## 29 1 1 1 1 ## 30 1 1 1 1 ## 31 1 1 1 1 ## 32 1 1 1 1 ## 33 1 1 1 1 ## 34 1 1 1 1 ## 35 1 1 1 1 ## 36 1 1 1 1 ## 37 1 1 1 1 ## 38 1 1 1 1 ## 39 1 1 1 1 ## 40 1 1 1 1 ## 41 1 1 1 1 ## 42 1 1 1 1 ## 43 1 1 1 1 ## 44 1 1 1 1 ## 46 1 1 1 1 ## 47 1 1 1 1 Subject 1 saw all items twice! There was apparently a bug in the code, or maybe two different subjects were mis-labeled as subject 1. Hard to know. This is one more reason why one should not load and analyze data immediately; always check that the design was correctly implemented. We could remove subject 1 from this data, leaving us with 43 subjects: df_fedorenko06<-subset(df_fedorenko06,subj!=1) sort(unique(df_fedorenko06$subj)) length(unique(df_fedorenko06$subj)) But we won’t do this yet. ## A full 2x2x2 ANOVA analysis: First, set up an ANOVA style coding for the data, in lmer: library(lme4) ## Loading required package: Matrix df_fedorenko06$noun<-ifelse(df_fedorenko06$nountype=="name",1/2,-1/2) df_fedorenko06$ld<-ifelse(df_fedorenko06$load=="hard",1/2,-1/2) df_fedorenko06$rc<-ifelse(df_fedorenko06$rctype=="obj",1/2,-1/2) df_fedorenko06$nounxld<-df_fedorenko06$noundf_fedorenko06$ld2 df_fedorenko06$nounxrc<-df_fedorenko06$noundf_fedorenko06$rc2 df_fedorenko06$ldxrc<-df_fedorenko06$lddf_fedorenko06$rc2 df_fedorenko06$nounxldxrc<-df_fedorenko06$noundf_fedorenko06$lddf_fedorenko06$rc4 I am using effect coding here ($\pm 1/2$), so that every estimate reflects the estimated difference in means for the particular comparison being made. When multiplying to get an interaction, I have to multiply with a factor of 2 or 4 to get the effect coding back; if I don’t do that I will get $\pm 1/4$ or $\pm 1/8$. I’m using effect coding because I was interested in getting an estimate of the relative clause effect for our Bayesian textbook’s chapter 6, on prior self-elicitation. The data frame now has new columns representing the contrast coding: head(df_fedorenko06) ## rctype nountype load item subj region RT noun ld rc nounxld nounxrc ## 2 obj name hard 16 1 2 4454 0.5 0.5 0.5 0.5 0.5 ## 6 subj name hard 4 1 2 5718 0.5 0.5 -0.5 0.5 -0.5 ## 10 subj name easy 10 1 2 2137 0.5 -0.5 -0.5 -0.5 -0.5 ## 14 subj name hard 20 1 2 903 0.5 0.5 -0.5 0.5 -0.5 ## 18 subj name hard 12 1 2 1837 0.5 0.5 -0.5 0.5 -0.5 ## 22 subj occ hard 19 1 2 1128 -0.5 0.5 -0.5 -0.5 0.5 ## ldxrc nounxldxrc ## 2 0.5 0.5 ## 6 -0.5 -0.5 ## 10 0.5 0.5 ## 14 -0.5 -0.5 ## 18 -0.5 -0.5 ## 22 -0.5 0.5 Display the estimated means in each condition, just to get some idea of what’s coming. round(with(df_fedorenko06,tapply(RT,nountype,mean))) ## name occ ## 1636 1754 round(with(df_fedorenko06,tapply(RT,load,mean))) ## easy hard ## 1608 1782 round(with(df_fedorenko06,tapply(RT,rctype,mean))) ## obj subj ## 1909 1482 The authors report raw and residual RTs, but we analyze raw RTs, and then analyze log RTs. There’s actually no point in computing residual RTs (I will discuss this at some point in a different blog post), so we ignore the residuals analysis, although nothing much will change in my main points below even if we were to use residual RTs. I fit the most complex model I could. If I were a Bayesian (which I am), I would have done the full Monty just because I can, but never mind. ## raw RTs m1<-lmer(RT~noun+ld+rc+nounxld + nounxrc + ldxrc + nounxldxrc+(1+ld+rc+ nounxrc + nounxldxrc\|\|subj) + (1+ld+rc+ nounxrc + nounxldxrc\|\|item),df_fedorenko06) #summary(m1) ## log RTs m2<-lmer(log(RT)~noun+ld+rc+nounxld + nounxrc + ldxrc + nounxldxrc+(1+ld+ nounxrc\|\|subj) + (1+ld+rc\|\|item),df_fedorenko06) #summary(m2) Here is the summary output from the two models: ## raw RTs: summary(m1)$coefficients ## Estimate Std. Error t value ## (Intercept) 1675.79757 125.88880 13.3117286 ## noun -116.07053 58.60273 -1.9806337 ## ld 155.14812 112.39971 1.3803249 ## rc 422.73102 77.84215 5.4306182 ## nounxld -103.39623 58.60364 -1.7643312 ## nounxrc -110.87780 61.73703 -1.7959692 ## ldxrc -23.47607 58.62151 -0.4004685 ## nounxldxrc -95.76489 66.95281 -1.4303340 ## log RTs: summary(m2)$coefficients ## Estimate Std. Error t value ## (Intercept) 7.20041271 0.06585075 109.3444271 ## noun -0.03127831 0.02457298 -1.2728741 ## ld 0.01532113 0.04530999 0.3381403 ## rc 0.21828914 0.02812620 7.7610591 ## nounxld -0.04645901 0.02457339 -1.8906222 ## nounxrc -0.02132791 0.02564043 -0.8318078 ## ldxrc -0.03704690 0.02457244 -1.5076607 ## nounxldxrc -0.01944255 0.02462971 -0.7893945 Both the models only show that object relatives are harder to process than subject relatives. No other main effect or interactions pan out. This (non-)result is actually consistent with what’s reported in the paper. For the story about the hard load conditions (discussed below) to hold up, a prerequisite is that the nounxldxrc interaction be significant, but it is not. ## Subset analyses using repeated measures ANOVA/paired t-test The Syn x noun interaction in the hard load conditions reported in the paper as the main result was computed by first subsetting the data such that we have only the hard load conditions’ data. This leads to a 2x2 design. Let’s see what happens when we do this analysis by subsetting the data. We have dropped half the data: dim(df_fedorenko06) ## [1] 1440 14 fed06hard<-subset(df_fedorenko06,load=="hard") dim(fed06hard) ## [1] 720 14 (1440-720)/1440 ## [1] 0.5 Let’s first do a repeated measures ANOVA (=paired t-tests) on raw RTs by subject and by item, because that is what Fedorenko et al did (well, they used residual RTs, but nothing will change if we use that dependent measure). ## aggregate by subject: fedsubjRC<-aggregate(RT~subj+rctype,mean,data=fed06hard) head(fedsubjRC) ## subj rctype RT ## 1 1 obj 3387.500 ## 2 2 obj 892.125 ## 3 3 obj 1373.000 ## 4 4 obj 1738.625 ## 5 5 obj 7074.000 ## 6 6 obj 3119.625 t.test(RT~rctype,paired=TRUE,fedsubjRC) ## ## Paired t-test ## ## data: RT by rctype ## t = 3.5883, df = 43, p-value = 0.0008469 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 176.7331 630.3151 ## sample estimates: ## mean of the differences ## 403.5241 fedsubjN<-aggregate(RT~subj+nountype,mean,data=fed06hard) head(fedsubjN) ## subj nountype RT ## 1 1 name 2971.500 ## 2 2 name 928.875 ## 3 3 name 1161.500 ## 4 4 name 1414.125 ## 5 5 name 4472.250 ## 6 6 name 1831.875 t.test(RT~nountype,paired=TRUE,fedsubjN) ## ## Paired t-test ## ## data: RT by nountype ## t = -2.6432, df = 43, p-value = 0.01141 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -383.91794 -51.61331 ## sample estimates: ## mean of the differences ## -217.7656 ## interaction: diff between obj and subj in name vs in occ conditions: RCocc<-aggregate(RT~subj+rctype,mean,data=subset(fed06hard,nountype=="occ")) diff_occ<-subset(RCocc,rctype=="obj")$RT-subset(RCocc,rctype=="subj")$RT RCname<-aggregate(RT~subj+rctype,mean,data=subset(fed06hard,nountype=="name")) diff_name<-subset(RCname,rctype=="obj")$RT-subset(RCname,rctype=="subj")$RT ## by-subject interaction: t.test(diff_occ,diff_name,paired=TRUE) ## ## Paired t-test ## ## data: diff_occ and diff_name ## t = 2.131, df = 43, p-value = 0.03885 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 21.71398 787.90534 ## sample estimates: ## mean of the differences ## 404.8097 One can of course do this with an ANOVA function: fedsubjANOVA<-aggregate(RT~subj+rctype+nountype,mean,data=fed06hard) head(fedsubjANOVA) ## subj rctype nountype RT ## 1 1 obj name 3030.25 ## 2 2 obj name 962.50 ## 3 3 obj name 1087.50 ## 4 4 obj name 1687.00 ## 5 5 obj name 5006.75 ## 6 6 obj name 1064.50 library(rstatix) ## ## Attaching package: 'rstatix' ## The following object is masked from 'package:stats': ## ## filter subj_anova_hard<-anova_test(data = fedsubjANOVA, dv = RT, wid = subj, within = c(rctype,nountype) ) get_anova_table(subj_anova_hard) ## ANOVA Table (type III tests) ## ## Effect DFn DFd F p p<.05 ges ## 1 rctype 1 43 12.876 0.000847 0.031 ## 2 nountype 1 43 6.986 0.011000 * 0.009 ## 3 rctype:nountype 1 43 4.541 0.039000 * 0.008 We see a significant interaction by subjects, as reported in the paper. By items: ## aggregate by item: feditemRC<-aggregate(RT~item+rctype,mean,data=fed06hard) head(feditemRC) ## item rctype RT ## 1 1 obj 1978.545 ## 2 2 obj 2076.667 ## 3 3 obj 3483.273 ## 4 4 obj 2489.000 ## 5 5 obj 1322.091 ## 6 6 obj 1268.800 t.test(RT~rctype,paired=TRUE,feditemRC) ## ## Paired t-test ## ## data: RT by rctype ## t = 3.9827, df = 31, p-value = 0.0003833 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 194.5260 602.8663 ## sample estimates: ## mean of the differences ## 398.6962 feditemN<-aggregate(RT~item+nountype,mean,data=fed06hard) head(feditemN) ## item nountype RT ## 1 1 name 1573.727 ## 2 2 name 1566.455 ## 3 3 name 1580.364 ## 4 4 name 2808.667 ## 5 5 name 1251.000 ## 6 6 name 1261.818 t.test(RT~nountype,paired=TRUE,feditemN) ## ## Paired t-test ## ## data: RT by nountype ## t = -1.6978, df = 31, p-value = 0.09957 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -477.39928 43.65591 ## sample estimates: ## mean of the differences ## -216.8717 ## interaction: diff between obj and item in name vs in occ conditions: RCocc<-aggregate(RT~item+rctype,mean,data=subset(fed06hard,nountype=="occ")) diff_occ<-subset(RCocc,rctype=="obj")$RT-subset(RCocc,rctype=="subj")$RT RCname<-aggregate(RT~item+rctype,mean,data=subset(fed06hard,nountype=="name")) diff_name<-subset(RCname,rctype=="obj")$RT-subset(RCname,rctype=="subj")$RT ## by-item interaction: t.test(diff_occ,diff_name,paired=TRUE) ## ## Paired t-test ## ## data: diff_occ and diff_name ## t = 2.0274, df = 31, p-value = 0.05129 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -2.45258 823.20437 ## sample estimates: ## mean of the differences ## 410.3759 Using ANOVA: feditemANOVA<-aggregate(RT~item+rctype+nountype,mean,data=fed06hard) head(feditemANOVA) ## item rctype nountype RT ## 1 1 obj name 2206.000 ## 2 2 obj name 1946.667 ## 3 3 obj name 1966.167 ## 4 4 obj name 2910.600 ## 5 5 obj name 1011.500 ## 6 6 obj name 1496.200 item_anova_hard<-anova_test(data = feditemANOVA, dv = RT, wid = item, within = c(rctype,nountype) ) get_anova_table(item_anova_hard) ## ANOVA Table (type III tests) ## ## Effect DFn DFd F p p<.05 ges ## 1 rctype 1 31 14.802 0.000557 * 0.085 ## 2 nountype 1 31 2.867 0.100000 0.028 ## 3 rctype:nountype 1 31 4.110 0.051000 0.023 Notice that even if the by-subjects and by-items analyses are significant, the MinF’ is not. As Clark points out in his paper, MinF’ is critical to compute in order to draw inferences about the hypothesis test. Since $F=t^2$, we can compute the MinF’. F1<-2.13^2 F2<-2.03^2 minf(F1,F2,43,31) ## [1] "minF(1, 71 )= 2.16 p-value= 0.15 crit= 4" We can do this MinF’ calculation with the ANOVA output as well: minf(4.541,4.11,43,31) ## [1] "minF(1, 71 )= 2.16 p-value= 0.15 crit= 4" We get exactly the same MinF’ in both the t-tests and ANOVA because they are doing the same test. ## The upshot of the ANOVA/t-test analyses using subsetted data The higher-order interaction is not significant. This diverges from the main claim in the paper. Now, let’s do the same by-subjects and items analyses using logRTs: fed06hard$logRT<-log(fed06hard$RT) ## aggregate by subject: fedsubjRC<-aggregate(logRT~subj+rctype,mean,data=fed06hard) head(fedsubjRC) ## subj rctype logRT ## 1 1 obj 7.980754 ## 2 2 obj 6.710152 ## 3 3 obj 7.125770 ## 4 4 obj 7.179976 ## 5 5 obj 8.625057 ## 6 6 obj 7.569662 t.test(logRT~rctype,paired=TRUE,fedsubjRC) ## ## Paired t-test ## ## data: logRT by rctype ## t = 4.6431, df = 43, p-value = 3.227e-05 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 0.1024094 0.2596785 ## sample estimates: ## mean of the differences ## 0.1810439 fedsubjN<-aggregate(logRT~subj+nountype,mean,data=fed06hard) head(fedsubjN) ## subj nountype logRT ## 1 1 name 7.835673 ## 2 2 name 6.740148 ## 3 3 name 6.988723 ## 4 4 name 7.167295 ## 5 5 name 8.249557 ## 6 6 name 7.142238 t.test(logRT~nountype,paired=TRUE,fedsubjN) ## ## Paired t-test ## ## data: logRT by nountype ## t = -2.2556, df = 43, p-value = 0.02924 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -0.151468895 -0.008471231 ## sample estimates: ## mean of the differences ## -0.07997006 ## interaction: diff between obj and subj in name vs in occ conditions: RCocc<-aggregate(logRT~subj+rctype,mean,data=subset(fed06hard,nountype=="occ")) diff_occ<-subset(RCocc,rctype=="obj")$logRT-subset(RCocc,rctype=="subj")$logRT RCname<-aggregate(logRT~subj+rctype,mean,data=subset(fed06hard,nountype=="name")) diff_name<-subset(RCname,rctype=="obj")$logRT-subset(RCname,rctype=="subj")$logRT ## by-subject interaction: t.test(diff_occ,diff_name,paired=TRUE) ## ## Paired t-test ## ## data: diff_occ and diff_name ## t = 1.0226, df = 43, p-value = 0.3122 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -0.07614228 0.23279035 ## sample estimates: ## mean of the differences ## 0.07832404 Now the by-subjects t-test for the interaction is nowhere near significant (cf the raw RTs analysis above). This means that there were some extreme values in the by-subjects data driving the effect. By items: ## aggregate by item: feditemRC<-aggregate(logRT~item+rctype,mean,data=fed06hard) head(feditemRC) ## item rctype logRT ## 1 1 obj 7.421371 ## 2 2 obj 7.339790 ## 3 3 obj 7.562565 ## 4 4 obj 7.590270 ## 5 5 obj 7.016257 ## 6 6 obj 7.031505 t.test(logRT~rctype,paired=TRUE,feditemRC) ## ## Paired t-test ## ## data: logRT by rctype ## t = 3.6988, df = 31, p-value = 0.0008374 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 0.08105886 0.28033104 ## sample estimates: ## mean of the differences ## 0.180695 feditemN<-aggregate(logRT~item+nountype,mean,data=fed06hard) head(feditemN) ## item nountype logRT ## 1 1 name 7.216759 ## 2 2 name 6.991090 ## 3 3 name 7.192081 ## 4 4 name 7.707697 ## 5 5 name 7.018699 ## 6 6 name 6.981104 t.test(logRT~nountype,paired=TRUE,feditemN) ## ## Paired t-test ## ## data: logRT by nountype ## t = -1.4239, df = 31, p-value = 0.1645 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -0.18854611 0.03351487 ## sample estimates: ## mean of the differences ## -0.07751562 ## interaction: diff between obj and item in name vs in occ conditions: RCocc<-aggregate(logRT~item+rctype,mean,data=subset(fed06hard,nountype=="occ")) diff_occ<-subset(RCocc,rctype=="obj")$logRT-subset(RCocc,rctype=="subj")$logRT RCname<-aggregate(logRT~item+rctype,mean,data=subset(fed06hard,nountype=="name")) diff_name<-subset(RCname,rctype=="obj")$logRT-subset(RCname,rctype=="subj")$logRT ## by-item interaction: t.test(diff_occ,diff_name,paired=TRUE) ## ## Paired t-test ## ## data: diff_occ and diff_name ## t = 0.93122, df = 31, p-value = 0.3589 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -0.09504043 0.25475215 ## sample estimates: ## mean of the differences ## 0.07985586 Here, too, the interaction by items is not significant. Conclusion: the MinF’ values show that there is no evidence for an interaction in the hard conditions, regardless of whether we use raw or log RTs. But there is actually more to this story, as I show below with linear mixed models analyses. ## Incorrect and correct lmer analysis Now, we do the same analyses as above, but in lmer instead of paired t-tests, and by subsetting the data as Fedorenko did. In order to compare this model (m4) with the correct analysis (model m6 below), we change the coding to be $\pm 1$ instead of $\pm 1/2$. ## 2x2 anova by doing subsetting: this is the main result in the paper fed06hard<-subset(df_fedorenko06,load=="hard") fed06hard$noun<-fed06hard$noun2 fed06hard$rc<-fed06hard$rc2 fed06hard$nounxrc<-fed06hard$nounxrc2 fed06hard<-fed06hard[,c("subj","item","RT","noun","rc","nounxrc")] head(fed06hard) ## subj item RT noun rc nounxrc ## 2 1 16 4454 1 1 1 ## 6 1 4 5718 1 -1 -1 ## 14 1 20 903 1 -1 -1 ## 18 1 12 1837 1 -1 -1 ## 22 1 19 1128 -1 -1 1 ## 26 1 31 2742 -1 1 -1 Here is the analysis using subsetting (this is incorrect): m4<-lmer(RT~noun+rc+nounxrc+(1+rc+nounxrc\|\|subj) + (1+noun+nounxrc\|\|item),fed06hard) summary(m4) ## Linear mixed model fit by REML ['lmerMod'] ## Formula: RT ~ noun + rc + nounxrc + ((1 \| subj) + (0 + rc \| subj) + (0 + ## nounxrc \| subj)) + ((1 \| item) + (0 + noun \| item) + (0 + ## nounxrc \| item)) ## Data: fed06hard ## ## REML criterion at convergence: 12401.9 ## ## Scaled residuals: ## Min 1Q Median 3Q Max ## -3.0317 -0.4256 -0.1341 0.2131 11.6983 ## ## Random effects: ## Groups Name Variance Std.Dev. ## subj (Intercept) 915329 956.73 ## subj.1 rc 42511 206.18 ## subj.2 nounxrc 1570 39.62 ## item (Intercept) 56296 237.27 ## item.1 noun 4123 64.21 ## item.2 nounxrc 11841 108.82 ## Residual 1540203 1241.05 ## Number of obs: 720, groups: subj, 44; item, 32 ## ## Fixed effects: ## Estimate Std. Error t value ## (Intercept) 1752.09 157.25 11.142 ## noun -110.39 47.65 -2.317 ## rc 200.80 55.86 3.595 ## nounxrc -103.48 50.56 -2.047 ## ## Correlation of Fixed Effects: ## (Intr) noun rc ## noun 0.000 ## rc 0.000 0.000 ## nounxrc 0.000 0.000 0.000 summary(m4)$coefficients ## Estimate Std. Error t value ## (Intercept) 1752.0945 157.24556 11.142410 ## noun -110.3885 47.64559 -2.316866 ## rc 200.8043 55.85609 3.595031 ## nounxrc -103.4786 50.56305 -2.046527 The above analysis shows a significant interaction. But this analysis is incorrect because it’s dropping half the data, which contributes to misestimating the variance components. Next, we will do the above analysis using all the data (no subsetting to load==hard conditions). We do this by defining nested contrasts The design is: ## Nested analysis: ## load e e e e h h h h ## noun o o n n o o n n ## rctyp s o s o s o s o # a b c d e f g h #--------------------------------------- #load -1 -1 -1 -1 1 1 1 1 #nn_easy -1 -1 1 1 0 0 0 0 #rc_easy -1 1 -1 1 0 0 0 0 #nn_hard 0 0 0 0 -1 -1 1 1 #rc_hard 0 0 0 0 -1 1 -1 1 Define the nested coding. This time I use $\pm 1$ coding; this is just in order to facilitate quick model specification by multiplying main effects in the lmer equation. I could have used $\pm 1/2$ and the only thing that would give me is that the estimated coefficients would reflect the difference in means, but I would need to define interactions as separate columsn (as done above). The statistical results will not change as a function of these two coding schemes. df_fedorenko06$ld<-ifelse(df_fedorenko06$load=="hard",1,-1) df_fedorenko06$nn_easy<-ifelse(df_fedorenko06$load=="easy" & df_fedorenko06$nountype=="occ",-1, ifelse(df_fedorenko06$load=="easy" & df_fedorenko06$nountype=="name",1,0)) df_fedorenko06$nn_hard<-ifelse(df_fedorenko06$load=="hard" & df_fedorenko06$nountype=="occ",-1, ifelse(df_fedorenko06$load=="hard" & df_fedorenko06$nountype=="name",1,0)) df_fedorenko06$rc_easy<-ifelse(df_fedorenko06$load=="easy" & df_fedorenko06$rctype=="subj",-1, ifelse(df_fedorenko06$load=="easy" & df_fedorenko06$rctype=="obj",1,0)) df_fedorenko06$rc_hard<-ifelse(df_fedorenko06$load=="hard" & df_fedorenko06$rctype=="subj",-1, ifelse(df_fedorenko06$load=="hard" & df_fedorenko06$rctype=="obj",1,0)) m6<-lmer(RT~ld + nn_hardrc_hard + nn_easyrc_easy + (1+ld + nn_hard:rc_hard \|\|subj) + (1+ld + nn_hard:rc_hard + nn_easy\|\|item),df_fedorenko06) summary(m6) ## Linear mixed model fit by REML ['lmerMod'] ## Formula: RT ~ ld + nn_hard rc_hard + nn_easy * rc_easy + ((1 \| subj) + ## (0 + ld \| subj) + (0 + nn_hard:rc_hard \| subj)) + ((1 \| item) + ## (0 + ld \| item) + (0 + nn_easy \| item) + (0 + nn_hard:rc_hard \| item)) ## Data: df_fedorenko06 ## ## REML criterion at convergence: 24455.1 ## ## Scaled residuals: ## Min 1Q Median 3Q Max ## -2.7050 -0.4408 -0.1223 0.2148 13.3538 ## ## Random effects: ## Groups Name Variance Std.Dev. ## subj (Intercept) 611484 781.97 ## subj.1 ld 79352 281.70 ## subj.2 nn_hard:rc_hard 20333 142.59 ## item (Intercept) 36175 190.20 ## item.1 ld 16443 128.23 ## item.2 nn_easy 1915 43.76 ## item.3 nn_hard:rc_hard 25223 158.82 ## Residual 1252751 1119.26 ## Number of obs: 1440, groups: subj, 44; item, 32 ## ## Fixed effects: ## Estimate Std. Error t value ## (Intercept) 1675.796 126.128 13.286 ## ld 77.641 56.535 1.373 ## nn_hard -110.046 41.745 -2.636 ## rc_hard 200.372 41.749 4.799 ## nn_easy -6.734 42.434 -0.159 ## rc_easy 224.046 41.749 5.366 ## nn_hard:rc_hard -103.208 54.818 -1.883 ## nn_easy:rc_easy -8.443 41.821 -0.202 ## ## Correlation of Fixed Effects: ## (Intr) ld nn_hrd rc_hrd nn_esy rc_esy nn_h:_ ## ld 0.000 ## nn_hard 0.000 0.000 ## rc_hard 0.000 0.000 0.000 ## nn_easy 0.000 0.000 0.000 0.000 ## rc_easy 0.000 0.000 0.000 0.000 0.000 ## nn_hrd:rc_h 0.000 0.000 0.000 0.000 0.000 0.000 ## nn_sy:rc_sy 0.000 0.000 0.000 0.000 0.000 0.000 0.000 #qqPlot(residuals(m6)) #summary(m6)$coefficients ## compare with a null model, without the ## nn_hard:rc_hard interaction: m6null<-lmer(RT~ld + nn_hard+rc_hard + nn_easyrc_easy +(1+ld + nn_hard:rc_hard \|\|subj) + (1+ld + nn_hard:rc_hard + nn_easy\|\|item),df_fedorenko06) anova(m6,m6null) ## refitting model(s) with ML (instead of REML) ## Data: df_fedorenko06 ## Models: ## m6null: RT ~ ld + nn_hard + rc_hard + nn_easy rc_easy + ((1 \| subj) + (0 + ld \| subj) + (0 + nn_hard:rc_hard \| subj)) + ((1 \| item) + (0 + ld \| item) + (0 + nn_easy \| item) + (0 + nn_hard:rc_hard \| item)) ## m6: RT ~ ld + nn_hard * rc_hard + nn_easy * rc_easy + ((1 \| subj) + (0 + ld \| subj) + (0 + nn_hard:rc_hard \| subj)) + ((1 \| item) + (0 + ld \| item) + (0 + nn_easy \| item) + (0 + nn_hard:rc_hard \| item)) ## npar AIC BIC logLik deviance Chisq Df Pr(>Chisq) ## m6null 15 24566 24645 -12268 24536 ## m6 16 24565 24649 -12266 24533 3.4482 1 0.06332 . ## --- ## Signif. codes: 0 '*' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 Removing subject 1 weakens the already non-significant interaction: m6a<-lmer(RT~ld + nn_hardrc_hard + nn_easyrc_easy +(1+ld + nn_hard:rc_hard \|\|subj) + (1+ld + nn_hard:rc_hard + nn_easy\|\|item),subset(df_fedorenko06,subj!