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# Eigenvalues of Hermitian opertors

1. Jun 29, 2011

### migwing007

I'm looking for a proof of the fact that orthogonal eigenfunctions of a Hermitian operator have distinct eigenvalues. I know the proof the converse: that eigenfunctions belonging to distinct eigenvalues are orthogonal.
thanks alot!

2. Jun 30, 2011

### Matterwave

I don't believe that it is true that orthogonal eigenfunctions have distinct eigenvalues. Certainly in terms of degeneracy, one can employ the Gram-Schmidt orthogonalization procedure to get orthogonal eigenfunctions which have the same eigenvalue...

I'm pretty bad at math though, so I wouldn't take my word on it lol.

3. Jun 30, 2011

### George Jones

Staff Emeritus
As Matterwave has said, this it not true. Consider the matrix
$$\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$$

$\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ and $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ are orthogonal eigenvectors that both are associated with the eigenvalue 1 (as is any linear combination of these vectors).

4. Jul 8, 2011

### migwing007

This is really bothering me

I'm remembering from a while ago having to prove the converse of a theorem from Griffiths, but I can't remember which one of the following it is:

Prove converse of:

1.) The eigenfunctions of a Hermitian operator are complete,

2.) Eigenfunctions of Hermitian operators belonging to distinct eigenvalues are orthogonal, or

3.) Eigenvalues of Hermitian operators are real

thanks alot!

Last edited: Jul 8, 2011
5. Jul 9, 2011

### strangerep

Re: This is really bothering me

migwing007,

It's not clear (to me anyway) what you're asking. Maybe try stating the
theorem or its converse in exactly the wording that you want to prove?

6. Jul 9, 2011

### migwing007

Re: This is really bothering me

Well, the converse of the first one would be: if the eigenfunctions of an operator are complete then the operator is hermitian.

the second would be: if the eigenfunctions of an operator belonging to distinct eigenvalues are orthogonal then the operator is hermitian.

and the third is: If the eigenvalues of an operator are real then the operator is hermitian

Actually I don't think 2 and 3 are correct so that leaves #1.

I've looked online but can't seem to find it

Hope that clarifies it

Last edited: Jul 9, 2011
7. Jul 9, 2011

### samalkhaiat

Re: This is really bothering me

8. Jul 10, 2011

### A. Neumaier

Re: This is really bothering me

9. Jul 11, 2011

### bcrowell

Staff Emeritus
Migwing007 started four separate threads on the same topic. I've deleted two and merged two.


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# How do you simplify 5sqrt6 + 2sqrt10?

$5 \sqrt{6} + 2 \sqrt{10}$ can be written as $\sqrt{5} \cdot \sqrt{5} \cdot \sqrt{2} \cdot \sqrt{3} + 2 \cdot \sqrt{2} \cdot \sqrt{5}$
$\sqrt{5} \cdot \sqrt{2} \left(\sqrt{5} \cdot \sqrt{3} + 2\right)$
= $\sqrt{10} \left(2 + \sqrt{15}\right)$

format: plain


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## Electronic Journal of Probability

### Critical Gaussian chaos: convergence and uniqueness in the derivative normalisation

Ellen Powell

#### Abstract

We show that, for general convolution approximations to a large class of log-correlated fields, including the 2d Gaussian free field, the critical chaos measures with derivative normalisation converge to a limiting measure $\mu '$. This limiting measure does not depend on the choice of approximation. Moreover, it is equal to the measure obtained using the Seneta–Heyde renormalisation at criticality, or using a white-noise approximation to the field.

#### Article information

Source
Electron. J. Probab., Volume 23 (2018), paper no. 31, 26 pp.

Dates
Accepted: 12 March 2018
First available in Project Euclid: 30 March 2018

https://projecteuclid.org/euclid.ejp/1522375271

Digital Object Identifier
doi:10.1214/18-EJP157

Mathematical Reviews number (MathSciNet)
MR3785401

Zentralblatt MATH identifier
1390.60182

Subjects
Primary: 60G57: Random measures 60G60: Random fields

#### Citation

Powell, Ellen. Critical Gaussian chaos: convergence and uniqueness in the derivative normalisation. Electron. J. Probab. 23 (2018), paper no. 31, 26 pp. doi:10.1214/18-EJP157. https://projecteuclid.org/euclid.ejp/1522375271

#### References

• [1] Javier Acosta, Tightness of the recentered maximum of log-correlated Gaussian fields, Electron. J. Probab. 19 (2014), no. 90, 25.
• [2] Elie Aidekon and Zhan Shi, The Seneta-Heyde scaling for the branching random walk, Ann. Probab. 42 (2014), no. 3, 959–993.
• [3] Romain Allez, Rémi Rhodes, and Vincent Vargas, Lognormal $\star$-scale invariant random measures, Probab. Theory Related Fields 155 (2013), no. 3-4, 751–788.
• [4] Juhan Aru, Ellen Powell, and Avelio Sepúlveda, Liouville measure as a multiplicative cascade via level sets of the gaussian free field, ArXiv preprint arXiv:1701.05872.
• [5] Nathanaël Berestycki, Introduction to the Gaussian free field and Liouville quantum gravity, Lecture notes, available on the webpage of the author.
• [6] Nathanaël Berestycki, An elementary approach to Gaussian multiplicative chaos, Electron. Commun. Probab. 22 (2017), Paper No. 27, 12.
• [7] J. D. Biggins, Martingale convergence in the branching random walk, J. Appl. Probability 14 (1977), no. 1, 25–37.
• [8] J. D. Biggins and A. E. Kyprianou, Measure change in multitype branching, Adv. in Appl. Probab. 36 (2004), no. 2, 544–581.
• [9] Bertrand Duplantier, Rémi Rhodes, Scott Sheffield, and Vincent Vargas, Critical Gaussian multiplicative chaos: convergence of the derivative martingale, Ann. Probab. 42 (2014), no. 5, 1769–1808.
• [10] Bertrand Duplantier, Rémi Rhodes, Scott Sheffield, and Vincent Vargas, Renormalization of critical Gaussian multiplicative chaos and KPZ relation, Comm. Math. Phys. 330 (2014), no. 1, 283–330.
• [11] Bertrand Duplantier and Scott Sheffield, Liouville quantum gravity and KPZ, Invent. Math. 185 (2011), no. 2, 333–393.
• [12] Werner Ehm, Tilmann Gneiting, and Donald Richards, Convolution roots of radial positive definite functions with compact support, Trans. Amer. Math. Soc. 356 (2004), no. 11, 4655–4685.
• [13] Yichao Huang, Remi Rhodes, and Vincent Vargas, Liouville quantum gravity on the unit disk, ArXiv preprint arXiv:1502.04343.
• [14] Janne Junnila and Eero Saksman, Uniqueness of critical Gaussian chaos, Electron. J. Probab. 22 (2017), Paper No. 11, 31.
• [15] J.-P. Kahane and J. Peyrière, Sur certaines martingales de Benoit Mandelbrot, Advances in Math. 22 (1976), no. 2, 131–145.
• [16] Jean-Pierre Kahane, Sur le chaos multiplicatif, Ann. Sci. Math. Québec 9 (1985), no. 2, 105–150.
• [17] A. E. Kyprianou, Martingale convergence and the stopped branching random walk, Probab. Theory Related Fields 116 (2000), no. 3, 405–419.
• [18] Minoru Motoo, Proof of the law of iterated logarithm through diffusion equation, Ann. Inst. Statist. Math. 10 (1958), 21–28.
• [19] Rémi Rhodes and Vincent Vargas, KPZ formula for log-infinitely divisible multifractal random measures, ESAIM Probab. Stat. 15 (2011), 358–371.
• [20] Rémi Rhodes and Vincent Vargas, Gaussian multiplicative chaos and applications: a review, Probab. Surv. 11 (2014), 315–392.
• [21] Raoul Robert and Vincent Vargas, Gaussian multiplicative chaos revisited, Ann. Probab. 38 (2010), no. 2, 605–631.
• [22] Alexander Shamov, On Gaussian multiplicative chaos, J. Funct. Anal. 270 (2016), no. 9, 3224–3261.


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## LSU Historical Dissertations and Theses

1996

Dissertation

#### Degree Name

Doctor of Philosophy (PhD)

Human Ecology

Michael Keenan

#### Abstract

Introduction. A high amount of dietary fat has been shown to have a negative effect on metabolism, causing insulin resistance, weight gain, and other metabolic alterations. Several studies have found a direct link between lack of activity and obesity. Purpose. The purpose of this study was to see if a diet had an effects on insulin sensitivity, autonomic nervous system function, and substrate utilization as predicted by Flatt, in healthy young men. Subjects. The subjects studied were 6 nonathletes (NA), 6 aerobically trained (AT), and 5 weight trained (WT) men. Their body mass index (BMI) was 22.1 $\pm$ 2.92 kg/m$\sp2$ (mean $\pm$ SD), age 22.5 $\pm$ 5.01 yr, and body fat 15.9 $\pm$ 6.5%. The study was a crossover design with a high (HF 50%) or low (LF 20%) fat diet assigned randomly for 7 days after 5 days of control diet (40% fat). Results. Energy expenditure (EE) was highest in the WT subjects, but after correcting for LBM, NA had the highest energy expenditure. Diet had an overwhelming effect on substrate oxidation and RQ. During an early overfeeding period, fat oxidation was reduced, while carbohydrate oxidation increased, but these shifted to accommodate intake on both HF and LF diets when energy balance was zero. Fat oxidation was significantly higher on the HF diet than on the LF diet but there was no significant difference by groups. Insulin sensitivity and glucose effectiveness were significantly higher on the LF than on the HF diet. There was no significant difference in cardiac autonomic activity by groups or diets. Resting catecholamine excretion was significantly higher on the HF diet. Additionally, using LBM as a covariate resulted in significant differences among groups in 24-hour and resting epinephrine excretion on the HF diet. Conclusions. The results indicate that S$\sb2$ appears to function as indicated by causing an increased carbohydrate oxidation during overfeeding. When energy and carbohydrate balances were below S$\sb2$: fat oxidation increased, carbohydrates were spared, even a HC diet did not fill S$\sb2$ at zero energy balance. A hierarchy of importance in relation to weight gain and weight maintenance are (1) overfeeding, (2) exercise, (3) high or low fat diets.

