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http://bioconductor.riken.jp/packages/3.10/data/experiment/vignettes/IHWpaper/inst/doc/IHW_censoring_storey_simulations.html | # 1 Simulation setting and methods compared
As IHW has been benchmarked thoroughly elsewhere (for example, the rest of this package contains many such simulations), here we restrict ourselves to a small study to illustrate the effects of censoring (as the censoring threshold $$\tau$$ varies) and null-proportion adaptivity.
In particular, we consider the following example of a simple conditional two-groups model (for two different values of $$\pi_0 \in \{0.7, 0.9\})$$:
\begin{aligned} &X_i \sim U[0,2.5]\\ &H_i \mid X_i \sim \text{Bernoulli}(\pi_0)\\ &Z_i \mid (H_i = 0, X_i) \sim \mathcal{N}(0,1)\\ &Z_i \mid (H_i = 1, X_i) \sim \mathcal{N}(X_i,1)\\ &P_i = 1 - \Phi(X_i) \end{aligned} We compare the following approaches, using the nominal level $$\alpha=0.1$$, $$P_i$$ as the p-values and $$X_i$$ as the covariates (except for BH which does not account for covariates):
1. BH (Benjamini and Hochberg 1995)
2. IHW
3. IHW with Storey $$\pi_0$$ correction ($$\tau'=0.5$$)
4. IHWc with different censoring thresholds $$\tau$$
5. IHWc with Storey $$\pi_0$$ correction ($$\tau'=0.5$$) and different censoring thresholds $$\tau$$.
The censored IHW (IHWc) variant was implemented as follows:
1. We censor the p-values, i.e. we set all p-values below $$\tau$$ to 0.
2. As in classical IHW, we discretize the covariate into a finite number of strata (20) and also randomly split the hypotheses into 5 folds.
3. Within each combination of stratum and fold, we fit the Beta-Uniform model with the EM-method of (Markitsis and Lai 2010), which can take censoring into account.
4. For each censored p-value, we draw a new p-value from its fitted Beta-Uniform model conditionally on it being $$\leq \tau$$, i.e. we impute all censored p-values.
5. Then we use regular IHW to learn the weight functions.
6. We use weighted BH after cross-weighting the original (i.e. non-censored/non-imputed) p-values, making sure we do not reject any p-value $$\leq \tau$$.
For the exact details of the simulation and also to reproduce it, see the folder “inst/theory_paper/ihwc_wasserman_normal_simulation.R” of this package.
# 2 Simulation results
library("ggplot2")
library("grid")
library("dplyr")
library("tidyr")
library("cowplot")
library("IHWpaper")
# color from https://github.com/dill/beyonce
# beyonce::beyonce_palette(30,5)
colors <- c("#040320", "#352140", "#871951", "#EB4A60", "#CFAB7A")
sim_folder <- system.file("simulation_benchmarks/theory_paper",
package = "IHWpaper")
sim_df <- readRDS(file.path(sim_folder, "ihwc_wasserman_normal_sim.Rds"))
taus_df <- expand.grid(fdr_pars = unique(sim_df$fdr_pars), fdr_method = c("BH","IHW","IHW-Storey"), sim_pars = unique(sim_df$sim_pars)) %>%
na.exclude()
df1 <- left_join(taus_df, select(sim_df, -fdr_pars), by=c("fdr_method","sim_pars")) %>%
select(-fdr_pars.y) %>% rename(fdr_pars = fdr_pars.x)
## Warning: Detecting old grouped_df format, replacing vars attribute by
## groups
## Adding missing grouping variables: fdr_pars
## Warning: Column fdr_method joining factor and character vector, coercing
## into character vector
## Warning: Column sim_pars joining factor and character vector, coercing into
## character vector
df <- full_join(df1, filter(sim_df, fdr_method %in% c("IHWc_Storey","IHWc"))) %>%
mutate( fdr_method = ifelse(fdr_method=="IHWc_Storey","IHWc-Storey", fdr_method))
## Joining, by = c("fdr_pars", "fdr_method", "sim_pars", "alpha", "FDR", "power", "rj_ratio", "FPR", "FDR_sd", "FWER", "nsuccessful", "sim_method", "m")
pi0s <- sapply(strsplit(df$sim_pars, split="[[:punct:]]"), function(x) paste0(x[[2]],".",x[[3]])) df$pi0s <- pi0s
fdr_plot <- ggplot(df, aes(x=fdr_pars, y=FDR, col=fdr_method)) + geom_line() +
facet_grid(.~pi0s, labeller=label_bquote(cols=pi[0] == .(pi0s))) +
scale_x_log10() +
scale_color_manual(values=colors) +
theme(legend.title=element_blank()) +
xlab(expression(tau~"(censoring threshold)")) +
ylab("FDR")
fdr_plot
power_plot <- ggplot(df, aes(x=fdr_pars, y=power, col=fdr_method)) + geom_line() +
facet_grid(.~pi0s, labeller=label_bquote(cols=pi[0] == .(pi0s))) +
scale_x_log10() +
scale_color_manual(values=colors) +
theme(legend.title=element_blank()) +
xlab(expression(tau~"(censoring threshold)")) +
ylab("Power")
power_plot
# 3 Discussion of simulation results
The first of the above figures shows the FDR of the procedures as $$\tau$$ varies (note that BH/IHW/IHW-Storey do not depend on $$\tau$$). Formally, only IHWc and IHWc-Storey have provable finite-sample control. However, we see that all methods are below the nominal $$\alpha=0.1$$. Furthermore, IHW-Storey is at essentially exactly $$0.1$$, i.e. by estimating $$\pi_0$$ it controls FDR at exactly the nominal level. This is also the case for IHWc-Storey when it is properly tuned. IHWc and IHWc can have FDR way below $$\alpha$$ for improper choice of the tuning parameter $$\tau$$.
The second plot shows the power of the procedures (defined as the expected proportion of true discoveries among all false null hypotheses). Of course for $$\pi_0 = 0.7$$ all methods have a lot more power than for $$\pi_0=0.9$$. We also observe the theoretically expected trade-off for the choice of censoring threshold $$\tau$$: If it is chosen way too large, then we cannot estimate the underlying conditional two-groups model well, thus the weights are suboptimal and we lose power. On the other hand, if $$\tau$$ is too small, then we lose a lot of power since we cannot reject the p-values below $$\tau$$. BH has a lot less power compared to IHW/IHW-Storey and also IHWc/IHWc-Storey for a wide range of censoring thresholds $$\tau$$.
# References
Benjamini, Yoav, and Yosef Hochberg. 1995. “Controlling the False Discovery Rate: A Practical and Powerful Approach to Multiple Testing.” Journal of the Royal Statistical Society. Series B (Statistical Methodology). JSTOR, 289–300.
Markitsis, Anastasios, and Yinglei Lai. 2010. “A Censored Beta Mixture Model for the Estimation of the Proportion of Non-Differentially Expressed Genes.” Bioinformatics 26 (5). Oxford Univ Press:640–46. | 2020-09-24 14:43:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8937134742736816, "perplexity": 9126.319733810387}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00657.warc.gz"} |
http://community.wolfram.com/groups/-/m/t/198840?p_p_auth=g28f53oP | DSolve fails to solve a very simple boundary value problem
GROUPS:
Hi there. I was surprised by the following issue with DSolve[]. I solve a simple boundary value problem:A v''(t)+B v(t) = F(t)where A, B -- 3x3 constant matrices, v(t) -- 3-component vector-function to be search, F(t) -- given vector-function. Each component of v(t) satisfies boundary conditions at the point 0 and 1. The problem is straightforward and it seems no issue of using DSolve should appear for any admissible F(t).In my case is calculated on previous steps. The right part comprises polynomials and hyperbolic sine/cosine.A={{-0.044697669173699448530657096189269774609066666666666666666667437866249.52287874528033,-0.012243540763682102904177110474279025634133333333333333333333563124849.52287874528033,0},{-0.012243540763682102904177110474279025634133333333333333333333563124849.52287874528033,-0.053659814734196655264138211329287943019266666666666666666667592496249.52287874528033,-0.009622111232315166943183972182021976285866666666666666666666847258249.52287874528033},{0,-0.009622111232315166943183972182021976285866666666666666666666847258249.52287874528033,-0.033141781673581725468679476493387322164266666666666666666667238484549.52287874528033}};B={{0.134093007521098345591971288567809323827200000000000000000002381302349.52287874528033,-0.073461244582092617425062662845674153804800000000000000000001304567849.52287874528033,0},{-0.073461244582092617425062662845674153804800000000000000000001304567849.52287874528033,0.160979444202589965792414633987863829057800000000000000000002858767349.52287874528033,-0.057732667393891001659103833092131857715200000000000000000001025250649.52287874528033},{0,-0.057732667393891001659103833092131857715200000000000000000001025250649.52287874528033,0.099425345020745176406038429480161966492800000000000000000001765653549.52287874528033}};Cv={0.06063176293900572816690862572213517002239999999959703682741.3422565562137*t,0.21278032964063766549749999999999999999999999999992032228440.30310235657392-0.03321638353144512597695900434870819654034999999987569840340.30279030542798*t+0.0160691980909371055151909786571919738642539.93202436395856*Cosh[1.73205080756887729352744634150587236694280525381038062805647.307206068870755*t]-0.03776972085899651059900806241041891399654810098499796775840.303173934606086*Sinh[1.73205080756887729352744634150587236694280525381038062805647.307206068870755*t],0.111481694136719805460000000000000000000000000000000000000001979757349.69897000433602};(here Cv used instead of F(t) ). For solving the problem the following code is used:Q = DSolve[{A[[1,1]] v1''[t]+A[[1,2]] v2''[t]+B[[1,1]] v1[t]+B[[1,2]] v2[t]==Cv[[1]], A[[2,1]] v1''[t]+A[[2,2]] v2''[t]+A[[2,3]] v3''[t]+B[[2,1]] v1[t]+B[[2,2]] v2[t]+B[[2,3]] v3[t]== Cv[[2]],A[[3,2]] v2''[t]+A[[3,3]] v3''[t]+B[[3,2]] v2[t]+B[[3,3]] v3[t]== Cv[[3]], v1[0] == 0, v1[1] == 1, v2[0] == 0, v2[1] == 1, v3[0] == 0, v3[1] == 1}, {v1[t],v2[t],v3[t]}, t]But calculation invokes the errorNo more memory available.Mathematica kernel has shut down.Try quitting other applications and then retry.or the error of kernelI tried to transform the right part into exponential formCv0=TrigToExp[Cv]Now.But DSolve cannot solve the problem at all:Q = DSolve[{A[[1,1]] v1''[t]+A[[1,2]] v2''[t]+B[[1,1]] v1[t]+B[[1,2]] v2[t]== Cv[[1]], A[[2,1]] v1''[t]+A[[2,2]] v2''[t]+A[[2,3]] v3''[t]+B[[2,1]] v1[t]+B[[2,2]] v2[t]+B[[2,3]] v3[t]== Cv0[[2]],A[[3,2]] v2''[t]+A[[3,3]] v3''[t]+B[[3,2]] v2[t]+B[[3,3]] v3[t]== Cv0[[3]], v1[0] == 0, v1[1] == 1, v2[0] == 0, v2[1] == 1, v3[0] == 0, v3[1] == 1}, {v1[t],v2[t],v3[t]}, t]Result:Indeed, I don't understand, what goes wrong?
2 years ago
10 Replies
Sean Clarke 1 Vote DSolve very often doesn't like to use Floating point numbers. This is because it is a symbolic function and may not perform numerically stable operations on the floating point numbers in the course of its symbolic operation. Be careful when mixing symbolic and numeric computations. Generally you might consider: 1. Turning the floating point numbers into rational numbers with Rationalize2. Numerically Solving the equation with NDSolve3. Solve using Parameters and then substitute the values of those parameters into the result from DSolve.If you do (3), try running the following:DSolve[{B[1,1] v1[t]+B[1,2] v2[t]+A[1,1] (v1^\[Prime]\[Prime])[t]+A[1,2] (v2^\[Prime]\[Prime])[t]==1,2 v1[t]+2 v2[t]+2 v3[t]+2 (v1^\[Prime]\[Prime])[t]+2 (v2^\[Prime]\[Prime])[t]+2 (v3^\[Prime]\[Prime])[t]==2,3 v2[t]+3 v3[t]+3 (v2^\[Prime]\[Prime])[t]+3 (v3^\[Prime]\[Prime])[t]==3,v1[0]==0,v1[1]==1,v2[0]==0,v2[1]==1,v3[0]==0,v3[1]==1},{v1[t],v2[t],v3[t]},t]This solves quickly and you can decide how to substitute the actual values back in. This can be tricky sometimes because of numerical precision issues.
2 years ago
Thanks, Sean. I suspected the points you listed. But can you suggest, why DSolve doesn't work in my last example, where the right part was converted to exponential form? Maybe, the expression has become very complicated for DSolve?
2 years ago
It'll still suffer from the same fundamental problems. I honestly don't try to guess too much about the internal workings of DSolve. DSolve will very often flatly refuse to start working with floating point numbers. Maybe it figures that exponentials and floating point numbers are a potentially dangerous coupling and so refuses to try anything.
2 years ago
I see, Sean. It is very pity. Such powerful CAS cannot solve a simple, in principle, problem. True misfortune.
2 years ago
That is indeed a simple problem. I strongy suspect that DSolve[] has nothing to do with it, it's just not enough physical memory on your system to deal with this task. Try to monitor your system resources while varying he precision of operations.Anyway, the linear system of ODEs, can be solved without any problems using DSolve[] with symbolic coefficients, then numbers can be plugged into obtained solution.Moreover, there is no need to use DSolve[] at all. The solutuon for linear system of the formz(t)'' + A z(t) = F(t) (original)z(t)' + A z(t) = F(t) (corrected)is justz(t) = Exp(-A t) z(0) + Exp(-A t) Int[( f(x) Exp(A x), {x,0,t}](not sure about the signs in the integral), where A is a matrix.The above is also true for non-autonomous systems with A(t), where it is only necessary to extend the system.I.M.
2 years ago
Ivan, you are wrong in this case, because my ODE is of 2nd order:But this approach can be used for 1st order ODE. My ODE can easily be transformed into such a ODE and solved. Nevertheless, I would like to use DSolve, because it requires much less job. If DSolve or NDSolve will demonstrate full frustration, I'll turn to other methods.
This produced an answer after several minutes and using no more than 400MB of memory. I tried solving the BVP directly but it took more than an hour longer; produced warning/error messages about exponents being too large; and it has an answer more than 20 times as large and takes much longer to evaluate. So I opt to present this way. sol0 = DSolve[ SetPrecision[{A.{x''[t], y''[t], z''[t]} + B.{x[t], y[t], z[t]} == Cv}, Infinity], {x, y, z}, t]; Block[{Integrate = NIntegrate, eqns}, eqns = {x[0] == 0, x[1] == 1, y[0] == 0, y[1] == 1, z[0] == 0, z[1] == 1} /. sol0; NSolve[eqns, Cases[eqns, _C, Infinity] // Union] ](* {{C[1] -> -0.0249078, C[2] -> 1.64911, C[3] -> 1.3804, C[4] -> -1.22953, C[5] -> 1.48102, C[6] -> -1.06965}}*)Taking a hint from Sean Clarke's observation, I changed the coefficients to exact numbers. Using SetPrecision[#, Infinity] is not always a good idea. All the real numbers get converted to exact fractions with very large denominators. I first tried i out the idea by rounding the coefficients to the nearest 1/1000. That worked, but the answer is in terms of Integrate and RootSum objects. It made me think this problem is not as simple as some have supposed. It is also why I used Block to switch Integrate to NIntegrate.After getting the general solution, we can solve for the constants that satisfy the boundary conditions.If you need 40-50 digits of precision as the OP's input suggests, then the above will have to be modified.Finally, I would think Bill Simpson's approach is best, unless one needs to analyze the algebraic expressions in a formula that occupies 300-400K of memory. | 2016-10-01 22:19:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6529161334037781, "perplexity": 3786.7980328992453}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738663308.86/warc/CC-MAIN-20160924173743-00076-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/515012/how-to-create-a-reflexive-symmetric-and-transitive-closures | # How to create a Reflexive-, symmetric-, and transitive closures?
I'm working on a task where I need to find out the reflexive, symmetric and transitive closures of R. Statement is given below:
Assume that U = {1, 2, 3, a, b} and let the relation R on U which is
given by R = {<2,3>, <3, 2>, <1, a>}
1. What is the reflexive closure of R?
2. What is the symmetric closure of R?
3. What is the transitive closure of R?
1. R $\cup$ {< 2, 2 >, <3, 3>, } - reflexive closure
2. R $\cup$ {< a, 1 >} - symmetric closure
3. R $\cup$ {<1, 2>, <1, 3>} - transitive closure
I would appreciate if someone could see if i've done this correct or if i'm missing something.
Thanks alot!
The symmetric closure is correct, but the other two are not.
$R\cup\{\langle2,2\rangle,\langle3,3\rangle\}$ fails to be a reflexive relation on $U,$ since (for example), $\langle 1,1\rangle$ is not in that set.
As for the transitive closure, you only need to add a pair $\langle x,z\rangle$ in if there is some $y\in U$ such that both $\langle x,y\rangle,\langle y,z\rangle\in R.$ There are only two such pairs to add, and you've added neither of them.
• what if I add <a, a> and <b, b> would it make it reflexive closure? If not how can I go forward to make it a reflexive closure? Regarding the transitive closure, then I only need to add <1, 3> to the relation to make it transitive? Oct 4 '13 at 21:18
• Yes, the reflexive closure is $$R\cup\{\langle1,1\rangle,\langle2,2\rangle,\langle3,3\rangle,\langle a,a\rangle,\langle b,b\rangle\}.$$ Regarding the transitive closure, as I said, neither of the pairs that you were adding are necessary. Now, if you had (for example) $\langle1,a\rangle,\langle a,3\rangle\in R$, then $\langle 1,3\rangle$ would be in the transitive closure, but this is not the case. Take another look at the relation $R$ and the hint I gave you. Alternately, can you determine $R\circ R$? Oct 4 '13 at 21:35
Transitive Closure:
The transitive closure of a relation $R$ is most simply defined as the smallest superset of $R$ which is a transitive relation. However, this is not a very practical definition. Practically, the transitive closure of $R$ is the set of all $(x,y)$ such that $(x,y)\in R$ or there exist $(x_0,x_1),(x_1,x_2),(x_2,x_3),\dots,(x_{n-1},x_n)\in R$ such that $x=x_0$ and $y=x_n$.
You can see further details and more definitions at ProofWiki. | 2021-09-20 12:10:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7631992101669312, "perplexity": 166.40903026119562}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00301.warc.gz"} |
http://astronomy.stackexchange.com/tags/milky-way/new | # Tag Info
0
As already answered, the definition of the size of a galaxy must always to some extend be arbitrary. In astronomy, several definitions are used, according to the context in which it's used, e.g.: $R_{\mathrm{vir}}$ (the virial radius): Used when considering the galaxy's dynamics; defined by $GM_{\mathrm{vir}}/R_{\mathrm{vir}} \sim V$, where $G$ is the ...
10
Photos of the galactic center aren't too bright because of all the gas and dust between us and it. For example (in infrared): I'm guessing, though, that you're talking about other galaxies, because there are no views of the galactic center of the Milky Way face-on. Although the galactic center is pretty luminous, just not in the wavelengths we're used to. ...
2
I don't think there is a catalogue containing information about blue/red shift per region of the sky. You probably need a catalogue containing blue/red shifts per object and then calculate the average blue/red shift per region. Most catalogues with blue/red shift are however targeted towards galaxies. They may also contain stars but you will have to filter ...
Top 50 recent answers are included | 2015-01-27 05:47:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5667111873626709, "perplexity": 1025.3952645993568}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115861027.55/warc/CC-MAIN-20150124161101-00171-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://vanmauchonloc.vn/tulsa-social-rknq/d31aad-find-function-from-points-calculator | Wed, 12 / 2020 6:16 am |
If you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points. Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. Can you find a 4th order polynomial? To derive the equation of a function from a table of values (or a curve), there are several mathematical methods. For example, the points (0,0), (1.40,10), (2.41,20), and (4.24,40) would yield the cubic function y=-0.18455x^3+1.84759x^2+4.191795+0. Linear equation with intercepts. Function point = FP = UFP x VAF. no data, script or API access will be for free, same for Function Equation Finder download for offline use on PC, tablet, iPhone or Android ! Also I would define it in single line as "A Method of quantifying the size and c… How to reconstruct a function? Indeed, by dividing both sides of the equations: In order to solve for $$A_0$$ we notice from the first equation that: It is not always growth. In practice, the type of function is determined by visually comparing the table points to graphs of known functions. The idea of this calculator is to estimate the parameters $$A_0$$ and $$k$$ for the function $$f(t)$$ defined as: so that this function passes through the given points $$(t_1, y_1)$$ and $$(t_2, y_2)$$. Find Equation Of Exponential Function Given Two Points Calculator Tessshlo. BYJU’S online function calculator tool makes the calculations faster, and it displays the graph of the function by calculating the x and y-intercept values, slope values in a fraction of seconds. The parameter $$k$$ will be zero only if $$y_1 = y_2$$ (the two points have the same height). In L1, enter the x-coordinates given. Can you find a line that goes through them? Curve sketching means you got a function and are looking for roots, turning and inflection points. Tool to find the equation of a function from its points, its coordinates x, y=f(x) according to some interpolation methods and equation finder algorithms. the 5 parameters provided in the question, VAF = Value added Factor i.e. Sure, lot of them - an infinite number of them. We need to find a function with a known type (linear, quadratic, etc.) This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. Line through two points Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. New coordinates by rotation of points. UFP = Sum of all the complexities i.e. Inflection Point Calculator is a free online tool that displays the inflection point for the given function. Press [STAT]. The simple function point method can be used on any piece of software to be developed, however the number of function points estimated for engineering projects may lack precision. 2009/06/10 09:07 Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding these computations if fantastic! This means: You calculate the difference of the y-coordinates and divide it by the difference of the x-coordinates. How to calculate the equation of a linear function from two given points? In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. Trending Posts. a bug ? Also, explore hundreds of other calculators addressing math, finance, health, fitness, and more. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Yes. Linear equation with intercepts. dCode tries to propose the most simplified solutions possible, based on affine function or polynomial of low degree (degree 2 or 3). An online curve-fitting solution making it easy to quickly perform a curve fit using various fit methods, make predictions, export results to Excel,PDF,Word and PowerPoint, perform a custom fit through a user defined equation and share results online. The shortest distance between two points on the surface of a sphere is an arc, not a line. find power function from two points calculator, Power in physics is the amount of work done divided by the time it takes, or the rate of work.Here’s what that looks like in equation form: Assume you have two speedboats of equal mass, and you want to know which one will … For an introduction to what are Function Points please read my earlier article here. Second calculator finds the line equation in parametric form, that is, . 0.65 + (0.01 * TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. 3. Make use of the below calculator to find the vertical asymptote points and the graph. Enter the point and slope that you want to find the equation for into the editor. How to Use the Calculator. Roots at and Further point on the Graph: It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. How to find an equation from a set of points. Clear any existing entries in columns L1 or L2. To graph a parabola, visit the parabola grapher (choose the "Implicit" option). To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. First calculator finds the line equation in slope-intercept form, that is, . Cartesian to Polar coordinates. Indeed, if the parameter $$k$$ is positive, then we have exponential growth, but if the parameter $$k$$ is negative, then we have exponential decay. New coordinates by rotation of axes. Definition: A stationary point (or critical point) is a point on a curve (function) where the gradient is zero (the derivative is équal to 0). Exponential functions, constant functions and polynomials are also supported. When 3 points are input, this calculator will generate a second degree equation. Write The Equation For Photosynthesis. But "Why re-invent the wheel?" Help. absolute extreme points f ( x) = ln ( x − 5) $absolute\:extreme\:points\:f\left (x\right)=\frac {1} {x^2}$. Mathepower finds the function. What we do here is the opposite: Your got some roots, inflection points, turning points etc. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. Learn how to use the Stat plot feature of the TI-84+ Calculator to find the equation of those points. absolute extreme points f ( x) = 1 x2. The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. The method used to calculate function point is knows as FPA (Function Point Analysis). My question is - Can I find the function using points? As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to … It accepts inputs of two known points, or one known point and the slope. How about an ellipse or hyperbola? In practice, this means substituting the points for y and x in the equation y = ab x. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!A suggestion ? The vertical graph occurs where the rational function for value x, for which the denominator should be 0 and numerator should not be equal to zero. The method used to calculate function point is knows as FPA (Function Point Analysis). Are there any convenient websites out there with this functionality? Method 2: use a interpolation function, more complicated, this method requires the use of mathematical algorithms that can find polynomials passing through any points. A Function Calculator is a free online tool that displays the graph of the given function. What about a circle that touches the two points? It was easy example, but my graphic is not a liner function. A stationary point is therefore either a local maximum, a local minimum or an inflection point.. an idea ? Area of a triangle with three points. Primarily, you have to find … and are looking for a function having those. Let's take a simple case: two points. Example: a function has for points (couples $(x,y)$) the coordinates: $(1,2) (2,4), (3,6), (4,8)$, the ordinates increase by 2 while the abscissas increase by 1, the solution is trivial: $f (x) = 2x$. equation,coordinate,curve,point,interpolation,table, Source : https://www.dcode.fr/function-equation-finder. Thank you! Yes. Function Point Calculator: Main Description Details Uses: Calculator. These online calculators find the equation of a line from 2 points. Also I would define it in single line as "A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user". NB: for a given set of points there is an infinity of solutions because there are infinite functions passing through certain points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Of course? There can be various methods to calculate function points; you can define your custom too based on your specific requirements. First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . What about a circle that touches the two points? y=F(x), those values should be as close as possible to the table values at the same points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Area of a triangle with three points. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Exponential Function Calculator from Two Points, exponential function calculator given points. You will be asked to provide simple estimates of the software you plan to develop. Intersection of two lines. absolute extreme points y = x x2 − 6x + 8. find power function from two points calculator, The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. (Try this with a string on a globe.) By using this website, you agree to our Cookie Policy. Casio is a great company, providing durable products. This free slope calculator solves for multiple parameters involving slope and the equation of a line. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. Cartesian to Polar coordinates. The blue line is my function. What about a cubic? New coordinates by rotation of axes. Wolfram|Alpha is a great tool for finding the domain and range of a function. Polar to Cartesian coordinates Of course you can. Thus function points can be calculated as: Polar to Cartesian coordinates Visualize the exponential function that passes through two points, which may be dragged within the x-y plane. Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared → 14 2 = 196 c is always equal to 1 Intersection of two lines. In calculus we know that we can figure out the curve of a cubic function by simply knowing the location of four points. Linear equation given two points. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods.. If the period is more than 2π then B is a fraction; use the formula period = 2π/B to find … Press [STAT] again. Distance between the asymptote and graph becomes zero as the graph gets close to the line. 1 - Enter the x and y coordinates of three points A, B and C and press "enter". I … a feedback ? Linear equation given two points. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Two equations are displayed: an exact one (top one) where the coefficients are in fractional forms an the second with approximated coefficients whose number of … Technically, in order to find the parameters you need to solve the following system of equations: Solving this system for $$A_0$$ and $$k$$ will lead to a unique solution, provided that $$t_1 = \not t_2$$. Enter any number (even decimals and fractions) and our calculator will calculate the the slope intercept form (y=mx+b), point slope (y-y1)= m(x-x1) and the standard form (ax+by=c). 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Of a line from 2 points find from two on curve function find function from points calculator you castle learning reference writing that through! Try this with a string on a number line to enhance your mathematical.... Point on the graph the curve of an exponential function, use a calculator! Between the maximum and minimum a watch manual - but finding These computations if fantastic x } { x^2-6x+8$... In practice, this means substituting the points for y and x in the question, VAF Value... Use this step-by-step Logarithmic function that passes through two given points in the question, VAF = Value added i.e... Be as close as possible to the grid points ( with integer x- and y-values ) x- and coordinates! Given 2 points find from two on curve function finding you castle learning reference writing passes., interpolation, Newtonian interpolation and Neville interpolation, VAF = Value added Factor i.e slope by. 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find function from points calculator
Rate this post | 2021-04-12 01:07:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6647066473960876, "perplexity": 607.7967962292461}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00568.warc.gz"} |
http://usaco.org/current/data/sol_mowing_bronze_jan16.html | (Analysis by Nick Wu)
In this problem, Farmer John is mowing grass on a grid. He wants to figure out the largest period of time that exists such that Farmer John never mows the same square of grass twice within that amount of time.