=1)) #summary(m6a) #qqPlot(residuals(m6)) #summary(m6)$coefficients ## compare with a null model, without the ## nn_hard:rc_hard interaction: m6anull<-lmer(RT~ld + nn_hard+rc_hard + nn_easyrc_easy +(1+ld + nn_hard:rc_hard \|\|subj) + (1+ld + nn_hard:rc_hard + nn_easy\|\|item),subset(df_fedorenko06,subj!=1)) anova(m6a,m6anull) ## refitting model(s) with ML (instead of REML) ## Data: subset(df_fedorenko06, subj != 1) ## Models: ## m6anull: RT ~ ld + nn_hard + rc_hard + nn_easy * rc_easy + ((1 \| subj) + (0 + ld \| subj) + (0 + nn_hard:rc_hard \| subj)) + ((1 \| item) + (0 + ld \| item) + (0 + nn_easy \| item) + (0 + nn_hard:rc_hard \| item)) ## m6a: RT ~ ld + nn_hard * rc_hard + nn_easy * rc_easy + ((1 \| subj) + (0 + ld \| subj) + (0 + nn_hard:rc_hard \| subj)) + ((1 \| item) + (0 + ld \| item) + (0 + nn_easy \| item) + (0 + nn_hard:rc_hard \| item)) ## npar AIC BIC logLik deviance Chisq Df Pr(>Chisq) ## m6anull 15 23385 23463 -11677 23355 ## m6a 16 23384 23467 -11676 23352 3.0927 1 0.07864 . ## --- ## Signif. codes: 0 '*' 0.001 '' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Notice that the interaction (nn_hard:rc_hard) is gone once we take the full data into account. ## subsetted data: summary(m4)$coefficients[4,] ## Estimate Std. Error t value ## -103.478646 50.563049 -2.046527 ## full data: summary(m6)$coefficients[7,] ## Estimate Std. Error t value ## -103.208049 54.817661 -1.882752 What has changed is the standard error of the interaction (in m4 it is 51 ms, in m6 it is 55 ms); the increase in the SE is due to the increased number of variance components in the full model with the nested contrast coding. The subsetted analysis (m4, and the paired t-tests) are hiding important sources of variance and giving a misleading result. Here are the variance components in model m4 (the subsetted data): Random effects: Groups Name Variance Std.Dev. subj (Intercept) 915329 956.73 subj.1 rc 42511 206.18 subj.2 nounxrc 1570 39.62 item (Intercept) 56296 237.27 item.1 noun 4123 64.21 item.2 nounxrc 11841 108.82 Residual 1540203 1241.05 Here are the variance components in the full model: Random effects: Groups Name Variance Std.Dev. subj (Intercept) 611472 781.97 subj.1 ld 317412 563.39 subj.2 nn_hard:rc_hard 20321 142.55 item (Intercept) 36172 190.19 item.1 ld 65756 256.43 item.2 nn_easy 1912 43.73 item.3 nn_hard:rc_hard 25229 158.84 Residual 1252759 1119.27 Notice that there are still some problems in these models: library(car) ## Loading required package: carData qqPlot(residuals(m6))	{}
# Small complex-regression example¶ Let’s just see how we can do a regression small CVNN. As usual, first import what is needed: import numpy as np import cvnn.layers as complex_layers import tensorflow as tf Let’s create random complex data input_shape = (4, 28, 28, 3) x = tf.cast(tf.random.normal(input_shape), tf.complex64) Now let’s create our network and compile it model = tf.keras.models.Sequential() This is it! Now you can train uwing the fit method or predict. You can check for example that the output of the model is still complex (as expected). y = model(x) assert y.dtype == np.complex64	{}
# ZERO DIVISOR GRAPHS OF SKEW GENERALIZED POWER SERIES RINGS • MOUSSAVI, AHMAD (Department of Pure Mathematics Faculty of Mathematical Sciences Tarbiat Modares University) ; • PAYKAN, KAMAL (Department of Pure Mathematics Faculty of Mathematical Sciences Tarbiat Modares University) • Published : 2015.10.31 #### Abstract Let R be a ring, (S,${\leq}$) a strictly ordered monoid and ${\omega}$ : S ${\rightarrow}$ End(R) a monoid homomorphism. The skew generalized power series ring R[[S,${\omega}$]] is a common generalization of (skew) polynomial rings, (skew) power series rings, (skew) Laurent polynomial rings, (skew) group rings, and Mal'cev-Neumann Laurent series rings. In this paper, we investigate the interplay between the ring-theoretical properties of R[[S,${\omega}$]] and the graph-theoretical properties of its zero-divisor graph ${\Gamma}$(R[[S,${\omega}$]]). 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# Contact electricity and photoelectric effect Most universities provide an experiment about the photoelectric effect to determine $h$ by measuring the stop voltage against the light frequency and calculating the slope $h/e$. But mostly they also talk about the contact voltage, which just changes the offset, but not the slope. I have several related questions regarding this contact voltage in combination with the photoelectric effect: • Why do they talk about contact voltages at all if it's irrelevant for the slope? • Why are the photo cells build with two different materials at all? The same material would give no contact voltage and everything would be fine. • Books and Wikipedia tell me contact voltage is the voltage between two materials that are connected directly. But the photo cell's electrodes are not contacted directly, but normaly through copper cables and with several devices inbetween. Doesn't this change the contact voltage inpredictable? • Are there papers or books or something that explains contact electricity from ground up? - ## 1 Answer • They talk about contact voltages because it affects “stop voltage” as measured by their instruments, but it doesn’t affect $\Delta U_{\rm stop} / \Delta\nu$ since aforementioned contact voltages are assumed independent of illumination conditions. • Because a photovoltaic cell made of a homogeneous piece of material won’t work. A piece of conductor will not produce any voltage, whereas you will be unable to extract the current from a dielectric. • No, it doesn’t. You can also read an interesting discussion at Ambiguity on the notion of potential in electrical circuits?. BTW, where does Wikipedia tell you about the contact voltage? https://en.wikipedia.org/wiki/Volta_potential , or… ? A posting that vaguely refers to “books and Wikipedia” without any specificity demonstrates a shortage in Internet communication skills, especially given that the people refers to several different things as to contact voltages. • I don’t known (I’m not an English speaker, usually). -	{}
CGAL 4.12.1 - 2D Arrangements CGAL::Arr_algebraic_segment_traits_2< Coefficient >::Curve_2 Class Reference #include <CGAL/Arr_algebraic_segment_traits_2.h> Definition Models the ArrangementTraits_2::Curve_2 concept. Represents algebraic curves. Internally, the type stores topological-geometric information about the particular curve. In order to use internal caching, instances should only be created using the Construct_curve_2 functor of the traits class. Modifiers Polynomial_2 polynomial () const returns the defining polynomial of the curve.	{}
6 of 7 make chain clearer. # Write a Shift Interpreter EDIT: As some of you suspected, there was a bug in the official interpreter: the order of composition in . was reversed. I had two versions of the interpreter, and used the wrong one here. The examples were also written for this incorrect version. I have fixed the interpreter in the repository, and the examples below. The description of > was also a bit ambiguous, so I've fixed that. Also, apologies for this taking so long, I was caught up in some real life stuff. EDIT2: My interpreter had a bug in the implementation of . which was reflected in the examples (they relied on undefined behavior). The issue is now fixed. # Introduction Shift is an esoteric functional programming language I made a couple of years ago, but published today. It is stack-based, but also has automatic currying like Haskell. # Specification There are two datatypes in Shift: • Functions, which have an arbitrary positive arity (number of inputs), and which return a list of outputs. For example, a function that duplicates its only input has arity 1, and a function that swaps its two inputs has arity 2. • Blanks, which are all identical and have no other purpose than not being functions. A Shift program consists of zero or more commands, each of which is a single ASCII character. There are 8 commands in total: • ! (apply) pops a function f and a value x from the stack, and applies f to x. If f has arity 1, the list f(x) is added to the front of the stack. If it has arity n > 1, a new (n-1)-ary function g is pushed to the stack. It takes inputs x1,x2,...,xn-1 and returns f(x,x1,x2,...,xn-1). • ? (blank) pushes a blank to the stack. • + (clone) pushes to the stack a unary function that duplicates its input: any value x is mapped to [x,x]. • > (shift) pushes to the stack a unary function that takes in an n-ary function f, and returns an (n+1)-ary function g that ignores its first argument x, calls f on the remaining ones, and tacks x in front of the result. For example, shift(clone) is a binary function that takes inputs a,b and returns [a,b,b]. • / (fork) pushes to the stack a ternary function that takes three inputs a,b,c, and returns [b] if a is a blank, and [c] otherwise. • $ (call) pushes to the stack a binary function that pops a function f and a value x, and applies f to x exactly as ! does. • . (chain) pushes to the stack a binary function that pops two functions f and g, and returns their composition: a function h that has the same arity as f, and which takes its inputs normally, applies f to them, and then fully applies g to the result (calls it as many times as its arity dictates), with unused items from the output of g remaining in the result of h. For example, suppose that f is a binary function that clones its second argument, and g is call. If the stack contains [f,g,a,b,c] and we do .!!, then it contains [chain(f,g),a,b,c]; if we do !! next, then f is first applied to a,b, producing [a,b,b], then g is applied to the first two elements of that since its arity is 2, producing [a(b),b], and the stack will finally be [a(b),b,c]. • @ (say) pushes a unary function that simply returns its input, and prints 0 if it was a blank, and 1 if it was a function. Note that all commands except ! simply push a value to the stack, there is no way to perform input, and the only way to output anything is to use @. A program is interpreted by evaluating the commands one by one, printing 0s or 1s whenever "say" is called, and exiting. Any behavior not described here (applying a blank, applying a stack of length 0 or 1, calling "chain" on a blank etc.) is undefined: the interpreter may crash, silently fail, ask for input, or whatever. # The Task Your task is to write an interpreter for Shift. It should take from STDIN, command line, or function argument a Shift program to be interpreted, and print to STDOUT or return the resulting (possibly infinite) output of 0s and 1s. If you write a function, you must be able to access the infinite-length outputs in some way (generator in Python, lazy list in Haskell, etc). Alternatively, you can take another input, a number n, and return at least n characters of the output if it is longer than n. The lowest byte count wins, and standard loopholes are disallowed. # Test Cases This Shift program prints 01: ?@!@@! Starting from the left: push a blank, push say, then apply the say to the blank. This outputs 0. Then, push say twice, and apply the second say to the first. This outputs 1. This program loops forever, producing no output: $+.!!+!! Push call and clone, then apply chain to them (we need two !s since chain is a binary function). Now the stack contains a function that takes one argument, duplicates it, and calls the first copy on the second. With +!!, we duplicate this function and call it on itself. This program prints 0010: ?@$.++>!.!!.!!.!!!!+?/!!!@!@>!!! Push a blank and say. Then, compose a binary function that copies its second argument b, then copies the first a and composes it with itself, then applies the composition to the copy of b, returning [a(a(b)),b]. Apply it to say and blank, then apply say to the two elements remaining on the stack. This program prints 0. For each !!! that you append to it, it prints an additional 0. ?@+$>!>!+>!///!!>!>!.!!.!!.!!+!!!! Push a blank and say. Then, compose a ternary function that takes f,g,x as inputs and returns [f,f,g,g(x)]. Clone that function, and apply it to itself, say, and the blank. This application does not change the stack, so we can apply the function again as many times as we want. This program prints the infinite sequence 001011011101111..., where the number of 1s always increases by one: @?/!@>!??/!!>!+.!!.!!.!!.+>!.!!+$>!>!$>!>!+>!\$>!>!>!+>!>!///!!>!>!>!.!!.!!.!!.!!.!!.!!.!!.!!.!!.!!+!!!!! The repository contains an annotated version.	{}
# Multilne captions neuwirthe shared this question 3 years ago Is ist possible to have a multiline caption for an object? I tried to insert line breaks in captions, but neither \n nor \\n did work 1 Use \cr or \\ The object caption needs to be in LaTeX format, that is, enclosed in \$ \$. Example: `\$line1 \cr line2\$` 1 Thanks, that solved my problem 1 One more question: LaTeX captions and plain text captions are using a different font, even when I use \textsf in LaTeX expressions. Is there a way to use the same font in LaTeX as in plain text captions. The visual appearance would be more consistent this way.	{}
Friday, July 18, 2014 IOI 2014 day 2 analysis I found day 2 much harder than day 1, and I still don't know how to solve all the problems (I am seriously impressed by those getting perfect scores). Here's what I've managed to figure out so far. Update: I've now solved everything (in theory), and the solutions are below. The official solutions are now also available on the IOI website. I'll try coding the solutions at some point if I get time. Gondola This was the easiest of the three. Firstly, what makes a valid gondola sequence? In all the subtasks of this problem, there will be two cases. If you see any of the numbers 1 to n, that immediately locks in the phase, and tells you the original gondola for every position. Otherwise, the phase is unknown. So, the constraints are that • if the phase is known, every gondola up to n must appear in the correct spot if it appears; • no two gondolas can have the same number. Now we can consider how to construct a replacement sequence (and also to count them), which also shows that these conditions are sufficient. If the phase is not locked, pick it arbitrarily. Now the "new gondola" column is simply the numbers from n+1 up to the largest gondola, so picking a replacement sequence is equivalent to deciding which gondola replaces each broken gondola. We can assign each gondola greater than n that we can't see to a position (one where the final gondola number is larger), and this will uniquely determine the replacement sequence. We'll call such gondolas hidden. For the middle set of subtasks, the simplest thing is to assign all hidden gondolas to one position, the one with the highest-numbered gondola in the final state. For counting the number of possible replacement sequences, each hidden gondola can be assigned independently, so we just multiply together the number of options, and also remember to multiply by n if the phase is unknown. In the last subtask there are too many hidden gondolas to deal with one at a time, but they can be handled in batches (those between two visible gondolas), using fast exponentiation. Friend This is a weighted maximum independent set problem. On a general graph this is NP-hard, so we will need to exploit the curious way in which the graph is constructed. I haven't figured out how to solve the whole problem, but let's work through the subtasks: 1. This is small enough to use brute force (consider all subsets and check whether they are independent). 2. The graph will be empty, so the sample can consist of everyone. 3. The graph will be complete, so only one person can be picked in a sample. Pick the best one. 4. The graph will be a tree. There is a fairly standard tree DP to handle this case: for every subtree, compute the best answer, either with the root excluded or included. If the root is included, add up the root-excluded answers for every subtree; otherwise add up the best of the two for every subtree. This takes linear time. 5. In this case the graph is bipartite and the vertices are unweighted. This is a standard problem which can be solved by finding the maximum bipartite matching. The relatively simple flow-based algorithm for this is theoretically $$O(n^3)$$, but it is one of those algorithms that tends to run much faster in most cases, so it may well be sufficient here. The final test-case clearly requires a different approach, since n can be much larger. I only managed to figure this out after getting a big hint from the SA team leader, who had seen the official solution. We will process the operations in reverse order. For each operation, we will transform the graph into one that omits the new person, but for which the optimal solution has the same score. Let's say that the last operation had A as the host and B as the invitee, and consider the different cases: • YourFriendsAreMyFriends: this is the simplest: any solution using B can also use A, and vice versa. So we can collapse the two vertices into one whose weight is the sum of the original weights, and use it to replace A. • WeAreYourFriends: this is almost the same, except now we can use at most one of A and B, and which one we take (if either) has no effect on the rest of the graph. So we can replace A with a single vertex having the larger of the two weights, and delete B. • IAmYourFriend: this is a bit trickier. Let's start with the assumption that B will form part of the sample, and add that to the output value before deleting it. However, if we later decide to use A, there will be a cost to remove B again; so A's weight decreases by the weight of B. If it ends up with negative weight, we can just clamp it to 0. Repeat this deletion process until only the original vertex is left; the answer will be the weight of this vertex, plus the saved-up weights from the IAmYourFriend steps. Holiday Consider the left-most and right-most cities that Jian-Jia visits. Regardless of where he stops, he will need to travel from the start city to one of the ends, and from there to the other end. There is no point in doing any other back-tracking, so we can tell how many days he spends travelling just from the end-points. This then tells us how many cities he has time to see attractions in, and obviously we will pick the best cities within the range. That's immediately sufficient to solve the first test case. To solve more, we can consider an incremental approach. Fix one end-point, and gradually extend the other end-point, keeping track of the best cities (and their sum) in a priority queue (with the worst of the best cities at the front). As the range is extended, the number of cities that can be visited shrinks, so items will need to be popped. Of course, the next city in the range needs to be added each time as well. Using a binary heap, this gives an $$O(n^2\log n)$$ algorithm: a factor of n for each endpoint, and the $$\log n$$ for the priority queue operations. That's sufficient for subtask 3. It's also good enough for subtask 2, because the left endpoint will be city 0, saving a factor of n. For subtask 4, it is clearly not possible to consider every pair of end-points. Let's try to break things up. Assume (without loss of generality) that we move first left, then back to the start, then right. Let's compute the optimal solution for the left part and the right part separately, then combine them. The catch is that we need to know how we are splitting up our time between the two sides. So we'll need to compute the answer for each side for all possible number of days spent within each side. This seems to leave us no better off, since we're still searching within a two-dimensional space (number of days and endpoint), but it allows us to do some things differently. We'll just consider the right-hand side. The left-hand side is similar, with minor changes because we need two days for travel (there and back) instead of one. Let f(d) be the optimal end-point if we have d days available. Then with a bit of work one can show that f is non-decreasing (provided one is allowed to pick amongst ties). If we find f(d) for d=1, 2, 3, ... in that order, it doesn't really help: we're only, on average, halving the search space. But we can do better by using a divide-and-conquer approach: if we need to find f for all $$d \in [0, D)$$ then we start with $$d = \frac{D}{2}$$ to subdivide the space, and then recursively process each half of the interval on disjoint subintervals of the cities. This reduces the search space to $$O(n\log n)$$. This still leaves the problem of efficiently finding the total number of attractions that can be visited for particular intervals and available days. The official solution uses one approach, based on a segment tree over the cities, sorted by number of attractions rather than position. The approach I found is, I think, simpler. Visualise the recursion described above as a tree; instead of working depth-first (i.e., recursively), we work breadth-first. We make $$O(\log n)$$ passes, and in each pass we compute f(d) where d is an odd multiple of $$2^i$$ (with $$i$$ decreasing with each pass). Each pass can be done in a single incremental process, similar to the way we tackled subpass 2. The difference is that each time we cross into the next subinterval, we need to increase $$d$$, and hence bring more cities into consideration. To do this, we need either a second priority queue of excluded cities, or we can replace the priority queue with a balanced binary tree. Within each pass, d can only be incremented $$O(n)$$ times, so the total running time will be $$O(n\log n)$$ per pass, or $$O(n\log n \log n)$$ overall. IOI 2014 day 1 Since there is no online judge, I haven't tried actually coding any of these. So these ideas are not validated yet. You can find the problems here. Rails I found this the most difficult of the three to figure out, although coding it will not be particularly challenging. Firstly, we can note that distances are symmetric: a route from A to B can be reflected in the two tracks to give a route from B to A. So having only $$\frac{n(n-1)}{2}$$ queries is not a limitation, as we can query all distances. This might be useful in tackling the first three subtasks, but I'll go directly to the hardest subtask. If we know the position and type of a station, there is one other that we can immediately locate: the closest one. It must have the opposite type and be reached by a direct route. Let station X be the closest to station 0. The other stations can be split into three groups: 1. d(X, Y) < d(X, 0): these are reached directly from station X and of type C, so we can locate them exactly. 