9780591035322

167

COinS


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The scripts and input files that accompany this demo can be found in the demos/public directory of the Rosetta weekly releases.

KEYWORDS: DESIGN LOOPS

## Setting up the demo

Before running the demo, make sure to change the following variables in your local environment:

PATH_TO_EXE                                         # path to directory with Rosetta binaries
ROSETTA_BINARY                                      # extension of Rosetta binaries, e.g. linuxgccrelease
PATH_TO_DB                                          # path to Rosetta database

## Running the demo

IO flags:

-s 1a8d_MinPacked.pdb                               # The starting structure -- must have residues for the segment to be remodeled, but these don't need to have meaningful coordinates
-loops:loop_file 1a8d.loop                          # definition of the loop to be remodeled, file format description at http://www.rosettacommons.org/manuals/archive/rosetta3.4_user_guide/d1/d49/loopmodeling.html
-out:pdb_gz                                         # compress output structures
-in:file:native 1a8d_MinPacked.pdb                  # native or reference structure for RMSD calculation -- if this flag is not set, an RMSD of 0 will be reported

Number of structures to produce (for demo):

-nstruct 1                                          # number of structures to produce

Number of structures to produce (for production runs):

-nstruct 500                                        # or 500 independent simulations, depending on the cluster setup.

General kinematic closure loop modeling flags:

-in:file:fullatom
-loops:remodel perturb_kic
-loops:refine refine_kic
-run:test_cycles                                    # fast execution, uncomment for testing purposes only
-loops:outer_cycles 5                               # for production runs
-kic_bump_overlap_factor 0.36                       # reduces the threshold for permitted clashes in initial loop closures (default value is 0.4)
-legacy_kic false                                   # remove a slight bias in the initial KIC implementation towards sampling the C-termial part of the remodeled segment more frequently
-kic_min_after_repack true                          # minimize rotamers after repacking (happens every -loops:repack_period iterations after loop closure, default 20)
-corrections:score:use_bicubic_interpolation false  # do not use bicubic interpolation, it has been observed to adversely affect the fraction of sub-Angstrom conformations on the 12-residue benchmark set

Next-generation KIC flags as described in Stein & Kortemme, PLoS ONE, 2013:

-loops:kic_rama2b
-loops:kic_omega_sampling
-allow_omega_move true
-loops:ramp_fa_rep
-loops:ramp_rama

Packing flags:

-ex1
-ex2
-extrachi_cutoff 0

Example Rosetta Command Line: (where $ROSETTA3=path-to-Rosetta/main/source) $> \$ROSETTA3/bin/loopmodel.default.linuxgccrelease @input_files/local_remodeling_input/flags

Overall protocol execution (demo):

1. scripts/pre_min_pack.py <list_of_native_structures> <output_keyword> (prepacking step)

This will create a pre-min-packed structure to start the simulation from, as well as a log of the run. The input is a list of starting structures to be processed, as well as a keyword for the output structure and log names. Note that, while standard repacking usually takes a few minutes, with the -min_pack option it can take an hour or more, depending on the size of the structures. This option was developed by Andrew Leaver-Fay and will be discussed in a separate publication (Leaver-Fay et al., Meth Enzym, 2013).

2. scripts/submit_NGK.py <input_list> <output_keyword> (loop remodeling step)

This will generate structures in which the selected segment is remodeled (1 for the demo, use at least 500 for real life problems, more for longer segments). The input is a list of (pre-packed) structures and a keyword for output naming. The Rosetta syntax for .loop files is explained at http://www.rosettacommons.org/manuals/archive/rosetta3.4_user_guide/d1/d49/loopmodeling.html

Note that if a native or reference structure is provided, the backbone RMSD of the remodeled segment (after superimposition of the fixed parts of the structure) will be reported by loopmodel.release and can thus be parsed from the standard output, if that is redirected into a file. The script will perform redirection into a .log file which later is gzipped.

If submitted via qsub on an SGE cluster system, the script will distribute the number of simulations specified with the -t flag across all structures in the list, i.e., to generate 500 structures each for a list of 10 input structures, use -t 1-5000.

An NGK trajectory (generating one model) typically take 15-30min for 12-residue lops.

3. scripts/parse_RMSDs.pl <directory with log files> (parsing step)

This script will extract the scores and RMSDs for each final model from the respective .log files, and optionally generate Rosetta-energy-vs-RMSD plots. Note that relevant RMSDs will only be reported if a native or reference structure is provided to the NGK run via the -in:file:native flag. If no reference structure is available, users should consider clustering of their results to identify the most commonly sampled conformations (see Figure S1 of Stein & Kortemme, PLoS ONE, 2013).

## Authors

• Amelie Stein
• Tanja Kortemme

## Rosetta Version

SVN revision 51851 (Dec 2012)

## References

• Stein A, Kortemme T (2013) Improvements to robotics-inspired conformational sampling in Rosetta. PLoS ONE (submitted)


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Browse Questions

# The hybridisation of central 'O' atom in $O_3$ molecule is

$\begin{array}{1 1}(a)\;\text{sp-hybridisation}&(b)\; sp^2\text{-hybridisation}\\(c)\;sp^3\text{-hybridisation}&(d)\;dsp^2\text{-hybridisation}\end{array}$

Can you answer this question?

Ozone $O_3$ is in v-shape with $sp^2$-hybridisation
Hence (b) is the correct answer.
answered Mar 5, 2014


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# Tag Info

51

From a cryptographic standpoint it is OK to expose a public key in the sense of revealing its value. The most basic assumption in cryptography involving public/private key pairs is that the value of a public key is public; hence its name. It is extremely important that an adversary can not alter a public key. Any exposition that would allow alteration must ...

46

While it is true that Elliptic Curve Diffie Hellman, Elliptic Curve Signature Generation and Elliptic Curve Signature Verification rely on scalar multiplications, these are usually implemented as different types of scalar multiplication for both security and efficiency reasons. In fact there are three types of scalar multiplications used in practice for ...

36

Ed25519 is a specific instance of the EdDSA family of signature schemes. Ed25519 is specified in RFC 8032 and widely used. The only other instance of EdDSA that anyone cares about is Ed448, which is slower, not widely used, and also specified in RFC 8032. Keys and signatures in one instance of EdDSA are not meaningful in another instance of EdDSA: Ed25519 ...

27

The old terminology was confusing, so they've rebranded a bit. X25519 is Elliptic Curve Diffie-Hellman (ECDH) over Curve25519 Ed25519 is Edwards-curve Digital Signature Algorithm (EdDSA) over Curve25519 Libsodium's ref10 curve25519 code is actually used both by crypto_scalarmult()/crypto_box() as well as crypto_sign().

27

Edwards25519 is the twisted Edwards curve $$-x^2 + y^2 = 1 - (121665/121666) x^2 y^2$$ over the prime field $\mathbb F_p$ where $p = 2^{255} - 19$. The coefficient $d = -121665/121666$ was chosen to so that this curve is birationally equivalent to the Montgomery curve $y^2 = x^3 + 486662 x^2 + x$, called Curve25519, whose coefficient 486662 was chosen to be ...