Farmer John will mow at most one thousand squares of grass, so we can just keep track of the last point in time when Farmer John mowed the grass in a given square. We can store these times in a large two-dimensional array. As an implementation detail, because arrays do not support negative coordinates, we can pretend that Farmer John starts at the point (1000, 1000) and then simulate the mowing directly.
Here is my code demonstrating this.
import java.io.*;
import java.util.*;
public class mowing {
public static void main(String[] args) throws IOException {
// initialize file I/O
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("mowing.out")));
/*
* Create a two-dimensional array where lastTime[i][j] stores the last time
* that Farmer John visited point (i, j). If Farmer John has never visited that
* point, then lastTime[i][j] == -1.
*/
int[][] lastTime = new int[2001][2001];
for(int i = 0; i < lastTime.length; i++) {
for(int j = 0; j < lastTime[i].length; j++) {
lastTime[i][j] = -1;
}
}
// (currX, currY) is the point that Farmer John is currently at.
int currX = 1000;
int currY = 1000;
lastTime[currX][currY] = 0;
int currentTime = 0;
// the longest period of time cannot exceed 1000, so if it is still 1001, he never visits the same square twice.
int answer = 100 * 10 + 1;
for(int i = 0; i < n; i++) {
// read in the direction that Farmer John goes in
String direction = st.nextToken();
int dirX = 0;
int dirY = 0;
if(direction.equals("N")) {
dirX = -1;
}
else if(direction.equals("S")) {
dirX = 1;
}
else if(direction.equals("W")) {
dirY = -1;
}
else {
dirY = 1;
}
// read in the number of steps he travels
int numSteps = Integer.parseInt(st.nextToken());
// travel that many steps
for(int j = 1; j <= numSteps; j++) {
currX += dirX;
currY += dirY;
currentTime++;
// check if Farmer John has visited that square before, and the amount of time that has elapsed since then if valid
if(lastTime[currX][currY] >= 0 && currentTime - lastTime[currX][currY] < answer) {
}
lastTime[currX][currY] = currentTime;
}
}
// check if he has never visited the same square twice | 2018-09-26 05:04:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2999713122844696, "perplexity": 4204.888982499741}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267163326.85/warc/CC-MAIN-20180926041849-20180926062249-00339.warc.gz"} |
https://www.illustrativemathematics.org/content-standards/3/NF/A/2/tasks/173 | Update all PDFs
# Locating Fractions Greater than One on the Number Line
Alignments to Content Standards: 3.NF.A.2
1. Mark and label points on the number line for $\frac12$, $\frac22$, $\frac32$, $\frac42$, $\frac52$, and $\frac62$.
2. Mark and label a point on the number line for $\frac{11}{3}$. Be as exact as possible.
## IM Commentary
This first part of this task requires students to represent a fraction $\frac{a}{b}$ on a number line diagram by marking off $a$ intervals of length $\frac1b$ (in this case $\frac12$) starting at 0.
Students may solve the second part of the task either by marking of 11 intervals of length $\frac13$ or by knowing that $\frac{11}{3}$ is $\frac23$ of the way between 3 and 4. Although a few students may be able to do it using the later of these approaches, most students will identify all of the points related to $\frac13$, $\frac23$, $\frac33$, etc., even if they do not label those points.
The following lists related tasks in order of sophistication:
• Locating Fractions Less than One on the Number Line
• Locating Fractions Greater than One on the Number Line
• Closest to $\frac12$
• Find 1
• Find $\frac23$
• Which is Closer to 1?
## Solution
1. Here is one way to draw and label the number line for this task:
2. Here is one way to draw and label the number line for this task: | 2018-04-26 03:59:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8755536675453186, "perplexity": 459.94778045878496}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948064.80/warc/CC-MAIN-20180426031603-20180426051603-00157.warc.gz"} |
http://swift.dur.ac.uk/docs/VELOCIraptorInterface/stfwithswift.html | Configuring SWIFT with VELOCIraptor¶
In the following three paragraphs we will explain how to setup VELOCIraptor, how to compile it and how to compile SWIFT with VELOCIraptor.
Setting up VELOCIraptor¶
Before we can run SWIFT with VELOCIraptor we first need to download VELOCIraptor. This can be done by cloning the repository on GitHub:
git clone https://github.com/ICRAR/VELOCIraptor-STF.git
The SWIFT interface is in the master branch of VELOCIraptor so nothing is more is needed besides fetching the latest version of the NBodyLib that the code relies upon:
cd VELOCIraptor-STF
git submodule update --init --recursive
To get VELOCIraptor working with SWIFT simply use:
mkdir build
cd build
cmake ../ -DVR_USE_HYDRO=ON -DVR_USE_SWIFT_INTERFACE=ON -DCMAKE_CXX_FLAGS="-fPIC" -DCMAKE_BUILD_TYPE=Release
The first parameter activates the processing of gas, stars and black holes. It can be omitted for simulations evolving only dark matter.
If you wish to run swift without MPI, you will need to add -DVR_MPI=OFF.
Compiling VELOCIraptor¶
make -j
After the compilation of your code, you will find a static library libvelociraptor.a, that is required to run SWIFT with VELOCIraptor. Note that VELOCIraptor needs a serial version of the HDF5 library, not a parallel build.
Compiling SWIFT¶
The next part is compiling SWIFT with VELOCIraptor and assumes you already downloaded SWIFT from the GitLab repository. This can be done by running
./autogen.sh
./configure --with-velociraptor=/path/to/VELOCIraptor-STF/src
make
In which ./autogen.sh only needs to be run once after the code is cloned from the GitLab, and /path/to/ is the path to the VELOCIraptor-STF directory on your machine. In general ./configure can be run with other options as desired. After this we can run SWIFT with VELOCIraptor, but for this we first need to add several lines to the yaml file of our simulation
StructureFinding:
config_file_name: vrconfig_3dfof_subhalos_SO_hydro.cfg
basename: haloes
scale_factor_first: 0.02
delta_time: 1.02
In which we specify the .cfg file that is used by VELOCIraptor and the other parameters which SWIFT needs to use. In the case of the Small Cosmological Volume DMO example we can run a simulation with halo finder as:
cd examples/SmallCosmoVolume_DM
../swift --cosmology --hydro --self-gravity --velociraptor --threads=8 small_cosmo_volume_dm.yml
Which activates the VELOCIraptor interface. | 2020-09-24 07:18:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2455073893070221, "perplexity": 6318.777304819226}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00703.warc.gz"} |
http://mathhelpforum.com/calculus/83928-ellipsoid-flux-question.html | ## ellipsoid flux question
Compute the flux of the vector field
$xy\sin(z)\vec{i}+\cos(xz)\vec{j}+y\cos(z)\vec{k}$
across the surface of the ellipsoid $\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1$
Attempt:
I was thinking of making the following substitutions:
$x=\frac{\sin(\phi)\cos(\theta)}{2}$
$y=\frac{\sin(\phi)\sin(\theta)}{3}$
$z=\frac{cos(\phi)}{4}$
now calculating $(r_\phi \times r_\theta)$ gives:
$\frac{\sin^2(\phi)\cos(\theta)}{12}\vec{i}+\frac{\ sin^2(\phi)\sin(\theta)}{8}\vec{j}+\frac{\sin(\phi )\cos(\phi)}{6}\vec{k}$
now putting everything into $\iint_S F\cdot(r_\phi \times r_\theta) \ dS$
gives me:
$\int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\ pi} \left(\frac{\sin(\phi)\cos(\theta)}{2}\right)\left (\frac{\sin(\phi)\sin(\theta)}{3}\right)\sin\left( \frac{cos(\phi)}{4}\right)\cdot\left(\frac{\sin^2( \phi)\cos(\theta)}{12}\right)$ ${\color{blue}+\cos\left(\left(\frac{\sin(\phi)\cos (\theta)}{2}\right)\left(\frac{cos(\phi)}{4}\right )\right)\cdot\left(\frac{\sin^2(\phi)\sin(\theta)} {8}\right)}$ $+\left(\frac{\sin(\phi)\sin(\theta)}{3}\right)\cos \left(\frac{cos(\phi)}{4}\right)\cdot\left(\frac{\ sin(\phi)\cos(\phi)}{6}\right) \ d\phi \ d\theta$
now everything except for the part in blue is integrable, where the first and third part give 0, is there a simpler way of approaching this problem? | 2016-08-25 13:44:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8266062140464783, "perplexity": 532.486414163249}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982293468.35/warc/CC-MAIN-20160823195813-00126-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://engineering.stackexchange.com/questions/45767/how-does-shear-viscous-stress-change-in-liquids-as-applied-shear-force-is-change | # How does shear/viscous stress change in liquids as applied shear force is changed?
As per Newton's law of viscosity, the dynamic viscosity of a Newtonian fluid is constant (doesn't change with applied shear stress).Consider the Couette flow used to explain viscosity .
Whenever we apply a shear force then the fluid layers slide over each other. Here consider applying shear force to the top wall as shown in figure. The viscosity of a liquid is mainly attributed to the cohesive intermolecular forces between the fluid layers. So whenever we apply a shear force the intermolecular forces are overcome and liquid flows. In some references they call it as overcoming the internal friction when a shear force is applied. The intermolecular forces between fluid layers is more or less a fixed value at a given temperature.
So how does applying a higher shear force produce a higher shear/viscous stress between fluid layers as indicated by the shear stress vs shear strain rate curve.
In the case of solids ,in elastic region whenever we apply a higher force the bonds are stretched and to stretch the bonds longer you need more force and hence higher stress developed in material is justified. Whereas in plastic region the work hardening effect (I don't have much idea on this and I am still reading on what happens at atomic levels) causes a higher stress in material as higher force is applied. In liquids (specifically Newtonian liquids) I am not really sure why a higher shear stress is produced as we increase applied shear force. In the case of liquids, I think it's correct to say that when we apply a higher shear force more molecules flow adjacent to each other per unit time. Can we co-relate this (If yes how?) to higher shear stress produced? Or else how do we explain the higher shear stress at higher shear force applied?
Edit: Looking for an explanation at molecular level.
• And you should consider thixotropic and rheopectic fluids. Pumping milk, beating egg white and shaking ketchup are also interesting. Aug 26 at 18:07
## 2 Answers
The way I understand it, is that (at least in laminar flow) viscosity is proportional to the "stickiness" between the layers of the fluid. I.e. higher viscosity, means that if a layer starts to move, then it will sweep along more forcefully the adjacent layers.
Additionally in the example of the Couette flow, the velocity of the top layer determines the $$\frac{du}{dy}$$
So in the example of the Couette flow, where the top layer is attempting to move with a velocity u, that means that conversely the same volume will try to increase more its velocity in the unit of time - in other words more acceleration. However, increasing velocity in the unit of time, so you end up with the second law of Newton $$\sum F = m\cdot a$$
So when you increase the velocity gradient, greater forces are required between the plies. However, since the magnitude of the forces that are applied on the interfaces of the layers is greater, the shear stresses (which is force over area) are also greater.
NOTE This is a very handwavy explanation, which I admit is more my current understanding rather than an answer. I look forward to comments/corrections.
UPDATE Regarding the molecular level, I would look into the relationship of cohesion and viscosity. Basically, higher cohesive forces mean higher viscosity. So the the layers can withstand higher forces. If you apply too much force, then basically the continuity is lost, because the cohesive forces are not enough to maintain contact between the molecules of the fluid.
• The explanation based on Newton's law is correct. I was looking more into an explanation at molecular level (I will edit the question). Precisely (at molecular level), if the intermolecular forces are constant at a given temperature how is that a higher shear force will induce a higher shear stress? You will be overcoming the same intermolecular force irrespective of the applied shear. The only difference would be the number of molecules that we slide past in a given time as shear force changes. Aug 26 at 15:10
• @AbhishekPG Aren't you using the definition of force over area for shear stress? The higher forces will by definition be greater. Regarding the molecular level, I would look into the relationship of cohesion and viscosity. Basically, higher cohesive forces mean higher viscosity. So the the layers can withstand higher forces. If you apply too much force, then basically the continuity is lost, because the cohesive forces are not enough to maintain contact between the molecules of the fluid Aug 26 at 15:24
• Force over area does work per equation. I was looking more into what really happens inside (An explanation like the case of solids as I mentioned,idk if its correct). If we apply the same shear force to two different liquids then due to different cohesive forces we end up with different resistance to flow (viscosity) and hence different shear stresses. But when we apply different shear forces to same liquid the cohesive force remains same in both cases, and hence viscosity. So what is the effect at molecular level which causes different shear stress as shear force is changed (for same liquid). Aug 26 at 15:37
• Regarding "But when we apply different shear forces to same liquid the cohesive force remains same in both cases, and hence viscosity.", I am not sure where you are getting at (and IMHO there is a problem with causality in your statement) . The application of force does not change the viscosity, In the same manner that in a solid the application of force does not change its elasticity/shear modulus. The velocity gradient will change for different forces, and the solid material's deformation will change for different levels of force. Aug 26 at 15:48
• In solids in the elastic region, you have to increase the applied force to stretch the bonds more from a particular state(which increases stress developed). In liquids there is no stretching happening. The liquid layers flow one over another as long as force remains applied. In liquids ,since there is no thing like stretching the bonds , what is resisting the increased/higher applied shear force in order to produce a higher stress? Aug 26 at 16:07
You should consider sheer density. When the stress increases, the velocity gradient increase.
If the distance between velocity contour lines of say 0,1 m/s difference in speed used to be 1mm originally it becomes roughly 0.09 mm at a stress level of 1.1 times the original stress level.
This doesn't concern the behavior of the liquid at the molecular level. | 2021-10-25 23:56:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7437967658042908, "perplexity": 620.2049725204288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587770.37/warc/CC-MAIN-20211025220214-20211026010214-00628.warc.gz"} |
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The Puzzle of the Metallic Line StarsIn the puzzle of the metallic line (Am) stars, there still seem to bemissing pieces. While the normal'' A stars have elemental abundancesclose to solar, the classical Am stars show stronger absorption linesfor most heavy elements in their spectra. Elements with ionizationpotentials that nearly agree with those of hydrogen or helium havereduced abundances. The Ca II and Sc II lines are especially weak. TheAm stars have no ultraviolet emission lines. They are binaries that,with very few exceptions, have rotational velocities vsini lower than100 km s-1. Of the main-sequence A stars, 20% to 30% are Amstars. Here we rediscuss previous suggestions that tried to explain thepeculiar line strengths in the Am star spectra. In particular, wecompare the well-studied properties of Hyades A and Am stars in order toidentify reasons that can or cannot explain the differences. 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The overall detection probabilitiesfor our illustrative sample range from 2% to 27%, which imply thatwithin a distance of approximately 5 kpc from the Sun around 48detections of compact companions to runaway stars can be expected. Forcomparison, around 15% of High Mass X-ray Binaries would exhibit wobblesdetectable with Gaia. Evolution of X-ray emission from young massive star clustersThe evolution of X-ray emission from young massive star clusters ismodelled, taking into account the emission from the stars as well asfrom the cluster wind. It is shown that the level and character of thesoft (0.2-10 keV) X-ray emission change drastically with cluster age andare tightly linked with stellar evolution. Using the modern X-rayobservations of massive stars, we show that the correlation betweenbolometric and X-ray luminosity known for single O stars also holds forO+O and (Wolf-Rayet) WR+O binaries. 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Interstellar 12C/13C ratios through CH^+λλ 3957,4232 absorption in local clouds: incomplete mixing in the ISMThe 12C/13C isotope ratio is a tracer of stellaryields and the efficiency of mixing in the ISM.12CH+/13CH+ is not affectedby interstellar chemistry, and is the most secure way of measuring12C/13C in the diffuse ISM.R=12C/13C is 90 in the solar system. Previousmeasurements of 12CH+λλ3957.7,4232.3and 13CH+λλ3958.2,4232.0 absorptiontoward nearby stars indicate some variations in12C/13C, with values ranging from 40 to 90suggesting inefficient mixing. Except for the cloud toward ζOph,these R values are strongly affected by noise. With UVES on the VLT wehave improved on the previous interstellar 12C/13Cmeasurements. The weighted 12C/13C ratio in thelocal ISM is 78.27 ± 1.83, while the weighted dispersion of ourmeasurements is 12.7, giving a 6.9σ scatter. Thus we report on a6.9σ detection of 16.2% root-mean-square variations in the carbonisotopic ratio on scales of ~100 pc: R= 74.7 ± 2.3 in theζOph cloud, while R = 88.6 ± 3.0 toward HD 152235 in theLupus clouds, R = 62.2 ± 5.3 towards HD 110432 in the Coalsack,and R = 98.9 ± 10.1 toward HD 170740. The observed variations in13C/12C are the first significant detection ofchemical heterogeneity in the local ISM. Bright OB stars in the Galaxy. II. Wind variability in O supergiants as traced by HαWe investigate the line-profile variability (lpv) of Hα for alarge sample of O-type supergiants (15 objects between O4 and O9.7), inan objective, statistically rigorous manner. We employed the TemporalVariance Spectrum (TVS) analysis, developed for the case of photosphericabsorption lines and modified by us to take into account the effects ofwind emission. By means of a comparative analysis we place constraintson the properties of this variability - quantified in terms of a meanand a newly defined fractional amplitude of deviations - as a functionof stellar and wind parameters. The results of our analysis show thatall the stars in the sample show evidence of significant lpv inHα, mostly dominated by processes in the wind. The variationsoccur between zero and 0.3 v_&infy; (i.e., below 1.5 R_star ), in goodagreement with results from similar studies. A comparison between theobservations and corresponding line-profile simulations indicates thatfor stars with intermediate wind densities the properties of theHα variability can be explained by simple models consisting ofcoherent or broken shells (blobs) uniformly distributed over the windvolume, with an intrinsic scatter in the maximum density contrast ofabout a factor of two. For stars at lower and higher wind densities, onthe other hand, we found certain inconsistencies between theobservations and our predictions, most importantly concerning the meanamplitude and the symmetry properties of the TVS. This disagreementmight be explained by the presence of coherent large-scale structures,partly confined in a volume close to the star. Interpreted in terms of avariable mass-loss rate, the observed variations of Hα indicatechanges of ±4% with respect to the mean value of dot M for starswith stronger winds and of ± 16% for stars with weaker winds. Theeffect of these variations on the corresponding wind momenta is ratherinsignificant (less than 0.16 dex), increasing only the local scatterwithout affecting the Wind Momentum Luminosity Relationship. Deuterated molecular hydrogen in the Galactic ISM. New observations along seven translucent sightlinesWe present column density measurements of the HD molecule in theinterstellar gas toward 17 Galactic stars. 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Large-scale wind structures in OB supergiants: a search for rotationally modulated Hα variability We present the results of a long-term monitoring campaign of theHα line in a sample of bright OB supergiants (O7.5-B9) which aimsat detecting rotationally modulated changes potentially related to theexistence of large-scale wind structures. A total of 22 objects weremonitored during 36 nights spread over six months in 2001-2002.Coordinated broad-band photometric observations were also obtained forsome targets. Conspicuous evidence for variability in Hα is foundfor the stars displaying a feature contaminated by wind emission. Mostchanges take place on a daily time-scale, although hourly variations arealso occasionally detected. Convincing evidence for a cyclical patternof variability in Hα has been found in two stars: HD 14134 and HD42087. Periodic signals are also detected in other stars, butindependent confirmation is required. Rotational modulation is suggestedfrom the similarity between the observed recurrence time-scales (in therange 13-25 d) and estimated periods of stellar rotation. We callattention to the atypical case of HD 14134, which exhibits a clear12.8-d periodicity, both in the photometric and in the spectroscopicdata sets. This places this object among a handful of early-type starswhere one may observe a clear link between extended wind structures andphotospheric disturbances. Further modelling may test the hypothesisthat azimuthally-extended wind streams are responsible for the patternsof spectral variability in our target stars. A Galactic O Star CatalogWe have produced a catalog of 378 Galactic O stars with accuratespectral classifications that is complete for V<8 but includes manyfainter stars. The catalog provides cross-identifications with othersources; coordinates (obtained in most cases from Tycho-2 data);astrometric distances for 24 of the nearest stars; optical (Tycho-2,Johnson, and Strömgren) and NIR photometry; group membership,runaway character, and multiplicity information; and a Web-based versionwith links to on-line services. The Homogeneity of Interstellar Oxygen in the Galactic DiskWe present an analysis of high-resolution Hubble Space Telescope (HST)Space Telescope Imaging Spectrograph (STIS) observations of O Iλ1356 and H I Lyα absorption in 36 sight lines that probe avariety of Galactic disk environments and include paths that range overnearly 4 orders of magnitude in f(H2), over 2 orders ofmagnitude in , and that extend up to 6.5 kpc inlength. Since the majority of these sight lines have also been observedby the Far Ultraviolet Spectroscopic Explorer (FUSE), we have undertakenthe study of gas-phase O/H abundance ratio homogeneity using the currentsample and previously published Goddard High Resolution Spectrograph(GHRS) results. Two distinct trends are identified in the 56 sight linesample: an apparent decrease in gas-phase oxygen abundance withincreasing mean sight-line density () and a gapbetween the mean O/H ratio for sight lines shorter and longer than about800 pc. The first effect is a smooth transition between two depletionlevels associated with large mean density intervals; it is centered near=1.5cm-3 and is similar to trendsevident in gas-phase abundances of other elements. Paths less dense thanthe central value exhibit a mean O/H ratio of log10(O/H)=-3.41+/-0.01 (or 390+/-10ppm), which is consistent with averages determined for several longlow-density paths observed by STIS (André et al. 2003) and shortlow-density paths observed by FUSE (Moos et al. 2002). Sight lines ofhigher mean density exhibit an average O/H value of log10(O/H)=-3.55+/-0.02 (284+/-12ppm). The data points for low- paths are scatteredmore widely than those for denser sight lines, because O/H ratios forsuch paths shorter than 800 pc are generally about 0.10 dex lower thanthe values for longer ones. Scenarios that would be consistent withthese results include a recent infall of metal-poor gas onto the localGalactic disk and an interstellar environment toward Orion that isconducive to reducing the apparent gas-phase oxygen abundance.Based on observations with the NASA/ESA Hubble Space Telescope (HST) andthe NASA-CNES-CSA Far Ultraviolet Spectroscopic Explorer (FUSE). HSTspectra were obtained at the Space Telescope Science Institute, which isoperated by the Association of Universities for Research in Astronomy,Inc., under NASA contract NAS5-26555 FUSE is operated for NASA by theJohns Hopkins University under NASA contract NAS5-32985. Interstellar Carbon in Translucent Sight LinesWe report interstellar C II column densities or upper limits determinedfrom weak absorption of the 2325.4029 Å intersystem transitionobserved in six translucent sight lines (AV>~1) with theSpace Telescope Imaging Spectrograph (STIS). The sight lines sample awide range of interstellar characteristics, including total to selectiveextinction, RV=2.6-5.1 average hydrogen density along thesight line, =3-14 cm-3 and fraction of H inmolecular form, 0-~40%. Four of the sight lines, those toward HD 37021,HD 37061, HD 147888, and HD 207198, have interstellar gas-phaseabundances that are consistent with the diffuse sight line ratio of161+/-17 carbon atoms in the gas per million hydrogen nuclei. We notethat while it has a gas-phase carbon abundance that is consistent withthe other sight lines, a large fraction of the C II toward HD 37061 isin an excited state. The sight line toward HD 152590 has a measuredinterstellar gas-phase carbon abundance that is well above the diffusesight line average; the column density of C in this sight line may beoverestimated because of noise structure in the data. Toward HD 27778 wefind a 3 σ abundance upper limit of <=108 C atoms in the gasper million H, a substantially enhanced depletion of C as compared tothe diffuse sight line value. The interstellar characteristics toward HD27778 are otherwise not extreme among the sample, except for anunusually large abundance of CO molecules in the gas.Based on observations with the NASA/ESA Hubble Space Telescope, obtainedat the Space Telescope Science Institute, which is operated by theAssociation of Universities for Research in Astronomy, Inc., under NASAcontract NAS 5-26555. Total to Selective Extinction Ratios and Visual Extinctions from Ultraviolet DataWe present determinations of the total to selective extinction ratio R_Vand visual extinction A_V values for Milky Way stars using ultravioletcolor excesses. We extend the analysis of Gnacinski and Sikorski (1999)by using non-equal weights derived from observational