2. d(0, X) + d(X, Y) = d(0, Y), but not of type 1: these are reached from station 0 via station X, so they lie to the left of station 0. 3. All other stations lie to the right of station X. Let's now consider just the stations to the right of X, and see how to place them. Let's take them in increasing distance from 0. This ensures that we encounter all the type D stations in order, and any type C station will be encountered at some point after the type D station used to reach it. Suppose Y is the right-most type D station already encountered, and consider the distances for a new station Z. Let $$z = d(0, Z) - d(0, Y) - d(Y, Z)$$. If Z is type C, then there must be a type D at distance $$\frac{z}{2}$$ to the left of Y. On the other hand, if Z is of type D (and lies to the right of Y), then there must be a type C station at distance $$\frac{z}{2}$$ to the left of Y. In the first case, we will already have encountered the station, so we can always distinguish the two cases, and hence determine the position and type of Z. The stations to the left of station zero can be handled similarly, using station X as the reference point instead of station 0. How many queries is this? Every station Z except 0 and X accounts for at most three queries: d(0, Z), d(X, Z) and d(Y, Z), where Y can be different for each Z. This gives $$3(n-2) + 1$$, which I think can be improved to $$3(n-2)$$ just by counting more carefully. Either way, it is sufficient to solve all the subtasks. Wall This is a fairly standard interval tree structure problem, similar to Mountain from IOI 2005 (but a little easier). Each node of the tree contains a range to which its children are clamped. To determine the value of any element of the wall, start at the leaf with a value of 0 and read up the tree, clamping the value to the range in each node in turn. Initially, each node has the range [0, inf). When applying a new instruction, it is done top-down, and clamps are pushed down the tree whenever recursion is necessary. An interesting aspect of the problem is that it is offline, in that only the final configuration is requested and all the information is provided up-front. This makes me think that there may be an alternative solution that processes the data in a different order, but I can't immediately see a nicer solution than the one above. Game I liked this problem, partly because I could reverse-engineer a solution from the assumption that it is always possible to win, and partly because it requires neither much algorithm/data-structure training (like Wall) nor tricky consideration of cases (like Rails). Suppose Mei-Yu knows that certain cities are connected. If there are any flights between the cities that she has not asked about, then she can win simply by saving one of these flights for last, since it will not affect whether the country is connected. It follows that for Jian-Jia to win, he must always answer no when asked about a flight between two components that Mei-Yu does not know to be connected, unless this is the last flight between these components? What if he always answers yes to the last flight between two components? In this case he will win. As long as there are at least two components left, there are uncertain edges between every pair of them, so Mei-Yu can't know whether any of them is connected any other. All edges within a component are known, so the number of components can only become one after the last question. What about complexity? We need to keep track of the number of edges between each pair of components, which takes $$O(N^2)$$ space. Most operations will just decrement one of these counts. There will be $$N - 1$$ component-merging operations, each of which requires a linear-time merging of these edge counts and updating a vertex-to-component table. Thus, the whole algorithm requires $$O(N^2)$$ time. This is optimal given that Mei-Yu will ask $$O(N^2)$$ questions. Monday, June 30, 2014 ICPC Problem H: Pachinko Problem A has been covered quite well elsewhere, so I won't discuss it. That leaves only problem H. I started by reading the Google translation of a Polish writeup by gawry. The automatic translation wasn't very good, but it gave me one or two ideas I borrowed. I don't know how my solution compares in length of code, but I consider it much simpler conceptually. This is a fairly typical Markov process, where a system is in some state, and each timestep it randomly selects one state as a function of the current state. One variation is that the process stops once the ball reaches a target, whereas Markov processes don't terminate. I was initially going to model that as the ball always moving from the target to itself, but things would have become slightly complicated. Gawry has a nice way of making this explicitly a matrix problem. Set up the matrix M as for a Markov process i.e., $$M_{i,j}$$ is the probability of a transition from state j to state i. However, for a target state j, we set $$M_{i,j}=0$$ for all i. Now if $$b$$ is our initial probability vector (equal probability for each empty spot in the first row), then $$M^t b$$ represents the probability of the ball being in each position (and the game not having previously finished) after $$t$$ timesteps. We can then say that the expected amount of time the ball spends in each position is given by $$\sum_{t=0}^{\infty} M^t b$$. The sum of the elements in this vector is the expected length of a game and we're told that it is less than $$10^9$$, so we don't need to worry about convergence. However, that doesn't mean that the matrix series itself converges: Gawry points out that if there are unreachable parts of the game with no targets, then the series won't converge. We fix that by doing an initial flood-fill to find all reachable cells and only use those in the matrix. Gawry then shows that under the right conditions, the series converges to $$(I - M)^{-1} b$$. This is where my solution differs. Gawry dismisses Gaussian elimination, because the matrix can be up to 200,000 square. However, this misses the fact that it is banded: by numbering cells left-to-right then top-to-bottom, we ensure that every non-zero entry in the matrix is at most W cells away from the main diagonal. Gaussian elimination (without pivoting) preserves this property. We can exploit this both to store the matrix compactly, and to perform Gaussian elimination in $$O(W^3H)$$ time. One concern is the "without pivoting" caveat. I was slightly surprised that my first submission passed. I think it is possible to prove correctness, however. Gaussian elimination without pivoting is known (and easily provable) to work on strictly column diagonally dominant matrices. In our case the diagonal dominance is weak: columns corresponding to empty cells have a sum of zero, those corresponding to targets have a 1 on the diagonal and zeros elsewhere. However, the matrix is also irreducible, which I think is enough to guarantee that there won't be any division by zero. EDIT: actually it's not guaranteed to be irreducible, because the probabilities can be zero and hence it's possible to get from A to B without being able to get from B to A. But I suspect that it's enough that one can reach a target from every state. Saturday, June 28, 2014 ICPC Problem L: Wires While this problem wasn't too conceptually difficult, it requires a lot of code (my solution is about 400 lines), and careful implementation of a number of geometric algorithms. A good chunk of the code comes from implementing a rational number class in order to precisely represent the intersection points of the wires. It is also very easy to suffer from overflow: I spent a long time banging my head against an assertion failure on the server until I upgraded my rational number class to use 128-bit integers everywhere, instead of just for comparisons. The wires will divide the space up into connected regions. The regions can be represented in a planar graph, with edges between regions that share an edge. The problem is then to find the shortest path between the regions containing the two new end-points. My solution works in a number of steps: 1. Find all the intersection points between wires, and the segments between intersection points. This just tests every wire against every other wire. The case of two parallel wires sharing an endpoint needs to be handled carefully. For each line, I sort the intersection points along the line. I used a dot produce for this, which is where my rational number class overflowed, but would probably have been safer to just sort lexicographically. More than two lines can meet at an intersection point, so I used a std::map to assign a unique ID to each intersection point (I'll call them vertices from here on). 2. Once the intersection points along a line have been sorted, one can identify the segments connecting them. I create two copies of each segment, one in each direction. With each vertex A I store a list of all segments A->B. Each pair is stored contiguously so that it is trivial to find its partner. Each segment is considered to belong to the region to its left as one travels A->B. 3. The segments emanating from each vertex are sorted by angle. These comparisons could easily cause overflows again, but one can use a handy trick: instead of using the vector for the segment in an angle comparison, one can use the vector for the entire wire. It has identical direction but has small integer coordinates. 4. Using the sorted lists from the previous step, each segment is given a pointer to its following segment from the same region. In other words, if one is tracing the boundary of the region and one has just traced A->B, the pointer will point to B->C. 5. I extract the contours of the regions. A region typically consists of an outer contour and optionally some holes. The outermost region lacks an outer contour (one could add a big square if one needed to, but I didn't). A contour is found by following the next pointers. A case that turns out to be inconvenient later is that some segments might be part of the contour but not enclose any area. This can make a contour disappear completely, in which case it is discarded. Any remaining contours have the property that two adjacent segments are not dual to each other, although it is still possible to both sides of an edge to belong to the same contour. 6. Each contour is identified as an outer contour or a hole. With integer coordinates I could just measure the signed area of the polygon, but that gets nasty with rational coordinates. Instead, I pick the lexicographically smallest vertex in the contour and examine the angle between the two incident segments (this is why it is important that there is a non-trivial angle between them). I also sort the contours by this lexicographically smallest vertex, which causes any contour to sort before any other contours it encloses. 7. For each segment I add an edge of weight 1 from its containing region to the containing region of its dual. 8. For each hole, I search backwards through the other contours to find the smallest non-hole that contains it. I take one vertex of the hole and do a point-in-polygon test. Once again, some care is needed to avoid overflows, and using the vectors for the original wires proves useful. One could then associate the outer contour and the holes it contains into a single region object, but instead I just added an edge to the graph to join them with weight 0. In other words, one can travel from the boundary of a region to the outside of a hole at no cost. 9. Finally, I identify the regions containing the endpoints of the new wire, using the same search as in the previous step. After all this, we still need to implement a shortest path search - but by this point that seems almost trivial in comparison. What is the complexity? There can be $$O(M^2)$$ intersections and hence also $$O(M^2)$$ contours, but only $$O(M)$$ of them can be holes (because two holes cannot be part of the same connected component). The slowest part is the fairly straightforward point-in-polygon test which tests each hole against each non-hole segment, giving $$O(M^3)$$ time. There are faster algorithms for doing point location queries, so it is probably theoretically possible to reduce this to $$O(M^2\log N)$$ or even $$O(M^2)$$, but certainly not necessary for this problem. ICPC Problem J: Skiing I'll start by mentioning that there is also some analysis discussion on the Topcoder forums, which includes alternative solutions that in some cases I think are much nicer than mine. So, on to problem J - which no team solved, and only one team attempted. I'm a little surprised by this, since it's the sort of problem where one can do most of the work on paper while someone else is using the machine. A possible reason is that it's difficult to determine the runtime: I just coded it up and hoped for the best. It is a slightly annoying problem for a few reasons. Firstly, there are a few special cases, because $$v_y$$ or $$a$$ can be zero. We also have to find the lexicographically smallest answer. Let's deal with those special cases first. If $$v_y = 0$$ then we can only reach targets with $$y = 0$$. If $$a \not= 0$$ then we can reach all of them in any order, otherwise we can only reach those with $$x = 0$$. To get the lexicographically smallest answer, just go through them in order and print out those that are reachable. I actually dealt with the $$v_y \not= 0$$, $$a = 0$$ case as part of my general solution, but is also easy enough to deal with. We can only reach targets with $$x = 0$$, and we can only reach them in increasing order of y. One catch is that targets are not guaranteed to have distinct coordinates, so if two targets have the same coordinates we must be careful to visit them in lexicographical order. I handled this just by using a stable sort. So, onwards to the general case. Before we can start doing any high-level optimisation, we need to start by answering this question: if we are at position P with X velocity $$v$$ and want to pass through position Q, what possible velocities can we arrive with? I won't prove it formally, but it's not hard to believe that to arrive with the smallest velocity, you should start by accelerating (at the maximum acceleration) for some time, then decelerate for the rest of the time. Let's say that the total time between P and Q is $$T$$, the X separation is $$x$$, the initial acceleration time is $$T - t$$ and the final deceleration time is $$t$$. Some basic integration then gives $x = v(T-t)+\tfrac{1}{2}a(T-t)^2 + (v+a(T-t))t - \tfrac{1}{2}at^2.$ This gives a quadratic in $$t$$ where the linear terms conveniently cancel, leaving $t = \sqrt{\frac{vT+\tfrac{1}{2}aT^2-x}{a}}$ The final velocity is just $$v+aT-2at$$. We can also find the largest possible arrival velocity simply by replacing $$a$$ with $$-a$$. Let's call these min and max velocity functions $$V_l(v, T, x)$$ and $$V_h(v, T, x)$$. More generally, if we can leave P with a given range of velocities $$[v_0, v_1]$$, with what velocities can we arrive at Q? We first need to clamp the start range to the range from which we can actually reach Q i.e., that satisfy $$\|vT - x\| \le \frac{1}{2}aT^2$$. A bit of calculus shows that $$V_l$$ and $$V_h$$ are decreasing functions of $$v$$, so the arrival range will be $$[V_l(v_1), V_h(v_0)]$$. Finally, we are ready to tackle the problem as a whole, with multiple targets. We will use dynamic programming to answer this question, starting from the last target (highest Y): if I start at target i and want to hit a total of j targets, what are the valid X velocities at i? The answer will in general be a set of intervals. We will make a pseudo-target at (0, 0), and the answer will then be the largest j for which 0.0 is a valid velocity at this target. Computing the DP is generally straightforward, except that the low-level queries we need are the reverse of those we discussed above i.e. knowing the arrival velocities we need to compute departure velocities. No problem, this sort of physics is time-reversible, so we just need to be careful about which signs get flipped. For each (i, j) we consider all options for the next target, back-propagate the set of intervals from that next target, and merge them into the answer for (i, j). Of course, for $$j = 1$$ we use the interval $$(-\infty, \infty)$$. The final inconvenience is that we must produce a lexicographical minimum output. Now we will see the advantage of doing the DP in the direction we chose. We build a sequence forwards, keeping track of the velocity interval for our current position. Initially the position will be (0, 0) and the velocity interval will be [0.0, 0.0]. To test whether a next position is valid, we forward-propagate the velocity interval to this next position, and check whether it has non-empty intersection with the set of velocities that would allow us to hit the required number of remaining targets. We then just take the next valid position with the lowest ID. What is the complexity? It could in theory be exponential, because every path through the targets could induce a separate interval in the interval set. However, in reality the intervals merge a lot, and I wouldn't be surprised if there is some way to prove that there can only be a polynomial number of intervals to consider. My solution still ran pretty close to the time limit, though. Friday, June 27, 2014 More ICPC analysis Now we're getting on to the harder problems. Today I've cover two that I didn't know how to solve. Some of the others I have ideas for, but I don't want to say anything until I've had a chance to code them up. Firstly, problem I. This is a maximum clique problem, which on a general graph is NP-complete. So we will need to use the geometry in some way. Misof has a nice set of slides showing how it is done: http://people.ksp.sk/~misof/share/wf_pres_I.pdf. I had the idea for the first step (picking the two points that are furthest apart), but it didn't occur to me that the resulting conflict graph would be bipartite. Now problem G. I discovered that this is a problem that has had a number of research papers published on the topic, one of which achieves $$O(N^3)$$. Fortunately, we don't need to be quite that efficient. Let's start by finding a polynomial-time solution. Let's suppose we've already decided the diameters of the two clusters, D(A) and D(B), and just want to find out whether this is actually possible. For each shipment we have a boolean variable that says whether it goes into part A (false) or part B (true). The constraints become boolean expressions: if d(i, j) > D(A) then we must have i \|\| j, and if d(i, j) > D(B) then we must have !i \|\| !j. Determining whether the variables can be satisfied is just 2-SAT, which can be solved in $$O(N^2)$$ time. Now, how do we decide which diameters to test? There are $$O(N^2)$$ choices for each, so the naive approach will take $$O(N^6)$$ time. We can reduce this to $$O(N^4\log N)$$ by noting that once we've chosen one, we can binary search the other (it's also possible to eliminate the log factor, but it's still too slow). So far, this is what I deduced during the contest. The trick (which I found in one of those research papers) is that one can eliminate most candidates for the larger diameter. If there is an odd-length cycle, at least one of the edges must be within a cluster, and so that is a lower bound for the larger diameter. What's more, we can ignore the shortest edge of an even cycle (with some tie-breaker), because if the shortest edge lies within a cluster, then so must at least one other edge. We can exploit this to generate an O(N)-sized set of candidates: process the edges from longest to shortest, adding each to a new bipartite graph (as for constructing a maximum spanning tree with Kruskal's algorithm). There are three cases: 1. The next edge connects two previously disconnected components. Add the edge to the graph (which will remain bipartite, since one can always re-colour one of the components. This edge length is a candidate diameter. 