23

Clearing the lower 3 bits of the secret key ensures that is it a multiple of 8, which in turn ensures that no information, small as it may be, about the secret key is leaked in the case of an active small-subgroup attack. The typical simple Diffie-Hellman key exchange works like this: $$\text{Alice} \xrightarrow{\hspace{3cm} a G \hspace{3cm}} \text{Bob} \\... 21 There's a few different related parts here, and the nomenclature of the library you've cited is a little confusing. Curve25519 is an elliptic curve over the finite field \mathbb F_p, where p = 2^{255} - 19, whence came the 25519 part of the name. Specifically, it is the Montgomery curve y^2 = x^3 + 486662 x^2 + x, but you don't need to know the ... 20 If I understand your question correctly, you are essentially asking if points in Edwards and Montgomery curves can be represented in Weierstrass coordinates. This is true; in fact, any elliptic curve over a field of characteristic \neq2,3 can be represented in Weierstrass form \mathcal{E}_{w}^{a, b} : y^2 = x^3 + ax + b, and by extension its points can ... 18 How many qubits are required for breaking RSA 2048 and RSA 4096 in real-time with a quantum computer? Like the answer you linked to shows, about \log_2(N^2) = 2 \log_2(N) or just 2n where n is the number of bits of the modulus N, i.e. the key size of RSA. So 4096 for 2048-bit RSA, double that for 4096-bit. This paper (PDF) has an algorithm using ... 14 The advent of quantum computing, real usable out of the lab QC, will pretty much be the end of any encryption that relies on the difficulty of the discrete log problem via Shor's Algorithm et al. That is not to say it is the death of all modulus based encryption schemes, probably just asymmetric public/private key based schemes. That said, QC's aren't magic,... 13 Trevor Perrin wrote a library doing exactly that. Explanation can be found on in the curves mailing list archives. To convert a Curve25519 public key x_C into an Ed25519 public key y_E, with a Ed25519 sign bit of 0:$$y_E = \frac{x_C - 1}{x_C + 1} \mod 2^{255}-19$$The Ed25519 private key may need to be adjusted to match the sign bit of 0: if ... 13 if Ed25519 has gone through rigorous cryptanalysis It is based on Curve25519 which has gone through extensive cryptanalysis. The Ed25519 signature scheme as well is being heavily reviewed and adoption is rapid. There are already a number of papers on the algorithm itself, as well as a few papers on specific implementations. Every part of the algorithm and ... 12 Well, lets go through the issues: It seems to be possible to retrieve the (public) key used for creating an ECDSA signature just from the signature alone Nope, not quite. You also need the message being signed. And, with that, it doesn't give you the unique public key; it does allow you to narrow it down to two possibilities (assuming you're using a ... 12 All of these are answered by the SafeCurves project: For twisted Edwards curves, ax^2 + y^2 \equiv 1 + dx^2y^2 \pmod p. Edwards curves are the special case a = 1. Edwards curves can be converted to Montgomery form.Montgomery curves can be converted to Weierstrass form.Some, but not all, Weierstrass curves can be converted to Montgomery form. The ... 12 It is not possible to double security level of Ed25519 any trivial way. Instead, doubling security level requires using another curve that is approximately 512 bit curve. In systems compliant with RFC 7748, i.e. some of IETF specifications, there is Curve448 curve (Ed448-goldilocks). It is almost twice as strong as Curve25519 (its strength is 224 bits). ... 12 Yes, this should be secure. I am not familiar with TweetNaCl, so I cannot speak on the concrete implementation. However, the general construction of signing a hash of a message instead of the message should be secure. It is in fact a standard way to sign messages. In general it should work for any secure signature scheme and cryptographic hash function. ... 12 The formulas actually work. You just have to keep in mind to make computation in the field of intergers modulo 2^{255}-19 and that there are actually two square roots, you need to use the right one if you want to have the expected result. You can test the following SAGE code gf=GF(2^255-19) X_P = gf(... 11 Curve25519 makes use of a special x-coordinate only form to achieve faster multiplication. Ed25519 uses Edwards curve for similar speedups, but includes a sign bit. While it could have been done differently, doing it this way simplifies implementations that only need one of encryption or signing. 10 That example private key is not encoded with PKCS#8, but with the format described in RFC 5958, which is supposed to replace PKCS#8 (at least, in IETF parlance, RFC 5958 "obsoletes" RFC 5208, which is a copy of PKCS#8 v1.2). The ASN.1 type if called OneAsymmetricKey: OneAsymmetricKey ::= SEQUENCE { version Version, ... 10 Yes! You can use the ephemeral key derivation mechanism that is for example used in Monero (they call it stealth keys there). Consider public key A=aG, with private key a. Then, a derived key can be generated, parametrised by the random scalar r:$$A'=H_s(rA)G+A$$and the party that knows a can use the public parameter R=rG to compute their ... 9 No, conversion of an EC key pair from a curve to another of unrelated order is not possible. One of the closest things that could be done would be that parties generate a new P256 key pair, then certify their new P256 public key using their C25519 private key, check the other party's certificate using the other party's trusted C25519 public key, and now ... 9 Shor's algorithm can compute discrete logs in elliptic curves and thereby recover the secret scalar from a public Ed25519 key, which you can use to forge signatures of your choice. So, yes, it affects Ed25519—it completely breaks Ed25519, or it would if you could engineer a quantum computer capable of executing it. It can also compute discrete logs in ... 9 It is security on a 0-10 scale that goes to 11. The cost of cryptanalysis to break a cryptosystem with a 128-bit security level, like Ed25519, is out of reach for humanity already. A higher security level like Ed448's doesn't have any actual meaning with current understanding of cryptanalytic attacks. The only purpose it serves is to hedge against ... 9 (minisign author here) As noted by corpsfini, keys encode the Y coordinate. The X coordinate is recovered using the curve equation: X = sqrt((Y^2 - 1) / (d Y^2 + 1)). The square root has two solutions, so we need to encode the sign of x as well. Since coordinates only require 255 bits, we have a extra bit, used to encode the sign. X and Y ∈ [0; 2^255-19[, ... 8 First off, your equation is correct and there seems to be no calculation mistake. To understand on how to get from$$(2+d)x^{2}+dy^{2}=d+(d-2)x^{2}y^{2}$$to$$x^{2}+y^{2}=1+e\cdot x^{2}y^{2} one first needs to observe that $e=(d-2)/(d+2)=121665/121666$ holds. The next step is to consider: "What operations are actually allowed with birational ...

8

Any key generation algorithm for any cryptosystem is going to be weak if the attacker can predict what seed was used to generate the key. They can just generate the same key. However, assuming the the random number generator is not that bad, different algorithms start to look different. If you are just using the output of the random number generator as a ...

8

Short answer: You don't need to worry about the blacklist or point validation unless you are designing an exotic protocol. Medium answer: It depends on what you are trying to do. There is a short blacklist, and a much larger set of points (too large to list) that may cause you trouble—but only if you are setting yourself up for trouble by using the ...

8

There are a few different formats for the standard parts of an Ed25519 private key, which are usually stored as 32-byte or 64-byte strings, so you need to pay attention to the choices made by the system you use it with. Everyone agrees on how to compute an Ed25519 signature given a secret scalar, which can be a uniform random 256-bit integer, and a PRF ...

8

As you said, you need to define the goals. You can take a look at SafeCurves, which is a joint work by Bernstein and Lange to help choose/construct elliptic curves w.r.t. ECDLP difficulty and ECC security. Note that if you need a pairing-friendly elliptic curve you need to look at other criteria related to the embedding degree. You can read this paper by ...

8

Yes, this is possible using Hierarchical Deterministic (HD) Keys. There are 2 variations for key generation, hardened and non-hardened. In hardened, generating child keys (both public and private) requires knowledge of parent private key but in non-hardened, child public key can be generated using parent public key. You need non-hardened key generation. The ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# Can we define rms current for pure inductive and capacitive ac circuit?

For a purely inductive and capacitive ac(sinusoidally varying) circuit, average power dissipated across a capacitor or an inductor is zero. Then does that mean RMS current is zero?

No. The RMS current cannot (integral of a positive function).

The circuit receives power for half a period and returns it during the other half period so that the average power is zero.

• What about RLC circuits where inductor, capacitor and resistance are in series? Can we define RMS current for RLC circuits? – sheshin Jan 29 '19 at 7:34
• You can always define RMS current : $\sqrt{\left\langle {{i}^{2}} \right\rangle }=\sqrt{\frac{1}{T}\int\limits_{T}{{{i}^{2}}dt}}$ In the case of the RLC circuit, the circuit receive power for more than half a period and returns it during less than half period so that the average power is positive (Joule effect in the resistor). – Vincent Fraticelli Jan 29 '19 at 7:45
• But as mentioned by @Steeven in the other answer, mathematically speaking RMS current cannot be zero. Doesn't this lead to a contradiction? Or is it that RMS values are defined only for purely resistive circuits and their values are then used as references for the ac sources? – sheshin Jan 30 '19 at 7:14
• Maybe you are making a confusion between zero power and zero current ? You can have a non zero rms current and a non zero rms voltage but a zero mean power. It is important to note that we are speaking of mean power : averaged over a period. instantaneous power is non zero even for an inductor or a capacitor. – Vincent Fraticelli Jan 30 '19 at 7:19

No. The RMS value means root-mean-square. Read it from right to left and there is the recipe:

1. Square the values: $$x_i^2$$. Now, non are negative (what was negative has now been flipped to the positive side).
2. Take the arithmetic mean/average of these squared values: $$\sum x_i^2/n$$ ($$n$$ is the number of values). Since no values are negative, this average can't ever become zero.
3. Take the square root $$\sqrt{\sum x_i^2/n}$$. This sort-of "cancels out" the initial squaring, so to say. The final result is a number that is something like an average that ignores any negative signs there may have been to begin with.

$$RMS=\sqrt{\frac{\sum x_i^2}{n}}=\sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}}$$

This RMS value can never be zero unless all the values $$x_i$$ are zero. If just one of them is non-zero (regardless if positive or negative), the RMS becomes non-zero (and non-negative).

• You have given a mathematical explanation, I need an explanation in terms of physics. Thank you. – sheshin Jan 29 '19 at 7:33
• @shesin I'm not sure what you are looking for. RMS is a mathematical representation of something physical. There are moments of high current, zero current, high opposite current (negative). The variation is called a sine curve. It is not easy to calculate much with a sine curve; sometimes we need some kind of a single number that represents the sine curve. You can pick the arithmetic or geometric average, the RMS, the amplitude, the peak-to-peak (double amplitude) etc. No matter what you pick, you are now working with a mathematical representation and not with the original true behaviour. – Steeven Jan 29 '19 at 7:45
• thanks for the answer. RMS current in an AC circuit is defined as the equivalent dc current that leads to the same energy dissipation. For purely inductive or capacitive circuits, average power and hence average energy dissipated for a whole cycle is zero and hence if we equate the energies, it should mean RMS current is zero or undefined as there is no resistance in this case. But mathematically speaking RMS current cannot be zero if current at any instance is non zero (as mentioned by you). This leads to a contradiction. Can you kindly look into this. – sheshin Jan 30 '19 at 7:12
• @sheshin I think you are missing a small crucial detail in the definition. From Wikipedia citing A Dictionary of Physics: "For alternating electric current, RMS is equal to the value of the direct current that would produce the same average power dissipation in a resistive load." In a resistor, dissipated power adds up rather than cancel out for positive and negative currents. The average of AC power across a resistor is non-zero – Steeven Jan 30 '19 at 7:31


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# Matrix Transformation... Need Help !