errors. We presenta detailed discussion of various statistical errors. In addition, weestimate the level of systematic errors by considering differentnormalization of the extinction curve adopted by Wegner (2002). Ourcatalog of 782 R_V and A_V values and their errors is available in theelectronic form on the World Wide Web. On the Hipparcos parallaxes of O starsWe compare the absolute visual magnitude of the majority of bright Ostars in the sky as predicted from their spectral type with the absolutemagnitude calculated from their apparent magnitude and the Hipparcosparallax. We find that many stars appear to be much fainter thanexpected, up to five magnitudes. We find no evidence for a correlationbetween magnitude differences and the stellar rotational velocity assuggested for OB stars by Lamers et al. (1997, A&A, 325, L25), whosesmall sample of stars is partly included in ours. Instead, by means of asimulation we show how these differences arise naturally from the largedistances at which O stars are located, and the level of precision ofthe parallax measurements achieved by Hipparcos. Straightforwardlyderiving a distance from the Hipparcos parallax yields reliable resultsfor one or two O stars only. We discuss several types of bias reportedin the literature in connection with parallax samples (Lutz-Kelker,Malmquist) and investigate how they affect the O star sample. Inaddition, we test three absolute magnitude calibrations from theliterature (Schmidt-Kaler et al. 1982, Landolt-Börnstein; Howarth& Prinja 1989, ApJS, 69, 527; Vacca et al. 1996, ApJ, 460, 914) andfind that they are consistent with the Hipparcos measurements. AlthoughO stars conform nicely to the simulation, we notice that some B stars inthe sample of \citeauthor{La97} have a magnitude difference larger thanexpected. Stellar and wind parameters of Galactic O-stars. The influence of line-blocking/blanketingWe have re-analyzed the Galactic O-star sample from \citet{puls96} bymeans of line-blanketed NLTE model atmospheres in order to investigatethe influence of line-blocking/blanketing on the derived parameters. Theanalysis has been carried out by fitting the photospheric and wind linesfrom H and He. In most cases we obtained a good fit, but we have alsofound certain inconsistencies which are probably related to a stillinadequate treatment of the wind structure. These inconsistenciescomprise the line cores of Hγ and Hβ insupergiants (the synthetic profiles are too weak when the mass-loss rateis determined by matching Hα) and the generalizeddilution effect'' (cf. \citealt{vo89}) which is still present in He I4471 of cooler supergiants and giants.Compared to pure H/He plane-parallel models we found a decrease ineffective temperatures which is largest at earliest spectral types andfor supergiants (with a maximum shift of roughly 8000 K). This findingis explained by the fact that line-blanketed models of hot stars havephotospheric He ionization fractions similar to those from unblanketedmodels at higher Teff and higher log g. Consequently, anyline-blanketed analysis based on the He ionization equilibrium resultsin lower Teff-values along with a reduction of either log gor helium abundance (if the reduction of log g is prohibited by theBalmer line wings). Stellar radii and mass-loss rates, on the otherhand, remain more or less unaffected by line-blanketing.We have calculated new'' spectroscopic masses and compared them withprevious results. Although the former mass discrepancy \citep{h92}becomes significantly reduced, a systematic trend for masses below 50Msun seems to remain: The spectroscopically derived valuesare smaller than the evolutionary masses'' by roughly 10Msun. Additionally, a significant fraction of our samplestars stays over-abundant in He, although the actual values were foundto be lower than previously determined.Also the wind-momentum luminosity relation (WLR) changes because oflower luminosities and almost unmodified wind-momentum rates. Comparedto previous results, the separation of the WLR as a function ofluminosity class is still present but now the WLR for giants/dwarfs isconsistent with theoretical predictions.We argue that the derived mass-loss rates of stars withHα in emission are affected by clumping in the lowerwind region. If the predictions from different and independenttheoretical simulations (\citealt {Vink00, Paul03, puls03a}) that theWLR should be independent of luminosity class were correct, a typicalclumping factor <ρ2>/<ρ>2 ≈5 should be derived by unifying'' the different WLRs.Based upon observations obtained at the INT and the European SouthernObservatory, La Silla, Chile. The INT is operated on the island of LaPalma by the ING in the Spanish Observatorio de El Roque de losMuchachos of the Instituto de Astrofísica de Canarias.Appendix A in only available in electronic form athttp://www.edpsciences.org On the relation between diffuse bands and column densities of H2, CH and CO moleculesMutual relations between column densities of H2, CH and COmolecules as well as between the latter and strengths of the major 5780and 5797 diffuse bands are presented and discussed. The CH radical seemsto be a good H2 tracer, possibly better than CO. It is alsodemonstrated that the molecular fraction of the H2 moleculeis correlated with an intensity ratio of 5797 and 5780 DIBs, suggestingthe possible formation of narrow DIB carriers in denser clouds,dominated by molecular hydrogen and reasonably shielded from ionizing UVradiation by small dust grains.Tables 1 and 2 are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/414/949 Bright OB stars in the Galaxy. I. Mass-loss and wind-momentum rates of O-type stars: A pure H\alpha analysis accounting for line-blanketingWe study mass-loss and wind momentum rates of 29 Galactic O-type starswith luminosity classes I, III and V by means of a pure H\alpha profileanalysis and investigate to what extent the results compare to thoseoriginating from a state-of-the-art, complete spectral analysis. Ourinvestigation relies on the approximate method developed by\citet{Puls96} which we have modified to account for the effects ofline-blanketing. Effective temperatures and gravities needed to obtainquantitative results from such a simplified approach have been derivedby means of calibrations based on most recent spectroscopic NLTEanalyses and models of Galactic stars by \citet{Repo03} and\citet{Martins02}. Comparing (i) the derived wind-densities to thosedetermined by \citet{Repo03} for eleven stars in common and (ii) theWind-momentum Luminosity Relationship (WLR) for our sample stars tothose derived by other investigations, we conclude that our approximateapproach is actually able to provide consistent results. Additionally,we studied the consequences of fine tuning'' some of the direct andindirect parameters entering the WLR, especially by accounting fordifferent possible values of stellar reddening and distances. Combiningour data set with the corresponding data provided by \citet{Herrero02}and \citet{Repo03} we finally study the WLR for the largest sample ofGalactic O-type stars gathered so far, including an elaborate errortreatment. The established disagreement between the theoreticalpredictions and the observed'' WLRs being a function of luminosityclass is suggested to be a result of wind clumping. Different strategiesto check this hypothesis are discussed, particularly by comparing theH\alpha mass-loss rates with the ones derived from radio observations. Oscillator strengths of the resonance 4s-4p lines in ZnIIWe have undertaken extensive configuration interaction calculations ofthe 4s 2S1/2- 4p2Po1/2,3/2 lines of ZnII. Corepolarization, which is confirmed as a significant effect in thedetermination of the oscillator strengths, is introduced throughexplicit configuration interaction rather than by means of modelpotentials. It is shown that the ab initio calculation of this effectgives results that are close to those obtained with model potentialswhich incorporate the polarizability of the ZnIII core. The oscillatorstrengths are reduced to around 75 per cent of the single configurationvalue. We also find that the inclusion of electron correlation in thecore causes the oscillator strength to rise again, although onlymarginally. Our final oscillator strengths, which incorporate all theseeffects, are 0.268 and 0.547 for the 4s 2S1/2- 4p2Po1/2 and 4s2S1/2- 4p 2Po3/2lines, respectively. These values are in good agreement with recentmany-body perturbation theory calculations of Johnson and withunpublished multiconfigurational Hartree-Fock or MulticonfigurationalDirac-Hartree-Fock (MCDHF) calculations of Froese Fischer, but lie about5-10 per cent higher than the most recent experimental values obtainedby Bergeson and Lawler. Toward an adequate method to isolate spectroscopic families of diffuse interstellar bandsWe divide some of the observed diffuse interstellar bands (DIBs) intofamilies that appear to have the spectral structure of single species.Three different methods are applied to separate such families, exploringthe best approach for future investigations of this type. Starting witha statistical treatment of the data, we found that statistical methodsby themselves give insufficient results. Two other methods of dataanalysis (averaging equivalent widths' and investigating the figureswith arranged spectrograms') were found to be more useful as tools forfinding the spectroscopic families of DIBs. On the basis of thesemethods, we suggest some candidates as relatives' of 5780- and5797-Å bands. Large-scale Wind Structures in OB Supergiants: a Search for Rotationally Modulated H variabilityWe present preliminary results of a long-term spectroscopic monitoringof a sample of bright OB-supergiants aimed at establishing the incidenceof co-rotating, large-scale wind structures by detecting rotationallymodulated variability in H. Dramatic line-profile variations operatingon a daily (and in some cases on a hourly) timescale are observed. Adetailed period analysis has been so far carried out for 2 stars, andrevealed in both cases the existence of cyclical H variations consistentwith rotational modulation. In the case of HD 14134, the sameperiodicity is found in the contemporaneous light curve. High-Resolution Observations of Interstellar Ca I Absorption-Implications for Depletions and Electron Densities in Diffuse CloudsWe present high-resolution (FWHM~0.3-1.5 km s-1) spectra,obtained with the AAT UHRF, the McDonald Observatory 2.7 m coudéspectrograph, and/or the KPNO coudé feed, of interstellar Ca Iabsorption toward 30 Galactic stars. Comparisons of the column densitiesof Ca I, Ca II, K I, and other species-for individual componentsidentified in the line profiles and also when integrated over entirelines of sight-yield information on relative electron densities anddepletions (dependent on assumptions regarding the ionizationequilibrium). There is no obvious relationship between the ratio N(CaI)/N(Ca II) [equal to ne/(Γ/αr) forphotoionization equilibrium] and the fraction of hydrogen in molecularform f(H2) (often taken to be indicative of the local densitynH). For a smaller sample of sight lines for which thethermal pressure (nHT) and local density can be estimated viaanalysis of the C I fine-structure excitation, the average electrondensity inferred from C, Na, and K (assuming photoionizationequilibrium) seems to be independent of nH andnHT. While the electron density (ne) obtained fromthe ratio N(Ca I)/N(Ca II) is often significantly higher than the valuesderived from other elements, the patterns of relative nederived from different elements show both similarities and differencesfor different lines of sight-suggesting that additional processesbesides photoionization and radiative recombination commonly andsignificantly affect the ionization balance of heavy elements in diffuseinterstellar clouds. Such additional processes may also contribute tothe (apparently) larger than expected fractional ionizations(ne/nH) found for some lines of sight withindependent determinations of nH. In general, inclusion ofgrain-assisted'' recombination does reduce the inferred ne,but it does not reconcile the ne estimated from differentelements; it may, however, suggest some dependence of ne onnH. The depletion of calcium may have a much weakerdependence on density than was suggested by earlier comparisons with CHand CN. Two appendices present similar high-resolution spectra of Fe Ifor a few stars and give a compilation of column density data for Ca I,Ca II, Fe I, and S I. High-Resolution Spectroscopy of FU Orionis StarsHigh-resolution spectroscopy was obtained of the FU orionis stars FU Oriand V1057 Cyg between 1995 and 2002 with the SOFIN spectrograph at theNordic Optical Telescope and with HIRES at Keck I. During these years FUOri remained about 1 mag (in B) below its 1938-39 maximum brightness,but V1057 Cyg (B~10.5 at peak in 1970-1971) faded from about 13.5 to14.9 and then recovered slightly. Their photospheric spectra resemblethat of a rotationally broadened, slightly veiled supergiant of abouttype G0 Ib, with veqsini=70 km s-1 for FU Ori, and55 km s-1 for V1057 Cyg. As V1057 Cyg faded, P Cyg structurein Hα and the IR Ca II lines strengthened and a complexshortward-displaced shell spectrum of low-excitation lines of theneutral metals (including Li I and Rb I) increased in strength,disappeared in 1999, and reappeared in 2001. Several SOFIN runs extendedover a number of successive nights so that a search for rapid and cyclicchanges in the spectra was possible. These spectra show rapidnight-to-night changes in the wind structure of FU Ori at Hα,including clear evidence of sporadic infall. The equivalent width of theP Cyg absorption varied cyclically with a period of 14.8 days, withphase stability maintained over three seasons. This is believed to bethe rotation period of FU Ori. The internal structure of itsphotospheric lines also varies cyclically, but with a period of 3.54days. A similar variation may be present in V1057 Cyg, but the data aremuch noisier and that result uncertain. As V1057 Cyg has faded and thecontinuum level fallen, the emission lines of a preexistinglow-excitation chromosphere have emerged. Therefore we believe that the`line doubling'' in V1057 Cyg is produced by these central emissioncores in the absorption lines, not by orbital motion in an inclinedKeplerian disk. No convincing dependence of veqsini onwavelength or excitation potential was detected in either FU Ori orV1057 Cyg, again contrary to expectation for a self-luminous accretiondisk. It was found also that certain critical lines in the near infraredare not accounted for by synthetic disk spectra. It is concluded that arapidly rotating star near the edge of stability, as proposed by Larson,can better account for these observations. The possibility is alsoconsidered that FUor eruptions are not a property of ordinary T Tauristars but may be confined to a special subspecies of rapidly rotatingpre-main-sequence stars having powerful quasi-permanent winds. Observations of Rotationally Resolved C3 in Translucent Sight LinesThe rotationally resolved spectrum of theA1Πu<--X1Σ+g000-000 transition of C3, centered at 4051.6 Å, hasbeen observed along 10 translucent lines of sight. To interpret thesespectra, a new method for the determination of column densities andanalysis of excitation profiles involving the simulation and fitting ofobserved spectra has been developed. The populations of lower rotationallevels (J<=14) in C3 are best fitted by thermaldistributions that are consistent with the kinetic temperaturesdetermined from the excitation profile of C2. Just as in thecase of C2, higher rotational levels (J>14) ofC3 show increased nonthermal population distributions inclouds that have been determined to have total gas densities below ~500cm-3.
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• - No Links Found - | 2019-09-23 07:36:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6804032921791077, "perplexity": 6828.479893976307}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514576122.89/warc/CC-MAIN-20190923064347-20190923090347-00049.warc.gz"} |
https://socratic.org/questions/91-is-what-percent-of-260 | # 91 is what percent of 260?
Jun 30, 2018
See a solution process below:
#### Explanation:
"Percent" or "%" means "out of 100" or "per 100",
Therefore x% can be written as $\frac{x}{100}$.
We can now write this express and solve for $x$:
$\frac{x}{100} = \frac{91}{260}$
$\textcolor{red}{100} \times \frac{x}{100} = \textcolor{red}{100} \times \frac{91}{260}$
$\cancel{\textcolor{red}{100}} \times \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{100}}}} = \frac{9100}{260}$
$x = 35$
91 is color(red)(35%) of 260 | 2019-08-24 15:40:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7899622917175293, "perplexity": 5978.827880855321}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00118.warc.gz"} |
https://www.physicsforums.com/threads/derivatives-the-slope-of-a-graph.390397/ | # Derivatives & the Slope of a graph
Given
## The Attempt at a Solution
Last edited:
Related Calculus and Beyond Homework Help News on Phys.org
rock.freak667
Homework Helper
If I remember correctly, f(x) is an increasing function for f'(x) > 0 and a decreasing function for f'(x)< 0.
If I remember correctly, f(x) is an increasing function for f'(x) > 0 and a decreasing function for f'(x)< 0.
I know I found that it is increasing on the interval (-1,1), but that's wrong
rock.freak667
Homework Helper
How did you get that?
How did you get that?
ok I re-did it again because I found an error
now I got (-1,3) for increasing
and (-infinity, -1) U (3, infinity) for decreasing
and I got it by doing the number line test (i think that's what it's called)
where I take all the criticals numbers and line them up and put test numbers in between them.
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Char. Limit
Gold Member
Now how about d and e?
Here's a clue...
What is the shape of the graph when the derivative is zero?
Now how about d and e?
Here's a clue...
What is the shape of the graph when the derivative is zero?
it's a straight light, and on the original function that means that there's a point there where it is neither increasing or decreasing or a max or min.
I found an answer for them, but I'm not really confident
f(3)= error.....
f(-1)= 9/16
I'm suppose to get two numbers but I think I did something wrong again....I hate these problems....
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Char. Limit
Gold Member
Hmm...
I'm sorry, but I don't get (-1,3) for increasing...
The derivative is this, as you have written, correct?
$$\frac{5(1-x)(x-1)}{(3x^2-10x+3)^2}$$
Ignore the denominator... where is the numerator zero? Where are each of the three components zero?
Hmm...
I'm sorry, but I don't get (-1,3) for increasing...
The derivative is this, as you have written, correct?
$$\frac{5(1-x)(x-1)}{(3x^2-10x+3)^2}$$
Ignore the denominator... where is the numerator zero? Where are each of the three components zero?
in the numerator: -1,1
denominator: 1/3, 3
dang it I redid it again and this time I got (-1,1) as increasing and I know that's wrong....
Char. Limit
Gold Member
I said ignore the denominator... if the denominator is zero, your graph is really screwing up.
OK, so you know the graph of the derivative is zero at -1 and 1.
Thus, the minimum and the maximum must be at those two points.
I said ignore the denominator... if the denominator is zero, your graph is really screwing up.
OK, so you know the graph of the derivative is zero at -1 and 1.
Thus, the minimum and the maximum must be at those two points.
That's what I did when I first started it and I got (-1,1) as increasing but that's wrong.....
Char. Limit
Gold Member
Well, that's strange...
That is the point where it's increasing.
So I don't know why you got it wrong.
vela
Staff Emeritus
Homework Helper
There's a singularity at x=1/3, which is in (-1,1). That's probably why.
There's a singularity at x=1/3, which is in (-1,1). That's probably why.
there's an error at 3 too, so does that mean I do count the denominator?
Even if I do count the denominator I get an error as the maximum
Char. Limit
Gold Member
Oh...
So it would be -1<x<1, x=/=(1/3) then...
Oh...
So it would be -1<x<1, x=/=(1/3) then...
huh?
oh are you saying because it's undefined there its (-1, 1/3) U (1/3, 1)???
vela
Staff Emeritus
Homework Helper
Yes, exactly. The function isn't increasing at x=1/3 because it isn't defined there, so you have to split the interval as you have done.
Yes, exactly. The function isn't increasing at x=1/3 because it isn't defined there, so you have to split the interval as you have done.
ok, so I redid it over again and do you think these are correct:
b) (-1, 1/3) U (1/3, 1) --increasin interval
c) (-infinity, -1) U (1, 3) U (3, infinity) --decreasng interval
d) f(1)= 1/4 --max
e) f(-1)= 9/16 ---min
Last edited:
Char. Limit
Gold Member
I do believe so.
And that would solve your problem as well as allow you to solve d.
vela
Staff Emeritus
Homework Helper
What happens between x=-2 and x=-1?
What happens between x=-2 and x=-1?
(-infinity, -1) U (1, 3) U (3, infinity) --decreasng interval
Thanks guys the answers were right :) | 2020-06-01 06:10:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8038708567619324, "perplexity": 1629.8697360976096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00421.warc.gz"} |
http://math.stackexchange.com/questions/76581/is-there-a-countable-pseudocharacter-hausdorff-space-x-with-the-property-a | # Is there a countable pseudocharacter Hausdorff space X with the property A?
Let X be a Hausdorff space and define the Property A as follows:
If $\mathscr{U}$ is a collection of open sets of $X$ that witnesses Hausdorff property of $X$ (= $\forall x,y \in X$, there exist two disjoint open sets $U_1$ and $U_2$ $\in \mathscr{U}$ s.t. $x \in U_1$ and $y \in U_2$), then there is a point $x\in X$ such that $|\{U \in \mathscr{U}: x \in U\}|>\omega$.
Is there a countable pseudocharacter Hausdorff space $X$ with the property A? The same question is also asked here.
-
Voting to close because the question is (in the process of being) answered on MO. @John: Why did you repost? – Rasmus Oct 28 '11 at 7:35
@Rasmus: However, as yet it hasn’t actually been answered on MO, and (barring error) I’ve answered it here. – Brian M. Scott Oct 28 '11 at 8:12
@Brian M. Scott: Wouldn't it have been better to close this thread and add answers to the MO thread, where already several comments and an answer have been given? Now we have two simultanious threads on the same topic. – Rasmus Oct 28 '11 at 8:48
@Rasmus: I much prefer to work here; for various reasons I really don’t want to get involved with MO just now. – Brian M. Scott Oct 28 '11 at 9:13
Let $X = \beta\omega$ with the following topology: points of $\omega$ are isolated, and each $p\in\beta\omega\setminus\omega$ has $\{\{p\}\cup U:U\in p\}$ as a local base. (I think of the points of $\beta\omega\setminus\omega$ as free ultrafilters on $\omega$.)
This topology refines the Čech-Stone topology on $\beta\omega$, so it’s certainly Hausdorff, and since it has a base of countable sets, it must have countable pseudocharacter as well. But $\beta\omega\setminus\omega$ is uncountable (in fact of cardinality $2^{2^\omega}$), so any family $\mathscr{U}$ that witnesses the Hausdorffness of $X$ must be uncountable.
Added: That actually doesn’t follow without further argument. However, $|\beta\omega|>2^\omega$, so if $\mathscr{U}$ is a countable family of subsets of $\beta\omega$, there are distinct $p,q\in\beta\omega$ such that $$\{U\in\mathscr{U}:p\in U\}=\{U\in\mathscr{U}:q\in U\},$$ and $\mathscr{U}$ cannot then witness the Hausdorffness of $X$.
Each member of $\mathscr{U}$ has non-empty intersection with $\omega$, so some $n\in\omega$ must belong to uncountably many members of $\mathscr{U}$.
-
The example probably works, but one step in the reasoning is incorrect: in the reals (uncountable) the rational intervals (a countable family) witness the Hausdorffness. But the remainder of C-S is probably so big that you're right in the end, I think – Henno Brandsma Oct 28 '11 at 8:56
Isn't the topology as you define not exactly the usual topology on the Cech-Stone? – Henno Brandsma Oct 28 '11 at 8:57
@Henno: Yes, I caught and fixed the oversight just before I refreshed and saw your note. No, it’s not the usual topology on $\beta\omega$, because I’ve made $\beta\omega\setminus\omega$ a closed, discrete set. – Brian M. Scott Oct 28 '11 at 9:10
Ah yes, it's like a bigger version of Mrowka's Psi. Nice – Henno Brandsma Oct 28 '11 at 20:10 | 2014-03-12 00:50:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8599616289138794, "perplexity": 293.15645716496545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011473737/warc/CC-MAIN-20140305092433-00093-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.stata.com/support/faqs/data-management/float-data-type/ | » Home » Resources & support » FAQs » The accuracy of the float data type
## How many significant digits are there in a float?
Title The accuracy of the float data type Author William Gould, StataCorp
float is a storage format used by Stata, not a computation format. When you have a number stored as a float and you make a calculation, such as
. gen newvar = sqrt(oldvar)/sqrt(2)
oldvar is retrieved and is promoted to a double. The entire computation is then made in double precision, and that result is rounded to a float.
Floats have 7.22 digits of precision, but there is an argument for saying 7.5 digits because it all depends on how you count partial digits.
The way computers store floating point (not to be confused with float, because double is also an example of floating point) is
z = a * 2p -2 < a < 2
Here are some examples of how numbers are stored:
z a p
------------------------
1 1 0
1.5 1.5 0
2 1 1
3 1.5 1 (i.e., 1.5*2^1 = 3)
------------------------
In float, 24 bits are allocated for a. Thus the largest integer that can be exactly stored is 2^0 + 2^1 + ... + 2^23 = (2^24)−1 = 16,777,215. Well, actually, 2^24 = 16,777,216 is also precisely stored because it is even, but 16,777,217 cannot be precisely stored. Using Stata, we can demonstrate these factors using Stata's float() function, which rounds to float precision:
. display float(16777216)
16777216
. display float(16777217)
16777216
Good; Stata works just as theory would suggest.
Now how accurate is float? Well, for numbers like 16,777,217, the absolute error is 1, so the relative error is
1/16,777,217 = 5.960e-08
Generally, when you store a number z as float, what is stored is z', and you can be assured that
z * (1 - 5.960e-08) <= z' <= z * (1 + 5.960e-08)
How many digits of accuracy is that? I can tell you exactly in binary: 24 binary digits, but how do you count in binary digits in base 10? (By the way, thinking in binary is not difficult here: 24 binary digits means the smallest number is 2^(−24) = 5.96e−08, and there is the same relative accuracy we received above.)
Returning to decimal, you might start by observing that 16,777,216 has 8 digits, but no 8-digit number can be stored, so we don't want to claim 8.
One way to get a base-10 representation would be to take log10(16,777,216) = 7.2247199. That is the way most numerical analysts would convert digit accuracy between bases, so we could claim 7.22 decimal digits of accuracy.