2. The next edge connects two vertices in the same component, but the graph remains bipartite. This edge is thus part of an even cycle, and so can be ignored. 3. The next edge connects two vertices, forming an odd cycle. This edge length is a candidate diameter, and the algorithm terminates. The edges selected in step 1 form a tree, so there are only O(N) of them. Wednesday, June 25, 2014 ACM ICPC 2014 ACM ICPC 2014 is over. The contest was incredibly tough: solving less than half the problems was enough to get a medal, and 20 teams didn't solve any of the problems (unfortunately including the UCT team, who got bogged down in problem K). I managed to solve 5 problems during the contest (in 4:30, since I didn't wake up early enough to be in at the start), and I've solved one more since then. Here are my solutions, in approximately easy-to-hard order. EDIT: note that I've made follow-up blog posts with more analysis. D: game strategy This was definitely the easiest one, and this is reflected in the scores. We can use DP to determine the set of states from which Alice can force a win within i moves. Let's call this w[i]. Of course, w[0] = {target}. To compute w[i+1], consider each state s that is not already a winning state. If one of Alice's options is the set S and $$S \subseteq w[i]$$, then Alice can win from this state in i+1 moves. We can repeat this N times (since no optimal winning sequence can have more than N steps), or just until we don't find any new winning states. We don't actually need to maintain all the values of w, just the latest one. K: surveillance Let's first construct a slower polynomial-time algorithm. Firstly, for each i, we can compute the longest contiguous sequence of walls we can cover starting at i. This can be done in a single pass. We sort all the possible cameras by their starting wall, and as we sweep through the walls we keep track of the largest-numbered ending wall amongst cameras whose starting wall we have passed. The camera with $$a > b$$ are a bit tricky: we can handle them by treating them as two cameras $$[a - N, b]$$ and $$[a, b + N]$$. Let's suppose that we know we will use a camera starting at wall i. Then we can use this jump table to determine the minimum number of cameras: cover as much wall starting from i as possible with one camera, then again starting from the next empty slot, and so on until we've wrapped all the way around and covered at least N walls. We don't know the initial i, but we can try all values of i. Unfortunately, this requires $$O(N^2)$$ time. To speed things up, we can compute accelerated versions of the jump table. If $$J_c[i]$$ is the number of walls we can cover starting from i by using c cameras, then we already have $$J_1$$, and we can easily calculate $$J_{a+b}$$ given $$J_a$$ and $$J_b$$. In particular, we can compute $$J_2, J_4, J_8$$ and so on, and them combine these powers of two to make any other number. This can then be used in a binary search to find the smallest c such that $$J_c$$ contains a value that is at least N. The whole algorithm thus requires $$O(N\log N)$$ time. One catch that broke my initial submission is that if the jump table entries are computed naively, they can become much bigger than N. This isn't a problem, until they overflow and become negative. Clamping them to N fixed the problem. C: crane balancing This is a reasonably straightforward geometry problem, requiring some basic mechanics. Let the mass of the crane be $$q$$, let the integral of the x position over the crane be $$p$$, let the base be the range $$[x_0, x_1]$$, and let the x position of the weight be $$x$$. The centre of mass of the crane is $$\frac p q$$, but with a weight of mass $$m$$, it will be $$\frac{p+mx}{q+m}$$. The crane is stable provided that this value lies in the interval $$[x_0, x_1]$$. This turns into two linear inequalities in $$m$$. Some care is then needed to deal with the different cases of the coefficient being positive, negative or zero. Computing the area and the integral of the x value is also reasonably straightforward: triangulate the crane by considering triangles with vertices at $$(0, i, i+1)$$ and computing the area (from the cross product) and centre of mass (average of the vertices) and then multiply the area by the x component of the centre of mass to get the integral. I prefer to avoid floating point issues whenever possible, so I multiplied all the X coordinates by 6 up front and worked entirely in integers. This does also cause the masses to be 6 times larger, so they have to be adjusted to compensate. E: maze reduction This is effectively the same problem I set in Topcoder SRM 378 (thanks to ffao on Topcoder for reminding me where I'd seen this). If you have an account on Topcoder you can read about an alternative solution in the match analysis, in which it is easier to compute worst-case bounds. The approach I used in this contest is similar in concept to D, in that you incrementally update a hypothesis about which things are distinguishable. We will assign a label to every door of each chamber. Two doors are given the same label if we do not (yet) know of a way to distinguish them. Initially, we just label each door with the degree of the room it is in, since there is nothing else we can tell by taking zero steps. Now we can add one inference step. Going clockwise from a given door, we can explore all the corridors leading from the current chamber and note down the labels on the matching doors at the opposite end of each corridor. This forms a signature for the door. Having done this for each door, we can now assign new labels, where each unique signature is assigned a corresponding label. We can repeat this process until we achieve stability, i.e., the number of labels does not change. We probably ought to use each door's original label in the signature too, but my solution passed without requiring this. Finally, two rooms are effectively identical if their sequence of door labels is the same up to rotation. I picked the lexicographically minimum rotation for each room and used this as a signature for the room. I'm not sure what the theoretical work-case performance is, but I suspect it would be quite large. For a start, by algorithm requires $$O(NE\log N)$$ time per iteration, which I suspect could be improved by using a hash table with some kind of rolling hash to rotation in O(1) time. I was surprised when my first submission ran in time. B: buffed buffet This one was deceptively sneaky, but the principles are reasonably simple. It's not too hard to guess that one will solve the continuous and discrete problems separately, and then consider all partitions between the two. Let's start with the continuous problem, since it is a little easier. I don't have a formal proof, but it shouldn't be too hard to convince yourself that a greedy strategy works. We start by eating the tastiest food. Once it degrades to being as tasty as the second-tastiest food, we then each both, in the proportion that makes them decay at the same rate. In fact, at this point we can treat them as a combined food with a combined (slower) decay rate. We continue eating this mix until tastiness decays to match the third-tastiest, and so on. There are some corner cases that need to be handled if there are foods that don't decay. Now what about the discrete case? Because the items have different weights, a greedy strategy won't work here, as with any knapsack problem. There is a reasonably straightforward $$O(DW^2)$$ dynamic programming, where you consider each possible number of serving of each discrete food type, but this will be too slow. But there is some structure in the problem, so let's try to exploit it. Let's say that $$f(i)$$ is the maximum tastiness for a weight of $$i$$ using only the foods we've already considered, and we're now computing an updated $$f'$$ by adding a new discrete food with weight $$w$$, initial tastiness $$t$$ and decay rate $$d$$. For a given i, $$f'(i)$$ clearly only depends on $$f(j)$$ where $$i - j$$ is a multiple of $$w$$, so let's split things up by the remainder modulo $$i$$. Fix a remainder $$r$$ and let $$g(i) = f(iw + r)$$ and $$g'(i) = f'(iw + r)$$. Now we can say that \begin{aligned} g'(i) &= \max_{0 \le j \le i}\big\{ g(j) + \sum_{n=1}^{i-j}(t - (n-1)d\big\}\\ &= \max_{0 \le j \le i}\big\{ g(j) + (i-j)t - \frac{(i-j-1)(i-j)}{2}\cdot d\big\}\\ &= \max_{0 \le j \le i}\big\{ g(j) + (i-j)t - \frac{i(i-1)+j(j+1)-2ij}{2}\cdot d\big\}\\ &= it - \frac{i(i-1)d}{2} + \max_{0 \le j \le i}\big\{ g(j)-\frac{j(j+1)d}{2} - jt + ijd\big\}\\ &= it - \frac{i(i-1)d}{2} + \max_{0 \le j \le i}\big\{ h(j) + ijd \big\} \end{aligned} Here we have defined $$h(j) = g(j)-\frac{j(j+1)d}{2} - jt$$, which can be computed in constant time per $$j$$. The key observation is that we have turned the expression inside the maximum into a linear function of $$j$$. Thus, if we plot $$h$$ on a graph, only the upper convex hull is worth considering. We can maintain this upper hull as we increase $$i$$ (remember, $$j \le i$$). The second observation is that the optimal choice of $$j$$ will increase as $$i$$ increases, because increasing $$i$$ will increase the $$ijd$$ more for larger values of $$j$$. It follows that we can incrementally find the optimal $$j$$ for each $$i$$ by increasing the previous $$j$$ along the upper hull until increasing it further would start decreasing the objective function. There are a few corner cases: the upper hull might not have any valid entries (because there might be no way to make up a weight of exactly $$jw+r$$), and we might have popped off the previous $$j$$ as part of updating the hull. These just need careful coding. Another snag to watch out for is that if all the foods decay very fast, the optimal result may in fact overload a 32-bit integer in the negative direction. The discrete part of the algorithm now requires O(DW), and the continuous part requires O(D\log D + W). F: messenger (solved after contest) This is a nasty problem mostly because of numerical stability problems. The problem itself is numerically unstable: a rounding error in measuring the total lengths of the paths is sufficient to change the answer between possible and impossible. It requires the black art of choosing good epsilons. I'll only describe the ideal case in which none of these nasty rounding problems occur. Suppose we pick a point on both Misha and Nadia's path. In general this won't work, because the courier will arrive too late or too early at this point to intercept Nadia. Let L(A, B) be the number of time units that the courier is late when travelling from A to B. L can be late if the courier is early. Valid solutions are those where L = 0. First, let's prove that L is an increasing function of A (where "increasing" means that A moves further along Misha's path). Suppose that the courier decided to, instead of moving straight to B, instead kept pace with Misha for a while, then went to B. Obviously this would mean arriving later (or at the same time). But this is equivalent to the arrival time if A increases. A similar argument shows that L is a decreasing function of B. In both cases, this is not strict. This means that there can be at most $$O(N)$$ pairs of segments for which L=0, rather than $$O(N^2)$$, because we cannot move forward on one path and backwards on another. We can iterate over the candidate pairs by successively advancing either one segment or the other, by measuring L at pairs of segment endpoints. Now we have reduced the problem to a single segment from each path. Given a point on one path, we can find the corresponding point on the other by solving what looks like a quadratic but which decays into a linear equation. Again, there are corner cases where the linear coefficient is also zero, in which case we have to break ties so as to minimise the travel time. Using this mapping function, we can identify the range of positions on Misha's path that correspond to valid positions on Nadia's path. I then used ternary search to find the optimal position along this segment. I didn't actually prove that the function is convex, but it seemed to work. The solution I implemented that passed is actually rather messier than what I've described, and ran close to the time limit. I tried implementing it as described above but it hits assertions in my code due to numeric instabilities, but I think it should still be possible to fix it. Monday, June 02, 2014 New website I've finally gotten around to replacing my ugly nineties-looking personal web page that was cobbled together with raw HTML and m4, with a much prettier version that nevertheless still contains roughly the same content. It is generated using Sphinx, with the individual pages written in reStructuredText. I've also merged my publications page into the main website instead of leaving it on the equally ugly but separately cobbled-together page I originally set up for my PhD.	{}
## Saturday, October 11, 2014 ### Illustrating Asymptotic Behaviour - Part I Learning the basics about the (large sample) asymptotic behaviour of estimators and test statistics is always a challenge. Teaching this material can be challenging too! So, in this post and in two more to follow, I'm going to talk about a small Monte Carlo experiment that illustrates some aspects of the asymptotic behaviour of the OLS estimator. I'll focus on three things: 1. The consistency of the OLS estimator in a situation where it's known to be biased in small samples. 2. The correct way to think about the asymptotic distribution of the OLS estimator. 3. A comparison of the OLS estimator and another estimator, in terms of asymptotic efficiency. The EViews program that I've written is on the code page for this blog, and it generates its own EViews workfile. The program will get extended for the subsequent posts, and re-linked. (If you're not an EViews user, you can read the program file with any text editor, and there are lots of comments in the code to explain what's going on.) The data-generating process (DGP) that's used for all three posts is: yt = β1 + β2 yt-1 + εt ; t = 2, 3, ....., n ; y= 0 . The error term, εt, is generated according to a uniform distribution on the interval (-1 , +1), so it has a mean of zero, and a variance of 0.08333. The errors have been chosen to be non-normal deliberately. One of the things that I want us to see is how the sampling distribution of the OLS estimator (which will be non-normal in finite samples in this case) eventually becomes normal as the sample size grows. My Monte Carlo experiment uses 5,000 replications, and the values of β1 and β2 have been set to 1.0 and 0.5 respectively in the DGP. In what follows the results will focus on the estimation of β2, but this is not especially important. In this post I'll be concentrating on just the first of the three numbered items above. The thing to keep in mind is that the OLS estimator is a random variable, so it has a probability distribution. Because the estimator is a sample statistic (a function of the random sample data), we call this distribution the "sampling distribution" of the estimator. The form that this distribution takes depends on the size of the sample that wee're using. This is important. Typically, the value of the point estimate will change as n grows. However, more fundamentally, the distribution of the (random) estimator will also change, and this is what we will be interested in here. To begin with, 5,000 different random samples (of size n = 20) have been generated, and in each case the model has then been estimated by OLS. The 5,000 point estimates of β2 have been saved. The distribution of these 5,000 values is a very close approximation to the true sampling distribution of the estimator. (The full sampling distribution would require that we do this an infinity of times, not 5,000, but we don't have time for that!) With n = 20, the sampling distribution of the OLS estimator of β2 looks like this: The mean of the sampling distribution is approximately 0.42, compared with the true value of β2, which is 0.5. So, the OLS estimator downward-biased as a result of having a lag of the dependent variable as a regressor. Notice that the standard deviation of the estimator (i.e., of its sampling distribution) is approximately 0.17. Finally, we can see from the Jarque-Bera test that the estimator's sampling distribution is non-normal. In particular, the distribution is substantially skewed to the left. Increasing the sample size to n = 100, and then to n = 250, we get the following results: We can see that as the sample size grows, the bias of the OLS estimator is decreasing - the expected value of the (sampling distribution of the) estimator is getting closer to the true value of the parameter, β2 = 0.5. In addition, the variability of the estimator is decreasing. The standard deviation of the sampling distribution falls from 0.17 when n = 10, to 0.08 and then to 0.05 when n = 250. However, even with the last of these sample sizes, the distribution of the OLS estimator is still non-normal. (When interpreting these graphs, and the ones below, keep an eye on the scale for the horizontal axis - typically it will be changing. This is why we might still see quite a lot of "spread" in the distributions, even though the standard deviation is declining.) Finally, when we set n = 1,000, and then n = 5,000, this is what we get: In this particular example, we need quite a large sample size before the sampling distribution of our OLS estimator is normal. In particular, there's a negative skewness to this distribution that's noticeable even for moderate sample sizes. The vanishing bias, and the decreasing variability of the estimator (as n grows) can be seen more directly in this graph: What can be observed here is the consistency of the OLS estimator. (For a discussion of different types of "consistency", see this earlier post.) If the sample size could be made infinitely large, the density for the sampling distribution would collapse to a degenerate "spike", centered at the true value of β2, and with negligible width (spread). At that final, hypothetical, stage the sampling distribution wouldn't really be a proper normal distribution. Also, as the "spike" would have no dispersion, it's not clear what it would mean to talk about the "asymptotic variance" of the estimator. Isn't it just zero, like the asymptotic bias? This last issue will be the topic for the next post in this sequence. 