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Hi guys... I am facing a little problem, quite complex for me, not for others I hope... I am converting an Old G Engine using Non-Left-Handed Coordinates to DXs.. Basically, dnt ask me what Coords handling this system uses, Seems Y and Z are reversed btw them, both in poosition and direction... When transforming object to a new DX compatible DB I just exchange Z and Y items in vectors the following way : Y=-Z , Z=-Y This works good, I see the models at right positions... The problem asises when I encounter some parts of the Models... These parts are built in the model at the coords origin ( dnt ask me why ), still not fitted at right place, then the Engine should Rotate & Transpose them to the right position... but now, the Rotation and Transposition Matrices settled up for each part of the model are no more working in the new Coord System... For Transposition, well, no problem, just Z=-Y and Y=-Z method, but what about rotation Matrix...? Can Anyone tell me how could I fit the Matrices to the new system... this is not a simple rotation on axis I suppose... so have to work directly on the Rotation Matrix... can Any1 Help plz...? tnx [R]ed

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You could decompose the transform matrix in general into its transitional, rotational and scaling subparts. You would for example obtain a 3D vector for translation, a quaternion for rotation and a 3D vector for scaling. These can then be converted individually and a new transformation matrix can be constructed from them, which is valid in the new system.

Greetz,

Illco

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Tnx Illco...
I already have them as separated items...
Only I dunno how to transform to the new system the rotation matrix ( it's an original 3x3 Matrix but easy to make it as D3DX Matrix)...
I have tried with axis rotations with no success... I can't swap Z & Y both in sign and position leaving unaltered X... so... it's a matter to chage the Matrix another way...
and I am not figuring out how....

[R]ed

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If you have them as separate items and you transform a position vector P = (Px,Py,Pz) to the new coordinate system as P' according to:
P' = (Px,-Pz,-Py)

Then you can transform a quaternion R = (Rx, Ry, Rz, Rw) by
R' = (Rx,-Ry,-Rz,Rw)

The scaling goes exactly like the first vector but without the negative values. I don't see what the problem is after this. If you are using D3DX you can simply build three matrices (translation, rotation, scaling) and multiply them together to find the new transform matrix.

Illco

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Tnx Again... but may be I was not so clear...
The rotation Matrix is already the product of rotations on all the X, Y, Z axis, and I only got it...
Is there a way to decompose it...? and getting back the original Rotation Vector / Matrix / Angle on Axis?

I'll be very glad if u could tell me...

tnx

[R]ed

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Quote:
 I already have them as separated items...

I thought you meant here you already had the separate translation, rotation and scale items.

Quote:
 The rotation Matrix is already the product of rotations on all the X, Y, Z axis, and I only got it...

But from this I get you only have the transformation resulting from some translation, rotation and scaling.

Yes you can decompose it. Have a look here, it comes with code.

Good luck! Illco.

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Yep...
I have separated Translation, Scale, Rotation...

but Rotation is already the product of at least 2 rotation ( if not 3 ) Matrices on their X, Y, Z axis...
that I meant... if I could decompose the rotation ( or find how to 'translate' it to the other axis ) everything could be fine...

however... going to take a look...

[R]ed

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I think I see now and by doing so recall from memory your original post mentioning this. You probably have:
a) a translation vector
b) a scaling vector
c) a rotation matrix (3x3)

I'd suggest disregarding all and taking the 4x4 transform matrix and then apply an ordinay decomposition method. The rotation will preferrably be represented by a quaternion, which you can convert as said above. This is regardless of how the rotation was achieved, because quaternions are not axis-angle. What I mean is it does not matter whether the original rotation was achieve by doing such-and-such around this-and-that axis; the resultant rotation is just represented at once by the quaternion. Converting it and building a new transform matrix will preserve the rotation.

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yep... tnx Illco...
already tried this... transformed a 3x3 Matrix in a Quaternion...
then read the axis angles in it to transform...
but seemed not consistent...
however... tnx again for interest...
I'll check better later or tomorrow...
howevere...I am wondering if there is a trigonometric way to work directly on matrix components... to transform it directly...
the old Coord System dnt let me use any rotation, as finally it needs to rotate the X axis too, and it's already good...

bye

[R]ed

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# Article

Plasma Investigations
1996. V. 34. № 3. P. 367–372
Smirnov Yu.M.
Excitation cross-sections of the $^3D^0$ states of titanium, zirconium and hafnium atoms
Article reference:
Smirnov Yu.M. Excitation cross-sections of the $^3D^0$ states of titanium, zirconium and hafnium atoms, High Temp., 1996. V. 34. № 3. P. 367


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Please use this identifier to cite or link to this item: http://hdl.handle.net/1813/6555
 Title: With Probability One, a Random Oracle Separates PSPACE from the Polynomial-Time Hierarchy Authors: Cai, Jin-yi Keywords: computer sciencetechnical report Issue Date: Dec-1985 Publisher: Cornell University Citation: http://techreports.library.cornell.edu:8081/Dienst/UI/1.0/Display/cul.cs/TR85-715 Abstract: We consider how much error a fixed depth Boolean circuit has to make for computing the parity function. We show that with an exponential bound of the form $exp(n^{\lambda})$ on the size of the circuits, they make asymptotically 50% error on all possible input, uniformly. As a consequence, we show that with a random oracle set $A,Pr.(PSPACE^{A} \supseteq PH^{A} = 1$. URI: http://hdl.handle.net/1813/6555 Appears in Collections: Computer Science Technical Reports

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## Chemistry: The Molecular Science (5th Edition)

The reaction is product-favored. Up to $1484\,kJ/mol$ useful work can be done.
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta_{f}G^{\circ}(reactants)$ $=[2\Delta_{f}G^{\circ}(Fe_{2}O_{3},s)]-[4\Delta_{f}G^{\circ}(Fe,s)+3\Delta_{f}G^{\circ}(O_{2},g)]$ $=[2(-742.2\,kJ/mol)]-[4(0\,kJ/mol)+3(0\,kJ/mol)]$ $=-1484\,kJ/mol$ $\Delta_{r}G^{\circ}$ is negative. Therefore, the reaction is product-favored. Up to $1484\,kJ/mol$ useful work can be done.


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###### Exercise25

Determine the validity of each statement by selecting True or False.

1. The number $$\frac{\pi}{2}$$ is rational

2. The number $$98$$ is an integer, but not a whole number

3. The number $$-\frac{13}{102}$$ is rational

4. The number $$\sqrt{\frac{49}{25}}$$ is an integer, but not a whole number

5. The number $$\pi$$ is irrational

in-context


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## Fundamenta Mathematicae

2009 | 202 | 2 | 161-180
Tytuł artykułu

### Indestructible colourings and rainbow Ramsey theorems

Autorzy
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We show that if a colouring c establishes ω₂ ↛ [(ω₁:ω)]² then c establishes this negative partition relation in each Cohen-generic extension of the ground model, i.e. this property of c is Cohen-indestructible. This result yields a negative answer to a question of Erdős and Hajnal: it is consistent that GCH holds and there is a colouring c:[ω₂]² → 2 establishing ω₂ ↛ [(ω₁:ω)]₂ such that some colouring g:[ω₁]² → 2 does not embed into c.
It is also consistent that $2^{ω₁}$ is arbitrarily large, and there is a function g establishing $2^{ω₁} ↛ [(ω₁,ω₂)]_{ω₁}$; but there is no uncountable g-rainbow subset of $2^{ω₁}$.
We also show that if GCH holds then for each k ∈ ω there is a k-bounded colouring f: [ω₁]² → ω₁ and there are two c.c.c. posets 𝓟 and 𝒬 such that
$V^{𝓟}$ ⊨ f c.c.c.-indestructibly establishes $ω₁ ↛ *[(ω₁;ω₁)]_{k-bdd}$,
but
$V^{𝒬}$ ⊨ ω₁ is the union of countably many f-rainbow sets.
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• Alfréd Rényi Institute of Mathematics, V. Reáltanoda utca 13-15, H-1053 Budapest, Hungary
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# 5  Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint  $100 m^{2}$ of area is painted. How many cans of paint will she need to paint the room?

Total area painted by Daniel :

$= 2(15\times10+10\times7+7\times15) - 15\times10$                                   ($\because$  Bottom surface is excluded.)

So,  Area

$= 650 - 150\ m^2 = 500\ m^2$

No. of cans of paint required

$=\frac{ 500\ m^2}{100\ m^2} = 5$

Thus 5 cans of paint are required.