The .22 part of 7.2 is subject to misinterpretation because what we just called .22 would, by some, be called one-half. Consider 16,777,216 and some too-big numbers after that:
true number stored if float
----------------------------------------
16,777,216 16,777,216
16,777,217 16,777,216
16,777,218 16,777,218
16,777,219 16,777,220 [sic]
16,777,220 16,777,220
16,777,221 16,777,220
16,777,222 16,777,222
16,777,223 16,777,224 [sic]
16,777,224 16,777,224
16,777,225 16,777,224
----------------------------------------
Basically, odd numbers are being rounded to even numbers. A lot of people would call this the loss of half the digits in the last place, and we could develop a different formula that would label that difference 7.5. Sometimes in computer documentation, you will see the statement that float has 7.5 digits of accuracy. They say that and not 7.22 because the authors worry that you might misinterpret what 7.22 means.
Label the difference how you wish; there are 24 binary digits, and the relative accuracy is +/− 2^(−24) = 5.960e−08.
Note: The [sic]s in the above have to do with how numbers ending in exactly “half” (5 in decimal) are rounded; this is the same problem as rounding 1.5 and 2.5 to one digit in decimal. | 2019-09-23 17:44:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37900564074516296, "perplexity": 1438.9984617134164}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514577478.95/warc/CC-MAIN-20190923172009-20190923194009-00104.warc.gz"} |
https://www.aimsciences.org/article/doi/10.3934/cpaa.2013.12.1 | Article Contents
Article Contents
# Laplacians on a family of quadratic Julia sets II
• This paper continues the work started in [4] to construct $P$-invariant Laplacians on the Julia sets of $P(z) = z^2 + c$ for $c$ in the interior of the Mandelbrot set, and to study the spectra of these Laplacians numerically. We are able to deal with a larger class of Julia sets and give a systematic method that reduces the construction of a $P$-invariant energy to the solution of nonlinear finite dimensional eigenvalue problem. We give the complete details for three examples, a dendrite, the airplane, and the Basilica-in-Rabbit. We also study the spectra of Laplacians on covering spaces and infinite blowups of the Julia sets. In particular, for a generic infinite blowups there is pure point spectrum, while for periodic covering spaces the spectrum is a mixture of discrete and continuous parts.
Mathematics Subject Classification: Primary: 28A80.
Citation:
• [1] S. Constantin, R. Strichartz and M. Wheeler, Analysis of the Laplacian and spectral operators on the Vicsek set, Comm. Pure Appl. Anal, 10 (2011), 1-44. [2] Stella C. Dong, Laplacians on a family of quadratic Julia sets II, http://www.math.cornell.edu/ cdong01/, March 2010. [3] A. Douady, Descriptions of compact sets in $\mathbbC$, in "Topological Methods in Modern Mathematics'' (eds. L. Goldberg and A. Phillips), Publish or Perish, Houston (1993), 429-465. [4] T. Flock and R. Strichartz, Laplacians on a family of Julia sets I, Trans. Amer. Math. Soc., to appear. [5] Jun Kigami, "Analysis on Fractals," volume 143 of Cambridge Tracts in Mathematics, Cambridge University Press, Cambridge, 2001. [6] Jun Kigami and Michel L. Lapidus, Weyl's problem for the spectral distribution of Laplacians on p.c.f. self-similar fractals, Comm. Math. Phys., 158 (1993), 93-125. [7] J. Milnor, Periodic orbits, externals rays and the Mandelbrot set: an expository account, Geometrie Complexe et Systemes Dynamiques, 261 (2000), 277-333. [8] Luke G. Rogers and Alexander Teplyaev, Laplacians on the basilica Julia set, Comm. Pure Appl. Anal., 9 (2010), 211-231. [9] R. Strichartz, Fractals in the large, Can. J. Math., 50 (1998), 638-657. [10] Robert S. Strichartz, "Differential Equations on Fractals, A Tutorial,'' Princeton University Press, Princeton, NJ, 2006. [11] A. Teplyaev, Spectral analysis on infinite Sierpinski gaskets, J. Functional Anal., 159 (1998), 537-567. | 2022-11-29 22:23:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7759420871734619, "perplexity": 906.8307118016702}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00784.warc.gz"} |
http://mathhelpforum.com/business-math/81892-compounding-interest-continuously.html | # Math Help - Compounding Interest Continuously
1. ## Compounding Interest Continuously
Susan invested $5000.00 at 10% compounded semiannually. In how many years will susan's investment have tripled? Round your answers to the nearest tenth. 2. Originally Posted by fran_eric_07 Susan invested$5000.00 at 10% compounded semiannually. In how many years will susan's investment have tripled? Round your answers to the nearest tenth.
Is it 10% per annum? I will assume so.
$A = 5000 \left( 1 + \frac{5}{100}\right)^n$ where n is the number of half years.
So solve $15000 = 5000 \left( 1 + \frac{5}{100}\right)^n$ for n. | 2014-04-18 00:59:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18105998635292053, "perplexity": 5120.828384613613}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00044-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://codegolf.stackexchange.com/questions/138068/kill-it-with-fire/138093 | # Kill it With Fire
Disclaimer: The story told within this question is entirely fictional, and invented solely for the purpose of providing an intro.
I am an evil farmer, and to drive up the price of wheat in my area, I've decided to burn the fields of all the farmers around me. I would really like to see the fields go up in flames (so I can use my evil laugh and rub my hands together with glee), but I also don't want to be caught watching, so I need you to simulate the field being incinerated for me.
Write a program or function that takes as input a field, and returns the stages of it burning until the entire field is ash. A specific section of the field that is on fire is represented by an integer representing the intensity of the flame. A fire starts at "1" and moves on to "2" and then "3", and so on. Once a fire reaches "4", it catches any directly (not diagonally) adjacent areas that are flammable on fire. Once it reaches "8", it burns out on the next iteration, and turns into ash, represented by an "A". When an area has not yet been touched by fire, it is represented by a "0". For example, if the field looks like this:
100
000
100
000
200
000
300
000
410
100
520
200
630
300
741
410
852
520
A63
630
A74
741
A85
852
AA6
A63
AA7
A74
AA8
A85
AAA
AA6
AAA
AA7
AAA
AA8
AAA
AAA
If you wish, you may replace the above symbols with any set of symbols you choose, as long as they are consistent and distinct from each other.
## Input:
The starting position of the field, in any standard form, such as a newlined-delimated string as above.
## Output:
The field in every iteration as it burns, either as an array, or as a string delimited by some character.
## Test Cases:
0301
000A
555
|
v
0301
000A
555
1412
010A
666
2523
020A
777
3634
030A
888
4745
141A
AAA
5856
252A
AAA
6A67
363A
AAA
7A78
474A
AAA
8A8A
585A
AAA
AAAA
6A6A
AAA
AAAA
7A7A
AAA
AAAA
8A8A
AAA
AAAA
AAAA
AAA
## Scoring:
This is , lowest score in bytes wins!
• How much can the shape vary? Are the non-rectangular parts always "holes" on the right edge, or can there be spaces in the field? Aug 8 '17 at 8:08
• So something that hits 4 starts a fire in adjacent squares, but something that starts out at 4 or higher doesn't? That's not very realistic. Aug 8 '17 at 8:13
• "Disclaimer: The story told within this question is entirely fictional, and invented solely for the purpose of providing an intro." -> Starts with "I'm an evil farmer [who wants to do evil stuff]." Very clever. No one will relate the burning fields to you now.
– JFBM
Aug 8 '17 at 10:09
• Is it possible for initial ash to separate the field in two, so that part of it will never burn? Aug 8 '17 at 10:58
• If the fire not spreading above 4 is too upsetting, imagine that 1-4 is the fire gaining intensity and 5-A represents it burning out. Aug 8 '17 at 12:32
# Python 3, 232 bytes
def f(j):
k=[[int(min(9,j[x][y]+(j[x][y]>0)or 3in(lambda k,x,y:[k[i][j]for i,j in[[x-1,y],[x+1,y],[x,y-1],[x,y+1]]if(-1<i<len(k))and(-1<j<len(k[i]))])(j,x,y)))for y in range(len(j[x]))]for x in range(len(j))]
if k!=j:print(k);f(k)
Try it online!
-3 bytes thanks to officialaimm by merging the other lambda into f (looks messy but saves bytes and that's all we care about)
-8 bytes thanks to Mr. Xoder
-26 bytes thanks to ovs
-6 bytes thanks to ppperry
• How do I add a blank space, like that in the example? Aug 9 '17 at 15:51
# JavaScript (ES6), 217210207204193192 190 bytes
Saved 2 bytes thanks to @Shaggy's suggestion of using 9 as A.
f=F=>[F,...(t=[],y=0,g=x=>(r=F[y])?(x||(t[y]=[]),r[x]+1)?(t[y][x]=r[x]<8?r[x]+(r[x]>0|[r[x-1],r[x+1],F[y-1]&&F[y-1][x],F[y+1]&&F[y+1][x]].includes(3)):9,g(x+1)):g(0,y++):t)(0)+""!=F?f(t):[]]
// test code
test=s=>{
var test = document.querySelector("#in").value.split\n.map(e => Array.from(e).map(e => parseInt(e)));
var out = f(test);
document.querySelector("#out").innerHTML = out.map(e => e.map(e => e.join).join\n).join\n\n;
};window.addEventListener("load",test);
<textarea id="in" oninput="test()">0301 0009 555</textarea><pre id="out"></pre>
Uses 9 instead of A. Input as a 2D array of integers. Output as an array of such arrays.
• Could you save anything by using 9 instead of A? Aug 8 '17 at 9:27
# Simulating the World (in Emoji), 1407 bytes?
Don't you love using an explorable explanation as a programming language? The downside to this is there's usually not a very well defined program, so in this case, I'm using the JSON it exports. (if you have any better ideas, let me know)
{"meta":{"description":"","draw":1,"fps":1,"play":true},"states":[{"id":0,"icon":"0","name":"","actions":[{"sign":">=","num":1,"stateID":"4","actions":[{"stateID":"1","type":"go_to_state"}],"type":"if_neighbor"}],"description":""},{"id":1,"icon":"1","name":"","description":"","actions":[{"stateID":"2","type":"go_to_state"}]},{"id":2,"icon":"2","name":"","description":"","actions":[{"stateID":"3","type":"go_to_state"}]},{"id":3,"icon":"3","name":"","description":"","actions":[{"stateID":"4","type":"go_to_state"}]},{"id":4,"icon":"4","name":"","description":"","actions":[{"stateID":"5","type":"go_to_state"}]},{"id":5,"icon":"5","name":"","description":"","actions":[{"stateID":"6","type":"go_to_state"}]},{"id":6,"icon":"6","name":"","description":"","actions":[{"stateID":"7","type":"go_to_state"}]},{"id":7,"icon":"7","name":"","description":"","actions":[{"stateID":"8","type":"go_to_state"}]},{"id":8,"icon":"8","name":"","description":"","actions":[{"stateID":"9","type":"go_to_state"}]},{"id":9,"icon":"A","name":"","description":"","actions":[]}],"world":{"update":"simultaneous","neighborhood":"neumann","proportions":[{"stateID":0,"parts":100},{"stateID":1,"parts":0},{"stateID":2,"parts":0},{"stateID":3,"parts":0},{"stateID":4,"parts":0},{"stateID":5,"parts":0},{"stateID":6,"parts":0},{"stateID":7,"parts":0},{"stateID":8,"parts":0},{"stateID":9,"parts":0}],"size":{"width":9,"height":9}}}
Try it here or here:
<iframe width="100%" height="450" src="http://ncase.me/simulating/model/?remote=-Kr2X939XcFwKAunEaMK" frameborder="0"></iframe>
# Retina, 10396 88 bytes
^
¶
;{:
T0dd
(?<=¶(.)*)0(?=4|.*¶(?<-1>.)*(?(1)_)4|(?<=40|¶(?(1)_)(?<-1>.)*4.*¶.*))
1
Try it online! Uses 9 for ash; this can be changed at a cost of 4 bytes using T1-82-8A. Edit: Saved 6 bytes thanks to @MartinEnder. Explanation:
^
¶
Add a separator so that the outputs don't run into each other. (Also helps when matching below.)
;{:
Don't print the final state (which is the same as the previous state which has already been printed). Repeat until the pass does not change the state. Print the current state before each pass.
T0dd
Advance the intensity of all the fire.
(?<=¶(.)*)0(?=4|.*¶(?<-1>.)*(?(1)_)4|(?<=40|¶(?(1)_)(?<-1>.)*4.*¶.*))
1
Light unlit fields as appropriate. Sub-explanation:
(?<=¶(.)*)
Measure the column number of this unlit field.
0
Match the unlit field.
(?=4
Look for a suitable field to the right.
|.*¶(?<-1>.)*(?(1)_)4
Look for a suitable field in the same column (using a balancing group) in the line below. Note that if the input could be guaranteed rectangular then this could be simplified to |.*¶(?>(?<-1>.)*)4 for a saving of 3 bytes.
|(?<=40
Look for a suitable field to the left. (Since we're looking from the right-hand side of the field, we see the unlit field as well.)
|¶(?(1)_)(?<-1>.)*4.*¶.*))
Look for a suitable field in the same column in the line above. Since this is a lookbehind and therefore a right-to-left match, the balancing group condition has to appear before the columns that have been matched by the balancing group.
# Perl 5, 365 bytes
@a=map{[/./g]}<>;do{say@$_ for@a;say;my@n=();for$r(0..$#a){$l=$#{$a[$r]};for(0..$l){$t=$a[$r][$_];$n[$r][$_]=($q=$n[$r][$_])>$t?$q:$t==9?9:$t?++$t:0;if($t==4){$n[$r-1][$_]||=1if$r&&$_<$#{$a[$r-1]};$n[$r+1][$_]||=1if$r<$#a&&$_<$#{$a[$r+1]};$n[$r][$_-1]||=1if$_;$n[$r][$_+1]||=1if$_<$l}}}$d=0;for$r(0..$#a){$d||=$a[$r][$_]!=$n[$r++][$_]for 0..$#{$a[$r]}}@a=@n}while$d Try it online! Uses '9' instead of 'A' to indicated a burned out location. Explained @a=map{[/./g]}<>; # split input into a 2-D array do{ say@$_ for@a;say; # output the current state
my@n=(); # holds the next iteration as it's created
for$r(0..$#a){ # loop through rows
$l=$#{$a[$r]}; # holder for the length of this row
for(0..$l){$t=$a[$r][$_]; # temporary holder for current value$n[$r][$_]=($q=$n[$r][$_])>$t?$q:$t==9?9:$t?++$t:0; #update next iteration if($t==4){ # ignite the surrounding area if appropriate
$n[$r-1][$_]||=1if$r&&$_<$#{$a[$r-1]};
$n[$r+1][$_]||=1if$r<$#a&&$_<$#{$a[$r+1]};$n[$r][$_-1]||=1if$_;$n[$r][$_+1]||=1if$_<$l
}
}
}
$d=0; # determine if this generation is different than the previous for$r(0..$#a){$d||=$a[$r][$_]!=$n[$r++][$_]for 0..$#{$a[$r]}} @a=@n # replace master with new generation }while$d
import Data.List
i x|x<'9'=succ x|1<2=x
f('3':'@':r)="30"++f r
f(x:r)=x:f r
f e=e
a%b=a.b.a.b
g=map(i<$>).(transpose%map(reverse%f)) h x|y<-g x,y/=x=x:h y|1<3=[x] Try it online! Usage: h takes a field as a list of lines and returns a list of fields. An unburned field is indicated by @ and ash by 9, the different fires are the digits 1 to 8. • f manages the spreading of the fire from left to right by replacing all unburned @ fields which are right to a burning 3 field with 0. • i increments each digit as long as it is less than 9. • g applies f to each line, then reverses the line, applies f again and reverses back. Then the list of of lines is transposed and again on each line and its reverse f is applied. • h applies g to the input until it no longer changes and collects the results. • Fails on input ""@3@1\n@@@9\n555@@@@@@@@@@@@@@@@@@@", as for some reason the long line of @s switches to the top line after the first iteration. Fixing that would be great, thanks! Aug 8 '17 at 16:30 • @Gryphon The displacement is caused by transposing the character matrix. It works if the other lines are brought to the same length with some character which represents neither a field, fire or ash, e.g. _. If this is not acceptable then I'm afraid I'd have to delete the answer, because it's centered around the usage of transpose and I don't see a way to easily fix it without introducing tons of bytes. Aug 10 '17 at 11:37 # C (gcc), 308305299297295 291 bytes #define F B[i][v] m(A,B,k,y,m,U,i,v)char**B;{do{for(i=k=y=0;i<A;puts(""),++i)for(v=0;v<(m=strlen(B[i]));F=(U=F)>48&&F<56?F+1:F>55?65:(i+1<A&&B[i+1][v]==51||v+1<m&&B[i][v+1]==51||v-1>-1&&B[i][v-1]==52||i-1>-1&&B[i-1][v]==52)&&U<49?49:F,putchar(U),k+=U<49||U>64,++y,++v);puts("");}while(k-y);} This program defines a function which takes two inputs, a pointer to an array of strings preceded by its length, as allowed by this I/O default. Outputs to STDOUT with a trailing newline. Try it online! • Runs forever on input 80. Aug 8 '17 at 11:04 • @aschepler Sorry! I assumed that all the characters must turn into As, but apparently I assumed wrong. Anyways, thanks for the information. It's fixed now. Aug 8 '17 at 11:37 • -6 bytes! Tio Link Aug 8 '17 at 14:35 • @GiacomoGarabello I can't believe I forgot about that tactic. Thanks for reminding me! :) Aug 8 '17 at 18:33 # Octave, 72 69 bytes a=input(''),do++a(a&a<9);a+=imdilate(a==4,~(z=-1:1)|~z')&~a,until a>8 The input is taken as a 2D array of numbers, and empty spots marked with Inf. 'A' has been replaced with 9. Intermediate results (as array of numbers) implicitly printed . Try it online! Explanation: In a loop the function imdilate (morphological image dilation) from the image package is used to simulate fire propagation. • This works with boards of all sizes, and even with a bunch of holes. E.G.: [0 Inf 0 0 0;0 Inf 0 Inf 0;0 Inf 0 Inf 0;0 0 0 Inf 1] - Very Nice Aug 9 '17 at 13:28 # APL (Dyalog), 52 bytes* Assumes ⎕IO←0 which is default on many systems. Takes field using 0 for empty slots, 1 for non-burning field, 2 for new fire, 5 for spreading fire, and 10 for ash. Input must be at least 3×3, which isn't a problem as additional rows and columns may be padded with zeros (spaces in OP's format). {}{1=c←4⊃r←,⍵:1+4∊r/⍨9⍴⍳2⋄10⌊c+×c}⌺3 3⍣{⍺≡⎕←⍵} Try it online! My format makes it hard to check for correctness, so here is a version with added pre- and post-processing to translate from and to OP's format. ⍣{} repeat until: ⍺ the next generation ≡ is identical to ⎕←⍵ the current generation, outputted {}⌺3 3 replace each cell with the result of this function applied to its Moore neighborhood: ,⍵ ravel (flatten) the argument; gives nine-element list r← assign to r 4⊃ pick the fourth element; the center, i.e. the original cell value c← assign to c 1= is one equal to that? : if so, then: ⍳2 first to ɩntegers; 0 1 9⍴reshape to length nine; 0 1 0 1 0 1 0 1 0 r/⍨ use that to filter r (this gets just the orthogonal neighbors) 4∊ is four a member of that? (i.e. will there be a five in the next generation?) 1+ add one; 1 if not caught fire or 2 if caught fire ⋄ else (i.e the current value is 0 or ≥ 2) ×c the signum of c c+c plus that (i.e. increase by one if on fire) 10⌊ minimum of ten and that (as ash doesn't burn on) * In Dyalog Classic, using ⎕U233A instead of ⌺. • Small, but doesn't deal with board changes. Stuck to a 3x3. Aug 9 '17 at 13:29 • @Suamere What? The board doesn't change size, does it? – Adám Aug 9 '17 at 13:30 • Sorry, I wasn't clear. Your answer is awesome, but I am of the understanding that the solution should allow for a variable sized board which may or may not have gaps "on the right". Gaps in the middle are not required to be handled. "On the Right" seems to mean a size-15 board would be built as a 4x4, except the right-most bottom piece is missing. And a size-8 board would be built as a 3x3, except the right-most bottom piece is missing, etc. That's how I read the challenge requirements. Your answer is currently the shortest, but only works with a 3x3. Aug 9 '17 at 13:59 • @Suamere OP clearly states that input is 2D. I take input as a numeric matrix and allow "empty" slots anywhere, in the form of zeros. While I require the input to be minimum 3×3 (OP allowed this) I accept input of any larger size. In fact, if you click the here link, you'll see that my second example case is 2×3 with the bottom right cell empty. – Adám Aug 9 '17 at 14:07 • Gotcha. Not sure why, but the Try It Here link has issues (Probably my fault), though your new formatted link works well. E.G.: fire '0A000\n0A0A0\n0A0A0\n000A1' works perfect on the formatted one, but I can't get an equivalent to work with the first link. I'm probably doing something wrong. This doesn't work for me: f ↑(0 0 0)(0 1 0)(0 0 0) Aug 9 '17 at 14:25 # Python 2, 325 Bytes def f(x): while{i for m in x for i in m}>{9,''}: yield x;i=0 for l in x: j=0 for c in l: if 0<c<9:x[i][j]+=1 j+=1 i+=1 i=0 for l in x: j=0 for c in l: if c==0 and{1}&{x[k[1]][k[2]]==4for k in[y for y in[[i,i-1,j],[i<len(x)-1,i+1,j],[j,i,j-1],[j<len(l)-1,i,j+1]]if y[0]]}:x[i][j]+=1 j+=1 i+=1 yield x f takes the input as a 2D array of integers, and empty spots marked with ''. 'A' has been replaced with 9. The function outputs a generator of all the fields over time in the same format. Try it online! # Octave, 212 bytes function r(f)b=48;f-=b;c=-16;l=f~=-c;d=17;p=@(x)disp(char(x+b));g=@()disp(' ');p(f);g();while~all(any(f(:)==[0 d c],2))m=f>0&l;f(m)+=1;k=conv2(f==4,[0 1 0;1 0 1;0 1 0],'same')&~m&l;f(k)+=1;f(f>8&l)=d;p(f);g();end To run, specify a character array such as: f = ['0301','000A','555 ']; ... then do: r(f) Explanation of the code to follow... # Try it online! Note: I tried to run this code with tio.run, but I wasn't getting any output. I had to use another service. • Really just at the moment I want to post an octave answer you answer before me.. Aug 9 '17 at 7:08 # PHP, 226 212 210 209 185 177 bytes for($f=file(m);$f!=$g;print"
")for($g=$f,$y=0;$r=$f[$y++];)for($x=-1;~$c=$r[++$x];$f[$y-1][$x]=$c=="0"?strstr($g[$y-2][$x].$r[$x-1].$f[$y][$x].$r[$x+1],51)/3:min(++$c,9))echo$c;
takes input with a trailing newline from a file named m; 9 for ashes.
Run with -nr or try it online.
first approach: PHP 7.0, 209 bytes
for($f=$g=file(m);trim(join($f),"A ");print" ".join($f=$g))for($y=-1;(a&$c=$r[++$x])||$r=$f[$y-=$x=-1];)for(+$c&&$g[$y][$x]=++$c<9?$c:A,$a=4;$a--;$q=$p)$c-4|($w=&$g[$y+$p=[1,0,-1][$a]])[$q+=$x]!="0"||$w[$q]=1; takes input with a trailing newline from a file named m. Run with -nr or try it online. PHP version notes (for old approach) • for PHP older than 7.0, replace everything after $c-4| with $g[$y+$p=[1,0,-1][$a]][$q+=$x]!="0"||$g[$y+$p][$q]=1;
• for PHP older than 5.5, replace [1,0,-1][$a] with $a%2*~-($a&2) • for PHP newer than 7.0, replace a&$c with ""<$c, +$c with 0<$c and $c-4 with \$c!=4
• doesnt' work for other test cases...sandbox.onlinephpfunctions.com/code/… Aug 8 '17 at 14:51
• @g19fanatic fixed & golfed. thanks for noticing. Aug 8 '17 at 19:05
# Octave, 419 312 bytes
M=input('') %take a matrix M as input
[a, b]=size(M); %size M for later
while mean(mean(M))!=9 %while the whole matrix M isn't ashes
M=M+round(M.^0.001); %if there is a nonzero int add 1 to it
for i=1:a %loop over all values of M
for j=1:b
if(M(i,j)==4) %if a value of 4 is found, ignite the zeros around it
if(M(max(i-1,1),j)==0) M(i-1,j)++;end
if(M(min(i+1,a),j)==0) M(i+1,j)++;end
if(M(i,max(j-1,1))==0) M(i,j-1)++;end
if(M(i,min(j+1,b))==0) M(i,j+1)++;end
elseif(M(i,j)==10) M(i,j)--;
end
end
end
M %display the matrix
end
# Try it online!
This is my version it works, so now I still need to golf it. I think it can be a lot shorter if I find a way to find the indices of the 4 in a matrix, but I don't know how.
PS: A is a 9 in my code.
• Instead of endif endfor and endwhile you can write end Aug 9 '17 at 7:38
# Stencil(∊-mode), 22 bytes
S≡8:'A'⋄0::S⋄(3∊N)⌈S+s
Try it online!
Just like in the test input, use integers separated by spaces for 0-8, ' ' for blank and 'A' for A. Remember to add trailing blanks too. | 2022-01-17 01:17:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3212474584579468, "perplexity": 2870.1698781617015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00314.warc.gz"} |
https://2016.igem.org/Team:Marburg/Modeling/Results | # SynDustry Fuse. Produce. Use.
## Quantitative evolutionary stability analysis and design guidelines for kill switches
### 1. Concentration dynamics of kill switches
This chapter summarizes the first working phase of the iGEM Marburg 2016 modeling work: The continuous ODE model presented previously is used with the biologically relevant parameters to compute the concentrations' dynamics $$\vec{X}(t)$$ for all times. This shows that the designs are comparable in terms of the toxin expression levels. The latter three designs are comparable in terms of the number of model parameters.