1. Hi, Thanks for this very helpful entry. In my class I also use a similar MC experiment to demonstrate the consistency of the OLS estimator of the AR(1) process. I use STATA and I thought it may be helpful for those are interested in STATA programming. So here is the code with the same DGP and parameter values. (PS: I admit that there may be more elegant ways of coding this :)) // Consistency of OLS estimator of AR(1) model clear capt prog drop arols program arols, rclass version 12 syntax [, N(int 50), burnin(int 100)] drop _all set obs n' gen double u = -1 + 2runiform() // random error, from U(-1,1) gen y = 0 replace y = 1 + 0.5y[_n-1] + u in 2/n' drop in 1/burnin' gen t = _n tsset t reg y l.y ret sca b1 =_b[_cons] ret sca b2 =_b[l.y] end graph drop _all glo numsim = 5000 local burnin = 200 foreach T of numlist 20 100 250 1000 5000 { local n = burnin' + T' simulate ARparT'=r(b2), /// reps(\$numsim) saving(OLSMCT', replace) nolegend nodots: /// arols, n(n') burnin(burnin') hist ARparT', normal name(nT') title("AR(1) Parameter: n=T'") } clear use OLSMC20 foreach T of numlist 100 250 1000 5000{ merge using OLSMC`T' drop _merge } sum tw (kdensity ARpar20) (kdensity ARpar100) (kdensity ARpar250) (kdensity ARpar1000) (kdensity ARpar5000) The code above draws the kernel density estimates similar to the last graph in your post. Also here are the summary statistics: Variable \| Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- ARpar20 \| 5000 .371518 .2144403 -.5469691 .9904804 ARpar100 \| 5000 .4751635 .0893891 .0822227 .7326534 ARpar250 \| 5000 .4906103 .0546967 .2765988 .6818358 ARpar1000 \| 5000 .4973098 .0272844 .3987602 .5796224 ARpar5000 \| 5000 .4998611 .0121114 .4362852 .5412318 2. very useful to understand the concept of consistency	{}
# Looping over files in C++ and changing their names to store data into a single file I have a lot of data in separate files generated from a simulation program. All data is clear and formatted via a single space to read in 4 columns. I can open and store data from each file one at a time, but that will get so impossible once I hit more than, say, 10 files of data with the same simulation data columns. I am just trying to find a better way to open files and store the data using an iterative method instead of typing, for example, "open file, read file data, append to file" over 1000 times. #include <iostream> #include <sstream> #include <fstream> #include <vector> using namespace std; int main () { // INITIALIZE int tmax = 80000; float time_step[tmax]; float LEx[tmax]; float LEy[tmax]; float LEz[tmax]; float LE[tmax]; float AvgLEx; float AvgLEy; float AvgLEz; float AvgLE; float Function[tmax]; // FILES ifstream infile; infile.open("File_1.prn"); // THIS INFILE is where I want to put a loop to change File_1 to File_2 to File_3 and so on and so on to append data from all those files into ONE single file if(infile.fail()) { cout << "error reading file" << endl; return 1; } for (int i = 0; i < tmax+1; ++i) { infile >> time_step[i] >> LEx[i] >> LEy[i] >> LEz[i] >> LE[i]; AvgLEx = AvgLEx + LEx[i]; AvgLEy = AvgLEy + LEy[i]; AvgLEz = AvgLEz + LEz[i]; AvgLE = AvgLE + LE[i]; } AvgLEx = AvgLEx/tmax; AvgLEy = AvgLEy/tmax; AvgLEz = AvgLEz/tmax; AvgLE = AvgLE/tmax; cout << "AvgLE: " << AvgLE << "\n"; cout << time_step[tmax] << " LEx " << LEx[tmax] << " LEy " << LEy[tmax] << " LEz: " << LEz[tmax] << " LE: " << LE[tmax] << endl; cout << "\n"; infile.close(); // FUNCTIONS for (int i = 0; i < tmax+1; ++i) { Function[i] = LE[i] - AvgLE; cout << LE[i] << " " << AvgLE << " " << Function[i] << "\n"; } cout << "\n Function set up \n"; // // CONVOLUTION float A_t; float A_dt; int dt, t; float max=0; for (int dt = 0; dt < 1 + tmax/10; ++dt) { //dt = 1; cout << "dt: " << dt << "\n"; //A_dt = 0; int window = tmax-dt; for (int t = 0; t < window; ++t) { if (t==0) { A_t = 0; cout << "A_t was set to zero \n"; } A_t+= Function[t] * Function[t + dt]; cout << "A_t is accumulation ...: " << A_t << "\n" ; } A_dt = A_t / window; if (A_dt > max) { max = A_dt; } // OUTPUT ofstream AutoCor; AutoCor.open ("AutoCor_25.prn", ios::app); AutoCor << dt << " " << A_dt << "\n"; AutoCor.close(); cout << "A_dt: " << A_dt << " " << "window used: " << window << "\n"; } cout << "max used: " << max << "\n"; // RETURN return A_dt; return 0; } • I'm sorry, but request regarding adding new implementation is off-topic. – Jamal Jul 14 '14 at 18:49 • I added all my working code. I just am trying to find a better way to open files and store the data using an iterative method instead of typing ex: open file, read file data, append to file. over 1000 times – Kitt Kat Jul 14 '14 at 18:55 • Okay. I'll replace the off-topic explanation with that, then reopen. – Jamal Jul 14 '14 at 18:56 • You can just use the already existing command cat to do this. – Martin York Jul 14 '14 at 20:05 • @Sarah You have to type the file names in somehwere! Unless they have a common prefix then you can use shell expansion cat file* > all.file or cat file{1..3} > file4 – Martin York Jul 16 '14 at 21:25 Creating Filenames If you know the naming format, you can use std::ostringstream or snprintf to create a filename: char filename[40]; for (unsigned int i = 0; i < 100; i++) { snprintf(filename, sizeof(filename), "file_%04d.prn"); std::ifstream infile(filename); //... } Another, platform specific, method is to iterate over each file. One algorithm is: 1. filename = Get_First_Filename(); 2. process file 3. filename = Get_Next_Filename(); 4. if (filename != NULL) goto to step 2 An array of structures vs. many arrays of variables Instead of declaring many arrays like this: float LEx[tmax]; float LEy[tmax]; float LEz[tmax]; float LE[tmax]; You may want to put them into a structure (record) and use an array of the structures: struct Record { float LEx; float LEy; float LEz; float LE; }; std::vector< Record > data(tmax); You may find that this change will speed up your program because all the relevant variables are next to each other instead of being separated by tmax elements. Use double not float In your programs prefer the double type over float. The double is more accurate and most floating point libraries are designed for double. The float should be used in constrained environments where memory space is restricted. In your program, you read a record of data then process it: for (int i = 0; i < tmax+1; ++i) { infile >> time_step[i] >> LEx[i] >> LEy[i] >> LEz[i] >> LE[i]; AvgLEx = AvgLEx + LEx[i]; AvgLEy = AvgLEy + LEy[i]; AvgLEz = AvgLEz + LEz[i]; AvgLE = AvgLE + LE[i]; } Most platforms are most efficient when the reading of data is continuous. This allows the hard drive to keep spinning rather than spinning down and stopping after a read, then spinning up again for the next read (yes, modern drives use caches to smooth this). A more efficient method is to read all or most of your data, then process it: struct Input_Record { double time_step; double LEx; double LEy; double LEz; }; std::vector< Input_Record > input_data(tmax); for (unsigned int i = 0; i < tmax; ++i) { infile >> input_data[i].time_step >> input_data[i].LEx >> input_data[i].LEy >> input_data[i].LEz; } for (unsigned int i = 0; i < tmax; i++) { AvgLEx += input_data[i].LEx; AvgLEy += input_data[i].LEy; AvgLEz += input_data[i].LEz; } • wow. So cool. This is such a clearer way to understand what is going on 'behind the scenes'. The use of structures is MOST definitely going to be helpful for me. – Kitt Kat Jul 14 '14 at 21:35 • I think that for the file part - I do know the file format because I create it myself in TCL to store the data when I run/call my simulation program. So if I understand what you wrote above correctly in the following -- char filename[40]; for (unsigned int i = 0; i < 100; i++) { snprintf(filename, sizeof(filename), "file_%04d.prn"); std::ifstream infile(filename); //... } .. you are using the "%04d" part to add the '1', '2', '3' ..etc changes to the file names so that it is then possible to loop over them and collect the data from each file? – Kitt Kat Jul 14 '14 at 21:37 • Yes, the snprint function is used to append numbers to a file name. So the first filename is "file_0000.prn", the second is "file_0001.prn", and so on. – Thomas Matthews Jul 15 '14 at 0:43 As you're new to C++, you should be aware that implementing an entire program in main() can make things difficult, especially when the code is already complex. As it appears, you have declare and initialize variables at the top, open the file, perform the operations, then output things. Even with these comments in place, you should still split this up into functions. 1. Do not maintain a list of variables as it could make maintenance more difficult, such as if variables are no longer being used. Instead. you should only declare or initialize variables as close to their intended use as possible. That way, if you no longer use that variable, you won't have to search for it at the top. This will also make your code easier to follow. 2. Consider just having main() do two things: attempt to open the file and call all the necessary functions afterwards. You can also open the file in an other function, but if you do it in main(), then you can easily return from there if the file cannot be opened. 3. Everything else, especially the file operations, should be in separate functions. Also keep in mind that a function should just have one primary purpose, in order to maintain the Single Responsible Principle (SRP). If a function needs to do something else that another function already does, then it should call that function. You should also name functions, in the verb form, based on what they're supposed to do. This will make the function's intent very clear to others. Example: void openFile() {} Miscellaneous: • Try not to use using namespace std in global scope as it could cause name-clashing bugs, especially in larger programs. See this post for more info on this. • This variable appears to be a constant: int tmax = 80000; If so, initialize it with const so that it cannot be modified: const int tmax = 80000; • For this approach at checking the file: infile.open("File_1.prn"); if(infile.fail()) { cout << "error reading file" << endl; return 1; } you should use std::ifstream, check for !infile(), print the error message to std::cerr, and then return EXIT_FAILURE: std::ifstream infile("File_1.prn"); if (!infile) { return EXIT_FAILURE; } • ok I will make those changes now and then repost a clearer code – Kitt Kat Jul 14 '14 at 19:44 • @Sarah: You may still wait for more answers. This is mostly a general review, and there may be more issues here. I haven't looked at every bit yet. – Jamal Jul 14 '14 at 19:45 • oh ok! I will check back then in a bit – Kitt Kat Jul 14 '14 at 19:51 Your current code looks buggy to me. The first (and most obvious) one that I saw was: for (int i = 0; i < tmax+1; ++i) ...then using i as an index into arrays defined as xxx[tmax]. With tmax as the size of an array, the allowable indices are 0 through tmax-1. Trying to read/write xxx[tmax] and xxx[tmax+1] leads to undefined behavior. Right now, the code also blindly assumes that each input file contains a pre-set number of rows of data (but the intent seems to be that it will then produce a file for which that assumption is false). Instead of defining a (fixed-size) array and basing your assumptions on each file containing a fixed number of items, I'd prefer to use std::vector, and read exactly as many items as the file contains. I agree with @Thomas Matthews' advice to create a struct to hold a single record, and create a vector (he said array, but see above) of those records. I'd go one more step though: I'd overload operator>> (and possibly also operator<<) for that record type to make reading the records convenient and easy as well. class record { float time_step; float LEx; float LEy; float LEz; float LE; friend std::istream &operator>>(std::istream &is, record &r) { return is >> time_step >> r.LEx >> r.LEy >> r.LEz >> r.LE; } }; Using this, reading one file of data into a vector of records can look like this: std::vector<record> records{std::istream_iterator<record>(infile), std::istream_iterator<record>()}; In your case (reading multiple files into a single vector) it's marginally less convenient, but only marginally so: std::vector<record> records; for (std::string const &filename : filenames) { std::istream infile(filename); std::copy(std::istream_iterator<record>(infile), std::istream_iterator<record>(), std::back_inserter(records)); } While it's possible to have the program synthesize the file names, it's generally easier to supply them on the command line, possibly including a wildcard, something like: process File_.prn In this case, your program can simply use argc/argv that are passed to main to process the file names: int main(int argc, char argv) { std::vector<std::string> filenames(argv+1, argv+argc-1); On Unix, passing a wildcard on the command line will normally be handled automatically by the shell. On Windows (at least with MSVC) you can link your file with setargv.obj to get the same effect: cl myfile.cpp setargv.obj. Alternatively, you could have your code enumerate the files fitting a specified pattern. If you're going to do this, I'd recommend using the Boost FileSystem library, so you can do the job portably across most major OSes--and better still, it's being used as the basis for work on an addition to the standard library, so the code you write now will probably work with the future standard library with little or no extra work. Although you didn't ask about them, I'd note that your FUNCTIONS and CONVOLUTION sections could be cleaned up quite a bit as well. It looks like std::accumulate could make computing averages quite a bit cleaner, and your: A_t+= Function[t] Function[t + dt]; ...looks like it could be implemented rather more cleanly with std::inner_product: A_t = std::inner_product(Function.begin(), Function.begin()+window, Function.begin()+dt, 0.0); (Note that this replaces the entire loop to calculate A_t, not just the one line quoted above). I guess I'll stop there (at least for now) since it's not really clear how much you care about cleaning up this part of the code. • No that is great ^^^ !!. Although I was mostly interested (desperate?) to find a solution to the file manipulation .. all these extra comments are so very much appreciated AND useful. I figured there was a lot about C++ syntax that I am clearly not exploiting/using/aware of and that was frustrating to me while I was writing the code initially. So thank you @JerryCoffin – Kitt Kat Jul 15 '14 at 17:22	{}
Power (physics) Last updated Power Common symbols P SI unit watt (W) In SI base units kgm 2s −3 Derivations from other quantities Dimension ${\displaystyle {\mathsf {L}}^{2}{\mathsf {M}}{\mathsf {T}}^{-3}}$ In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. [1] [2] [3] Power is a scalar quantity. Contents Power is related to other quantities; for example, the power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft. Likewise, the power dissipated in an electrical element of a circuit is the product of the current flowing through the element and of the voltage across the element. [4] [5] Definition Power is the rate with respect to time at which work is done; it is the time derivative of work: ${\displaystyle P={\frac {dW}{dt}}}$ where P is power, W is work, and t is time. If a constant force F is applied throughout a distance x, the work done is defined as ${\displaystyle W=\mathbf {F} \cdot \mathbf {x} }$. In this case, power can be written as: ${\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\left(\mathbf {F} \cdot \mathbf {x} \right)=\mathbf {F} \cdot {\frac {d\mathbf {x} }{dt}}=\mathbf {F} \cdot \mathbf {v} }$ If instead the force is variable over a three-dimensional curve C, then the work is expressed in terms of the line integral: ${\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {r} =\int _{\Delta t}\mathbf {F} \cdot {\frac {d\mathbf {r} }{dt}}\ dt=\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt}$ From the fundamental theorem of calculus, we know that ${\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt=\mathbf {F} \cdot \mathbf {v} .}$ Hence the formula is valid for any general situation. Units The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second. Other common and traditional measures are horsepower (hp), comparing to the power of a horse; one mechanical horsepower equals about 745.7 watts. Other units of power include ergs per second (erg/s), foot-pounds per minute, dBm, a logarithmic measure relative to a reference of 1 milliwatt, calories per hour, BTU per hour (BTU/h), and tons of refrigeration. Average power As a simple example, burning one kilogram of coal releases much more energy than detonating a kilogram of TNT, [6] but because the TNT reaction releases energy much more quickly, it delivers far more power than the coal. If ΔW is the amount of work performed during a period of time of duration Δt, the average power Pavg over that period is given by the formula: ${\displaystyle P_{\mathrm {avg} }={\frac {\Delta W}{\Delta t}}}$ It is the average amount of work done or energy converted per unit of time. The average power is often simply called "power" when the context makes it clear. The instantaneous power is then the limiting value of the average power as the time interval Δt approaches zero. ${\displaystyle P=\lim _{\Delta t\to 0}P_{\mathrm {avg} }=\lim _{\Delta t\to 0}{\frac {\Delta W}{\Delta t}}={\frac {dW}{dt}}}$ In the case of constant power P, the amount of work performed during a period of duration t is given by: ${\displaystyle W=Pt}$ In the context of energy conversion, it is more customary to use the symbol E rather than W. Mechanical power Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity, or the product of a torque on a shaft and the shaft's angular velocity. Mechanical power is also described as the time derivative of work. In mechanics, the work done by a force F on an object that travels along a curve C is given by the line integral: ${\displaystyle W_{C}=\int _{C}\mathbf {F} \cdot \mathbf {v} \,dt=\int _{C}\mathbf {F} \cdot d\mathbf {x} }$ where x defines the path C and v is the velocity along this path. If the force F is derivable from a potential (conservative), then applying the gradient theorem (and remembering that force is the negative of the gradient of the potential energy) yields: ${\displaystyle W_{C}=U(A)-U(B)}$ where A and B are the beginning and end of the path along which the work was done. The power at any point along the curve C is the time derivative: ${\displaystyle P(t)={\frac {dW}{dt}}=\mathbf {F} \cdot \mathbf {v} =-{\frac {dU}{dt}}}$ In one dimension, this can be simplified to: ${\displaystyle P(t)=F\cdot v}$ In rotational systems, power is the product of the torque τ and angular velocity ω, ${\displaystyle P(t)={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }}}$ where ω measured in radians per second. The ${\displaystyle \cdot }$ represents scalar product. In fluid power systems such as hydraulic actuators, power is given by ${\displaystyle P(t)=pQ}$ where p is pressure in pascals, or N/m2 and Q is volumetric flow rate in m3/s in SI units. If a mechanical system has no losses, then the input power must equal the output power. This provides a simple formula for the mechanical advantage of the system. Let the input power to a device be a force FA acting on a point that moves with velocity vA and the output power be a force FB acts on a point that moves with velocity vB. If there are no losses in the system, then ${\displaystyle P=F_{\text{B}}v_{\text{B}}=F_{\text{A}}v_{\text{A}},}$ and the mechanical advantage of the system (output force per input force) is given by ${\displaystyle \mathrm {MA} ={\frac {F_{\text{B}}}{F_{\text{A}}}}={\frac {v_{\text{A}}}{v_{\text{B}}}}.}$ The similar relationship is obtained for rotating systems, where TA and ωA are the torque and angular velocity of the input and TB and ωB are the torque and angular velocity of the output. If there are no losses in the system, then ${\displaystyle P=T_{\text{A}}\omega _{\text{A}}=T_{\text{B}}\omega _{\text{B}},}$ ${\displaystyle \mathrm {MA} ={\frac {T_{\text{B}}}{T_{\text{A}}}}={\frac {\omega _{\text{A}}}{\omega _{\text{B}}}}.}$ These relations are important because they define the maximum performance of a device in terms of velocity ratios determined by its physical dimensions. See for example gear ratios. Electrical power The instantaneous electrical power P delivered to a component is given by ${\displaystyle P(t)=I(t)\cdot V(t)}$ where • ${\displaystyle P(t)}$ is the instantaneous power, measured in watts (joules per second) • ${\displaystyle V(t)}$ is the potential difference (or voltage drop) across the component, measured in volts • ${\displaystyle I(t)}$ is the current through it, measured in amperes If the component is a resistor with time-invariant voltage to current ratio, then: ${\displaystyle P=I\cdot V=I^{2}\cdot R={\frac {V^{2}}{R}}}$ where ${\displaystyle R={\frac {V}{I}}}$ is the electrical resistance, measured in ohms. Peak power and duty cycle In the case of a periodic signal ${\displaystyle s(t)}$ of period ${\displaystyle T}$, like a train of identical pulses, the instantaneous power ${\textstyle p(t)=\|s(t)\|^{2}}$ is also a periodic function of period ${\displaystyle T}$. The peak power is simply defined by: ${\displaystyle P_{0}=\max[p(t)]}$ The peak power is not always readily measurable, however, and the measurement of the average power ${\displaystyle P_{\mathrm {avg} }}$ is more commonly performed by an instrument. If one defines the energy per pulse as: ${\displaystyle \varepsilon _{\mathrm {pulse} }=\int _{0}^{T}p(t)\,dt}$ then the average power is: ${\displaystyle P_{\mathrm {avg} }={\frac {1}{T}}\int _{0}^{T}p(t)\,dt={\frac {\varepsilon _{\mathrm {pulse} }}{T}}}$ One may define the pulse length ${\displaystyle \tau }$ such that ${\displaystyle P_{0}\tau =\varepsilon _{\mathrm {pulse} }}$ so that the ratios ${\displaystyle {\frac {P_{\mathrm {avg} }}{P_{0}}}={\frac {\tau }{T}}}$ are equal. These ratios are called the duty cycle of the pulse train. Power is related to intensity at a radius ${\displaystyle r}$; the power emitted by a source can be written as:[ citation needed ] ${\displaystyle P(r)=I(4\pi r^{2})}$ Related Research Articles In mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object. The magnitude of an object's acceleration, as described by Newton's Second Law, is the combined effect of two causes: Continuum mechanics is a branch of mechanics that deals with the mechanical behavior of materials modeled as a continuous mass rather than as discrete particles. The French mathematician Augustin-Louis Cauchy was the first to formulate such models in the 19th century. A centripetal force is a force that makes a body follow a curved path. Its direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path. Isaac Newton described it as "a force by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre". In Newtonian mechanics, gravity provides the centripetal force causing astronomical orbits. In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors. In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. It represents the capability of a force to produce change in the rotational motion of the body. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Torque is defined as the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically , the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted by M. In physics, the Navier–Stokes equations are certain partial differential equations which describe the motion of viscous fluid substances, named after French engineer and physicist Claude-Louis Navier and Anglo-Irish physicist and mathematician George Gabriel Stokes. They were developed over several decades of progressively building the theories, from 1822 (Navier) to 1842–1850 (Stokes). Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies without considering the forces that cause them to move. Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system. Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics. In physics, angular velocity or rotational velocity, also known as angular frequency vector, is a pseudovector representation of how fast the angular position or orientation of an object changes with time. The magnitude of the pseudovector represents the angular speed, the rate at which the object rotates or revolves, and its direction is normal to the instantaneous plane of rotation or angular displacement. The orientation of angular velocity is conventionally specified by the right-hand rule. In physics, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force. The moment of inertia, otherwise known as the mass moment of inertia, angular mass, second moment of mass, or most accurately, rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis, akin to how mass determines the force needed for a desired acceleration. It depends on the body's mass distribution and the axis chosen, with larger moments requiring more torque to change the body's rate of rotation. In the physical science of dynamics, rigid-body dynamics studies the movement of systems of interconnected bodies under the action of external forces. The assumption that the bodies are rigid simplifies analysis, by reducing the parameters that describe the configuration of the system to the translation and rotation of reference frames attached to each body. This excludes bodies that display fluid, highly elastic, and plastic behavior. Particle velocity is the velocity of a particle in a medium as it transmits a wave. The SI unit of particle velocity is the metre per second (m/s). In many cases this is a longitudinal wave of pressure as with sound, but it can also be a transverse wave as with the vibration of a taut string. Particle displacement or displacement amplitude is a measurement of distance of the movement of a sound particle from its equilibrium position in a medium as it transmits a sound wave. The SI unit of particle displacement is the metre (m). In most cases this is a longitudinal wave of pressure, but it can also be a transverse wave, such as the vibration of a taut string. In the case of a sound wave travelling through air, the particle displacement is evident in the oscillations of air molecules with, and against, the direction in which the sound wave is travelling. In mechanics, virtual work arises in the application of the principle of least action to the study of forces and movement of a mechanical system. The work of a force acting on a particle as it moves along a displacement is different for different displacements. Among all the possible displacements that a particle may follow, called virtual displacements, one will minimize the action. This displacement is therefore the displacement followed by the particle according to the principle of least action. The work of a force on a particle along a virtual displacement is known as the virtual work. In a compressible sound transmission medium - mainly air - air particles get an accelerated motion: the particle acceleration or sound acceleration with the symbol a in metre/second2. In acoustics or physics, acceleration is defined as the rate of change of velocity. It is thus a vector quantity with dimension length/time2. In SI units, this is m/s2. In calculus, the Leibniz integral rule for differentiation under the integral sign, named after Gottfried Leibniz, states that for an integral of the form In differential calculus, the Reynolds transport theorem, or simply the Reynolds theorem, named after Osborne Reynolds (1842–1912), is a three-dimensional generalization of the Leibniz integral rule. It is used to recast time derivatives of integrated quantities and is useful in formulating the basic equations of continuum mechanics. In electromagnetism, current density is the amount of charge per unit time that flows through a unit area of a chosen cross section. The current density vector is defined as a vector whose magnitude is the electric current per cross-sectional area at a given point in space, its direction being that of the motion of the positive charges at this point. In SI base units, the electric current density is measured in amperes per square metre. Linear motion, also called rectilinear motion, is one-dimensional motion along a straight line, and can therefore be described mathematically using only one spatial dimension. The linear motion can be of two types: uniform linear motion with constant velocity or zero acceleration; and non-uniform linear motion with variable velocity or non-zero acceleration. The motion of a particle along a line can be described by its position , which varies with (time). An example of linear motion is an athlete running 100m along a straight track. References 1. Fowle, Frederick E., ed. (1921). Smithsonian Physical Tables (7th revised ed.). Washington, D.C.: Smithsonian Institution. OCLC 1142734534. Archived from the original on 23 April 2020. Power or Activity is the time rate of doing work, or if W represents work and P power, P = dw/dt. (p. xxviii) ... ACTIVITY. Power or rate of doing work; unit, the watt. (p. 435) 2. Heron, C. A. (1906). "Electrical Calculations for Rallway Motors". Purdue Eng. Rev. (2): 77–93. Archived from the original on 23 April 2020. Retrieved 23 April 2020. The activity of a motor is the work done per second, ... Where the joule is employed as the unit of work, the international unit of activity is the joule-per-second, or, as it is commonly called, the watt. (p. 78) 3. "Societies and Academies". Nature. 66 (1700): 118–120. 1902. Bibcode:1902Natur..66R.118.. doi:. If the watt is assumed as unit of activity... 4. Halliday and Resnick (1974). "6. Power". Fundamentals of Physics.{{cite book}}: CS1 maint: uses authors parameter (link) 5. Chapter 13, § 3, pp 13-2,3 The Feynman Lectures on Physics Volume I, 1963 6. Burning coal produces around 15-30 megajoules per kilogram, while detonating TNT produces about 4.7 megajoules per kilogram. For the coal value, see Fisher, Juliya (2003). "Energy Density of Coal". The Physics Factbook. Retrieved 30 May 2011. For the TNT value, see the article TNT equivalent. Neither value includes the weight of oxygen from the air used during combustion.	{}
## Margin Percentage Calculator The margin calculator provides a simple percentage calculation of the required Leverage (also known as Margin Level) for each tradable instrument offered on our platform. With Fortrade’s Margin Calculator you can calculate exactly how much margin is required in order to guarantee a position that you would like to open, and adjust its size accordingly, if necessary. #### How to use the Margin Percentage Calculator? 2. Choose the currency pair for which you would like to calculate the margin percentage. 3. Define the margin ratio from the predefined ratios in the dropdown list. 4. Type the amount you would like to calculate, using numbers only. 5. Click on the CONVERT button, and the result will appear underneath. • Didn't find what you were looking for?	{}
When we learn how to multiply polynomials, we often encounter special cases that occur very often. These special cases involve multiplying two or more binomials. Instead of grinding out the work each time, we can memorize a formula in each case, and then apply the formula to our given problem. These types of problems are known as special products or special polynomial products. Test Objectives • Demonstrate an understanding of the rules of exponents • Demonstrate the ability to multiply conjugates • Demonstrate the ability to square a binomial • Demonstrate the ability to cube a binomial Special Polynomial Products Practice Test: #1: Instructions: Find each product. $$a)\hspace{.2em}(3x + 7)(3x - 7)$$ $$b)\hspace{.2em}(5x - 6)(5x + 6)$$ #2: Instructions: Find each product. $$a)\hspace{.2em}(6x - 1)^2$$ $$b)\hspace{.2em}(x + 6)^2$$ #3: Instructions: Find each product. $$a)\hspace{.2em}(7 + 6x)(7 - 6x)$$ $$b)\hspace{.2em}(2x^2 + 3)^2$$ #4: Instructions: Find each product. $$a)\hspace{.2em}(8x - 2)^3$$ $$b)\hspace{.2em}(5x + 1)^3$$ #5: Instructions: Find each product. $$a)\hspace{.2em}(3x^2 - y^2)^3$$ $$b)\hspace{.2em}(5x^2 + 3y^4)^3$$ Written Solutions: #1: Solutions: $$a)\hspace{.2em}9x^2 - 49$$ $$b)\hspace{.2em}25x^2 - 36$$ #2: Solutions: $$a)\hspace{.2em}36x^2 - 12x + 1$$ $$b)\hspace{.2em}x^2 + 12x + 36$$ #3: Solutions: $$a)\hspace{.2em}49 - 36x^2$$ $$b)\hspace{.2em}4x^4 + 12x^2 + 9$$ #4: Solutions: $$a)\hspace{.2em}512x^3 - 384x^2 + 96x - 8$$ $$b)\hspace{.2em}125x^3 + 75x^2 + 15x + 1$$ #5: Solutions: $$a)\hspace{.2em}27x^6 - 27x^4y^2 + 9x^2y^4 - y^6$$ $$b)\hspace{.2em}27y^{12}+ 135x^2y^8 + 225x^4y^4 + 125x^6$$	{}
# Why does maximal entropy imply equilibrium? From a purely thermodynamical point of view, why does that entropy have to be a maximum at equilibrium? Say there is equilibrium, i.e. no net heat flow, why can the entropy not be sitting at a non-maximal value? From the second law of thermodynamics, it follows that $S$ never gets smaller and of course I know that for an isolated system there are many statements involving $\text{d}S=\frac{\delta Q}{T}$, which say what might happen for processes. But if I have equilibrium, then no relevant processes are going on. Some proofs in thermodynamics involve arguments how if we don't have maximal entropy, then we can do something which raises entropy. But why is that relevant or related equilibrium, i.e. to the positions for the termodynamic parameters, which don't change with time? One could argue that probably the energy $U$ doesn't sit at the minimal value, but in thermodynamics, without microscopic forces, the statement that the energy changes towards equilibrium and seeks its minimal value seems to be derived from maximal entropy. And how can I conclude the converse? Why does equilibrium follow from $\text{d}S=0$? (edit: I see the question was just bumped on the front page, and as it's a year old now, I guess my interest in the topic has changed in so far as I'm not particularly happy with the formulation of the initial question anymore. That is, I guess without proper stating the definitions, it might be difficult to give a good answer - the thing I still don't "like", and that might be kind of a language problem, is that "to maximize" implies that there is a family of values it could take - but it takes the maximum - while at the same time, you often consider a deviation away from a thermodynamical state to be a transition into a configuration where thermodynamics doesn't apply anymore. Hence, if you go away from the "maximal value", you might lose the concept of there being a temperature at all, but since this is what makes the parameter space with respect you use the word "maximal" entropy, you get into language problems. But at least I do see it's use in explainaing how the entropy/energy evolves once you take the extensive parameters into your control and change the system quasistatically. In any case, I am and was only only interested in a non-statistical mechanics answer here. Clearly, I can make sense of the physics using the microscopic picture anyway, but was purely interested in the formulation of the thermodynamical theory here.) - Sometimes "entropy is maximum in equilibrium" is just put into the formulation of the second law. So what is the exact formulation of the second law you are using? – Yrogirg Jan 30 '12 at 17:41 The Clausius or Kelvin formulations at the beginning of the wikipedia page would suffice. The mathematical formulations involving $S$ are good as well. It's just that I don't see how a statement involving equilibrium, which is a concept involving time and time evolution, can be characterized in classical thermodynamics in terms or $dS$ and so on. – NikolajK Jan 30 '12 at 18:02 @Nick Kidman Comparative statics has a concept of equilibrium but no dynamical theory. This reminds me of Joan Robinson's famous critique of neo-neo-classical economic theory's use of comparative statics. – joseph f. johnson Jun 8 '13 at 16:20 @josephf.johnson: erm.. – NikolajK Jun 8 '13 at 17:33 well, there's these things called "thermodynamic potentials"---I'm thinking about an answer to your question. – joseph f. johnson Jun 8 '13 at 18:22 First, if ${\rm d}S\neq 0$, then the entropy will change, and because something is changing, it's obviously not an equilibrium. If the physical system doesn't maximize the entropy and it's composed of many parts that may interact with each other, directly or indirectly (it's not disconnected), then any path that allows the entropy to be increased (given fixed values of conserved quantities such as energy) will be realized, so you will be away from the equilibrium because something will change. If a system is composed of two or more decoupled, non-interacting components – like a bottle of blue lemonade and a bottle of red lemonade – they may be in equilibrium even if the entropy isn't maximized. One could increase the entropy by mixing the liquids but because they're not in contact, they won't be mixed. On the contrary, if the entropy is already maximized, the only way how the system may evolve is to evolve into another state with the same, maximum value of entropy: there's no higher allowed value and the second law of thermodynamics prohibits a decreasing entropy. This is atypical because when we maximize entropy among all states with the same conserved quantities, the state of maximum entropy is typically unique. For example, if there are also movable macroscopic bodies that may create heat by friction, the entropy is maximized only when the friction stops the macroscopic motion and converts its energy to heat. In all these discussions, one has to be careful on whether or not we're maximizing the entropy among all states or just the states with the same value of energy (and other conserved quantities). If we allow the energy to change arbitrarily, the entropy isn't really bounded from above (and discussions about its maximization are rendered meaningless) because any body may be heated to pretty much arbitrarily high temperature (or it may collapse into a black hole with an ever greater mass and therefore an ever greater entropy, too). - Okay, the second part basically says that if the maximum is unique, then the entropy has nowhere to go, so equilirbium. I still don't see why $\text{S}\ne 0$ implies that the entropy will change. Just becasue $S(U,V)$ is such that it grows with $U$ and $V$, why does this imply that $S$ will change with time on it's own? In classical mechanics, if $F=-\nabla U$ isn't $0$, then I understand that something will happen. However, I don't see how what in thermodynamics says that "if $\text{d}S\ne 0$, then the entropy will change". What says such a $(U,V),(T,P),...$ or whatever point is not stable? – NikolajK Jan 29 '12 at 18:36 Dear Nick, as I said, I think that your statement that "something will change" is only true if the entropy is (locally) non-maximal among states with the same energy. Then it is guaranteed that the system will evolve in the direction in which the entropy increases - and free energy decreases. The force is really $-\nabla \Phi$, a gradient of the free energy, when one does it right, so this increases with the gradient of the entropy. – Luboš Motl Jan 30 '12 at 7:06 But where in the laws of thermodynamics is this movement away from non extremal values encoded? I don't see that if thermodynamics is a theory which makes statements about equilibrium, how it handles points down at the math, which "are not yet" at equilibrium and tells them where to go. – NikolajK Jan 30 '12 at 7:15 Dear Nick, the increase of entropy is the second law of thermodynamics. When the entropy isn't maximized, the second law may really be interpreted as a "strict inequality": the entropy strictly increases, so something has to be changing about the system because $S$, a function of the variables, goes up. That's why it's not an equilibrium. The actual rate by which $S$ increases depends on the situation but as I indicated, the force trying to move the system in the higher-entropy direction is really given by $F=-\nabla (E-TS)$, a gradient of free energy. It includes $+T\cdot\nabla S$. – Luboš Motl Jan 30 '12 at 9:38 @Motl: I still don't see why the laws of thermodynamics can make statements about a change in time. I principally don't see how there can be a theorem about what will happen, if there is no time involved in the theory based on three axioms. The force relation $mx''(t)=F=-\nabla...$ isn't part of classical thermodynamics, is it? – NikolajK Jan 31 '12 at 22:08 Let us focus on the case of a simple homogeneous substance that does not dissociate or undergo other chemical reactions. Then a thermodynamic state is any state where the substance has a definite pressure, equal in all its parts, a definite temperature, without any temperature variations in its different parts, and a fixed volume (it is not busy expanding or contracting). As usual we also assume the mass is fixed and so the density is also definitely related to the volume: so nothing more will be said about mass or density. A bit of, e.g., gas cannot be in equilibrium if different parts have different temperatures, this is an observational fact: the temperatures tend to equalise. But we just make it part of the definition of a 'state' that the system has a definite temperature. It is only in more advanced, not so classical, thermodynamics that one relaxes to some extent this assumption. This is not the definition of equilibrium, it is the definition of 'state'. Now further, the definition of equilibrium is that if there is any thermal contact with the environment, then the environment must also be at a definite temperature, the same as that of the system. If the system can interact with the environment through doing work by expanding its volume, then the pressure of the system must be equal to the pressure of the environment. This is the definition of equilibrium. Note carefully: if there is no interaction with the environment, then every thermodynamic state is a state of equilibrium, because there is no condition that it has to satisfy. All of these remarks are logically prior to either the First or the Second Law of Thermodynamics, prior, too, to notions of reversibility or irreversibility, even to the notion of 'path'. For that reason, they are sometimes called 'The Zero'th Law of Thermodynamics'. All of this makes sense even if one does not know that temperature or pressure is the result of the motion of molecules: it was formulated before the modern theory of heat was understood, it was sometimes thought that heat was a fluid. Comparative Statics does not have a true dynamics, but it is meant as a shortcut for use when the true dynamical evolution of the system is either unknown or too complicated to analyse. One defines `equilibrium' by the balance of forces, by having the net sum of the forces acting at any point be zero. In thermodynamics, the two main principles are equalisation of temperature, thought of as the result of transferring heat from the hotter body to the colder, and equalisation of pressure, which occurs by expansion or contraction of the volume. It is assumed that if one studies the properties of a system at one equilibrium point (i.e., the equilibrium state which would subsist under one set of values of the parameters such as the temperature of the environment and the pressure of the environment), and then studies the properties of that same system at a different equilibrium point which would subsist under different values of the parameters, then one has a good deal of information even about the process of transformation from the first point to the second point. In thermodynamics this is true provided the transformation proceeds slowly enough. Equilibrium is meant to embody the notion that the system is in a stable state and will not evolve. Even so, the notion in thermodynamics of a reversible path is that a system passes through several different equilibrium