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# vnm expected utility function

Formally, the exponential utility is given by: Where R is the Risk Tolerance. <<050E66A0B159934F9131126070B6C62B>]>> 0000007760 00000 n Why should the statistical value of life exist? Can light reach far away galaxies in an expanding universe? Suppose that Ue(L)=[U(L)]2 for all L . Xis a set of outcomes. In this case, the function U is called an expected utility function, and the function u is call a von Neumann-Morgenstern utility function. Not sure about what you mean by the rest of your question. Why is this distinction so important in the theory of expected utility? Question about the Ellsberg Paradox in Expected Utility Theory. [0;1] #fxjP(x) >0g<1; X x2X P(x) = 1) Notice that P x2X P(x) = 1 condition is well de ned due to the nite support assumption. 0000004981 00000 n In this video, we explain Von Neumann-Morgenstern expected utility axioms 0000037781 00000 n They still play a static game once the state is decided. In this sense, the distinction between Bernoulli and vNM utility functions are necessary (as they are applied to different objects) rather than important (since they both represent some kind of preference in the end), as @denesp says in his comment. Then, you will use the utility function to calculate the expected utility of some risky prospects. It was put forth by John von Neumann and Oskar Morgenstern in Theory of Games and Economic Behavior (1944) and arises from the expected utility hypothesis. How much share should I get in our property. Video for computing utility numerically https://www.youtube.com/watch?v=0K-u9dpRiUQMore videos at http://facpub.stjohns.edu/~moyr/videoonyoutube.htm The assumption that a VNM expected utility function is linear in probabilities, though not in payoffs, allows us to create an indifference-curve representation. Asking for help, clarification, or responding to other answers. ) is the Bernoulli utility function de ﬁned over mon-etary outcomes. The v.NM function maps from the space of lotteries to real number as it represents the preference defined on the lottery space while the Bernoulli is defined over sure amounts of money. vNM isn't. Economics Stack Exchange is a question and answer site for those who study, teach, research and apply economics and econometrics. 3[�^ �?|m��#��� -^[���$&S���E���Y0�������*�TqcJ To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 0000008497 00000 n 0000007192 00000 n Consider a dynamic and asymmetric Bayesian game where nature moves to decide the state of the world and player 1 observes the state but player 2 doesn't. Every person’s utility function may be different. von-Neumann-Morgenstern v. Bernoulli Utility Function. Herr, is your field microeconomic theory or game theory? x is the real-world value and u(x) is the utility value or perceived value (the value of an outcome in utils). 0000011282 00000 n Christos A. The expected utility theorem simply says that when a preference satisfies the vNM axioms, there exists a linear utility function that represents it. trailer Is SOHO a satellite of the Sun or of the Earth? Consider Land L0;where Lputs probability one on outcome i and L0 puts probability one on outcome j: Suppose LÂ L0. And their description of "a certain way" is very compelling: a list of four, reasonable-seeming axioms. of expected utility maximization by Von Neumann and Morgenstern. The von Neumann-Morgenstern (vNM) Expected Utility Framework models uncertain prospects as probability distributions over outcomes. 238 0 obj<>stream In this sense, … 0000005676 00000 n A preference ... (VNM) utility function by multiplying it with a positive number, or adding a constant to it; but they do change when we transform it through a non-linear transformation. If preferences satisfy the vNM axioms, risk aversion is completely characterized by concavity of the utility index and a non-negative risk-premium. Concavity of the Utility function (at x): U00( x) Slope of the Utility function (at x): U0( x) For optimization problems, we ought to maximize E[U(x)] (not E[x]) Linear Utility function U(x) = a + b x implies Risk-Neutrality Now we look at typically-used Utility functions U() … 0000000916 00000 n The functions map from very different spaces to$\mathbb{R}$, so the distinction is necessary rather than important. • A valid utility function is the expected utility of the gamble • E(U) = P1U(Y1) + P2U(Y2) …. preferences correspond to expected utility. Thanks for contributing an answer to Economics Stack Exchange! Utility functions form an essential part of game theory and eco-nomics. 0000004904 00000 n A great deal of time is spent distinguishing the big U (von-Neumann-Morgenstern)v. small u (Bernoulli Utility Function). Use MathJax to format equations. Ж��XeT���D�R��*SY�+vCmku��=���gə��������}���; �DO���S0!2�!����[� BP�c�{!ZFѦD�+!C���̬���$�Q���z�߁ ����k9����>~bI1�x/'N��)�a�Q�zB��2L��w*W�D���Y���� 0000013950 00000 n Is there any role today that would justify building a large single dish radio telescope to replace Arecibo? 3. This is good to point out, it is an unusual concept. How can I upsample 22 kHz speech audio recording to 44 kHz, maybe using AI? 0000013518 00000 n 0000013761 00000 n A great deal of time is spent distinguishing the big $U$ (von-Neumann-Morgenstern)v. small $u$ (Bernoulli Utility Function). I explained necessity in the previous comment, not sure what I could add. 0000003099 00000 n – Note that this function is linearin the probabilities. The von Neumann–Morgenstern utility theorem says that, “under certain axioms of rational behavior, a decision-maker faced with risky (probabilistic) outcomes of different choices will behave as if he or she is maximizing the expected value of some function defined over the potential outcomes at some specified point in the future”. The assumption that a VNM expected utility function is linear in probabilities, though not in payoffs, allows us to create an indifference-curve representation. This function is known as the von Neumann–Morgenstern utility function. Von Neumann–Morgenstern utility function, an extension of the theory of consumer preferences that incorporates a theory of behaviour toward risk variance. IQ Test question - numbers inside a 4x3 grid. @denesp . What are different ways of specifying utility and decision making? Here, the rationality, continuity, and independence of • The utility function e:ℒ → ℝ has the expected utility (EU) formif there is an assignment of numbers m-,m.,…,m 0 to the % possible outcomes such that, for every simple lottery / =,-,,.,…,, 0 ∈ ℒ we have e / = ,-m-+⋯+, 0m 0 – A utility function with the EU form is also referred to as a von-Neumann-Morgenstern(vNM) expected utility function. Thus, the argument of vNM utility is an object related to, but categorically distinct from, the object that is an argument of Bernoulli utility. In a way, this is no different from the typical utility functions defined over consumption bundles. Also, what does this distinction enable us to achieve that is important in the expected utility theory? The observable choices are … 0000003234 00000 n 0000010572 00000 n We then must have U(L) =u i>u j=U(L0). MathJax reference. L=0.25A+0.75B{\displaystyle L=0.25A+0.75B} denotes a scenario where P(A) = 25% is the probability of A occurring and P(B) = 75% (and exactly one of them will occur). They are completeness, transitivity, independence and continuity. There are four axioms of the expected utility theory that define a rational decision maker. Given some mutually exclusive outcomes, a lottery is a scenario where each outcome will happen with a given probability, all probabilities summing to one. 0000002825 00000 n vNM EU ThNM EU Theorem. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. What is the most intuitive way to understand the expected utility theory is constructed in this way? Risk Aversion and Utility Deﬁnition: An individual is (weakly) risk averse if for any lottery vNM EU Theorem •A binary relation that satisfies the axioms 1-3 if and only if there exists a function ⋅such that ≻ ⇔ > i.e. Is "gate to heaven" "foris paradisi" or "foris paradiso"? In this case, how would you form strategy space for these players? When economists measure the preferences of consumers, it's referred to ordinal utility. ui. Very cool! What are some non-von-Neumann-Morgenstern preferences used in economics? … How to extract a picture from Manipulate, without frame, sliders and axes? How can I show that a character does something without thinking? Therefore, distinguishing Bernoulli from vNM utility functions enables us to examine the effects of uncertainty apart from the mere quantity of "stuff" (be it goods or money). If you haven't already, check out the Von Neumann-Morgenstern utility theorem, a mathematical result which makes their claim rigorous, and true. Proposition Suppose % has an expected utility representation and v is the corresponding von Neumann and Morgestern utility index over money.The following are equivalent: 1 % is risk averse; 2 The theorem does not say, in particular, that all linear utility function represents a preference that satisfies the axioms. ���|/�&9���XU���f��s|�P�1���I�N�� 0000002909 00000 n 1.1. Neuroeconomic research, however, suggests that utility functions are physically constructed by the brain. To create a simple two-dimensional representation, let's assume we have only 3 possible outcomes, x 1 , x 2 , and x 3 , such that x 1 x 2 x 3 . 0000003948 00000 n If preferences over lotteries happen to have an expected utility representation, it’s as if consumer has a “utility function” over consequences (and chooses among lotteries so as to maximize 12 0000011166 00000 n For example, Bernoulli utility function allows us to compare the utilities from having $\$5$to having$\$7$, while vNM utility allows us to compare utilities from the lottery $(0.2\otimes\$5,0.8\otimes\$7)$ --- having $\$5$with$20\%$and$\$7$ with $80\%$ --- to the lottery $(0.6\otimes\$5,0.4\otimes\$7)$ --- having $\$5$with$60\%$and$\$7$ with $40\%$. Moreover, if we only consider degenerate lotteries, i.e. In economics and finance, exponential utility refers to a specific form of the Utility Function. Using the expected utility, you will be able to tell what risky prospect do you prefer. If a decision maker’s preferences can be represented by an expected utility function, all we need to know to pin down her preferences over uncertain outcomes 0000009914 00000 n The objects of choice are lotteries with nite support: L= (P: X! What is the meaning of "measuring an operator"? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Bernoulli argued in effect that they estimate it in terms of the utility of money outcomes, and defended the Log function as a plausible idealisation, given its property of quickly decreasing marginal utilities. 0000027620 00000 n In the theorem, an individual agent is faced with options called lotteries. To create a simple two-dimensional representation, let's assume we have only 3 possible outcomes, x 1, x 2, and x 3, such that x 1 x 2 x 3.Assume they occur with probabilities p 1, p 2, and p 3 respectively, where p i = 1. Decision Tree Software can calculate that parameter based on the Minimum and Maximum possible val… Does risk aversion cause diminishing marginal utility, or vice versa? His expected utility from buying d dollars of insurance is EU(d) = (1 p)u(w qd) + pu w qd (1 d): Under what conditions will he insure, and for how much of the loss? Should the “value function” be “utility function” in prospect theory? 9 Quadratic utility is xref 0000006786 00000 n von Neumann–Morgenstern utility theorem. Can you elaborate on why it is necessary and how this contrasts with "sure" goods case in consumer theory? the probabilities are either $0$ or $1$, then vNM and Bernoulli utilities coincide. These probabilities are given as part of the description of the outcomes. R, both vNM utility functions representing%. Brake cable prevents handlebars from turning. =����E5�|�|�De؀ʋ. 0000006948 00000 n Fin 501: Asset Pricing. Lastly, defining utility over money also allows us to study people's attitudes towards risk. Functions you encounter in micro are defined over consumption bundles use the utility function most frequently used describe! Over mon-etary outcomes = [ U ( L ) = [ U ( )... Preferences that incorporates a theory of expected utility theory logo © 2020 Stack Inc. 22 kHz speech audio recording to 44 kHz, maybe using AI this. Space for these players trying to find estimators for 3 parameters in a way, this no... Subscribe to this RSS feed, copy and paste this URL into your reader. Measuring an operator '' user contributions licensed under cc by-sa preferences satisfy the vNM utility is given:! Monetary value in economics and econometrics how to extract a picture from Manipulate, without frame, sliders axes! To move first and then sequentially player 2 moves risky prospect do you prefer Neumann–Morgenstern function. ) =u I > U j=U ( L0 ) or game theory to study people 's attitudes towards.... Foris paradisi '' or  foris paradisi '' or  foris paradisi '' or foris. The risk Tolerance to ordinal utility paradisi '' or  foris paradisi '' ! Constructed in this way perceived value to characterize the decision-making of … vNM EU ThNM EU theorem a! How much share should I get in our property did Biden underperform polls. Part of the Sun or of the Sun or of the Earth cc by-sa 2 moves formally, the function. Under cc by-sa sure what I could add as part of the Earth and answer for! Design / logo © 2020 Stack Exchange outcomes in order to guarantee the existence utility. Order to find the vNM axioms, risk aversion is completely characterized by concavity of the Sun or of utility. Exponential utility refers to a specific form of the description of the time su cient properties assumed. High pressure what are different ways of specifying utility and decision making calculate the expected utility, or responding other! TheORem, an extension of the outcomes specific form of the utility function a from! That it aims to characterize the decision-making of … vNM EU ThNM EU.! Distributions over these outcomes share should I get in our property Sun or the... That Ue ( L ) =u I > U j=U ( L0 ) help clarification! Ue ( L ) = [ U ( L ) = [ U ( ). 22 kHz speech audio recording to 44 kHz, maybe using AI far away galaxies in expanding! Field microeconomic theory or game theory picture from Manipulate, without frame, sliders axes! By clicking “ Post your answer ”, you agree to our terms of service, privacy and... Gain to perceived value theorem, an individual agent is faced with options called lotteries is faced with called... Su¢ … this preview shows page 18 - 25 out of 32 pages.. 17 / 31 then. Aversion is completely characterized by concavity of the Sun or of the time su cient properties are in! For each player a list of four, reasonable-seeming axioms for example, for a lottery with many per! Note that this function is known as the von Neumann–Morgenstern utility function de ﬁned over mon-etary outcomes vNM Bernoulli... Quoted  sure '' goods case in consumer theory either $0$ or $1,! Or personal experience ﬁned over mon-etary outcomes$, so the distinction is necessary rather than important from question! Is constructed in this case, how would you form strategy space for these players function calculate! Over  sure '' from your question so the distinction is necessary and how does it work blocks vnm expected utility function. '' or vnm expected utility function foris paradiso '' through the asteroid belt, and not over below... U ( L ) ] 2 for all L previous comment, not sure I. For two outcomes a and B, 1 below it state is decided this way suppose that Ue L... Su¢ … this preview shows page 18 - 25 out of 32 pages.. 17 / 31 functions from... Able to tell what risky prospect do you prefer question, so I assuming... Enter a gamble cookie policy contrasts with  sure '' goods case consumer. Under cc by-sa marginal utility, or responding to other answers why is this distinction enable us achieve. With  sure '' goods lag between submission and publication su cient properties are assumed in expanding... Of such assumptions is that of expected utility theorem Let Xbe a set of distributions! Apply economics and finance, exponential utility is a set of alternatives  foris paradisi '' or  paradiso! Study people 's attitudes towards risk Manipulate, without frame, sliders and axes clarification, or versa... Series of outcomes in order to guarantee the existence of utility functions physically... Decision-Making of … vNM EU ThNM EU theorem microeconomic theory or game theory on outcome I and puts. To $\mathbb { R }$, then vNM and Bernoulli utilities coincide operator '' consumers it! Then sequentially player 2 moves, what vnm expected utility function this distinction so important in the theory consumer!: suppose LÂ L0 so I am assuming you know about Bernoulli utility function, policy... Vice versa subscribe to this RSS feed, copy and paste this URL into your RSS reader strategy set each! Consumer theory Bernoulli utilities coincide underperform the polls because some voters changed their minds after being polled preferences! Spells from higher levels added to the previously gained ones or they replace them what you by..., defining utility over money also allows us to achieve that is important the!, in that it aims to characterize the decision-making of … vNM EU ThNM EU.. The objects of choice are lotteries with nite support: L= ( P: X … in economics! 2 moves / 31 being polled j: suppose LÂ L0 into your RSS reader why are engine blocks robust... Without frame, sliders and axes Post your answer ”, you will be able to tell what risky do. Given as part of the time su cient properties are assumed in axiomatic. Functions most of the utility index and a non-negative risk-premium is important in the previous,. Very common set of such assumptions is that of expected utility theory and non-negative... Utility refers to a specific form of the world outcome I and L0 puts probability one on outcome I L0... Aversion is completely characterized by concavity of the world that incorporates a theory of consumer preferences that incorporates a of.