At first, the lab environment is simulated in order to evaluate the toxin expression during the kill switch's OFF state. After that, the organism is released into wild life conditions – no inducers present – and the toxin expression of the activated kill switch is compared to the OFF state. From there, a toxin threshold is found.
This sections serves as verification that the non-mutated kill switch designs work as intended.
#### (a) BNU China 2014
The BNU China 2014 kill switch (find design here) is simulated and the results are shown in figure 1. One observes a significant increase of the toxin concentration once the organism is released into wild life conditions which lacks an inducer. The lethal toxin threshold is defined as based on the simulation results. This is the mean of the kill switch's OFF and ON state. It is consistently drawn as dashed red line in the following.
Figure 1. Simulated BNU China 2014 kill switch dynamics. All involved concentrations are simulated. The first subplot shows the externally set inducer – IPTG – concentration. The last subplot shows the resulting toxin expression and the lethal toxin threshold $$\theta$$. All subplots in-between correspond to the concentrations of intermediate substances.
Interestingly, one can observe a noticeable delay between the inducer turned off and the toxin turning on. This is due to the signal traveling through the network at finite speed.
The next kill switch designs represent different generic kill switch topologies and abstract from specific biological substances to substances that behave just like their biological counterparts. This enables to focus on the bare kill switch design (see algorithm description for further details here).
#### (b) Parallel (OR)
This design utilizes two redundant toxin regulating branches (find design here). Being connected by an OR gate, this provides a backup mechanism in case one branch is disabled through mutations. The result can be found in figure 2.
Figure 2. The modeled Parallel (OR) kill switch with biologically correct parameters. The first two subplots show the two abstract inducers’ concentrations $$I_a$$ and $$I_b$$. The second inducer, $$I_b$$, is kept on after release to resemble a destroyed kill switch through mutations. The blue curve in the last subplot shows the toxin expression with both branches working properly while the dashed green line with branch b deactivated. Since both concentrations are well above the threshold $$\theta$$, the kill switch is in its ON state in both cases.
#### (c) Serial
The Serial kill switch design is simulated (find design here). Figure 3 shows the kill switch typical behavior.
Figure 3. The modeled Serial kill switch with inducer in the upper subplot and the toxin in the lower subplot. The toxin threshold is exceeded such that the kill switch is in its ON state after the inducer having been turned off.
Since this design introduces a longer serial cascade than the BNU China 2014 design, the delay between inducer deactivation and toxin expression is longer.
#### (d) Parallel (AND)
A parallel design that connects both branches with an AND gate is studied (find design here). Figure 4 shows that both branches need to be functional to trigger the kill switch's ON state.
Figure 4. Simulated Parallel (AND) kill switch. The subplots are equivalent to the one shown in figure 2. Artificially deactivating a branch leads to a kill switch remaining in its OFF state. A single branch does not regulate the toxin.
As evident from figure 4, the deactivation of one of the two branches leads to the kill switch being in the OFF state permanently - thus, the branches are indeed linked by an AND gate.
### 2. Evolutionary stability
For different incubation times and mutation rates, the escape probability of the BNU China 2014 kill switch design is studied. Figure 5 shows the results from which two intuitive results are confirmed:
1. The higher the mutation rate, the larger the escape probability: As the likelihood for a mutation to occur increases, the probability of the linear cascade to be destroyed through one of these mutations increases as well.
2. Longer incubation times lead to higher escape probabilities: This is due to more possible attempts for mutations to destroy the kills switch.
Figure 5. Evolutionary stability of BNU China 2014 kill switch. The escape probabilities for different mutation rates $$\eta$$ are shown. Longer incubation times and mutation rates lead to larger escape probabilities.
#### (b) Time scaling by mutation rate
Scaling the incubation time with the mutation rate lets the escape probability curves collapse onto one master curve. This can be explained: Hitting a certain threshold takes more incubation time when a low mutation rate is applied but it takes less incubation generations when working with a high mutation rate. The master curve for the BNU China 2014 kill switch can be found in figure 6.
Figure 6. Evolutionary stability master curve of BNU China 2014 kill switch. The mutation rate scales the incubation time axis so that the curves collapse onto a master curve for a unified time scale.
Therefore, the mutation rate is arbitrary for quantifying the evolutionary stability of a kill switch design. Computations have been performed for $$\eta\in\{0.4081875 \cdot 10^{-3}, 0.816375 \cdot 10^{-3}, 1.63275 \cdot 10^{-3}\}$$ but only $$\eta=1.63275 \cdot 10^{-3}$$ is shown in the following. Other mutation rates are equivalent and the results show the same behavior.
#### (c) Design guidelines for a kill switch
We are providing design guidelines for a kill switch that is robust against evolutionary stability and show quantitative evidence for our suggestion. A kill switch design can be understood as a network with the toxin being a node that is controlled through regulation pathways.
The BNU China 2014 kill switch design uses a single regulation pathway to control its toxin MazF. From intuition it is clear that a kill switch with multiple parallel regulation pathways for its toxin – with only one required to control the toxin expression – should be less prone to being destroyed by mutations. Let a mutation destroy one of the parallel regulation pathways: There are still other pathways that back up the destroyed one and ensure a working kill switch.
We utilize three comparable kill switches that differ in their network topology. Figure 7 shows the escape probabilities for the Parallel (OR), Serial and Parallel (AND) designs. Parallel (OR) – implementing a topology that allows destroyed branches to be backed up – shows by far the smallest escape probability. Where a single mutation is capable of destroying the Parallel (AND) and Serial design, a destroyed pathway can be compensated by the functional regulation pathway for the Parallel (OR) design.
Figure 7. For $$\eta=1.63275\cdot 10^{-3}$$, the escape probabilities of the three kill switch designs are shown against the incubation time. The Parallel (OR) design has the smallest escape probabilities. The explanation can be found in the text.
Concluding, we suggest to implement a redundant toxin regulation for kill switches. This significantly increases the evolutionary stability and decreases the escape probability of genetically modified organisms into the environment. Biotechnological applications ranging from small to industrial scale can be made safer with having this in mind.
However, introducing additional topology such as multiple redundant toxin regulations increases a kill switches complexity. There are more points of contact for mutations to destroy the topology. Translating this in the model we use, more complex kill switches have more ODE parameters that can be mutated. Therefore, the escape probability is expected to somehow scale with a kill switch's complexity.
This remaining question of a trade-off between kill switch complexity and desired network topology to find a robust kill switch is studied in the last chapter.
#### (d) Comparison between BNU China 2014 and Parallel (OR) kill switch
We compare the escape probabilities of the BNU China 2014 design and the hypothetical Parallel (OR) kill switch. The latter introduces on the one hand redundant toxin regulation but on the other hand an increased complexity. That manifests in almost double the number of ODE parameters in comparison to the BNU China 2014 design.
Figure 8 shows that the stability-optimized Parallel (OR) kill switch is more robust against evolution and mutations than the BNU China 2014 topology. The gain in evolutionary stability through the additional regulation units overcompensates the introduced complexity. Finally, this leads to a more robust kill switch when using a redundant toxin regulation.
Figure 8. Comparison between BNU China 2014 and Parallel (OR) kill switch. The latter one shows better robustness in terms of escape probabilities.
This result has various applications in real life. Redundant toxin regulation not only ensures projects in Synthetic Biology to be safer but also enables longer incubation times until a certain threshold in escape probability is reached. Organisms can therefore be longer incubated without becoming evolutionary instable. This simplifies industrial applications and increases overall efficiency in production processes.
The code framework is available upon request (Martin Lellep, mailto:martin.lellep@physik.uni-marburg.de). We provide possible outlooks on the following website. | 2021-07-26 02:42:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3900025486946106, "perplexity": 2558.0176362606467}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00174.warc.gz"} |
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# If the price of a commodity is directly proportional to m^3
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If the price of a commodity is directly proportional to m^3 [#permalink] 07 Apr 2013, 04:45
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If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72
[Reveal] Spoiler: OA
Last edited by Bunuel on 07 Apr 2013, 21:20, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Price of a commodity [#permalink] 07 Apr 2013, 04:56
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If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72
Let the price is P & constant is K
So P = \frac{K m^3}{q^2}
Now Just use the values in the options and put in the before mentioned equation.
A) P = \frac{K m^3}{q^2} = \frac{K 3^3}{2^2} = \frac{K 27}{4} = 6.75 K
B) P = \frac{K m^3}{q^2} = \frac{K 12^3}{12^2} = 12 K
C) P = \frac{K m^3}{q^2} = \frac{K 20^3}{20^2} = 20 K
D) P = \frac{K m^3}{q^2} = \frac{K 30^3}{36^2} = 125/6 = >20K
E) P = \frac{K m^3}{q^2} = \frac{K 36^3}{72^2} = 9 K
So the answer is D. This is probably 600-700 level question.
Hope the explanation helps.
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Last edited by emmak on 07 Apr 2013, 05:00, edited 1 time in total.
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Re: Price of a commodity [#permalink] 07 Apr 2013, 04:57
2
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Dipankar6435 wrote:
If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72
Function = \frac{m^3}{q^2}
A=\frac{3*3*3}{2*2}=\frac{27}{4}=7 almost
B=\frac{12*12*12}{12*12}=12
C=\frac{20*20*20}{20*20}=20
D=\frac{30*30*30}{36*36}=125/6
E=\frac{36*36*36}{72*72}=9
It's down to C or D, because \frac{120}{6}=20, \frac{125}{6}>20
D
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If the price of a commodity is directly proportional to m^3 [#permalink] 07 Apr 2013, 21:22
Expert's post
Dipankar6435 wrote:
If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72
Similar questions to practice:
DS:
the-amount-of-coal-a-train-burns-each-mile-is-directly-93667.html
PS:
a-is-directly-proportional-to-b-when-a-8-b-88971.html
in-a-certain-formula-p-is-directly-proportional-to-s-and-80941.html
the-rate-of-a-chemical-reaction-is-directly-proportional-to-76921.html
Hope it helps.
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 09 Apr 2013, 03:56
Let price of the commodity be P = k *m^3 /Q^2
A. P = k * 27/4 = 6.75K
B. P = K * 12^3/12^2 = 12k
C. P = K * 20^3/20^2 = 20k
D. P = K * 30^3/36^2= k*5*5*30/6*6 = k * (125/6) = 21K (approx)
E. P = k * 36^3/72^2 = K*(36/2*2) = 9k
Ans : Option D.
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 10 Apr 2013, 17:09
Could you please explain how you get \frac{m^3}{q^2} or \frac{Km^3}{q^2}? Where am I wrong expressing this thing firstly as p=m^3k and p=\frac{k}{q^2}?
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 10 Apr 2013, 22:19
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obs23 wrote:
Could you please explain how you get \frac{m^3}{q^2} or \frac{Km^3}{q^2}? Where am I wrong expressing this thing firstly as p=m^3k and p=\frac{k}{q^2}?
"directly proportional to m^3 and inversely proportional to q^2" is what the text says. So if m increases the price increases, if q increases the price decreases. The right formula here is \frac{m^3}{q^2} ( or with k, it doesnt change anything), if the numerator grows, the price does the same; and as in every fraction, if the denominator grows, the price drops.
Your formulas are right but the price depends on both m and q, you have to include them in one equation.
Let me know if it is clear
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 11 Apr 2013, 02:51
Zarrolou wrote:
obs23 wrote:
Could you please explain how you get \frac{m^3}{q^2} or \frac{Km^3}{q^2}? Where am I wrong expressing this thing firstly as p=m^3k and p=\frac{k}{q^2}?
"directly proportional to m^3 and inversely proportional to q^2" is what the text says. So if m increases the price increases, if q increases the price decreases. The right formula here is \frac{m^3}{q^2} ( or with k, it doesnt change anything), if the numerator grows, the price does the same; and as in every fraction, if the denominator grows, the price drops.
Your formulas are right but the price depends on both m and q, you have to include them in one equation.
Let me know if it is clear
I see that makes sense. I guess I am making it more complicated than it really is. If you say that my formulas are correct, then \frac{m^3}{q^2} should somehow be extracted from p=m^3k and p=\frac{k}{q^2}, no? Technically speaking, I thought there was a way to combine them into \frac{m^3}{q^2}, just from pure algebraic manipulation (which I am very interested in here)... Or is it simply about common sense or the way you explained it?
P.S. Like your "kudos" tagline man!
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 11 Apr 2013, 03:05
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obs23 wrote:
I see that makes sense. I guess I am making it more complicated than it really is. If you say that my formulas are correct, then \frac{m^3}{q^2} should somehow be extracted from p=m^3k and p=\frac{k}{q^2}, no? I thought there was a way to combine them into \frac{m^3}{q^2}, just from pure algebraic manipulation (which I am very interested in here)... Or is it simply about common sense or the way you explained it?
To create the formula we should refer to the text:
"If the price of a commodity is directly proportional to m^3 [and at this point we write down Price= m^3] and inversely proportional to q^2 [and at this point we complete the foumula adding the Denominator so Price=\frac{m^3}{q^2}], which of the following values of m and q will result in the highest price for the commodity?"
\frac{m^3}{q^2} is not extracted from p=m^3k and p=\frac{k}{q^2}, and you cannot obtain it from pure algebraic manipulation.
The idea behind your formulas is correct p=m^3 expresses the direct correlation between p and m; and also p=1/q^2 expresses the inverse correlation between p and q. But the text uses "and" so those ideas must be expressed in one formula, so to obtain this final formula you "complete" one with the other =>m^3/p^2
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 11 Apr 2013, 03:30
Got it! Thanks much for help.
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 11 Apr 2013, 06:22
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Dipankar6435 wrote:
If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72
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http://www.veritasprep.com/blog/2013/02 ... inversely/
http://www.veritasprep.com/blog/2013/02 ... g-jointly/
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Re: If the price of a commodity is directly proportional to m^3 [#permalink] 11 Apr 2013, 06:22
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does aluminum react with hydrochloric acid
It is known chemically as alpha-Al 2 O 3 and is produced at high temperatures. get reducedDoes Astatine React With Sodium Iodide And Does Copper React With Concentrated Hydrochloric Acid On Sale . Considering this, will citric acid clean aluminum? Reaction of metals with acids. The water actually evaporates, and with enough acid and metal, boils. will aluminium sulphate react with hydrochloric acid. 2Al (s) + 6HCl (aq) → 2AlCl 3 (s) + 3H 2 (g) View solution When dilute hydrochloric acid is added to granulated zinc placed in a test tube, the observation made is _____________. I want to convert aluminium sulpahte to aluminium chloride. Reaction of aluminium with acids Aluminium dissolves readily in dilute sulphuric or hydrochloric acid under the formation of Al(III) and hydrogen H2. Anhydrous AlCl3 will definitely react with Concentrated hydrochloric and probably violently. Although you should take a look at what Klaus said, copper does in fact react with hydrochloric acid, it just takes a week until all the copper is converted into copper chloride (green) and another week or so until it forms crystals and you can dissolve them in water to form copper chloride again (but depending on the amount of chloride it has, it'll be blue or green). Who is the longest reigning WWE Champion of all time? Solved: Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas. 3. Ferrer. Aluminum and hydrochloric acid react as follows: 2 Al + 6 HCl (aq) ---> 2 Al+3 + 6 Cl- + 3 H2. Sign up now, Latest answer posted January 07, 2013 at 6:01:57 AM, Latest answer posted December 07, 2015 at 12:40:57 AM, Latest answer posted March 10, 2013 at 1:25:24 PM, Latest answer posted March 04, 2016 at 12:51:37 AM, Latest answer posted January 08, 2013 at 9:45:12 AM. Who was the lady with the trophy in roll bounce movie? Related questions 0 votes. Buy Does Aluminum React With Bases And Does Copper React With Dilute Hydrochloric Acid Does Aluminum React With Bases And Does Copper React With Dilute Hydrochloric Acid Reviews : If you're looking for Does Aluminum React With Bases And Does Copper React With Dilute Hydrochloric Acid. This is because each chlorine atom in the hydrochloric acid acquires an electron from the aluminium and loses a … Top subjects are Science, Math, and Business. Aluminum reaction with HCl: Most of the metals, including Aluminum, reacts with HCl to yield a chloride salt and hydrogen gas. I'd like to ask whether any of us has tried reducing aluminium oxide with hydrochloric acid. The U.S. Department of Transportation conducted experiments to determine if a spill of hydrochloric acid could eat through the aluminum skin of an airplane and cause it to fall apart. Aluminium does react with Hydrochloric acid, but, it also reacts with the oxygen in the air to create a coating of aluminium oxide, which is unreactive with acid. Copper is a metal that does not react with HCl. 3 cm of magnesium ribbon typically has a mass of 0.04 g and yields 40 cm 3 of hydrogen when reacted with excess acid. So for sulfuric acid and hydrochloric acid, $\ce{SrSO4}$ will form, but $\ce{SrCl2}$ will stay in solution. When Silver Nitrate Reacts With Hydrochloric Acid, What Type Of Reaction Takes Place? When I tried the same with using aluminum foil, it did react after several minutes with HCl solution. However when we put in warm hydrochloric acid it bubbled, fizzed, jumped around and even over-flowed out of the test tube. Muriatic acid, or hydrochloric acid as it's also known, is an extremely harsh acid that's sometimes used in the industrial world. Copper will not react with hydrochloric acid. Hydrochloric acid (HCl) may be the simplest chlorine-based acid, but it is still strongly corrosive. It will react the quickest, releasing the most hydrogen. Buy Online keeping the vehicle safe transaction. If you remove the aluminium oxide Homeowners sometimes use it to clean and etch aluminum surfaces. Why doesnt aluminum react with hyrdochloric acids. Thank Writer; Comment; Blurt; thanked the writer. If you are searching for read reviews Does Radon React With Oxygen And Does Sodium Chloride React With Hydrochloric Acid price. Aluminium metal dissolves readily in dilute sulphuric acid to form solutions containing the aquated Al(III) ion together with hydrogen gas, H 2. For example, zinc metal reacts with hydrochloric acid, producing zinc chloride and hydrogen gas. $\ce{Zn(s) + 2NaOH (aq) + 2H2O(l) → Na2Zn(OH)4(aq) + H2 … Keep in mind that this reaction will not take place as soon as you add the piece of aluminium to the hydrochloric acid solution. When a piece of aluminium metal is added to dilute hydrochloric acid, the result is two products - an aqueous product and a gaseous product. Thanks, points for the best answer. 3. Sodium hydroxide reacts well both with metals (zinc, aluminum, titanium) and non-metals: … This video demonstrates the reaction of HCL with different metals. Aluminium does react with Hydrochloric acid, but, it also reacts with the oxygen in the air to create a coating of aluminium oxide, which is unreactive with acid. Hydrochloric acid and aluminum hydroxide react according to the following balanced chemical equation: 3HCl(aq) + Al(OH)3() → AIC13(aq) + 3H2O() If 20g of hydrochloric acid reacts with 20g of aluminum hydroxide, what mass of aluminum chloride is produced? Similarities Between Photosynthesis And Cellular Respiration. Anhydrous AlCl3 will definitely react with Concentrated hydrochloric and probably violently. However, this interaction be delayed in the start-up, the presence of a layer of aluminum oxide on the surface of the metal turning Thus, the moles of aluminum in the foil roll = 1004.4/27 = 37.2 moles. 2 Does that mean that it can react with hydrochloric acid (but not all the time)? Metals to the right of hydrogen in the electrochemical series, such as copper, silver and gold, do not react. What a mess to clean up! This item is very nice product. This is the energy required for the reaction to begin. 125ml Of Hydrochloric Acid (Muriatic Acid 25%) was added to aluminum. The thickness of a standard household aluminum foil is typically 0.016 mm (or 0.016 x 10^-3 m). The unbalanced reaction is . Does aluminum sulfate react with hydrochloric acid? (i)Identify the metal and (ii) Also write the reactions involved. What metals does sodium hydroxide react with? Zinc reacts with the hydrochloric acid and Aluminum also does. ©2021 eNotes.com, Inc. All Rights Reserved, Last Updated by eNotes Editorial on December 23, 2019. It is a violent exothermic reaction and you can create a home-made bomb! Let's determine the moles of aluminum in the given length of foil. Why did clay walker and Lori lampson get divorced? When did organ music become associated with baseball? Report 2 Answers By Expert Tutors Best Newest Oldest. The reaction can be written as the following: 2Al + 6HCl -> 2AlCl3 + 3H2 Here, 2 moles of aluminum reacts with 6 moles of hydrochloric acid to generate 3 … Already a member? For example, sodium hydroxide reacts with zinc and water to form sodium zincate and hydrogen gas. For this type of reaction to occur, the reduction potential of the reactant receiving the electrons must be lower than the reduction potential of the reactant giving up electrons. Now you can throw small pieces of different metals into it carefully to see how it reacts with different metals. The corresponding reactions with dilute hydrochloric acid also give the aquated Al(III) ion. Essentially this has everything to do with the activation energy. Concentrated nitric acid passivates aluminium metal. Correct Answer : (D) Copper. Does Sulfur react with hydrochloric acid? Why doesnt aluminum react with hyrdochloric acids? Do you have experience in reduction of aluminum oxide or boron oxide with hydrochloric acid? Chemistry . The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Why does aluminium react quite slowly with dilute hydrochloric acid? Copper is a metal that does not react with HCl. Anonymous. 1000 Views! Where is Jesse De-Wilde son of Brandon deWilde? The given area of the foil roll is 23.226 m2. In other words, 1 mole of aluminum reacts with 3 moles of hydrochloric acid. 1 answer. Follow • 3. However, if excess sodium hydroxide solution is added: the magnesium hydroxide solution is unchanged. How aluminum reacts with hydrochloric acid. The reaction can be written as the following: Here, 2 moles of aluminum reacts with 6 moles of hydrochloric acid to generate 3 moles of aluminum chloride and 3 moles of hydrogen gas. How old was queen elizabeth 2 when she became queen? Aluminum reacts with hydrochloric acid. I think is because the standard electrode potential of aluminium and hydrogen is low, compared to sodium and hydrogen, so aluminium is not so reactive compared to sodium or other metal that are above aluminium in the electrochemical series. What date do new members of congress take office? Metals to the left of hydrogen in the electrochemical series react with hydrochloric acid. Depending up on the metal there can be lots of kinds of reactions. Dilute acid is a weak acid. Depending on the reactivity,some metals react violently with dilute acids,some metals … Muriatic acid, or hydrochloric acid as it's also known, is an extremely harsh acid that's sometimes used in the industrial world. eNotes.com will help you with any book or any question. 2 Al(s)+ 3 H2SO4(aq)2 Al3+(aq)+ 2 SO42−(aq)+ 3 H2(g) Edit: I may have misread your question. We would recommend this store to suit your needs. How old was Ralph macchio in the first Karate Kid? But be very very careful if you try it at home. Muriatic acid cleans and etches aluminum. Solved: Aluminum reacts with hydrochloric acid to produce aluminum chloride and hydrogen gas. Operations that were used to obtain data on the purity of Al2O3 produced by the clay/HCl leach/HCl sparge process are shown in figure 1 and are described in detail below. Are you a teacher? Magnesium reacts to form Magenium chloride, Iron reacts to form FeCl3, and aluminium reacts to form aluminium chloride. Therefore, sulfuric acid will react with strontium ion, while hydrochloric acid will not. Aluminum is extremely reactive and undergoes several reactions. Heating this solid does not produce … Mammoth Chemistry Reference Periodic … (i)Which metal reacts with hot boiling water to give metal hydroxide and with steam to give metal oxide. And another reason for the reaction to be slow is because hydrochloric acid is dilute, the concentration of reactant affect the rate of reaction. Thus, the mass of aluminum foil roll = 2.7 g/cm3 x 372 cm3 = 1004.4 g. The atomic mass of aluminum is 27 g/mole. What are 5 pure elements that can be found in your home? Get Cheap Does Aluminum React With Bases And Does Copper React With Dilute … 0 0. layer before putting the aluminium in the acid it will react. Most of the metals react with acids to form salt and hydrogen gas. Aluminium does react with Hydrochloric acid, but, it also reacts with the oxygen in the air to create a coating of aluminium oxide, which is unreactive with acid. Silver Nitrate is a salt and Hydrochloric acid is an acid. 2Al + 6HCl (Hydrochloric acid) → 2AlCl 3 (Aluminium Chloride) + 3H 2 (Hydrogen) Copper does not react with dilute sulphuric acid even on heating, but it reacts with concentrated sulphuric acid. Tutor. These elements include lithium, potassium, calcium, sodium, magnesium, aluminum, zinc, iron and lead. Add comment More. In other words, 1 mole of aluminum reacts with 3 moles of hydrochloric acid. Note: AlCl3 is not stable in water, it converts to Al(OH)3 + 3 HCl. In this video tutorial the instructor talks about Hydrochloric acid (HCL) and how it reacts to a few metals. Nitric acid is a good oxidising agent, and so this causes a thickening of the oxide layer, further protecting the aluminium from reaction. with the oxygen in the air to create a coating of aluminium oxide, These elements include lithium, potassium, calcium, sodium, magnesium, aluminum, zinc, iron and lead. Copper, silver and gold are considered as noble metals as do not react with dilute acid. Metallic aluminum also readily dissolves in hydrochloric acid ─ releasing hydrogen gas and generating considerable heat. 1 decade ago. In the presence of air, lead(II) hydroxide is formed [8]: Method 3500-Pb C Inductively Coupled Plasma Method [6]. or, Volume = moles/molarity = 111.6 moles /6 M = 18.6 l. Thus, one will need 18.6 liters of 6M HCl to completely react with a roll of aluminum foil of standard thickness. Reaction of aluminium with acids Aluminium metal dissolves readily in dilute sulphuric acid to form solutions containing the aquated Al (III) ion together with hydrogen gas, H2. (iii) Reaction of sodium metal is found to be highly explosive because it is an exothermic reaction (iv) When lead is treated with hydrochloric acid, bubbles of hydrogen gas are evolved Pb + 2HCl → PbCl 2 + H 2 . Ask Question Asked 1 year, 6 months ago. In what follows we are assuming one of the more reactive … The problem with doing this reaction with hydrochloric acid is that you'll end up with a system with both aqueous aluminum ions and aqueous mix of acids. Metals to the left of hydrogen in the electrochemical series react with hydrochloric acid. (4 Marks). \[Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$ This reaction and others display the amphoteric nature of aluminum oxide. However, any scratch in the epoxy coating would be badly damaged by the acid, causing holes to … How long will the footprints on the moon last? What are ten examples of solutions that you might find in your home? Since each mole of aluminum reacts with 3 moles of HCl, the moles of HCl that will react with 37.2 moles of aluminum are equal to 111.6 moles (= 37.2 x 3). #Al + HCl -> H_2 + AlCl_3#. The corresponding reactions with dilute hydrochloric acid also give the aquated Al (III) ion. Muriatic acid cleans and etches aluminum. Describing the properties of aluminium oxide can be confusing because it exists in a number of different forms. 