states under outside influence (namely, a change in the external parameters which means that what was formerly an equilibrium state is no longer in equilibrium with the environment). Although it may violate strict logic, this is a very useful idealisation of reality. (In reality, only irreversible processes can take place.) All of Classical Thermodynamics is Equilibrium Thermodynamics (to a first approximation), but it studies heat engines and refrigerators, in which the system does pass through different states, changing its pressure and volume. The theory uses comparative statics to be able to make statements about the end results of these processes, without being able to say anything about how the process happens exactly, at what rate, or anything truly dynamical. If the system is in the state given by an equilibrium point, it will not change its state unless the external conditions are changed. If the external conditions, such as the temperature of the environment, go through a finite non-zero change, the shock will lead to an irreversible transformation through disequilibrium states, e.g., the part of the system closer to the boundary will heat up before the rest of the system, and this means the system is not in a thermodynamic state at all. Eventually the system will reach a new equilibrium state, but the amount of work done or heat absorbed during this transition period is not exactly calculable using Classical Equilibrium Thermodynamics. (At least, not in the elementary part.) If the non-zero change is very small, it will be a reasonable approximation to treat it as the result of an infinite number of infinitesimal changes. An infinitesimal change does not shock the system out of equilibrium, but coaxes it gently to a neighbouring state of equilibrium. Hence the system traverses a reversible path of equilibrium states. Infinitely slowly. In reality this is not possible. Logically, if a path is reversible, then there is no reason for the system to pick which direction in which to traverse it, so the system will stay put. But this is an idealisation which is a useful approximation to the real world of irreversible transformations through disequilibrium states. More importantly, it gives us a theoretical bound on the efficiency of an irreversible process, an efficiency which can never be improved on. The Second Law of Thermodynamics will be taken in the formulation due to Lord Kelvin and preferred by Nobel laureate Max Planck. It is impossible to find a cycle (of paths) whose only effects on the environment are to produce work and take heat from the environment. Note by contrast that the Carnot cycle both takes in heat and puts out heat. In fact, it has to have two heat reservoirs, at two different temperatures. It takes in heat from the hot one, and puts out heat to the cold one. The point of the Second Law is that it is impossible to turn into work all the heat taken from the hot reservoir. Some of the heat has to be 'passed on' to a colder reservoir. Our strategy will be to first consider reversible cycles. A reversible path is one where we infinitesimally nudge one equilibrium state to a nearby one and there are only two ways to do this: we can put it into thermal contact with an inifitesimally hotter (or colder) heat reservoir. It will absorb (or lose) a tiny amount of heat so slowly that the equilibrium is not disturbed, and any other adjustments allowed by the setup will also happen infinitely slowly and delicately. We can adjust its volume infinitesimally: we can alter, very delicately and infinitesimally, the pressure of the environment which, if we allow the system to be in the kind of setup where it feels the pressure and can alter its volume, it then contracts (or expands) until pressure is equalised. There are only two ways to reversibly move the equilibrium state around, any reversible path is really a combination of adiabatic transformations and isothermal transformations. If we allow thermal contact with the environment, that is of course an isothermal transformation. If we do not allow thermal interaction with the environment, then of course no heat can be exchanged, and that by definition is an adiabatic transformation, and that is the only way the temperature can be altered. The first point is to deduce from Lord Kelvin's formulation of the Second Law of Thermodynamics that no cycle (heat engine) can be more efficient than a reversible cycle (heat engine). The basic idea of this deduction is that if C were a cycle (and, for simplicity, assume it operates between two heat reservoirs only, 1 is 'hot' and 2 is 'cold') that was more efficient than a reversible cycle R operating between the same two reservoirs, then the Second Law would be violated. For we could run the engine C and use its external work to drive R in reverse, as a refrigerator, since R is reversible. By scaling R up or down, if necessary, we can assume that R requires the same amount of heat input from the cold reservoir as is output there by C. Since C is more efficient than R, C produces more work than R really needs to operate. Hence, the combination of running C then R is that the cold heat reservoir is unchanged, heat has been absorbed from the hot reservoir only, and net work has been produced (since not all the work produced by C was needed by R). But this violates the Second Law of Thermodynamics. The second point is to note that therefore, every reversible cycle operating between the same two temperatures has the same efficiency. For we can apply the above argument to two reversible cycles, and then neither of them can be more efficient than the other. We are about to use a little algebra, so we have to formalise the notation for the intuitive argument we already presented in words. Let $W$ be the total work done in one cycle by a given engine. Since it is a cycle, there is no change in the internal energy $U$, so the work done is $W = \Delta Q_1 - \Delta Q_2$, where $\Delta Q_1$ is the heat absorbed at the hot reservoir and $\Delta Q_2$ is the wasted heat, expelled to the cold reservoir. The efficiency, by definition, is $W \over \Delta Q_1$ and this is equal to $1-{\Delta Q_2 \over \Delta Q_1}$. Hence, all reversible cycles operating between the same two reservoirs have the same ratio $\Delta Q_2 \over \Delta Q_1$. This fact is used to define the temperature of the reservoirs by $${\Delta Q^R_2 \over \Delta Q^R_1 } = {T_2 \over T_1} .$$ (I wrote $R$ this time to emphasise that this is only for reversible cycles.) This defines the temperature up to a scaling factor. It also immediately gives us $${\Delta Q^R_2 \over T_2 } = {\Delta Q^R_1 \over T_1} .$$ Now, redefine $\Delta Q_2$ to mean, just as with the other reservoir, the heat absorbed by the system from the cold reservoir. This is more rational, less prejudiced against refrigerators. We now get for all reversible cycles, $${\Delta Q^R_2 \over T_2 } + {\Delta Q^R_1 \over T_1} = 0$$ and for an arbitrary cycle, $${\Delta Q_2 \over T_2 } + {\Delta Q_1 \over T_1} \leq 0 .$$ By approximating an arbitrary path by segments like these and using lots of reservoirs, two for each segment, we get the curved versions: $$\int_{\mbox{closed path}} {dQ \over T} \leq 0,$$ with equality whenever the path is completely reversible. This is Clausius's inequality, and it follows from the Second Law by the line of reasoning we followed involving comparing an arbitrary cycle to a reversible cycle being run in reverse. Arranging it to cancel out the effect of the given cycle on one heat reservoir, and getting a contradiction to the Second Law. If we now confine ourselves to reversible transformations between equilibrium thermodynamic states only, then the fact that this path integral is zero means we can define a potential function $S$ (well-defined up to an additive constant) such that $$S(\mbox{state two}) - S(\mbox{state one}) = \int_1^2 {dQ \over T}.$$ This is pure mathematics. It's like an antiderivative (except in two dimensions). This is the basic definition of entropy in Classical Thermodynamics (as distinguished from the informational definition of entropy in Stats or in Stat Mech, due to Boltzmann, Sir Ronald Fisher, and Shannon). But from the Clausius inequality, it follows that for an irreversible path, $$\int_1^2 {dQ \over t} < S_2 - S_1 .$$ Now, for an isolated system, $dQ$ is always zero since no heat can be exchanged with the environment. So $$S_2 - S_1 > 0.$$ I.e., along an irreversible path, the entropy of an isolated system increases, and along a reversible path, the entropy cannot change. There is something illogical about this. Entropy has only been defined for equilibrium states, and if a system is isolated, there is no way to disturb the system, and so, no processes will take place. But as an idealisation, to give us simplified approximations to real processes provided they take place very slowly, it is logical in its own way. Clausius's inequality does not tell us how fast the dynamics will go along a path, but it will tell us which direction is possible and which direction is impossible. Because $$\int_1^2 {dQ \over t} = - \int_2^1 {dQ \over t} ,$$ so if one direction yields a negative value of the integral, the other direction will yield a positive value, which is forbidden by Clausius's inequality so that other direction is impossible. - This is now Part II. PART II entropy and disequilibrium I will explain the ideas from a footnote in Fermi, Thermodynamics, p. 50, on entropy in disequilibrium. There is the idea, an imperfect one, that 'entropy is additive'. If two systems each undergo a change in entropy $\Delta S_i$, then the combined system undergoes a change in entropy $\Delta S_1 + \Delta S_2$. Fermi comments that this is not always valid, but is if 'the energy of the system is the sum of the energies of all the parts and if the work performed by the system during a transformation is equal to the sum of the amounts of work performed by all the parts. Notice that these conditions are not quite obvious and that in some cases they may not be fulfilled.' I already remarked in Part I that if a substance does not possess one uniform temperature throughout its volume, it is not even in a thermodynamic state. But there is something we can do about this provided there is no turbulence: provided the variation in temperature in space is smooth, we can divide the substance into many small subsystems and make the approximation that the subsytems are each at a uniform temperature and pressure. Can we define the entropy of each small subsystem? Assume further that we can treat the entropy as additive. To do this, we have to assume each small piece is in equilibrium, so it cannot be in thermal contact with the other subsystems. So imagine that each small volume is divided from the rest of the region by little non-conducting walls, which are also immovable. For simplicity, let us consider only two smaller regions. Imagine two boxes of one cubic light year each are filled with hydrogen gas and placed side by side. The left box is at equilibrium at temperature $T_1$, the right one at the higher temperature $T_2$, and each box contains the same mass of gas. Each box is isolated from the environment and from the other box. Then each one is in equilibrium and the combined system is also in equilibrium. We can measure changes in entropy from this state as a starting point, state one in the integral formulas from Part I. If we replace the dividing wall by a perfect thermal conductor, the states are no longer in equilibrium and an irreversible transformation will take place until the temperatures are equalised. We will call this state two. But in order to use our definition of entropy, we have to think of how to connect state one with state two by a reversible transformation that takes place through equilibrium states. We can do this. It is impossible, but still physically meaningful. Replace the dividing wall by a perfect thermal conductor wall only for an infinitesimal period of time, then put the insulating wall back. This is a reversible transformation to a state where the left box is in equilibrium at temperature $T_1 + dT_1$, and the right box has temperature $T_2 + dT_2$. The pressures will have changed a well since the volume has been constant. If these are perfect gases, one can show that $dT_1 = - dT_2$ by conservation of energy, but never mind. This is not really a dynamical analysis since time as a variable is not being considered. Note that this reversible transformation could also be brought about with two external heat reservoirs instead, leaving the dividing insulating wall in place. In Part I we proved that the exact details of how a reversible transformation between two equilibrium states is brought about make no difference to the value of the entropy change. Repeat this process of replacing the insulating wall by a conducting wall briefly, allowing the equilibrium states of the two boxes to be nudged to neighbouring equilibrium states with closer temperatures. Continue until no further change occurs. Obviously by symmetry, this will be for the average temperature $T_1+T_2\over 2$. Or conservation of energy since no work is done, the change in heat is the same as the change in internal energy $U$, and $dQ=c_vdT$; for a perfect gas $c_v$ is a constant. This is a reversible transformation. We can return to state one. We use two heat reservoirs. Leave the dividing insulating wall in place, put the left in contact with an infinitesimally cooler reservoir and the right in contact with an infinitesimally hotter reservoir, repeat as needed until state one is reached. Because the dividing wall was non-conducting, each state traversed is a state of equilibrium. Therfore we can calculate the change in entropy of each box, add them up. The formulas from Part I are $$S_2-S_1 = \int_{T_1}^{T_1 + T_2 \over 2} {dQ_1\over T_1} + \int_{T_2}^{T_1 + T_2 \over 2} {dQ_2\over T_2}.$$ In general, $dQ = dU + dW$ where $U$ is the internal energy and $W$ is the work done. Here, no work is done, and by energy conservation, $dU_1 = - dU_2$. In fact, by the perfect gas law, $dQ = c_vdT$ so we get $$S_2-S_1 = \int_{T_1}^{T_1 + T_2 \over 2} {c_v dT_1\over T_1} + \int_{T_2}^{T_1 + T_2 \over 2} {c_v dT_2\over T_2}.$$ Performing the integrations, this becomes $$S_2-S_1 = c_v\log{T_1 + T_2 \over 2T_1} - c_v\log{T_1 + T_2 \over 2T_2}.$$ Now this transformation took place in thermal isolation from the environment. Planck proves the theorem that 'If a system of perfect gases pass in any way from one state to another, and no changes remain in surrounding bodies, the entropy of the system is certainly not smaller, but either greater than, or, in the limit, equal to that of the initial state; in other words, the total change of entropy $\geq$ 0. The sign of inequality corresponds to an irreversible process, the sign of equality to a reversible process.' One can indeed check that the sign of this change in entropy is positive if $${(T_1+T_2)^2\over 4T_1T_2} > 1,$$ i.e., if $$(T_1+T_2)^2 > 4T_1T_2,\,\, \mbox{, i.e., if }\,\, (T_1-T_2)^2 > 0$$ which is elementary. Its minimum is when $T_1 = T_2$, that is the only case where the change in entropy is zero, hence that is the only case when the entropy is maximal among all these equilibrium states with constant energy, connected to state one by a reversible transformation. In order to use the definition of entropy we had to show how to reverse the transformation from state one to state two but we did not have to reverse it while remaining in isolation. If we remain in isolation, all we can manipulate is the dividing wall, we cannot use heat reservoirs. While the system is in isolation, then, it moves through equilibrium states in a manner which is reversible in one sense, but irreversible in another sense. Therefore the entropy has a maximum when the temperatures are equalised, which is the only stable equilibrium. It is the only state which will not evolve further no matter what we do with the dividing wall. The other equilibrium states become non-equilibrium when the dividing wall is allowed to conduct heat. This is the sense of the slogan, 'equilibrium is when entropy is maximised.' It is also true that even this stable equilibrium state would become non-equilibrium if put into contact with an external heat reservoir, which is how we showed we could reverse the transformation and get back to state one. But then the system is no longer isolated, but there is no law that entropy increases for a non-isolated system. - Thank you for your answer, I'll study it in detail when I get to it! – NikolajK Jun 23 '13 at 21:35 "entropy is maximized" does not make sense as a dynamical statement in thermodynamics (in the usual definition). Entropy is constant in td at the equilibrium for fixed boundary conditions (dS=/=0 makes sense in td, for slow processes under a change of boundary conditions, but that is not directly related to the "entropy is maximized" statement). Here is a way to make sense of the "entropy is maximized" statement: 1) Consider a set of fixed td states without transitions. E.g. in the previous lemonade example, consider a sample of fixed td states, with different mixtures in the two bottles, such that there is equilibrium in each bottle. The total entropy is maximal on perfectly mixed configurations (you could measure the difference by a process like in 2)). So entropy is maximal for the perfect mixings on the set so defined. But the system stays in any td state with lower entropy as well. 2) Now allow a dynamical transition between the configurations in 1). The non-perfectly mixed states will develop to the perfectly mixed states, which are the equilibrium states. Entropy is maximized dynamically on the set of n-e states. The entropy function which is maximized here is strictly speaking no longer the td entropy, as the latter is not defined except for equilibrium states. It is well-defined in statistical physics, as logarithm of the multiplicity of the n-e state. 1) is not a dynamical statement, 2) takes one outside of td. There is no dynamical maximization of the td entropy, but of a minor generalization of an entropy function defined on n-e states. - thanks for the response. – NikolajK Nov 12 '12 at 9:05 Thermodynamics is a science that happens to be concerned with extremely complicated systems that can be analyzed with only a few variables. For example, the gas in a pressurized bottle consists of something like $10^{23}$ atoms, each with a position (3 variables) and a velocity / momentum (3 more variables). That's a lot of variables, but thermodynamics gives us a way of analyzing the system (I'll use the energy formulation) with just volume, total energy, and number of atoms. From this we get the extensive values pressure, temperature and chemical potential (this last is not needed for the fixed amount of gas in a bottle), and other things such as heat capacity, etc. In making an analysis of these systems, we find that there is another variable that is very useful to know, the "entropy". From a microscopic point of view, the entropy is the logarithm of the number of states that are possible for the system. By "number of states" we mean the number of possible positions and momenta for those $10^{23}$ atoms. The reason entropy is a useful thing is that it tells us how "easy" it is to set up the atoms in a particular situation. If there is only one way to do it (for example, all the atoms sitting next to each other in a crystalline solid at the bottom of the bottle), then that will be very difficult to achieve. On the other hand, if there are "billions and billions" (never mind exactly how many) ways of assembling that situation (as far as total energy, volume, and number of atoms goes), then that situation will be easy to achieve. Sometimes you can get a system where the entropy is very small, compared to its maximal value. An example is a bottle with all the (ideal) gas on one side of the container. Such a situation is not in equilibrium because there are so many ways it could reorganize itself that it is doomed to change to a situation of higher entropy. Hence S must me maximized. Now let me give an example of a situation where the numbers are extremely small but you'll still get an idea how the math works. Suppose we had a handful of 6-sided dice, say $N$ dice. A given situation is that each die shows a number in {1,2,3,4,5,6}. A thermodynamic variable for the handful is "the total of all the numbers showing on the dice." We suppose that the dice have an interaction which causes them to randomly change their orientation (and hence their numbers). For $N$ dice, the average (or expectation value) for this thermodynamic variable is $N$ times the average number on a die which happens to be 7/2. Thus, on average, the dies will add up to $7N/2$. Suppose the dies happened to be in a situation where their numbers summed up to $N$ (or $6N$). Such a situation has only one way of being achieved -- all the dies have to show the same number, i.e. "1" (or "6"). Such a situation has an entropy of log(1) as there is only one way it can be achieved, and the entropy is $\ln(1) = 0$. This has an unnaturally low entropy and is not an equilibrium situation. On the other hand, there are many ways of getting the dies to sum up to $7N/2$, (at least supposing you have an even number of dice!) Thus this situation is one with high entropy. So a handful of jiggling dice, tends to approach thermodynamic equilibrium by exhibiting a total value that is equal to the sum of their average possible values. - Thanks for the response. The point is that I'm rather looing for an explaination "from a purely thermodynamical point of view". A systematic derivation from the laws of thermodynamics would suffice, however I guess a clear definition of equilirbium will have to be stated as well. – NikolajK Jan 29 '12 at 19:03 Although the previous answers are well articulated i would like to give an alternative and more intutitive viewpoint very shortly from two inter-connected angles: 1. By the 2nd Law, the entropy of a system increases or becomes a maximum i.e $\Delta{S} \ge 0$. This means that once $\Delta{S}$ reaches a stationary point (maximum entropy) the system is in equilibrium. 