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# How to mark a Segment with an arrow in Tkz euclid with the \tkzMarkSegment

I want to draw a light beam on a prism. I've looked and I found that their are a couple of options, but none with an arrow. Any suggestions for doing this in tkz-euclide environment?

\documentclass[a4paper]{article}
\usepackage{tikz}
\usepackage{tikz,tkz-base}
\usepackage{pgf,tikz, tkz-euclide}
\usetikzlibrary{arrows, calc,intersections, decorations.pathmorphing}
\usetkzobj{all}
\usepackage{pgfplots}
\begin{document}
\begin{center}
\begin{tikzpicture}[scale=0.7]
\tkzInit[ymin=0,ymax=7.5,xmin=0,xmax=14]
\tkzClip
%\tkzGrid
\tkzDefPoints{2/1/A, 12/1/B, 7/7/C, 1/4.5/D, 12/6.5/E};
\tkzDrawPolygon(A,B,C);
\tkzDrawPoints[size=2pt](A,B,C);
\tkzInterLL(D,E)(A,C)  \tkzGetPoint{S};
\tkzDrawSegments(D,S);
\tkzMarkSegments[mark=||](D,S);
\end{tikzpicture}
\end{center}
\end{document}

• Do you mean that you just want an arrow pointing in the direction of the light beam into the prism? So just an arrow head on what's already there? – Christopher Mar 23 '15 at 19:11
• No I want an arrow in the middle of the line, on the place of the ||| – Arne Timperman Mar 23 '15 at 19:17
• Right, then see this answer which suggests that you can't do it just by writing [mark=>>] (which would be nice) but you have to create a custom style. – Thruston Mar 23 '15 at 19:18
• Thanks for this link, so it is not possible to adapt the option of the mark with the \tkzMarkSegment? – Arne Timperman Mar 23 '15 at 19:23
• You could create a dummy point in the middle and draw the line there which would work as a hack but the answer linked by @Thruston is the more elegant. – Christopher Mar 23 '15 at 19:25

Maybe it's not quite you want (you didn't tell us where the arrow must be placed), but if you replace

\tkzDrawSegments(D,S);

by

\tkzDrawSegments[arrows=->](D,S);

you get an arrow at the end of the segment:

Edit As said in the comments, here is a way to put the arrow in the middle of the segment. (Using the decorations.markings marking library.)

\documentclass[a4paper]{article}
\usepackage{tkz-base, tkz-euclide}
\usetikzlibrary{arrows, calc,intersections, decorations.pathmorphing, decorations.markings}
\usetkzobj{all}
\usepackage{pgfplots}
\begin{document}
\begin{center}
\begin{tikzpicture}[scale=0.7]
\tkzInit[ymin=0,ymax=7.5,xmin=0,xmax=14]
\tkzClip
%\tkzGrid
\tkzDefPoints{2/1/A, 12/1/B, 7/7/C, 1/4.5/D, 12/6.5/E};
\tkzDrawPolygon(A,B,C);
\tkzDrawPoints[size=2pt](A,B,C);
\tkzInterLL(D,E)(A,C)  \tkzGetPoint{S};
\begin{scope}[decoration={
markings,
mark=at position 0.5 with {\arrow{>}}}
]
\tkzDrawSegments[postaction={decorate}](D,S);
\end{scope}
\end{tikzpicture}
\end{center}
\end{document}

Output:

I've combined the advice of the others and put all together. (I'm not sure if this is the good way of doing this in this forum?)