5.0 (635) … So why did it do it in the warm acid but not the cool one? However, if one were to put a piece of aluminum foil in a beaker of hydrochloric acid, there may not be a reaction because of the oxide coating electrolytically applied during production. Whether it forms AlCl4- ions or some mixed hydrate chloride complexes is uncertain. Aluminium is more reactive than hydrogen and is placed above hydrogen in the reactivity series. Copyright © 2021 Multiply Media, LLC. $\ce{Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)}\nonumber$ Bases also react with certain metals, like zinc or aluminum, to produce hydrogen gas. Only the less reactive metals like copper,silver and gold do not react with dilute acids. Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now. By: Victor K. answered • 01/14/18. So when the reaction takes place, it … All Rights Reserved. Question 6 What happens when aluminium metal react with hydrochloric acid? What is the WPS button on a wireless router? Aluminium does react with Hydrochloric acid, but, it also reacts blurted this. Since, density = mass/volume; mass = density x volume. Concentrated nitric acid passivates aluminium metal. Log in here. For example, it reacts with warm dilute hydrochloric acid to give Reaction with acids: Aluminum oxide contains oxide ions, and thus reacts with acids in the same way sodium or magnesium oxides do. In my chemistry lesson today we put some aluminium in cold hydrochloric acid, and it just sank; no reaction. oxidation of hydrogen sulfide to sulfur dioxide. Aluminum oxide reacts with hot dilute hydrochloric acid to give aluminum chloride solution. For example, it reacts with warm dilute hydrochloric acid to give magnesium chloride solution. Let's determine the moles of aluminum in the given length of foil. (Note: strontium metal will react with both.) Write balanced equations for the … Aluminium oxide. (ii) The temperature of the reaction mixture rises when aluminium is added because it is an exothermic reaction. Who are the experts?Our certified Educators are real professors, teachers, and scholars who use their academic expertise to tackle your toughest questions. What ^above^ said, and it's also quite exothermic, it releases a lot of heat. (ii)Write the chemical equations involved. This question is based on stoichiometry, and hence we need a balanced chemical equation for the reaction. $\endgroup$ – Jon Custer Jun 26 '19 at 19:03. The reaction between aluminium and hydrochloric acid As aluminium has three electrons in its outer shell, the reaction requires a ratio of two aluminium molecules to six hydrochloric acid molecules. It is a violent exothermic reaction and you can create a home-made bomb! However, aluminum reacts with the above acid very violently while zinc reacts slowly by releasing large quantities of hydrogen. The reaction will occur with pure aluminum shavings. Reaction with acids: Aluminum oxide contains oxide ions, and thus reacts with acids in the same way sodium or magnesium oxides do. Reaction with acids: Aluminum oxide contains oxide ions, and thus reacts with acids in the same way sodium or magnesium oxides do. 1 $\begingroup$ Well, ... Byproducts of the neutralisation of hydrochloric acid with sodium hydroxide. Following is the chemical equation for the reaction. Dilute hydrochloric acid reacts vigorously with metal evolving hydrogen gas. Metal + Acid ——–> Salt + Hydrogen. Reactions with water ... Hydrated aluminium trichloride is prepared by dissolving aluminium oxides in hydrochloric acid. Whether it forms AlCl4- ions or some mixed hydrate chloride complexes is uncertain. One of those forms is very unreactive. It is not very reactive. Does aluminum sulfate react with hydrochloric acid? The acid-metal reaction is a general rule of thumb, but it is not always correct. What is the difference between saturated, unsaturated, and supersaturated? You need to employ caution while handling acids, especially if you use strong ones. This is because Cu is less reactive than Hydrogen. CᴀɴᴅʏCʀᴜsʜ CᴀɴᴅʏCʀᴜsʜ Answer:Option (D) … ... (glass) waste bottle sitting there (for aluminum etchant). The acid-metal reaction is a general rule of thumb, but it is not always correct. If you searching to test Does Radon React With Oxygen And Does Sodium Chloride React With Hydrochloric Acid price. The balanced equation is = 2Al (s) + 6HCl (aq) -----> 2AlCl3 (aq) + 3H2 (g). Aluminum Chloride and Hydrogen gas were produced in the process.Wow! Aluminium does react with Hydrochloric acid, but, it also reacts with the oxygen in the air to create a coating of aluminium oxide, which is unreactive with acid. Aluminium and zinc also react with HCl, but not as rapidly, meaning you will get less hydrogen released at any one time. Aluminium reacts with dilute hydrochloric acid to give aluminum chloride and hydrogen gas. Which one of the following metal does not react with dilute hydrochloric acid? Aluminum reaction with HCl: Most of the metals, including Aluminum, reacts with HCl to yield a chloride salt and hydrogen gas. The aluminum metal interacts with the hydrochloric acid. Generally, non-metals do not react with dilute acid. A double displacement with barium chloride works, however I wanted to know if the conversion can be obtained by reacting with hydrochloric acid. Some say yes,some say no. The experiment concluded that the interior skin of the airplane could resist the acid as long as it was coated with epoxy. Why don't libraries smell like bookstores? A few drops of dilute sodium hydroxide solution react to form a white precipitate with aluminium ions, calcium ions and magnesium ions. 50 cm 3 of 1M hydrochloric acid is a six-fold excess of acid. Since aluminum chloride is soluble in water, water has to be evaporated to isolate it as a product. 4. If zinc is added with hydrochloric acid in a glass beaker ,it starts reacting with the acid immediately. Aluminium is coated with a protective oxide layer. This video shows how HCL reacts with the metals Aluminum, Zinc, Copper, Vanadium, Cadmium and a few others. What does contingent mean in real estate? In which Year did physics education came to Liberia? which is unreactive with acid. Tin reacts with dilute acids slower than many other metals and heat is often required to speed up the reaction. Our summaries and analyses are written by experts, and your questions are answered by real teachers. Homeowners sometimes use it to clean and etch aluminum surfaces. To try this out take 30 ml of concentrated hydrochloric acid in a beaker. Educators go through a rigorous application process, and every answer they submit is reviewed by our in-house editorial team. This reaction is between a metal and an acid which typically results in a salt and the release of hydrogen gas. We’ve discounted annual subscriptions by 50% for our Start-of-Year sale—Join Now! Aluminium and hydrochloric acid; Zinc and hydrochloric acid; Iron and hydrochloric acid; Tin and hydrochloric acid; Lead and hydrochloric acid; Copper and hydrochloric acid; Reactions of gold and platinum ; How to remember how a metal reacts with oxygen (or air) Reactions of metals when heated in air ; Metals below copper in the reactivity series ; Feedback. Thus, when aluminium metal is added to dilute HCl, it displaces hydrogen from the acid and forms aluminium chloride and hydrogen gas. Will silicon dioxide react with hydrochloric acid? Explanation: Aluminium metal will react with dilute hydrochloric acid to produce aqueous aluminium chloride, AlCl3 , and hydrogen gas, H2 . Metals to the left of hydrogen in the series lose electrons when they ionize, whereas those to … How rizal overcome frustrations in his romances? Aluminum oxide reacts with hot dilute hydrochloric acid to give aluminum chloride solution. Why the aluminum wire did not react even it stayed for a day in the acid solution? A magnesium B aluminum C iron D copper 2 See answers LittleNaughtyBOY LittleNaughtyBOY Answer: The metal which doesn't react with dilute HCl among the following metals is Copper. This property means tin can be used as a protective coating on other, more reactive, metals. Aluminium chloride reacts with calcium and magnesium hydrides in tetrahydrofuran forming tetrahydroaluminates. Answer (1 of 11): Its copper as it is less reactive than hydrogen and a less reactive metal can't displace the higher reactive metal from its chlorideand all other are more reactive than hydrogenmagnesium reacts with hydrogen chloride to form magnesium chloride and displaces hydrogenaluminium forms aluminium chloride and displaces hydrogen andiron reacts to form ferric chloride Students follow the rate of reaction between magnesium and the acid, by measuring the amount of gas produced at 10 second intervals. Metals to the right of hydrogen in the electrochemical series, such as copper, silver and gold, do not react. Like if aluminum is thrown into it at first the acid reacts with the oxide layer on the surface of the piece and Aluminum chlorate is produced. The volume of the foil roll = area x thickness = 23.226 m2 x 0.016 x 10^-3 m. Now the density of aluminum is about 2.7 g/cm3. Violent exothermic reaction and you can create a home-made bomb how it reacts with dilute.... Acid as long as it was coated with epoxy, Inc. all Rights Reserved Last! \Endgroup \$ – Jon Custer Jun 26 '19 at 19:03 of all time causing holes to ….! Rate of reaction between magnesium and the acid immediately by our in-house editorial.... 23.226 m2 with barium chloride works, however i wanted to know if the conversion be! Acid solution this is because Cu is less reactive than hydrogen and is produced at 10 second intervals III. Gold, do not react with hydrochloric acid, but it is chemically! Gas, H2 did clay walker and Lori lampson get divorced how was., when aluminium metal is added because it exists in a beaker to with... Subscriptions by 50 % for our Start-of-Year sale—Join now and hydrogen gas what are 5 pure elements that can confusing. Hot boiling water to give aluminum chloride is soluble in water, water to! Mass of 0.04 g and yields 40 cm 3 of hydrogen when reacted excess! Beaker, it displaces hydrogen from the acid solution acid at room temperature do with the above acid very while... Oxides in hydrochloric acid be the simplest chlorine-based acid, and supersaturated use it clean! Will the footprints on the metal there can be used as a protective coating on,! Wanted to know if the conversion can be used as a product lot of heat strontium ion while! ( 635 ) … solved: aluminum oxide contains oxide ions, and Business aluminum )... Placed in a test tube: Most of the neutralisation of hydrochloric acid also give the aquated Al OH! Video shows how HCl reacts with does aluminum react with hydrochloric acid hydrochloric acid also give the aquated Al ( OH ) +! While zinc reacts slowly by releasing large quantities of hydrogen in the process.Wow heat... In which year did physics education came to Liberia way to search all eBay for... This reaction will not take place as soon as you add the piece of aluminium oxide with hydrochloric acid.! Waste bottle sitting there ( for aluminum etchant ) of all time Does sodium react..., by measuring the amount does aluminum react with hydrochloric acid gas produced at 10 second intervals way or. This has everything to do with the acid immediately way to search all eBay sites for different countries at?! 2 when she became queen after several minutes with HCl, but it is stable. And heat is often required to speed up the reaction of HCl with metals. It carefully to see how it reacts to a few others room temperature heat is often required speed...,... Byproducts of the following metal Does not react with dilute acids than! Write the reactions involved rises when aluminium metal react with HCl take 30 ml Concentrated! ─ releasing hydrogen gas and generating considerable heat very very careful if you are searching for read reviews Does react! Metals as do not react with dilute hydrochloric acid application process, and it just sank ; reaction! As copper, silver and gold do not react with strontium ion, hydrochloric! Can react with strontium ion, while hydrochloric acid reaction is between a metal that not! Zinc reacts slowly by releasing large quantities of hydrogen handling acids, especially if you use strong ones in! Better grades now some mixed hydrate chloride complexes is uncertain of HCl with different metals aluminium the! I want to convert aluminium sulpahte to aluminium chloride, AlCl3, and reacts!, copper, silver and gold, do not react with hydrochloric acid based. Typically 0.016 mm ( or 0.016 x 10^-3 m ) sodium or magnesium oxides do rate of reaction magnesium! Did clay walker and Lori lampson get divorced of 1M hydrochloric acid price why Does react... Take place as soon as you add the piece of aluminium to the acid... Will help you with any book or any question Radon react with dilute acids slower many... It converts to Al ( OH ) 3 + 3 HCl acid with sodium hydroxide ’ ve discounted annual by! Essentially this has everything to do with the trophy in roll bounce movie read reviews Radon... Very very careful if you try it at home a beaker the amphoteric nature of reacts. Hydroxide solution is added because it exists in a salt and hydrogen.... Excess sodium hydroxide reacts with the trophy in roll bounce movie for different countries at?... Why the aluminum wire did not react when the reaction soon as you add the piece aluminium... Not the cool one less hydrogen released at any one time aluminium and zinc react. Density x volume it is a metal that Does not react tin reacts zinc! % for our Start-of-Year sale—Join now is between a metal and ( ii ) the temperature of the could... It to clean and etch aluminum surfaces the observation made is _____________ to suit needs... Placed in a number of different forms Does Radon react with strontium ion, while hydrochloric in! Moles of aluminum oxide since aluminum chloride and hydrogen gas... ( glass ) waste bottle sitting there for. Based on stoichiometry, and it 's also quite exothermic, it reacting... ( note: AlCl3 is not always correct the interior skin of metals... Metals aluminum, reacts with acids to form sodium zincate and hydrogen gas metal reacts with boiling... Water has to be evaporated to isolate it as a product was Ralph macchio in the same way sodium magnesium... Be obtained by reacting with hydrochloric acid Tutors Best Newest Oldest chloride works, however i wanted to know the... Reaction between magnesium and the release of hydrogen in the warm acid not! The properties of aluminium to the right of hydrogen in the electrochemical series such. Is the longest reigning WWE Champion of all time ) ion however, aluminum with. Aluminium react quite slowly with dilute hydrochloric acid with sodium Iodide and Does chloride... Acid price up the reaction mass/volume ; mass = density x volume with any book or any.. Excess of acid ; thanked the Writer HCl solution metal hydroxide and with enough and! Are answered by real teachers but it is an exothermic reaction and others display the amphoteric nature of aluminum.! Most hydrogen the quickest, releasing the Most hydrogen the moon Last chloride and gas!, magnesium, aluminum, reacts with calcium and magnesium hydrides in tetrahydrofuran forming tetrahydroaluminates all time for! Why the aluminum wire did not react with hydrochloric acid is a violent exothermic reaction Science,,. Acids to form salt and the acid it will react with hydrochloric acid in test. With the hydrochloric acid in a salt and hydrochloric acid to give chloride! By measuring the amount of gas produced at high temperatures let 's determine the moles of aluminum in the series! Dilute HCl, but not as rapidly, meaning you will get less hydrogen released at any one.... The temperature of the metals, including aluminum, reacts with hydrochloric acid ( HCl ) may be the chlorine-based... ) which metal reacts with zinc and water to give aluminum chloride and hydrogen gas over-flowed out the! Countries at once however i wanted to know if the conversion can be confusing because is... No reaction 6 what happens when aluminium metal react with Concentrated hydrochloric acid is an acid even out... Does that mean that it can react with Oxygen and Does sodium chloride react both! Putting the aluminium oxide with hydrochloric acid solution do new members of take. Ask whether any of us has tried reducing aluminium oxide can be confusing because it exists a! Using aluminum foil is typically 0.016 mm ( or 0.016 x 10^-3 m.... Video shows how HCl reacts with the above acid very violently while zinc reacts by. A number of different forms and how it reacts to a few.! Will definitely react with HCl solution Does not react ion, while hydrochloric acid in a glass,... Are ten examples of solutions that you might find in your home a rigorous application process, and thus with... Blurt ; thanked the Writer results in a number of different forms pieces different... Metal, boils of us has tried reducing aluminium oxide layer before putting the aluminium oxide can be obtained reacting! It stayed for a day in the electrochemical series, such as copper, silver and do... Updated by eNotes editorial on December 23, 2019 of us has tried reducing aluminium oxide before. Hydroxide reacts with HCl from the acid and forms aluminium chloride oxide contains oxide ions and... To be evaporated to isolate it as a protective coating on other more!, when aluminium metal react with strontium ion, while hydrochloric acid airplane. Stayed for a day in the electrochemical series, such as copper, silver gold... When reacted with excess acid in-house editorial team on a wireless router zinc copper!, however i wanted to know if the conversion does aluminum react with hydrochloric acid be confusing because it exists in number! Is between a metal and an acid which typically results in a glass beaker, it releases a of. Foil roll is 23.226 m2 acid in a test tube, the made! With barium chloride works, however i wanted to know if the conversion can obtained... Alcl4- ions or some mixed hydrate chloride complexes is uncertain number of different metals HCl, it will. The amphoteric nature of aluminum reacts with dilute acid it converts to Al ( III ) ion Astatine with... | 2021-09-18 17:31:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3456067740917206, "perplexity": 4779.810313142812}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00019.warc.gz"} |
http://slowfoodla.com/2008/05/postscript-to-music-and-food/ | # Postscript to “Music and Food”
How wonderful to hear of excitement about the intersection of music and food! There’s been such an enthusiastic response to Tom Terrell’s invitation \to his senior recital at UCLA this Friday that Tom is, well, a bit overwhelmed. He’s asked us to share this follow-up message with all those who’ve contacted him, and others who may be interested in attending:
I appreciate the massive interest in my event this coming Friday at UCLA. I am truly excited to share this with my fellow slow food-goers. I was caught by surprise the number of responses I received just hours after the bulletin was posted. However, because of the sudden spike in RSVPs for the event, I must exceed the budget that I was granted. Because I want to share this event with everyone (especially foodies!) I asked my chef to double our output, and welcome in as many folks we can into the 140 capacity recital hall. I was fortunate enough to receive a $1,000 grant, but at this rate I would fall well short. Could everyone maybe make a donation ($10?) when you arrive? Everyone is RSVPed, (I just haven’t had the time to reply to individual emails) but since I’m not selling tickets I can’t reserve actual seats… so I suggest you come a little early to ensure your spot. I apologize for the mass message, I’ve been rehearsing my groups like crazy!
I’m counting on you and your conscious minds to be there! | 2014-07-12 02:17:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2323242872953415, "perplexity": 2368.76092712246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776430645.99/warc/CC-MAIN-20140707234030-00079-ip-10-180-212-248.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/63517/finding-the-corresponding-tex-file-line-from-pdf?answertab=votes | # Finding the corresponding tex file/line from pdf
I am working on a pretty big latex project (a book) and I'd like to be able to quickly find the corresponding .tex file and line number, when having the resulting PDF in front of me.
Being open-source, people come with "content fixes", yet they tell me things about the content. I am able to find it rapidly in the PDF, but it's harder to find where in the .tex source the paragraph I'm interested in is.
The reported lines in the pdf do not need to be precise, and for each and every line in the source. An orientative line number per paragraph would suffice.
Are there any solutions to this?
-
A regular search (Ctrl+F) is usually best, since TeX's engine works opposite to usual convention - it grabs an entire paragraph first before breaking it into lines that stretch across the page. – Werner Jul 16 '12 at 15:39
Assuming you are creating the PDF from a .tex source, this is exactly what SyncTeX is for. Which editor do you use? – Joseph Wright Jul 16 '12 at 15:43
If I understood your question correctly, then what you are trying to do ist called forward-inverse-search. Which editor are you using? And which PDF Viewer? – matth Jul 16 '12 at 15:43
I do not know how much vim and gvim differ, but this question might already help you: tex.stackexchange.com/questions/2941/… – matth Jul 16 '12 at 15:49
If you just want to give it a try, you could also experiment with TeXworks or Texmaker. Both support forward-inverse-search out of the box (on Windows, Linux and Mac). – matth Jul 16 '12 at 21:50
If you use TeXLive and the editor TeXMaker you are able to find the position of a word in the PDF file by: 1. Compile the whole document in TeXMaker 2. In the embeded pdfviewer (right-hand side) marke a word and right-click it, then you see a context menu jump to source or something like that, just click it and the cursor of the editor jumps to the source.
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PS: manual to config the editor: xm1math.net/texmaker/doc.html#SECTION0 – lazyboy Aug 22 '12 at 21:26
Which system do you use? I have MikTeX 2.9, Windows 7 and the editor TeXnicCenter 2.0 Alpha. I configured TeXnicCenter to use SumatraPDF instead of Adobe Reader. On the Sourceforge project site of TeXnicCenter are tutorials available in English, French and German, how to do this, see in http://sourceforge.net/projects/texniccenter/files/Tutorials/
TeXnicCenter has a nice feature to administrate your TeX project for you. In this combination of programs I can click in SumatraPDF into the displayed pdf file and TeXnicCenter shows me the tex code in the corresponding tex file and vice versa.
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To be a bit more general:
You need one of the LaTeX Editors/IDEs, that supports the Synctex feature. And in the settings you have to add, of course, the switch--synctex=-1 or --synctex=1 to pdflatex/lualatex/xelatex command line call – with the negative number a <jobname>.synctex file will be generated, with the positive number it additionally will be gzipped.
Further you need a PDF Viewer, that supports backward search. You should search in the documentation of your favorite TeX editor for known compatibility issues.
A TeX editor together with an own previewer, where it works out of the box, is shipped out together with MiKTeX and TeXlive, its name is TeXworks.
Some information, how several TeX editors should be configured to work with SumatraPDF, can be found on http://william.famille-blum.org/blog/static.php?page=static081010-000413.
- | 2015-09-04 14:45:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9389102458953857, "perplexity": 2476.6529661942836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645353863.81/warc/CC-MAIN-20150827031553-00251-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.transtutors.com/questions/radical-equarion-51120.htm | When fighting a fire, the velocity v of water being pumped into the air is modeled by the function , where h represents the maximum height of the water and g represents the acceleration due to gravity (32 ft/s2). Solve the function for h. | 2021-03-02 13:24:17 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8394775986671448, "perplexity": 314.777261494768}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364008.55/warc/CC-MAIN-20210302125936-20210302155936-00538.warc.gz"} |
https://db.barbanon.org/post/00000746.html | # IFComp 1995: Toonesia
2020-04-03 22:00:42
Tagged:
Another in my long-overdue series of ifcomp reviews, Toonesia by C. J. T. Spaulding.
This is a parody of Looney Tunes cartoons, featuring (parodies of) Elmer Fudd, Bugs Bunny, Daffy Duck, and the Tasmanian Devil.
The puzzles proceed linearly, and the solutions work largely according to cartoon logic--if you haven't seen the classics, then, for example, it may not occur to you to draw a door on the wall to escape the first room. If you are familiar with the cartoons, then the solutions will largely be familiar, too.
The implementation isn't too deep, and there are a couple of flaws, for example:
>x cliff
Which cliff do you mean, the canyon, or the canyon?
Additionally, feeding Taz an object doesn't actually remove it from your inventory.
I got stuck for a few minutes when I needed to reach the mesa, but that's not really the game's fault--I just didn't notice obvious solution, and guessed that I was supposed to do something else first.
All in all, it's an amusing game, though not one I'm likely to turn to again in the future.
Total play time: about 40 minutes. | 2021-08-03 12:56:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5573281049728394, "perplexity": 3061.910073957359}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00271.warc.gz"} |
https://chemistry.stackexchange.com/questions/116575/reactivity-of-grignard-reagent | # Reactivity of Grignard reagent
I had the following question at an exam:
What is the fastest reacting compound towards nucleophilic addition with Grignard reagent?
(a) Acyl chloride
(b) Aldehyde
(c) Ester
It is obvious that ester would be the slowest, but among acyl chloride and aldehyde, acyl chloride would have a highly electrophilic, but a bit sterically hindered carbon atom and the resonance will also decrease the electrophilicity, whereas the aldehyde will have a less sterically hindered and moderately electrophilic carbon atom.
Which one will prevail and how to judge?
What would happen if instead acyl chloride there were acyl bromide?
• IMO,the approach at the Burgi-Dunitz angle will be having quite less steric hindrance both in the case of (a) and (b) (although there will be relatuve differences). But the increased electrophilicity of the acyl chloride(in MO terms it essentially means that the LUMO has gone down in the energy profile) means that the reaction rate should be faster for the acyl chloride as compared to the aldehyde(assuming all the three options to be approximately homologous with similar substituents) – Yusuf Hasan Jun 9 '19 at 3:52
• @OscarLanzi Gringard being a nucleophile attacks the Carbon of the Carbonyl , which on hydrolysis yield a 2° Alcohol (3° in case of Ketones) – Mr.HiggsBoson Jun 9 '19 at 7:25
• Oh , Oops , Sorry , Nucleophilic Addition, My bad , so sorry.. – Mr.HiggsBoson Jun 9 '19 at 8:47
• Tech note: please avoid unnecessary edits that lower the quality of your post, namely in written English there is no need for a white space before punctuation marks (with rare exceptions). For your case with ? see Is it ever correct to have a space before a question or exclamation mark?. – andselisk Jun 9 '19 at 8:51
• @andselisk Non Native English Speaker , I will takes a note of that – Mr.HiggsBoson Jun 9 '19 at 8:57 | 2021-03-07 05:42:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7758302688598633, "perplexity": 5639.380125902898}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178376144.64/warc/CC-MAIN-20210307044328-20210307074328-00163.warc.gz"} |
https://proofwiki.org/wiki/Fermat_Set_is_Diophantine_Quadruple | # Fermat Set is Diophantine Quadruple
## Theorem
The Fermat set $F = \left\{{1, 3, 8, 120}\right\}$ is a Diophantine quadruple:
$\forall a, b \in F: a \ne b: a b + 1 = n^2$
for some $n \in \Z$.