2. By Boltzmann's (aka statistical mechanics) relation, entropy is analogous to the number of configurations ($S = k_BlogW$) in that state of energy. This implies: maximum entropy $\implies$ maximum number of configurations. This means that if the system is (slightly) perturbed from this state, it has a larger amount of available configurations to cope for this perturbation (or is more likely to be stable). This can be related to similar results of KAM theory for hamiltonian systems. Regarding the relation between maximum entropy state and equilibrium state of a dynamical system. Consider the equilibrium state as a state of a system where no more work can be extracted (this is again a generality in the spirit of the question), as such it is also a maximum entropy state. Conversely in maximum entropy state, no more work can be extracted, thus it is also an equilibrium state. Consider a dynamical system which can achieve a number of configurations at a specific energy. In the maximum entropy state, all the available system configurations are already used by the system and this represents the cancellation of any potential differences with the system environment. Thus (at that energy state of the system) no more work can be extracted and is in (dynamical in general) equilibrium (with environment). Conversely, a dynamical system at equilibrium (with environment) (at a specific energy), has null potential differences (with environment) thus no more work can be extracted and this is in maximum entropy. The analogy is very similar to the least action principle of mechanics (which actually is not necesarily a least action but a stationary action). Why should least action give a dynamical law? (the answer is somewhat similar to both questions, although a deeper investigation can make things more clear and inter-connected) - Thanks for the response. Though my comment to Carl Brannens answer applies here too: In the first sentence of the question I say I consider the question from the point of view of pure thermodynamics, not arguing using the statistical physics model of it. – NikolajK Jun 21 at 21:29 @NikolajK, using a thermodynamics picture is the same (as the 1st argument), i.e system will evolve dynamically towards a state of maximum entropy. Using stat. mechanics gives another formulation. Also temperature can also be used. Take a look at Generalized Thermodynamics book by Keenan, Hatzopoulos for generalisations of the 2nd law even for non-equilibrium physics – Nikos M. Jun 22 at 10:42 @NikolajK, i fully understand the question, since i have similar questions regarding 1st principles understanding. The thing is that thermodynamics (although having thermo- prefix), is extremely general, do not asssume anything (a-priori) about tha constitution of the system, or the nature of particles (if any) or constraints etc.. In this sense 2nd law is a (general) dynamical law (it states the way or state sth will evolve). The thermo- prefix actually signifies that this investigation was triggered by heat engines, however it is stated and related to dynamics in general terms – Nikos M. Jun 22 at 10:53 @NikolajK, edited answer to reflect the comment – Nikos M. Jun 23 at 2:10	{}
CryptoDB Paper: Efficient Verifiably Encrypted Signature and Partially Blind Signature from Bilinear Pairings Authors: Fangguo Zhang Reihaneh Safavi-Naini Willy Susilo URL: http://eprint.iacr.org/2004/004 Search ePrint Search Google Verifiably encrypted signatures are used when Alice wants to sign a message for Bob but does not want Bob to possess her signature on the message until a later date. Such signatures are used in optimistic contact signing to provide fair exchange. Partially blind signature schemes are an extension of blind signature schemes that allows a signer to sign a partially blinded message that include pre-agreed information such as expiry date or collateral conditions in unblinded form. These signatures are used in applications such as electronic cash (e-cash) where the signer requires part of the message to be of certain form. In this paper, we propose a new verifiably encrypted signature scheme and a partially blind signature scheme, both based on bilinear pairings. We analyze the security and efficiency of these schemes and show that they are more efficient than the previous schemes of their kinds. BibTeX @misc{eprint-2004-11980, title={Efficient Verifiably Encrypted Signature and Partially Blind Signature from Bilinear Pairings}, booktitle={IACR Eprint archive}, keywords={public-key cryptography / Verifiably encrypted signature, partially blind signature, Bilinear pairings}, url={http://eprint.iacr.org/2004/004}, note={Indocrypt 2003, LNCS 2904, pp. 191-204, Springer-Verlag. fangguo@uow.edu.au 12475 received 6 Jan 2004, last revised 26 Feb 2004}, author={Fangguo Zhang and Reihaneh Safavi-Naini and Willy Susilo}, year=2004 }	{}
0 RESEARCH PAPERS # Fatigue Crack Modeling and Simulation Based on Continuum Damage Mechanics [+] Author and Article Information Masakazu Takagaki Institute of Industrial Science, The University of Tokyo, 4-6-1 Komaba, Meguro, Tokyo, 153-8505 Japanmtak@iis.u-tokyo.ac.jp Toshiya Nakamura1 Aviation Program Group, Japan Aerospace Exploration Agency, 6-13-1 Osawa, Mitaka, Tokyo, 181-0015 Japannakamt@chofu.jaxa.jp Square root of a tensor: $A=X1∕2$ or $A2=X$ is defined for a positive semi-definite tensor $X$(15-16). 1 Corresponding author. J. Pressure Vessel Technol 129(1), 96-102 (Mar 10, 2006) (7 pages) doi:10.1115/1.2388993 History: Received April 20, 2005; Revised March 10, 2006 ## Abstract Numerical simulation of fatigue crack propagation based on fracture mechanics and the conventional finite element method requires a huge amount of computational resources when the cracked structure shows a complicated condition such as the multiple site damage or thermal fatigue. The objective of the present study is to develop a simulation technique for fatigue crack propagation that can be applied to complex situations by employing the continuum damage mechanics (CDM). An anisotropic damage tensor is defined to model a macroscopic fatigue crack. The validity of the present theory is examined by comparing the elastic stress distributions around the crack tip with those obtained by a conventional method. Combined with a nonlinear elasto-plastic constitutive equation, numerical simulations are conducted for low cycle fatigue crack propagation in a plate with one or two cracks. The results show good agreement with the experiments. Finally, propagations of multiply distributed cracks under low cycle fatigue loading are simulated to demonstrate the potential application of the present method. <> ## Figures Figure 13 Propagation of two cracks (H0W15, N=2000) Figure 14 Propagation of two cracks (H10W0, N=2000) Figure 15 Simulation of crack propagation for distributed cracks Figure 1 Two-dimensional crack Figure 2 FE model for elastic analyses (angle of the crack θ=30deg) Figure 3 Contour of stress σy (angle of crack θ=30deg) Figure 4 Comparison of σy distribution (angle of crack θ=30deg) Figure 5 Yield surface (plane stress) Figure 6 Assumption of damage evolution Figure 7 Specimen straight section and analysis model with location of initial cracks (displacement controlled push-pull: Δδ∕L=0.5%) Figure 8 Nominal stress-strain response Figure 9 Analysis result of crack propagation (H0W0, single crack) Figure 10 Analysis of crack propagation Figure 11 Propagation of two cracks (H2.5W15, N=2000) Figure 12 Propagation of two cracks (H10W20, N=1200) ## Discussions Some tools below are only available to our subscribers or users with an online account. ### Related Content Customize your page view by dragging and repositioning the boxes below. Related Journal Articles Related Proceedings Articles Related eBook Content Topic Collections	{}
Lemma 10.137.15. Let $R$ be a ring. Let $S = S' \times S''$ be a product of $R$-algebras. Then $S$ is smooth over $R$ if and only if both $S'$ and $S''$ are smooth over $R$. Proof. Omitted. Hints: By Lemma 10.137.13 we can check smoothness one prime at a time. Since $\mathop{\mathrm{Spec}}(S)$ is the disjoint union of $\mathop{\mathrm{Spec}}(S')$ and $\mathop{\mathrm{Spec}}(S'')$ by Lemma 10.21.2 we find that smoothness of $R \to S$ at $\mathfrak q$ corresponds to either smoothness of $R \to S'$ at the corresponding prime or smoothness of $R \to S''$ at the corresponding prime. $\square$ In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).	{}
# Questions & Answers of Network Solution Methods: Nodal and Mesh Analysis ## Weightage of Network Solution Methods: Nodal and Mesh Analysis Total 2 Questions have been asked from Network Solution Methods: Nodal and Mesh Analysis topic of Networks subject in previous GATE papers. Average marks 1.50. Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k> 0, the elements of the corresponding star equivalent will be scaled by a factor of	{}
## Cryptology ePrint Archive: Report 2014/688 White-Box AES Implementation Revisited Chung Hun Baek and Jung Hee Cheon, and Hyunsook Hong Abstract: White-box cryptography is an obfuscation technique for protecting secret keys in software implementations even if an adversary has full access to the implementation of the encryption algorithm and full control over its execution platforms. This concept was presented by Chow et al. with white-box implementations of DES and AES in 2002. The strategy used in the implementations has become a design principle for subsequent white-box implementations. However, despite its practical importance, progress has not been substantial. In fact, it is repeated that as a proposal for a white-box implementation is reported, an attack of lower complexity is soon announced. This is mainly because most cryptanalytic methods target specific implementations, and there is no general attack tool for white-box cryptography. In this paper, we present an analytic toolbox on white-box implementations in this design framework and show how to reveal the secret information obfuscated in the implementation using this. For a substitution-linear transformation cipher on $n$ bits with S-boxes on $m$ bits, if $m_Q$-bit nonlinear encodings are used to obfuscate output values in the implementation, our attack tool can remove the nonlinear encodings with complexity $O(\frac{n}{m_Q}2^{3m_Q})$. We should increase $m_Q$ to obtain higher security, but it yields exponential storage blowing up and so there are limits to increase the security using the nonlinear encoding. If the inverse of the encoded round function $F$ on $n$ bits is given, the affine encoding $A$ can be recovered in $O(\frac{n}{m}\cdot{m_A}^32^{3m})$ time using our specialized affine equivalence algorithm, where $m_A$ is the smallest integer $p$ such that $A$ (or its similar matrix obtained by permuting rows and columns) is a block-diagonal matrix with $p\times p$ matrix blocks. According to our toolbox, a white-box implementation in the Chow et al.'s framework has complexity at most $O\left(\min\left\{ \tfrac{2^{2m}}{m}\cdot n^{m+4}, n\log n \cdot 2^{n/2}\right\}\right)$ within reasonable storage, which is much less than $2^n$. To overcome this, we introduce an idea that obfuscates two AES-128 ciphers at once with input/output encoding on 256 bits. To reduce storage, we use a sparse unsplit input encoding. As a result, our white-box AES implementation has up to 110-bit security against our toolbox, close to that of the original cipher. More generally, we may consider a white-box implementation on the concatenation of $t$ ciphertexts to increase security. Category / Keywords: white-box cryptography, white-box implementation, specialized affine equivalence algorithm, AES, block cipher Date: received 2 Sep 2014, last revised 20 May 2016 Contact author: hongsuk07 at snu ac kr Available format(s): PDF \| BibTeX Citation Note: This paper will be to appear in the Journal of Communications and Networks. Short URL: ia.cr/2014/688 [ Cryptology ePrint archive ]	{}
16/32=x/14 Simple and best practice solution for 16/32=x/14 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework. If it's not what You are looking for type in the equation solver your own equation and let us solve it. Solution for 16/32=x/14 equation: x in (-oo:+oo)16/32 = x/14 / - x/1416/32-(x/14) = 01/2-1/14x = 0 / - 1/2-1/14x = -1/2 / : -1/14x = -1/2/(-1/14)x = 7x = 7` Related pages 2xy x yroman numerals for 1965lcm of 5 72y 3x 4square root of 288 simplifiedmath equations solver2x 4y 8what is the square root of 2704850isfactor 4x 2-12x-5y 0 find slopederive sin 2xwhat is the prime factorization of 52sina sinb2x 3y 12 graph5yz1.4fwhat is the square root of 9618.25 x 20what is 10 percent of 2000.00calculator to solve fractionsfind common multiples calculator84 prime factorizationwhat is the prime factorization of 854000 dollars in rupeeswhat is prime factorization of 7535cm to inchessolve 2x x 31978 in roman numeralsprime factorization of 128solving binomial equations calculatorgreatest common factor of two monomials calculatorsin 2xcos 2xtwo step equations calculatorfactoring calculator with solution800 349 9756square root of 940960 x3what is the greatest prime factor of 964x 2y 7write 0.4 as a fractionstep by step system of equations solver33.333 as a fraction285.9calculator for two step equationsyi8 gamesrx 270xleast common multiples calculatorgcf calculator15 000 euros to dollarswhat is the gcf of 84webwork ttu1.2m incheswhat is the greatest common factor of 32 and 72graph 5x y 3simplify the square root of 162greatest common factor of 120prime factorization 67differentiate e cos xx ln x derivative2logyx52cos 2x-1x senxcos2 pisubtracting fractions calculator mixed numbersderivative of cos squaredsolve 2048derivative of tan 3xhow to write 1075 on a checksolve 5x 25common multiples of 153x 2 5x 2 factor3x2 5x117t15000 pesos to dollars	{}
Volume 35 • Number 2A • 2012 • On the Commutant of Certain Operators in the Bergman Space Ali Abkar Abstract. We study the commutant of analytic multiplication operator $M_{z^2}$ on the weighted Bergman space $A^2_\alpha$. According to a result of Zhu [Reducing subspaces for a class of multiplication operators, J. London Math. Soc. (2) 62 (2000), no.2, 553-568], a bounded linear operator $T$ defined on the standard Bergman space $A^2$ commutes with $M_{z^2}$ if and only if there exist two bounded analytic functions $\phi$ and $\psi$ such that $Tf=\phi f_e+\psi f_o/z$ where $f=f_e+f_o$ is the even-odd decomposition of $f$. We shall prove that this statement holds true in the weighted Bergman space as well. 2010 Mathematics Subject Classification: 47B38 (46E20, 30H20). Full text: PDF	{}
# zbMATH — the first resource for mathematics ## Obaya, Rafael Compute Distance To: Author ID: obaya.rafael Published as: Obaya, R.; Obaya, Rafael External Links: ORCID · dblp Documents Indexed: 80 Publications since 1992, including 1 Book all top 5 #### Co-Authors 0 single-authored 31 Novo, Sylvia 26 Núñez, Carmen A. 23 Sanz, Ana M. 9 Alonso, Ana Isabel 8 Hong, Jialin 7 Johnson, Russell Allan 4 Villarragut, Víctor M. 3 Campos, Juan 3 Fabbri, Roberta 3 Langa, Jose’ Antonio 3 Longo, Iacopo P. 3 Tarallo, Massimo 2 Caraballo Garrido, Tomás 2 de la Llave, Rafael 2 Maroto, Ismael 1 Arratia, Oscar 1 Calzada, Juan A. 1 Cardoso, C. A. E. N. 1 Faria, Teresa 1 Gil, A. S. 1 Jorba, Àngel 1 Mierczyński, Janusz 1 Muñoz-Villarragut, Víctor 1 Ortega, Rafael 1 Paramio, Miguel 1 Pituk, Mihály 1 Sansaturio, María-Eugenia 1 Tatjer, Joan Carles all top 5 #### Serials 14 Journal of Differential Equations 10 Journal of Dynamics and Differential Equations 5 Discrete and Continuous Dynamical Systems 5 Discrete and Continuous Dynamical Systems. Series B 4 Nonlinearity 3 Israel Journal of Mathematics 3 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 3 Applied Mathematics Letters 3 Discrete and Continuous Dynamical Systems. Series S 2 Studia Mathematica 2 Proceedings of the American Mathematical Society 2 Proceedings of the Royal Society of Edinburgh. Section A. Mathematics 2 SIAM Journal on Mathematical Analysis 2 International Journal of Bifurcation and Chaos in Applied Sciences and Engineering 2 Communications on Pure and Applied Analysis 1 Computers & Mathematics with Applications 1 Journal of Mathematical Analysis and Applications 1 Duke Mathematical Journal 1 Illinois Journal of Mathematics 1 Transactions of the American Mathematical Society 1 Physica D 1 Comptes Rendus de l’Académie des Sciences. Série I 1 Journal of Difference Equations and Applications 1 Proceedings of the Royal Society of London. Series A. Mathematical, Physical and Engineering Sciences 1 Comptes Rendus. Mathématique. Académie des Sciences, Paris 1 Developments in Mathematics 1 Boletín de la Sociedad Española de Matemática Aplicada. S$$\vec{\text{e}}$$MA all top 5 #### Fields 57 Dynamical systems and ergodic theory (37-XX) 56 Ordinary differential equations (34-XX) 7 Biology and other natural sciences (92-XX) 6 Measure and integration (28-XX) 6 Partial differential equations (35-XX) 5 Difference and functional equations (39-XX) 4 Calculus of variations and optimal control; optimization (49-XX) 4 General topology (54-XX) 3 Systems theory; control (93-XX) 2 Global analysis, analysis on manifolds (58-XX) 1 General and overarching topics; collections (00-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Real functions (26-XX) 1 Functional analysis (46-XX) 1 Operator theory (47-XX) 1 Manifolds and cell complexes (57-XX) 1 Probability theory and stochastic processes (60-XX) 1 Mechanics of particles and systems (70-XX) 1 Mechanics of deformable solids (74-XX) 1 Quantum theory (81-XX)	{}
# What is the best loss function for convolution neural network and autoencoder? What is the best choice for loss function in Convolution Neural Network and in Autoencoder in particular - and why? I understand that the MSE is probably not the best choice, because little difference in lighting can cause a big difference in end loss. What about Binary cross-entropy? As I understand, this should be used when target vector is composed as 1 at one place and 0 at all others, so you compare only class that should be correct (and ignore others),... But this is an image (although the values are converted in 0-1 values,...) • Hi. Can you just focus on one model? Please, edit your question and ask just about one model, otherwise, this question might become too broad. – nbro Jul 11 '19 at 17:12 There is no right answer to this. Finding the right loss function is a tough and difficult problem. So your goal as the architect is to try to find one that best suits your needs. So lets think about your needs. You mention that you dont want lighting shifts to cause large error, so ill take a leap and assume you care more about the shapes and style of the image more than the coloring. To deal with this, maybe consider using difference of the gram matrices (this is considered common place in style transfer literature: A Neural Algorithm of Artistic Style) Note that you could use the encoder to get the representation of the output as well for the loss, $$L(x) = D(Gram(Enc(x)), Gram(Enc(\hat x))$$ where $$D$$ would be some distance metric like euclidian distance. Maybe the outline is all you care about. You could use some known edge detector filter and compare those, ex: $$D(Edge(x), Edge(\hat x))$$ Maybe you just dont care about color shifts, you could do $$D(x - \mu_x, \hat x - \mu_{\hat x}$$). Note that you can play around with whatever distance metric you use, whether it be with MSE, RMSE, MAE, etc... Each has their own small pros/cons based on the loss manifolds they create. In your case i dont think the difference there will be night and day though, but you never know. Also mixing and matching is always nice: ex: $$L(x) = \lambda_1 D(x, \hat x) + \lambda_2D(Gram(Enc(x)), Gram(Enc(\hat x)) + ...$$ Takeaway: MSE might actually be fine, but it really depends on what you prioritize, and once you figure that out you can start getting clever and design the loss that fits your needs and problem	{}

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