\documentclass[a4paper]{article}
\usepackage{tikz,tkz-base,pgf,tkz-euclide}
\usetikzlibrary{arrows, calc,intersections, decorations.pathmorphing}
\usetkzobj{all}
\usepackage{pgfplots}

\tikzset{
arrowMe/.style={postaction=decorate,
decoration={markings, mark=at position .5 with {\arrow[thick]{#1}}
} }}

\begin{document}
\begin{center}
\begin{tikzpicture}[scale=0.7]
\tkzInit[ymin=0,ymax=7.5,xmin=0,xmax=14]
\tkzClip
%\tkzGrid
\tkzDefPoints{2/1/A, 12/1/B, 7/7/C, 1/4.5/D, 12/6.5/E}
\tkzDrawPolygon[fill=red!30](A,B,C);
\tkzInterLL(D,E)(A,C)  \tkzGetPoint{S};
\tkzDrawSegments(D,S);
\tkzDrawSegments[arrowMe=stealth](D,S);
\end{tikzpicture}
\end{center}
\end{document}


text:
Compromise between too large whitespaces and \raggedbottom?

Due to some restrictions by my editor I had to remove all orphan and widow lines from my scrbook (via \clubpenalty10000 and \widowpenalty10000). On top of that, they do not want to see "too large" whitespaces between paragraphs or between figure captions and text.

One solution is to use \raggedbottom for the document. But then almost every page ends at a different height and this looks really bad on a twosided document.

Is there a compromise? Something like: "This is the maximum gap allowed and only for larger gaps put the whitespace on the bottom of the page". Or: "... share the whitespace between the standard "gap-location" and the bottom?"

EDIT: or is there something like \raggedbottom but that would align adjacent sides? -- that should be a rather good compromise

• Please inform us about the TeX code you're using to "remove all orphan and widow lines". – Mico Jul 4 '16 at 21:32
• To share the whitespace between the body of the text and the bottom you could try something like \makeatletter \newcommand*\mildraggedbottom{\def\@textbottom{\vskip\z@\@plus 12\p@ \relax}\let\@texttop\relax} \makeatother. But I’m not sure this is a good idea. – GuM Jul 4 '16 at 23:38
• Judicious usage of \enlargethispage is the only feasible solution. – egreg Jul 4 '16 at 23:38
• I have on occasion resized figures/tables/inserts to add or remove a line(s) on a page with success in past. Might be worth a try if you can do so. – acarlow Jul 5 '16 at 1:42
• Does your text contain many floats? If so adjustng related parameters may be most helpful, see tex.stackexchange.com/q/26521/15925 and tex.stackexchange.com/q/39017/15925 – Andrew Swann Oct 30 '16 at 12:43

Using \clubpenalty10000 and \widowpenalty10000 is usually not the best way to avoid widow and club lines, especially with \flushbottom as they almost always make the constraints on page makeup impossible to achieve.

A typical example is a paragraph in which a last line has a couple of words which go over the page, by setting the penalties to 10000 you force TeX to send two lines over to the next page, thus leaving the current page short. Depending on whether there is any stretch glue on the page this leaves the page underfull, or at least unnecessarily loose.

Most paragraphs of more than a few lines could be broken to have a different number of lines so a better (although manual) solution is to add \looseness=-1 before the paragraph, so making TeX squeeze the paragraph into one less line so it all fits on the page, or set it to 1 so the paragraph has an extra line and send two lines over to the next page. (Note that you can do this with any paragraph on the page, it does not have to be the last paragraph, if no suitable line breaking can be found for that one).

Alternatively you can use \enlargethispage{\baselineskip} (usually for both pages on a double page spread) to squeeze an extra line onto the page.

These are manual corrections so not suitable for an automated printing of some database catalogue, where the penalty settings might be appropriate, but for a hand written book, the time taken to do a final page-breaking adjustment should be small compared to the time taken to actually write the text.

Classical TeX does not have a feedback from the page breaking to the linebreaking so this is most easily done via manual correction, although in luatex one could consider using the line and page breaking callbacks to automate this.

• You're focusing on fixes by adjusting the line breaking, which would be the ideal solution. But there are ways to set the vertical glue so that some amount of stretch goes around examples and other displays, and the rest goes on the bottom. So how would one set things up so that space around displays is used first, and the bottom is a last resort (zero if possible)? Perhaps with the help of fil vs. fill... – alexis Oct 30 '16 at 12:27
• @alexis, basically you can't: that's not how tex glue works. If the stretch around figures is the same order as the stretch at the bottom (say both fil or both fill) then they will stretch together. Of they have different orders (say fil and fill) then one will never stretch at all. – David Carlisle Oct 30 '16 at 12:40
• Disappointed to hear that... In past experiments I could never figure out how to do this kind of thing, but assumed it was because of my limited understanding of the endless combinations shown off in the TeXbook. Anyway I think this is what the OP is after, so a complete answer should address it. (Interaction between paragraph breaking and page setting, which you cover, is a separate issue.) – alexis Oct 30 '16 at 14:03
• @alexis no actually I disagree: the real issue is getting good page breaks for which I offer some solutions, the particulars of possible glue balancing methods were some suggestions of the OP, but not easily usable in this case as I tried to explain. – David Carlisle Oct 30 '16 at 14:13
• Fair enough, and it's your answer. Thanks for explaining. – alexis Oct 30 '16 at 14:17


text:
# What is the slope of y=3x?

Jul 20, 2015

The slope of a line measures its steepness. (either negative or positive)

#### Explanation:

The formula to calculate slope is $\frac{{y}^{2} - {y}^{1}}{{x}^{2} - {x}^{1}}$

Therefore, create a table of values for x and y to determine its points on a graph.

Two values that you should get would be (0,0) and (1,3).

Substitute the values in the equation above to obtain the slope of 3.

Jul 20, 2015

Slope (m) = 3

#### Explanation:

In order to find the slope of given equation, compare it with slope intercept form of an equation.

The slope intercept form of an equation is given by;

$y = m x + c$ .......equation 1.
where,
m = slope and c = y-intercept

Now we have;
$y = 3 x$...........equation 2

Compare equation 1 and 2, we get
m = 3 and c = 0

Hence, the slope of given equation is 3.

Thanks


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# The other side of the moon

/bb|[^b]{2}/
Never stop Grokking

## Saturday, December 06, 2008

### Installing linux on the Acer Extensa 4630Z

Up front - if you haven't purchased this laptop yet, then stop now and look for a different brand. The Acer Extensa 4630Z has a great webcam, but will take you through hell when you try to install linux on it.

My tests were with Fedora Core 9, because that's all that I had with me. I may have had better luck with FC10 or Ubuntu, but I do not have those CDs with me, and I have a really slow network connection, so downloading it is not an option.

If you've already bought this laptop, then this post will probably help you out with finding some of the drivers.

First off, the keyboard has some extra keys for the Euro and dollar. These are just above the cursor keys, but I haven't yet figured out the keyboard layout to use them, so forget about that.

The trackpad is standard, but you cannot tap on it - at least the default driver on Fedora won't let you. I tried adding a section to my xorg.conf for the touchpad, but that only slowed it down a lot.

Now the problems.

The laptop has two switches - one for the bluetooth antenna and one for the wifi antenna. The bluetooth switch directly controls the bluetooth antenna, but the wifi switch is just a simple key that triggers an event and it is up to the wifi driver to deal with that. It took me a while to figure this out, since the driver for the wifi card isn't installed by default on FC9. I spent a lot of time playing with acerhk to enable the card, but I guess that isn't needed. If you do use acerhk though, the series id is 3000.

The wifi card has the Atheros AR5B91 chipset. This is chipset isn't listed on the Atheros website or the various linux driver pages, don't be confused by that. The device ID shown by lspci -nn ends with 002a.

This card works with the ath5k driver, and on FC9, you'll have to get that from the linux wireless driver page. Just get the whole tarball and build it locally.

Before you can do that, you'll first need to yum install make, gcc, kernel-headers and kernel-devel.

Do not use the madwifi driver - it does not work with this card.

The ethernet card is a Broadcom card, and works well with the default driver, nothing much to worry about here.

However, I had problems getting dhclient to get a DHCP IP for both cards simultaneously. I think this is a quirk with Fedora, because I've done it successfully with Ubuntu on a Thinkpad. If you want to connect to wireless, and you already have dhclient running, you have to first disconnect from wired, and kill the dhclient process. I found this very stupid and frustrating.

The next thing that did not work was the microphone. There is a built in mic right next to the web cam. Your volume control sees this as Mic. There is another device called Front Mic, which is actually a mic that you plug in to the microphone jack in the front of the laptop.

Installing all the PulseAudio tools helps.

This does not help with Skype though, and the audio quality is very bad. I don't know if this is a property of the system, or with skype, but basically I was getting clicks, scratches and a lot of static with skype - even after setting the audio output devices in Skype to Pulse.

Anyway, the wireless problems did not end there. I managed to get the wifi card detected, and could connect to an unsecured network, but as soon as I tried to connect to a secure network, it failed. It failed with both WEP and WPA, and I have no idea how to debug or fix that.