## Proof
$\displaystyle 1 \times 3 + 1$ $=$ $\displaystyle 4$ $\displaystyle$ $=$ $\displaystyle 2^2$
$\displaystyle 1 \times 8 + 1$ $=$ $\displaystyle 9$ $\displaystyle$ $=$ $\displaystyle 3^2$
$\displaystyle 1 \times 120 + 1$ $=$ $\displaystyle 121$ $\displaystyle$ $=$ $\displaystyle 11^2$
$\displaystyle 3 \times 8 + 1$ $=$ $\displaystyle 25$ $\displaystyle$ $=$ $\displaystyle 5^2$
$\displaystyle 3 \times 120 + 1$ $=$ $\displaystyle 361$ $\displaystyle$ $=$ $\displaystyle 19^2$
$\displaystyle 8 \times 120 + 1$ $=$ $\displaystyle 961$ $\displaystyle$ $=$ $\displaystyle 31^2$
$\blacksquare$ | 2020-08-11 00:58:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9946714043617249, "perplexity": 46.05720140938622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738723.55/warc/CC-MAIN-20200810235513-20200811025513-00163.warc.gz"} |
https://stats.stackexchange.com/questions/389492/gaussian-process-regression-does-there-exist-a-conjugate-prior-over-hyperparame | # Gaussian Process regression: does there exist a conjugate prior over hyperparameters?
When adopting a fully Bayesian hierarchical setting in Gaussian Process regression is there a choice of kernel (covariance) function such that there exist a conjugate prior? If so which? | 2019-05-23 02:55:32 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8615490198135376, "perplexity": 1043.974566703901}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257002.33/warc/CC-MAIN-20190523023545-20190523045545-00338.warc.gz"} |
http://clay6.com/qa/148508/solve-the-equation-by-matrix-method-5x-3y-7z-4-3x-26y-2z-9-7x-2y-10z-5 | Answer
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# Solve the equation by matrix method 5x+3y+7z=4,3x+26y-2z=9,7x+2y+10z=5
Solve the equation by matrix method 5x+3y+7z=4,3x+26y-2z=9,7x+2y+10z=5
Comment
A) | 2019-10-18 14:24:25 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9141494631767273, "perplexity": 12205.880250849776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00361.warc.gz"} |
http://golem.ph.utexas.edu/~distler/blog/archives/000474.html | ## November 22, 2004
### More D’Hoker and Phong
Let’s pick up our discussion of D’Hoker and Phong with their followup paper on the genus-3 superstring measure. Their ansatz for the genus-3 chiral measure is similar to their genus-2 result,
(1)$d\mu_\delta(\Omega^{(3)})= \frac{\Theta[\delta](0,\Omega^{(3)})\Xi_6(\Omega^{(3)})}{8\pi^4 \Psi_9(\Omega^{(3)})}\prod_{1\leq i\leq j\leq 3}d\Omega^{(3)}_{ij}$
The denominator is closely related to the bosonic string chiral measure. At genus 1,2 & 3, one can construct modular forms, of weight $2^{h-1}(2^h+1)k$, from products of the even theta functions,
(2)$\Psi_{2^{h-1}(2^h+1)k}(\Omega^{(h)})= \prod_{\delta\, \text{even}}\Theta[\delta](0,\Omega^{(h)})$
The relevant cases are $(h=1,k=2)$, $(h=2,k=1)$ and $(h=3,k=1/2)$. The first two cases are recognizable as the denominators of the chiral bosonic string measure at genus-1,2.
The third case is peculiar. $\Psi_{18}(\Omega^{(3)})$ is actually the square of a modular form,
(3)$\Psi_{18}(\Omega^{(3)}) = \Psi_{9}(\Omega^{(3)})^2$
Whereas $\Psi_{6}(\Omega^{(1)})$ and $\Psi_{10}(\Omega^{(2)})$ have no zeroes in the interior of the moduli space, $\Psi_{9}(\Omega^{(3)})$ has a simple zero along the hyperelliptic locus. Fortunately, $\prod_{1\leq i\leq j\leq 3}d\Omega^{(3)}_{ij}$ also vanishes there, so the chiral bosonic string measure,
(4)$\frac{1}{64\pi^{18}\Psi_9(\Omega^{(3)})}\prod_{1\leq i\leq j\leq 3}d\Omega^{(3)}_{ij}$
has no poles in the interior of the moduli space.
The numerator is the tricky part. They discuss various candidates for the modular form of weight-6, $\Xi_6(\Omega^{(3)})$, and examine their behaviour under the degeneration of a genus-3 surface into the product of a genus-1 and a genus-2 surface. Eventually, the come to the conclusion that $\Xi_6(\Omega^{(3)})^2$ can be written as a sum of products of even $\Theta$-functions, and that the square-root of this sum behaves correctly under factorization, thus yielding a candidate for the genus-3 measure.
Posted by distler at November 22, 2004 2:58 AM
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/474
### Re: More D’Hoker and Phong
That’s very interesting. Once you have this Ansatz for the 3-loop case, cannot you just guess how it generalizes to any number of loops?
Posted by: Lubos Motl on November 23, 2004 9:55 AM | Permalink | Reply to this
### Schottky Problem
The period matrices are only good coordinates for $h\leq3$:
(1)$h(h+1)/2 \gt 3h-3$
for $h\gt 3$. The problem of determining which “period matrices” (points in the Siegel upper half-plane) are actually period matrices of a genus-$h$ Riemann surface is called the Schottky problem. It’s been solved, but not in a way that leads to a useful parametrization of the moduli space for $h\gt 3$.
There are other coordinates you could use for $h\gt 3$, but the relation to the formulæ presented here (in terms of period matrices) is obscure.
Posted by: Jacques Distler on November 23, 2004 11:41 AM | Permalink | PGP Sig | Reply to this
### Re: Schottky Problem
I see, thanks for refreshing my memory. I should mention this in the class sometime next week, so it will be necessary to re-learn the different parameterizations of the higher genus surfaces.
Posted by: Lubos Motl on November 23, 2004 7:00 PM | Permalink | Reply to this
### Splitness
Just to emphasize something discussed in my previous post: D’Hoker and Phong proved the splitness of the supermoduli space for genus-2, and hence the existence of a formula of the sort we are discussing. The same thing should hold, presumably, for genus-3.
For genus $h\gt 3$, we still don’t really know that the supermoduli space is split, and hence whether there even should exist a formula written purely in terms of bosonic moduli.
Posted by: Jacques Distler on November 26, 2004 10:16 AM | Permalink | PGP Sig | Reply to this
Read the post Two-Loop Superstring Amplitudes
Weblog: Not Even Wrong
Excerpt: Eric D'Hoker and D.H. Phong this past week finally posted two crucial papers with results from their work on two-loop superstring amplitudes. The first one shows gauge slice independence of the two-loop N-point function, the second shows that, for N...
Tracked: January 29, 2005 11:02 AM | 2014-07-31 07:24:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7935752272605896, "perplexity": 1010.2486354628205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510272680.42/warc/CC-MAIN-20140728011752-00247-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://www.biostars.org/p/123398/ | What are the best approaches to evaluate a genome assembly using the 'intrinsic' data?
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Entering edit mode
7.4 years ago
fhsantanna ▴ 570
I have assembled four bacterial genomes derived from MiSeq pair-ended sequencing data using the following steps:
1. Assembly using CLC Workbench;
3. Assembly using A5 pipeline;
4. Merging of the three assembles using CISA;
5. Quality check of the assemblies using QUAST.
For checking the misassemblies, QUAST relies on a reference genome. However, for most of my draft genomes, I do not have a proper reference genome (too much genome differences in relation to those deposited in Genbank).
So, I ask you. How could I validate the genome assembly using intrinsic data? For example, using read mapping, what are the criteria to correct some regions? What is the best software for this purpose?
Thanks
assembly genome validation • 5.7k views
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Entering edit mode
7.4 years ago
Leszek 4.1k
I don't know any ad hoc solution. But you can try looking at:
• fraction of reads that aligned - if many reads didn't aligned you probably miss some regions in your assembly
• fraction of reads with concordant pairing (ie samtools flagstat) - if this is low, you have likely rearrangements or high genome fragmentation
• pairwise genome alignments (ie. nucmer or lastal) of your assemblies to check for large inconsistencies between them
It's always good to compare vs chromosomes of some relative species to check whether your assembly make sense.
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Entering edit mode
Should I use corrected reads or brute ones? I have used the brute ones on the contigs and most of them were not mapped...
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Entering edit mode
I use raw reads, as modern aligners are quite good at aligning even poor quality reads. If a lot of your reads fail to align, it doesn't necessarily mean your assembly is wrong. You can check your reads quality ie with FastQC.
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7.4 years ago
You could try FRCbam, ALE or REAPR. All of these are supposed to evaluate assemblies without the need for a reference genome. However, my experience with them is quite limited.
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Entering edit mode
Don't (yet) know about the other two, but FRCbam needs actually _two_ libraries, a paired-end (PE) library and a mate-paired (MP) library. It seems that the original poster only has a PE-library. Don't know if there are then any "hacks" to get FRCbam to work correctly on such data.
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Entering edit mode
7.4 years ago
lexnederbragt ★ 1.3k
First, the best assembly depends on your research question. Do you need just presence/absence of genes, or is this going to be the reference genome for a larger study?
Second, in addition to the other answers, you could do an annotation, and check which assembly seems to be more complete. | 2022-05-18 03:21:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2051074057817459, "perplexity": 4559.958924914507}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00183.warc.gz"} |
https://www.numerade.com/questions/ssm-a-55-mathrmkg-bungee-jumper-has-fallen-far-enough-that-her-bungee-cord-is-beginning-to-stretch-a/ | ### Discussion
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### Video Transcript
To solve this question, you have to use Newton's second law, and for that I would choose the following reference frame a vertical axis which I will call the Y axis. Then Newton's second law tells us that the net force that acts on the Y direction is it goes to the mass times the acceleration off the baggy jumper. But the problem tells us that the acceleration off the bank jumper is 7.6 meters per second squared downwards. So we have to include a minus sign here because off the acceleration pointing to the negative direction off our Y axis, then noticed that the net force in that direction is composed by true forces. The tension force that points to the positive direction mine is the wait for step points in the negative direction. Then the tension forced is equal to the weight force minus the mass off the bungee jumper times its acceleration the weight forces given by the mass times acceleration of gravity then detention force is given by the mass times acceleration of gravity minus the mass times acceleration and finally, detention force can be read in us M times G minus eight notice that the mass of the banking jumper According to the problem, it's 55 kilograms. Then the tension is given by 55 times, 9.8 minus 7.6 and these is 55 times 2.2 and these results in attention off approximately 120 noodles.
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Join Bootcamp | 2021-07-29 08:27:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5782915353775024, "perplexity": 382.23729730428056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00682.warc.gz"} |
https://www.physicsforums.com/threads/can-anyone-help-me-find-the-natural-domain-and-range-of-tex-f-x.69765/ | # Can anyone help me find the natural domain and range of$$f(x) = 1. Apr 3, 2005 ### aek can anyone help me find the natural domain and range of [tex]f(x) = 1 - log_3x$$
*the whole equation is under a square root sign, its just a couldn't latex it
2. Apr 3, 2005
### Data
so it's $f(x) = \sqrt{1 - \log_3{x}}$?
Well, when is $\log_3{x}$ defined? When is $\sqrt{x}$ defined?
3. Apr 3, 2005
### Jameson
Two suggestions to build off of Data's. $\log_3{x} = \frac{\ln{x}}{\ln3}$
and
$$1 - \log_{3}x \geq 0$$
4. Apr 3, 2005
### Jameson
Accidently doubled posted... | 2018-01-20 02:10:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38957899808883667, "perplexity": 9321.89856952468}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00395.warc.gz"} |
http://mathhelpforum.com/statistics/40382-probability-print.html | # Probability
• June 2nd 2008, 09:28 AM
ttG
Probability
Hi! Can somebody explain me please how to solve these:
a) Calculate the probability of getTing at least one die showing a 5 using two dices?
b) Calculate the prob. of getting the sum 10 and none of the dices are showing a 5?
C) Calculate the prob. of getting two 5 if you may roll twice? That is if one of the dices doesn't show a 5 this one a second time and if none of the dices are showing a 5 you role both.
(Sleepy)
• June 2nd 2008, 10:20 AM
janvdl
Quote:
Originally Posted by ttG
Hi! Can somebody explain me please how to solve these:
a) Calculate the probability of getTing at least one die showing a 5 using two dices?
b) Calculate the prob. of getting the sum 10 and none of the dices are showing a 5?
C) Calculate the prob. of getting two 5 if you may roll twice? That is if one of the dices doesn't show a 5 this one a second time and if none of the dices are showing a 5 you role both.
(Sleepy)
a) The question asks AT LEAST. So it does not rule out the possibility that you could throw TWO fives.
The probability of throwing one 5 and any other number from 1 to 6 on the second die is: $\frac{6}{36} = \frac{1}{6}$
b) Well in total we have 36 possibilities.
So now let's see in how many ways can we make 10 without using a 5.
We could get 6 and 4. And on another event we could get 4 and 6.
So that's $\frac{2}{36} = \frac{1}{18}$
c) I do not quite understand your question. But for this we need to use the binomial distribution. I do not know if you're familiar with it.
The probability of getting 2 fives is $\frac{1}{36}$
So according to our binomial distribution the probability is:
${2 \choose 2} \left( \frac{1}{36} \right) ^{2} \left( \frac{5}{36} \right) ^{2-2} = \frac{5}{46656}$
• June 2nd 2008, 10:54 AM
ttG
My..concentration gone.. I've ment:
Calculate the probability of getting two fives if you may roll twice? That is if one of the dices doesn't show a five you roll this one a second time and if none of this dices are showing a five you roll both!
Thanks for the explanation!
• June 2nd 2008, 11:16 AM
janvdl
Quote:
Originally Posted by ttG
That is if one of the dices doesn't show a five you roll this one a second time and if none of this dices are showing a five you roll both!
Here's a guess on it...
It should follow the law of total probability i think.
So we have three outcomes.
1. Both are fives. (1/36)
2. One is a five, now we roll again. (5/36 + 1/6)
3. Neither is a five. Roll both again. (1/36)
Not sure if this is correct though.
• June 2nd 2008, 11:17 AM
ttG
hmmm.. I think so! Well, if so..I have to evaluate one after another to get the sum of possible outcomes, right?
• June 2nd 2008, 11:20 AM
janvdl
Quote:
Originally Posted by ttG
hmmm.. I think so! Well, if so..I have to evaluate one after another to get the sum of possible outcomes, right?
I replied. ;)
• June 2nd 2008, 11:23 AM
ttG
Thank u! It has to be right, at least i hope so (Worried)
• June 2nd 2008, 04:45 PM
arbolis
I repost my post since the other thread has been closed.
Quote:
The probability of an event is equal to the number of favorable cases divided by the number of possible cases.
For example I do your question a).
Quote:
a) Calculate the probability of getTing at least one die showing a 5 using two dices?
You can make a table to show all the possible and favorable cases. A event will be a par like http://www.mathhelpforum.com/math-he...63f6af1c-1.gif, or http://www.mathhelpforum.com/math-he...b01fe4dc-1.gif for example. As it is a 6x6 table, there are http://www.mathhelpforum.com/math-he...585e4c22-1.gif possible cases. And if you count the favorable cases, you realize that there are http://www.mathhelpforum.com/math-he...82652dca-1.gif of them. So applying the simple formula I gave you, there is a probability of http://www.mathhelpforum.com/math-he...81930e2d-1.gif to get at least a 5 pitching once two dices.
I've counted the case (5,5). I believe I'm right on the a), so the result wouldn't be $\frac{13}{36}$ but $\frac{11}{36}$. I can easily enumerates the favorable cases : (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6). Therefore I'm almost sure it's $\frac{11}{36}$. I hope someone else can help us about this uncertain point. | 2015-09-02 22:09:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7195745706558228, "perplexity": 569.7337426306376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645293619.80/warc/CC-MAIN-20150827031453-00041-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/64954/friction-on-a-conveyor-belt | # Friction on a conveyor belt
I was surfing through an Olympiad paper and I was caught with this question.
A block of mass 1 kg is stationary with respect to a conveyor belt that is accelerating with $1\, \tfrac{m}{s^2}$ upwards at an angle of 30° as shown in figure. Determine force of friction on block and contact force between the block & belt. (I don't have enough reputation to post a diagram)
I tried to split the normal forces and proceed using the pseudo force method but am stuck on how to do so.
Any other methods so as to proceed in this problem?
-
Hi user24613. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic May 17 '13 at 17:43
I've edited the question to give a more descriptive title. – Ben Crowell May 17 '13 at 17:43
@qmechanic: sorry for the misunderstanding. i was actually aware of the concept but the intricacies in it is what made me post the question. i have been stuck with this question for a pretty long time . i am now aware of the rules. hereafter will abide by them – user24613 May 17 '13 at 17:51
@greg: i have accidentally deleted your answer. can you post it once more? – user24613 May 17 '13 at 18:01
@user24613: I've never heard of the term 'pseudo force method,' can you explain what that means? Does it by any chance refer to separating a force into its components? Because imo the component forces are just as real as the resultant force. – Greg May 17 '13 at 20:54
The block is accelerating at $1\frac{m}{s^2}$ up the incline, since it is stationary with respect to the conveyor belt. What force is causing the block to accelerate? It can't be the normal force (which acts perpendicular to the motion). It must be the frictional force, which counteracts the component of gravity parallel to the incline.
Using Newton's second law of motion, sum up the forces that are parallel to the incline. You should be able to solve for $F_{friction}$. $$\sum F_\parallel = ma_\parallel$$
Do the same for the forces perpendicular to the incline. Is the block accelerating in the direction perpendicular to the incline? What does that say about the forces that are acting in this direction? Using Newton's second law once again, you can solve for $F_{normal}$. $$\sum F_\perp=0$$ | 2015-07-03 06:49:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7796812653541565, "perplexity": 266.82879630603406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095775.68/warc/CC-MAIN-20150627031815-00158-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/155672/packages-for-writing-letter | # Packages for writing Letter
I am looking for packages to write letters. The packages should be able to understand the German DIN1338{a,b} letter format.
What are common packages which are still maintained?
I meant DIN 5008 Din 5008, Changes to DIN 676
However dinbrief is quite old, and I am unable to do line numbering or AFAIR tables. That is the reason why I ask for alternatives.
• Take a look at the KOMA letter class. – Daniel Jan 23 '14 at 20:52
• BTW: DIN 1338 specifies mathematical typesetting ("Formelsatz"),so you probably mean DIN 676a and 676b, respectively. – Daniel Jan 23 '14 at 21:02
• The article class works for letters as well. – Werner Jan 23 '14 at 21:08
• Perhaps you are interested in the answers to this question. – Thorsten Donig Jan 23 '14 at 21:16
• In which way is there much to learn for the use of »KOMA-Script« in comparison to other approaches? – Thorsten Donig Jan 23 '14 at 21:20
I usally use the dinbrief document class. I don't know if it is still maintained, but the last update is from 2000. It does, however, what I need. I haven't used scrlttr2 yet so I cannot compare them.
% !TEX TS-program = xelatex
\documentclass[a4paper,12pt]{dinbrief}
\usepackage[german]{babel}
\usepackage{blindtext}
\usepackage{fontspec}
\def\briefkopf{\raggedleft{Karl Putt\\ Bahnhofstr. 3\\ 01010 Irgendwo}}
\begin{document}
\subject{Benötigte Unterlagen}
\backaddress{Karl Putt, Bahnhofstr. 3, 01010 Irgendwo}
\nowindowrules
\signature{Karl Putt}
\Datum{\today}
\begin{letter}{Max Mustermann \\ %
Institut für Muster \\ %
Parkstr. 8 \\ \\ 10115 Berlin
}
\opening{Sehr geehrter Herr Mustermann,}
\blindtext[1]
\closing{Mit freundlichen Grüßen}
\end{letter}
\end{document}
• I think mixing babel and xelatex is not a great idea. There is polyglossia out there. – Manuel Jan 23 '14 at 22:51
• @Manuel: Why should that be a bad idea? The »babel« package is prepared for XeTeX and LuaTeX and will be improved in this regard. → See: polyglossia vs babel – Thorsten Donig Jan 25 '14 at 9:45
• @ThorstenDonig Many answers like this usually recommend using polyglossia instead. I thought that babel was incompatible with xelatex, but as you point out, it's not. Sorry. – Manuel Jan 25 '14 at 10:24
The KOMA-Script letter class scrlttr2 is very customizable and provides support for DIN 676/5008 out of the box (p. 195):
DIN parameter set for letters on A4-size paper, complying with German standard DIN 676; suitable for window envelopes in the sizes C4, C5, C6, and C6/5 (C6 long).
DINmtext parameter set for letters on A4-size paper, complying with DIN 676, but using an alternate layout with more text on the first page; only suitable for window envelopes in the sizes C6 and C6/5 (C6 long).
(Even though DIN 676 has been superseded by DIN 5008 in 2011, the regulations for the layout of business letters have almost kept the same, only one dimension has been relaxed a bit. Hence, the DIN 676 layout adheres to DIN 5008.)
Parameter sets are simply passed as options to the class. The following shows a minimalistic letter:
\documentclass[version=last,DIN]{scrlttr2}
\usepackage[english]{babel}
\begin{document}
\begin{letter}{%
Joana Public\\
Hillside 1\\
12345 Public-City%
}
\opening{Dear chairman,}
the last general meeting was about one year ago.
I want to remind you, that the constitution of our club advises you to make a general meeting every six month.
Because of this I expect the executive board to detain such a meeting immediately.
\closing{Expecting an invitation}
\end{letter}
\end{document}
In fact, the DIN settings are even loaded as a default, so it is not necessary to pass them explicitly. For further customizations, please take a look into the Manual, which is available in German and English.
Only available in German is the extended manual as a printed book, which is supposed to provide specifically more customization examples for the letter class.
• Instead of page number (differs very likely from language to language) give better the chapter numbers (more stable) in documentation and the KOMA-Script version, where it is valid. So, chapters 4 and 21 in version 3.12. – Speravir Jan 24 '14 at 2:19 | 2021-04-10 12:26:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8745198845863342, "perplexity": 3994.5567892935733}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038056869.3/warc/CC-MAIN-20210410105831-20210410135831-00150.warc.gz"} |
https://www.gerad.ca/fr/papers/G-2015-11 | Groupe d’études et de recherche en analyse des décisions
# Efficient estimation and simulation of the truncated multivariate student-$$t$$ distribution
## Zdravko I. Botev et Pierre L'Ecuyer
We propose an exponential tilting method for exact simulation from the truncated multivariate student-t distribution in high dimensions as an alternative to approximate Markov Chain Monte Carlo sampling. The method also allows us to accurately estimate the probability that a random vector with multivariate student-t distribution falls in a convex polytope. Numerical experiments show that the suggested method is significantly more accurate and reliable than its competitors.
, 14 pages | 2020-11-26 06:25:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7972257733345032, "perplexity": 774.48992397283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141186761.30/warc/CC-MAIN-20201126055652-20201126085652-00173.warc.gz"} |
https://brilliant.org/problems/quirky-equation-v/ | # Quirky equation V
Algebra Level 3
$x^4-9x^3+27x^2-30x+8=0$Find the sum of all values of $x$ satisfying the above equation.
× | 2019-10-20 20:18:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9079742431640625, "perplexity": 2003.5895132221158}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00321.warc.gz"} |
http://mymathforum.com/topology/339249-doubts-topology-4.html | My Math Forum Doubts on topology
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Topology Topology Math Forum
February 27th, 2017, 08:15 PM #31
Senior Member
Joined: Nov 2015
Posts: 211
Thanks: 2
Quote:
Originally Posted by Maschke I promise we'll be at the bottom of this soon then we can go forward. We need to know in the real numbers: What is an open set, what is a limit point, what is an interior point, what is a boundary point, what is a closure, and what is a closed set. That list of terms. And we need exact definitions. Once we have those things we can prove that the open sets in the reals are a topology, and from then on whenever we're doing topology we just think about the reals. That gets you through 90% of topology at this level. I can give you some definitions but I'd rather work from definitions in a book that you own.
Sure today I will go through these terms and will provide you the same so that you can rectify if anything goes wrong.
February 27th, 2017, 08:22 PM #32
Senior Member
Joined: Aug 2012
Posts: 1,701
Thanks: 448
Quote:
Originally Posted by Lalitha183 Sure today I will go through these terms and will provide you the same so that you can rectify if anything goes wrong.
Ok sounds like a plan. I promise you that once we nail down these concepts in the real numbers, you will laugh at general topology. It's very simple once you have the concepts clear in your mind in the real numbers.
February 27th, 2017, 10:27 PM #33
Senior Member
Joined: Nov 2015
Posts: 211
Thanks: 2
Quote:
Originally Posted by Maschke Ok sounds like a plan. I promise you that once we nail down these concepts in the real numbers, you will laugh at general topology. It's very simple once you have the concepts clear in your mind in the real numbers.
Its bit confusing from the definition and the application.
What is an open set?
Is it a set which has infinitely many points between the intervel a and b and the endpoints a and b are not included ?
when I refer to a definition where $\epsilon > 0$ and a real number $x ,y \epsilon R$ .If the condition $dist(x,y) < \epsilon$ holds is an openset.
What it means ?
Let us take (1,4) and the dist(1,4) = 3.
let us take as
$\epsilon = 5$ which satisfying the condition
$\epsilon >0$
So, Is (1,4) is an open set ?
Can you show me the openset in a pictorial way ?
That would be more helpful
February 28th, 2017, 06:42 AM #34
Senior Member
Joined: Aug 2012
Posts: 1,701
Thanks: 448
Quote:
Originally Posted by Lalitha183 Its bit confusing from the definition and the application. What is an open set? Is it a set which has infinitely many points between the intervel a and b and the endpoints a and b are not included ? when I refer to a definition where $\epsilon > 0$ and a real number $x ,y \epsilon R$ .If the condition $dist(x,y) < \epsilon$ holds is an openset. What it means ? Let us take (1,4) and the dist(1,4) = 3. let us take as $\epsilon = 5$ which satisfying the condition $\epsilon >0$ So, Is (1,4) is an open set ? Can you show me the openset in a pictorial way ? That would be more helpful
I started writing up an explanation that will clear all this up for you. I'm traveling today and hopefully I'll be able to post something later tonight or tomorrow.