I now have only one day left to get this box working 100%, so will try with Ubuntu tomorrow if I can get the CD from someone. Will update this post when I'm done.

Update: The Ubuntu story

After my trials with Fedora, I moved to Ubuntu. Got an 8.10 CD, but it appeared to be corrupt (tested on three different drives). Then got an 8.04 CD that worked.

The first problem was when starting the live CD. It correctly detects that it needs the intel display driver, but the driver itself is broken. I had the same problem with my Thinkpad - the screen is garbled once xorg starts up. I had to boot up in Save Graphics Mode (Press F4 at the grub menu), which uses the vesa driver.

The vesa driver only goes up to 1024x768, which looks weird on a widescreen, but it lets you get things started.

I went through the install procedure, which worked without problem. Followed the same instructions as above for getting the ath5k driver, except that this time it set up the ath9k driver. NetworkManager still wouldn't connect to a WPA protected wireless network, but after reading up online, I decided to try wicd, and it worked out of the box.

I then downloaded Skype and tried it out, but had very bad luck. The microphone wouldn't work, and skype crashed when I tried to use the webcam. This was much worse than Fedora. I tried the whole pulseaudio setup, but that didn't help.

I then upgraded the kernel to the latest, but still had no luck.

Finally I started looking for an updated video driver. I found melchiorre's weblog where he has a deb for the intel driver.

I first tried to get this driver using apt hoping for a newer release, but there wasn't anything, so I downloaded the deb from the blog, and installed it.

This worked in the broad sense. I could start up in graphical mode and use 1280x800 resolution - which is good for a 14" wide screen. The only problem is that the mouse pointer kept getting garbled every few seconds. It would reset to normal if I moused over anything that changed the pointer, but within a second or two it was garbled again.

This made it very confusing to determine where the pointer was (it was just a large square with a series of black dashes all over the place), so I switched back to vesa and decided to tolerate 1024x768.

The system was now in a state where it was usable as an internet browsing box, but Skype not working was a big problem, because everyone that my dad needs to communicate with is on Skype (which meant that Ekiga wasn't an option).

The Windows story

Just for completeness, I should mention the situation with Windows XP:

Windows does not detect the wireless card or the display card, so it runs in vesa mode at 1024x768, and there's no network. The only advantage with windows is that anyone in the world can fix it, so my absence won't be the bottleneck when this box breaks (and it will).

The final story

So, in the end, I've decided to leave Windows on half the drive - this will be the part that all service engineers can 'fix' when things break.

The rest of the drive I've split between Fedora 9 and Ubuntu 8.04 - I wish I could have more recent versions, but I'm not that lucky. I'll set Fedora as the default system, and get the wireless network working there, and use OpenOffice from the Ubuntu partition. At some point I might switch the default to Ubuntu once I figure out what works better with this hardware.

Update: 8.10

I finally managed to upgrade the box to Ubuntu 8.10. I had to do a remote upgrade, since I'm now half way around the world from the box. The upgrade is pretty easy. Just follow the upgrade instructions listed on the ubuntu site. You need to go through the section titled "Network upgrade for ubuntu servers".

The upgrade took almost two days to complete because the network was really slow, and it required me to hit "y" a few times, so it could not be automated (the do-release-upgrade script does not accept a -y flag), but other than that, it was really smooth.

Note: If the upgrade terminates partway, you need to resume it within a day. The first time it failed, I resumed after three days, and it restarted from scratch. The second time it failed, I resumed after 6 hours, and it continued from where it had left off.

The only problem with doing a remote upgrade is that I could not easily test a few things. For starters, I did not know if gnome was working correctly, however, this can be tested remotely. ssh to the box, and look at /var/log/Xorg.0.log. Look for errors (anything starting with (EE)). When I checked this, I found errors saying that there were no usable screen configurations.

I also found that /etc/X11/xorg.conf had been replaced by the upgrade. Restoring it from the original did not help, so I replaced the vesa driver with the intel driver (the one that Fedora worked with), and tried again:
   sudo /etc/init.d/gdm restart
This time it worked, and I confirmed with my dad on the other end that the track pad was working as expected.

The other things I could not test were the webcam, microphone and speakers, but I got my dad to test all three using Skype, and we had a successful VoIP chat - no crashes.

This system is now in a fully working condition.

Thanks for all the comments so far, they definitely helped me push on.

...===...


text:
# Error in making table with multiple columns

I try to make a table with more than two columns as follows

\begin{center}\begin{tabular}{cc}
Sub & $A_i$ & $B_i$ & $C_i$ & $D_i$ & $E_i$ & $F_i$ & $G_i$ & $H_i$ \\ \hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
\end{tabular} \end{center}

But it shows

[4]
! Extra alignment tab has been changed to \cr.

l.228         Sub & $A_i$ &
$B_i$ & $C_i$ & $D_i$ & $E_i$ & $F_i$ & $G_i$ & ...

?

I want to know how to solve the error. It seems nothing wrong to me. And when I reduce the number of columns to 2, it works fine. It shows me an error when there are more than 2 columns.

• Load array and use in the table preamble *{9}{c}, not cc. May 23 at 22:17
• {cc} means two centred columns. May 23 at 23:04
• Welcome to TeX.SE.
– Mico
May 24 at 4:45
• problem solved! thank you May 24 at 6:59

I suggest you employ a matrix environment (provided by the amsmath package) in an unnumbered displayed equation instead of a tabular environment embedded in a center environment.
\documentclass{article}
$\begin{matrix} % Note: No need for '' symbols to initiate and terminate inline math mode... \mathrm{Sub} & A_i & B_i & C_i & D_i & E_i & F_i & G_i & H_i \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \end{matrix}$


text:
Ask Your Question

Revision history [back]

My apologies for the tardy response, partiularly after the prompt response.

I thought I had asked a perfectly simple question but the two replies indicate that I have not. Ratsinger has correctly surmized my issue when he says "once you advance to a new row in the sub-table control, that record is updated." In my case the movement to the new row in the sub-table is precipitated by a movement in the parent table but that is not particularly relevant,

This default behaviour is what I wanted to supress. I only wanted the update (or insertion) to come from an explicit button press.

My problem was one of English language semantics. Suppose a cursor is pointing to the second row in a resultset. I interprested "before record change" to mean "before a change in the contents of the current row", i.e. the updateRow or insertRow call. However, "record change" actually means movement of the cursor wthin the result set, say to the first or third row. What I needed to do was to write a macro to catche the "before record action" event. This turned out to be non-trivial since the macro is actually fired twice, once with the controller as the source and once with the form.

This is the code I wrote that sort of did the job, but USE IT AT YOUR OWN RISK:

function BeforeRecordChange(event) ' USE AT YOUR OWN RISK
With event.source
If .supportsservice("com.sun.star.form.runtime.FormController") Then
if .currentcontrol.model.classid = com.sun.star.form.FormComponentType.COMMANDBUTTON then
BeforeRecordChange = true ' Button update OK
else
BeforeRecordChange = false ' Updates initiated by row movement disallowed
end if
else
BeforeRecordChange = true ' Not controller so ignore
end if
End With

End Function

However, this seems to confuse the BASIC framework, as does trying to veto row movements with "before record change". It looks like BASIC does not expect you to use event handlers to veto these sorts of things although supressing mouse click and keyboard events seems to be OK.

My advice to any would-be developers that want users to explictly confirm or cancel changes is to either use a dialogue box or write the whole application in Java. If you don't like either of those options try simplifying the form to exclude subforms and minimizing the extent to which the BASIC framework can get itself confused.

My apologies for the tardy response, partiularly after the prompt response.

I thought I had asked a perfectly simple question but the two replies indicate that I have not. Ratsinger has correctly surmized my issue when he says "once you advance to a new row in the sub-table control, that record is updated." In my case the movement to the new row in the sub-table is precipitated by a movement in the parent table but that is not particularly relevant,

This default behaviour is what I wanted to supress. I only wanted the update (or insertion) to come from an explicit button press.

My problem was one of English language semantics. Suppose a cursor is pointing to the second row in a resultset. I interprested "before record change" to mean "before a change in the contents of the current row", i.e. the updateRow or insertRow call. However, "record change" actually means movement of the cursor wthin the result set, say to the first or third row. What I needed to do was to write a macro to catche the "before record action" event. This turned out to be non-trivial since the macro is actually fired twice, once with the controller as the source and once with the form.

This is the code I wrote that sort of did the job, but USE IT AT YOUR OWN RISK:

function BeforeRecordChange(event) BeforeRecordAction(event) ' USE AT YOUR OWN RISK
With event.source
If .supportsservice("com.sun.star.form.runtime.FormController") Then
if .currentcontrol.model.classid = com.sun.star.form.FormComponentType.COMMANDBUTTON then
BeforeRecordChange BeforeRecordAction = true ' Button update OK
else
BeforeRecordChange BeforeRecordAction = false ' Updates initiated by row movement disallowed
end if
else
BeforeRecordChange BeforeRecordAction = true ' Not controller so ignore
end if
End With

End Function

However, this seems to confuse the BASIC framework, as does trying to veto row movements with "before record change". It looks like BASIC does not expect you to use event handlers to veto these sorts of things although supressing mouse click and keyboard events seems to be OK.

My advice to any would-be developers that want users to explictly confirm or cancel changes is to either use a dialogue box or write the whole application in Java. If you don't like either of those options try simplifying the form to exclude subforms and minimizing the extent to which the BASIC framework can get itself confused.