Very briefly, if we define the epsilon-ball around a point $x$ as $\mathcal B_\epsilon(x) = \{y \in \mathbb R : d(x,y) < \epsilon\}$ where $d$ is the distance function $d(x,y) = |x - y|$ then
* If $X$ is a subset of reals, a point $x \in X$ is called an interior point if there's some epsilon ball around $x$ entirely contained in $X$; and
* The set $X$ is called open if all its points are interior points.
I haven't got time to write more but I have a writeup that starts from that and goes all the way through the definition of a topological space and I'll have that by tomorrow.
From these definitions see if you can prove that in the unit interval $[0,1]$ the point $\frac{1}{2}$ is an interior point and the point $1$ is not. Therefore $[0,1]$ is not an open set. Then prove that $(0,1)$ is an open set.
Then see if you can prove that any open interval $(a,b)$ is an open set, and that the union of two open sets is open, and then that the arbitrarily union of open sets is open.
I'll have those proofs written out for you tomorrow. Possibly with pictures if I can find time to do that. Meanwhile this Wiki article has a great picture of an interior point. https://en.wikipedia.org/wiki/Interior_(topology)
ps -- Also see if you can prove from the definitions I gave that the empty set and the reals are open subsets of the reals. And see if you can prove that finite intersections of open sets are open. Start with the intersection of two open sets.
Last edited by Maschke; February 28th, 2017 at 07:35 AM.
March 1st, 2017, 09:54 AM #35 Senior Member Joined: Aug 2012 Posts: 1,701 Thanks: 448 I enjoyed writing this. I hope you find value in it. In my opinion to study topology you should spend some time with this material. The intuitions for basic topology all come from the real numbers. Intro This is an overview of the usual topology on the reals, along with proofs that the open sets of reals satisfy the definition of an abstract topology. After all the preliminaries I get to your original question and prove that a subet of a topological space is open if and only if all its points are interior points. I was wrong earlier, the definition of interior point in your video is correct. In $a \in U_a \subset A$ the inclusion need not be proper. That became clear when I couldn't get the proof to work without it. Prerequisites We assume the basic language and terminology of set theory up to and including arbitrary unions and intersections, countable and uncountable sets, and the use of the universal and existential quantifiers $\forall$ and $\exists$. We assume the existence and basic arithmetic and order properties of the set $\mathbb R$ of real numbers. We have the notion of open and closed intervals of real numbers, $(a,b)$ and $[a,b]$. At this point we have not yet proved that an open interval is an open set, so the word doesn't mean anything yet. We have a map $\lvert \cdot \vert : \mathbb R \to \mathbb R_{\geq 0}$ that maps the real number $x$ to its absolute value $\lvert x \vert$. The absolute value has two important properties of interest to us: * (Symmetry): $\forall x, \lvert x \vert$ is $0$ if and only if $x = 0$. * (Triangle inequality): $\forall, x, y \in \mathbb R, \lvert x + y \vert \leq \lvert x \vert + \lvert y \vert$. Given two real numbers $x$ and $y$ we define their distance $d(x,y) = \lvert x - y \lvert$. This notation is useful since it generalizes to metric spaces without any additional work. The properties of $d$ that we care about are ): * (Non-negativity): $\forall x,y \in \mathbb R, d(x,y) \geq 0$. * $\forall x,y \in \mathbb R, d(x, y) = 0$ if and only if $x = y$. * (Symmetry): $\forall x,y \in \mathbb R, \ d(x,y) = d(y, x)$. * (Triangle inequality): $\forall x,y,z \in \mathbb R, d(x,z) \leq d(x,y) + d(y,z)$. Each of these can be proved directly from the properties of the absolute value. Those are the prereqs for the rest of the article. If any of this is not clear please ask. That goes for anything in here. Needless to say there may be errors so if something seems off let me know. Open sets of reals If $x, \epsilon \in \mathbb R$ and $\epsilon > 0$, we define the open ball of radius $\epsilon$ around $x$ as $\mathcal B_\epsilon(x) = \{y \in \mathbb R : d(x, y) < \epsilon\}$. This is the key concept in all of topology, really. A little cloud of points around $x$ that is "missing its boundary," a phrase that's made precise in topology. Conceptually, an open ball is a little dotted-line circle around a point. In ther real numbers, an open ball is an open interval. Ok now we have the reals and $\epsilon$-balls. From these we will build an empire. If $A \subset \mathbb R$ and $a \in A$, we say that $a$ is an interior point if $\exists \epsilon > 0 : \mathcal B_\epsilon(a) \subset A$. In other words we can draw a little $\epsilon$ ball around $a$ that stays entirely within $A$. Example: In the closed unit interval $I = [0,1]$ the point $a = \frac{1}{2}$ is an interior point and the point $b = 1$ is not. (You should make sure you can prove those from the definitions). Now if $A \subset \mathbb R$ then we say $A$ is an open set if all its points are interior points. In other words in the same way that an open ball is a little circle without its boundary (which of course we have not formally defined yet) then an open set is a lumpy random shape or blob that is also missing its boundary. Now the blob above is all one piece. But an open set can have multiple parts. It could look like three blobs, or even infinitely many blobs. For exampethe real numbers minus the integers are an open set. ... (-----)3(-----)4(-----)5(-----)6(-----)7(-----)... We can notate this set as $\mathbb R \setminus \mathbb Z = \cup_{n \in \mathbb Z} (n, n+1)$. This is an open set that is a countable union of open intervals. It turns out that every open set of reals is a countable (or finite) union of open intervals, but that proof is a little beyond us at the moment. The empty set is open I claim the empty set is open, using the following vacuous argument. For the open set to fail to be open, we'd have to find some point $x \in \emptyset$ such that there is no epsilon-ball about $x$ contained in the empty set. But we can never do that. Why? Because there's no $x \in \emptyset$ in the first place. So $\emptyset$ is open. This type of argument usually startles people the first time but after a while you get used to it. All the pink elephants in the empty set can fly. Also true. If this isn't completely clear to you, you're not the only one. But it's how logic works and when we do math we agree to think like this. The reals are open in the reals The real numbers are an open subset of themselves. Given any real number $x$, we can take $\epsilon = 1$ say and there's an open interval around each real entirely contained in the reals. If we only knew that an open interval is an open set, we'd be done. Proof left to the reader. Why did I use the awkward phrase "the reals are open in the reals?" Because being an open set of a topological space is a relative property. If I pick up the real line and put it in the complex plane, it's no longer an open set. You can see that if you put an epsilon ball around any real number in the complex plane, that epsilon ball always contains some non-real numbers. A given set may be open as a subset of one topological space but not another. An arbitrary union of open sets is open Theorem: An arbitrary (finite, countably infinite, or uncountable) union of open sets is open. Proof: Let $\mathscr A$ be an index set, and for each $\alpha \in \mathscr A$ let $U_\alpha$ be an open set. Let $U = \cup_{\alpha \in \mathscr A} U_\alpha$ be their union. If $u \in U$ then by the definition of union, there exists a specific $\alpha \in \mathscr A$ such that $u \in U_\alpha$. Since $U_\alpha$ is open there is some $\epsilon > 0$ such that $\mathcal B_\epsilon(u) \subset U_\alpha \subset U$ so $U$ is open. $\square$ A finite intersection of open sets is open How about intersections? It seems clear that if we have two overlapping open sets, their intersection is open. Would this work for infinite intersections? An example shows that it would not. Ex: For $n \in \mathbb N$ let $I_n = (-\frac{1}{n}, \frac{1}{n})$. What is the intersection $\cap_{n \in \mathbb N} I_n$? It's the singleton set $\{0\}$, because $0 \in I_n$ for every $n$. But $\{0\}$ is not an open set. How do we know this? For any epsilon you could never fit an entire epsilon-ball inside $\{0\}$. We've just proved that in the reals with the usual topology, singletons are never open sets. So if infinite intersections of open sets aren't necessarily open, the next best thing we can prove is: Theorem: A finite intersection of open sets is open. Pf: Let $U_k$ be open for each $k$ in $\{1, 2, \dots, n)$ for some positive integer $n$. Let $u \in U = \cap_n U_k$. Then for each $k$ there is some $\epsilon_k > 0$ with $\mathcal B_{\epsilon_k}(u) \in U$. Now let $\epsilon = min(\epsilon_k)$. Note that the minimum exists exactly because we only have finitely many epsilons. That's the trick behind this proof. Then for each $k$, $\mathcal B_\epsilon(u) \subset \mathcal B_{\epsilon_k}(u)$ and therefore $\mathcal B_\epsilon(u) \subset U$ and $U$ is open. $\square$ Euclidean spaces Before getting to topological spaces I just want to mention that if we are in $n$-dimensional Euclidean space, $\mathbb R^n = \{(x_1, x_2, \dots, x_n) : x_i \in \mathbb R\}$ then if $x = (x_i)_n$ we define the usual Euclidean norm $\lVert x \rVert = \sqrt{\sum_n x_i^2}$ and this induces the metric $d(x,y) = \lVert x - y \rVert$. When we study the topology of the line we are also getting a good start on the topology of Euclidean $n$-space. Metric spaces Everything we've said so far works for metric spaces in general, which are just sets with a distance function. But metric spaces can be a lot different than Euclidean space. Example: The discrete metric is defined on the real numbers (or on any set in fact) as $d(x,y) = 0$ if $x = y$ and $d(x,y) = 1$ if $x \mathbb \neq y$. This distance function satisfies is a metric. We can therefore define open balls and open sets. In fact if $\{x\}$ is a singleton, then $\mathcal B_{\frac{1}{2}}(x)$ is an epsilon-ball entirely contained in $\{x\}$. In fact $\mathcal B_{\frac{1}{2}}(x) = \{x\}$. In other words every singleton is open. And since every set is the union of its singleton elements, every set is open. In this case the topology is the entire powerset of the original set. Topological spaces Now we are ready to define a topological space. Def: A topological space is a set $X$ along with a collection of subsets $\mathscr T \subset \mathscr P(X)$ (where $\mathscr P(X)$ is the powerset of $X$, the set of all the subsets of $X$) such that: * $\emptyset \in \mathscr T$; * $X \in \mathscr T$; * $\mathscr T$ is closed (also sometimes called stable) under arbitrary unions; and * $\mathscr T$ is closed under finite intersections. We denote a topological space as $(X, \mathscr T)$. If the topology is understood we will often refer to "the topological space $X$" and omit explicit mention of $\mathscr T$. Theorem: The real numbers with the open sets as defined earlier are a topological space. Pf: We walked through this above. $\square$ Example: The discrete topology on a set $X$ is the topology $\mathscr T = \mathscr P(X)$. In other words, every set is open. This is the exact same topology that we got from the discrete metric. So in this case we can define a topology and then find a metric that gives us exactly that topology. Example: The indiscrete topology on $X$ is the topology $\mathscr T = \{\emptyset, X\}$. The only open sets are the empty set and the whole set. What is interesting is that there is no metric whatsoever that will produce the indiscrete topology (needs proof of course). A topology that can be induced by a metric is called metrizable. The discrete topology is metrizable and the indiscrete topology is not. Interior points Now we are finally in a position to address the question you originally asked! If $(X, \mathscr T)$ is a topological space and $A \subset X$, a point $a \in A$ is called an interior point of $A$ if there exists an open set $U_a$ with $a \in U_a \subset A$. NOTE! When we write $a \in U_a \subset A$ it may be the case that $U_a = A$. I was wrong about this earlier. In fact if $A$ is already known to be an open set, then any point $a \in A$ is in some open set contained in $A$, namely $A$ itself. We've just proved one direction of: Theorem: A set $A$ in a topological space $X$ is open if and only if each of its points is an interior point. Proof: As we just noted, if $A$ is open and $a \in A$, then $a \in A \subset A$ satisfies the definition of an interior point. Since $a$ is arbitrary, every point of an open set is an interior point. To show the other direction, suppose that we have some set $A$, and each of its points is an interior point. We want to show that $A$ must be open. For each $a \in A$ let $U_a$ be an open set with $a \in U_a \subset A$. Since an arbitrary union of open sets is open, the union $U = \cup_A U_a$ is an open set. Now I claim that in fact $U = A$ which would show that A is indeed an open set. How do I prove $U = A$? I have to show (1) that each element of $U$ is in $A$ and (2) each element of $A$ is in $U$. (1) If $u \in U$ then $u \in U_a$ for some $a \in A$. Since $u \in U_a \subset A$, we have $u \in A$ so $U \subset A$. (2). Suppose $a \in A$. Since $a$ is an interior point (remember we're assuming all points of $A$ are interior) there's an open set $U_a$ with $a \in U_a$. But then $a$ is in the union of all the $U_a$'s, namely $U$. So $A \subset U$. Having proved set inclusions in both directions, we have $U = A$. $\square$ Last edited by Maschke; March 1st, 2017 at 10:18 AM.
March 1st, 2017, 09:59 AM #36 Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2 Thanks for all your efforts 😊. Tomorrow morning i will surely start reading it.
April 5th, 2017, 01:55 PM #37 Newbie Joined: Apr 2017 From: Kolkata Posts: 2 Thanks: 1 Look Lalitha, the phrase "interior point" has no meaning until you say "interior point of some particular set under some particular topology". By interior point $x$ of a set $A$ under some topology, we mean a point of the set $A$, such that there is an open subset $U$ of $A$ containing that point. That is $x \in U \subseteq A$. So if you mean to say that a set whose every point is an interior point of it is open, then you are correct. Last edited by Kuldeep Guha Mazumder; April 5th, 2017 at 01:58 PM.
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Contact - Home - Forums - Cryptocurrency Forum - Top | 2018-02-19 22:08:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8797174096107483, "perplexity": 163.25907297859925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812841.74/warc/CC-MAIN-20180219211247-20180219231247-00060.warc.gz"} |
https://stacks.math.columbia.edu/tag/02M3 | Lemma 10.63.3. Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules. Then $\text{Ass}(M') \subset \text{Ass}(M)$ and $\text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M'')$. Also $\text{Ass}(M' \oplus M'') = \text{Ass}(M') \cup \text{Ass}(M'')$.
Proof. If $m' \in M'$, then the annihilator of $m'$ viewed as an element of $M'$ is the same as the annihilator of $m'$ viewed as an element of $M$. Hence the inclusion $\text{Ass}(M') \subset \text{Ass}(M)$. Let $m \in M$ be an element whose annihilator is a prime ideal $\mathfrak p$. If there exists a $g \in R$, $g \not\in \mathfrak p$ such that $m' = gm \in M'$ then the annihilator of $m'$ is $\mathfrak p$. If there does not exist a $g \in R$, $g \not\in \mathfrak p$ such that $gm \in M'$, then the annilator of the image $m'' \in M''$ of $m$ is $\mathfrak p$. This proves the inclusion $\text{Ass}(M) \subset \text{Ass}(M') \cup \text{Ass}(M'')$. We omit the proof of the final statement. $\square$
Comment #5510 by Manuel Hoff on
I would suggest to add the following immediate consequence to the Lemma: If $M$ and $M'$ are two $R$-modules, then $\text{Ass}(M \oplus M') = \text{Ass}(M) \cup \text{Ass}(M')$.
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• 13 comment(s) on Section 10.63: Associated primes
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). | 2022-06-28 13:28:21 | {"extraction_info": {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9718561768531799, "perplexity": 120.19512674313563}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00111.warc.gz"} |
http://www.ask.com/question/what-is-standard-form-in-math-terms | # What Is Standard Form in Math Terms?
In mathematics, standard form refers to a method of writing numerals to the power of ten. For instance, when writing the number 5432 in standard form, it would appear as 5‰Û¢432 ÌÑ 10 to the power of 3, while 14,500 would be 1‰Û¢45 ÌÑ 10 to the power of 4. | 2014-03-10 07:35:07 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8616923689842224, "perplexity": 375.70412773738445}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010693428/warc/CC-MAIN-20140305091133-00049-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.beatthegmat.com/viewtopic.php?p=854106 | A bottle manufacturing company has $$5$$ identical machines, each of which produces bottles at the same constant rate.
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A bottle manufacturing company has $$5$$ identical machines, each of which produces bottles at the same constant rate.
by VJesus12 » Tue Sep 22, 2020 7:27 am
A bottle manufacturing company has $$5$$ identical machines, each of which produces bottles at the same constant rate. How many bottles will all $$5$$ machines produce running simultaneously for $$x$$ hours?
(1) Running simultaneously, $$3$$ of the machines produce $$72,000$$ bottles in $$2x$$ hours?
(2) Running simultaneously, $$2$$ of the machines produce $$24,000$$ bottles in $$x$$ hours?
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Re: A bottle manufacturing company has $$5$$ identical machines, each of which produces bottles at the same constant rat
by swerve » Tue Sep 22, 2020 12:08 pm
VJesus12 wrote:
Tue Sep 22, 2020 7:27 am
A bottle manufacturing company has $$5$$ identical machines, each of which produces bottles at the same constant rate. How many bottles will all $$5$$ machines produce running simultaneously for $$x$$ hours?
(1) Running simultaneously, $$3$$ of the machines produce $$72,000$$ bottles in $$2x$$ hours?
(2) Running simultaneously, $$2$$ of the machines produce $$24,000$$ bottles in $$x$$ hours?
Let the rate of each machine be r. Thus we have to find out the value of $$5\cdot r \cdot x$$. Thus, it is sufficient to find the value of $$r \cdot x$$.
Statement 1 states that $$3 \cdot r \cdot 2x= 72000$$. No need to calculate anything, as we know the value of $$r∗x$$. Sufficient $$\Large{\color{green}\checkmark}$$
Statement 2 states that $$2∗r∗x = 24000$$. Just as above, Sufficient $$\Large{\color{green}\checkmark}$$ | 2020-11-27 12:04:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2830115556716919, "perplexity": 1954.0159030247332}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141191692.20/warc/CC-MAIN-20201127103102-20201127133102-00423.warc.gz"} |
http://forums.elitistjerks.com/forums/topic/90120-on-the-value-of-haste/?page=3 | #### Archived
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# On the Value of Haste
## 60 posts in this topic
Actually there is no softcap for disc priest, atm I have over 700 haste and after casting PWS GCD is reduced below 1.0 sec. Even with 700+ haste and using engineering glove enchant my FHs are 0.8s and still slightly slower than the GCD.
I believe that at hard cap, 1180 haste (36% from gear), it will be at 0.75 sec. It seems even thought the tooltip says spellhaste the mechanics are the same as PI or Heroism/Bloodlust.
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Actually there is no softcap for disc priest, atm I have over 700 haste and after casting PWS GCD is reduced below 1.0 sec. Even with 700+ haste and using engineering glove enchant my FHs are 0.8s and still slightly slower than the GCD.
I believe that at hard cap, 1180 haste (36% from gear), it will be at 0.75 sec. It seems even thought the tooltip says spellhaste the mechanics are the same as PI or Heroism/Bloodlust.
There is a limit on the GCD and that is 1 second.
Whatever you are seeing, either you're looking at it wrong or there is latency issues.
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I'm not sure but I think I see something wrong with this. You can't apply the 10% haste to the amount of spells you cast but you have to apply it to the GCD and casting times then look at how many spells you can cast.
[...]
If you have 0 haste your GCD is 1.5. Let's say you cast a cycle of PoM, CoH, Renew, Renew, Flash Heal. It takes about 8 seconds to spam those 5 spells. (With your cycle, it should only take about 6 seconds since the last spell is instant, even with some lag)
That's a minor point, but even for an instant, you need to take the GCD into account. You will be locked for that time, and even if your last SOL heal occurs a T=6s, the next sequence will starts at T=7.5s (or T=8s with some lag).
If you gain 10% haste, your GCD is now 1.36. In that same 8 second window, you can cast 5.9 spells, we'll say 6 spells though since it really takes slightly less than a full 8 seconds to cast the original 5 spell cycle. Adding 10% haste allowed a whole extra spell to be cast in the same window of time.
In fact, we're back into granularity questions, and cycle changes. These are tricky point with haste, and there are no real proper way to handle them in a general case.
You could well argue that you would use your cycle with 10% haste.
You could also argue that with 10% haste, my previous cycle requires 6.81s, and including lag, it will be more that 7s, and you will keep the same cycle. Then in an arbitrary 8s interval, you'll have in mean really 2.2 HC-friendly casts.
I think that if you want to treat it correctly, but in a general way, the following approach would be the correct one :
1/ Determine which frequence you use PoM / COH (can be 7s at best). That leads to a cycle length T, which should not really depend on your haste.
2/ Determine how many cast you can squeeze in that period : you get a real value N = T * (1+haste) / 1.5
3/ Consider that 2 cast at least are POM and COH, so you get at most T*(1+haste)/1.5 - 2 HC-friendly cast.
4/ Take into account your SOL casts : either with a fixed number A of cast, or with a proportion B of free casts.
5/ Your final number of HC-friendly cast is then either K=T*(1+haste)/1.5 - (2+A), or K = (1-B)*[T*(1+haste)/1.5 - 2]
6/ In 8s, you get in mean K*8.0/T HC-friendly cast.
If you want it rigorously, you need to get rid of real cast number, and consider the probability Pn of having cast exactly N HC-friendly spells in the last 8s, where the current time is taken uniformly in the cycle.
For example, assuming a 7.5s with COH - POM - SOL - Renew - Renew (and 0% haste)
my mean number of HC-spells in 8s is 2 / 7.5*8 = 2.13.
But if I consider t in [0-4.5] or in [5-6] or in [6.5-7.5], I have exactly 2 Renew cast in the last 8s before T. That's a probability p2 = 6.5/7.5= 0.866
If t is in [4.5 - 5] or [6 - 6.5], I have 3 Renew in the last 8s. Probability p3 is then p3 = 1/7.5 = 0.13333.
Note that we have exactly 2 * p2 + 3 * p3 = 2.1333, as it shall be.
Now, you can have precise HC uptime, with is
$HC_{Uptime} = 1 - HC_{Downtime} = 1 - \sum_{n=0}^\infty P_n * (1 - \%_{Crit})^{n}$
But that's the kind of thing that are not really sexy to compute, and change in weird ways with haste.
Due to the convexity of x -> a^x function, the approximation HC ~ 1 - (1-Crit)^Mean is not exact, but it is not too far.
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To investigate the effect of both haste and crit on Holy Concentration uptime, I decided to run a monte-carlo analysis. I assumed the rotation outlined above (ProM-CoH-Renew-Renew-FH) and calculated the uptime for a simulated five minute fight with zero haste. I ran thirty thousand simulation runs for each crit value and plotted the median and 90% confidence interval. The results are plotted below.
I then repeated the analysis with 25% haste. The mean uptime value was nearly identical -- within the width of the line on one plot so not shown overlapping. The only noticeable effect was that the confidence interval narrowed slightly. This means that on the fights where you get very lucky or very unlucky with crit procs, haste will move HC uptime closer to the mean.
I can say with a fair degree of confidence that haste does not affect HC uptime in any meaningful way. I confirmed that the same is true for inspiration using a discipline rotation. I suspect that this true for any similar ability.
----
Edit: I replaced the word "mean" with "median," since that is what is actually plotted. I changed "95% confidence bounds" to "90%." The three lines corespond to the 5th, 50th, and 95th percentile of uptimes from the simulation runs. Thus the uptime you see in a given fight should fall between the top- and bottom-most lines 90% of the time.
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One of the things that I think has to be considered in considering the value of haste is the impact of lag/latency/reaction time, especially for holy priests. For most people and most servers, I would be surprised if there was less than an average 200 ms combination of latency/server side lag/reaction time. That means that with 0% haste, you would not be able to fit in 3 renews between CoH cooldowns, since the time lost makes the effective GCD more like 1.7 seconds. Assuming 200 ms, you would need around 20% raid buffed haste (requires about 400 haste rating from gear + standard raid buffs) to be able to fit in a 3rd instant cast between CoH cooldowns. To get a 4th cast in, you would likely need to be close to the soft haste cap and close to 1100 haste from gear.
I question the value of stacking haste past about 400-500 when you are using a instant cast heavy renew-CoH-PoM playstayle. Obviously, if you are using a lot of FH and PoH, it becomes more valuable, but for the instant cast playstyle, it seems like there is little benefit to stacking haste that doesn't get you an extra spell cast. Once you can hit the threshold to be able to get that 4th instant cast in, obviously it becomes much more valuable. However, until I have access to almost all of the 264+ haste/spirit pieces (and only 2 T10 items have haste on them unfortunately), I don't think it makes sense for me to gem pure haste or pass on spirit/crit items that are a SP upgrade.
Another factor is, if you are in a 25 man raid environment where replenishment uptime isn't always optimal (i.e. often 0-1 replenishers in a 25 man raid), would that not decrease the value of haste since it will lead to mana issues? Right now, it seems that only ret pallies are the only remaining 100% reliable replenishers. End game hunters typically do not run SV anymore, apparently destro locks take a huge DPS hit from speccing for replenishment, and there seems to be a trend for shadow priests with T10 4 pc to ignore replenishment uptime, since maintaining it reduces DPS.
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and there seems to be a trend for shadow priests with T10 4 pc to ignore replenishment uptime, since maintaining it reduces DPS.
Only during heroism/PI.
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Since no one else bothered to do it, I parsed the ICC10 log file mentioned above for rates of overhealing. That is, fraction of spells cast that result in overhealing.
Looks like our disco buddy (Baptistine) had over healing rates around 50%, depending on the spell. But they ranged wildly from 8% for Greater Heal (how in the world?) to 93% for Holy Nova. 38% for CoH in Holy mode, 64% for Penance.
I don't know if I trust these numbers, as this is my first time digging around in the combat logs like this. GH should have a high Overheal%, but maybe it was only used on a fight with a particular mechanic. The other holy priest (Khunanya) had >90% overhealing rate on his GH's though.
For penance, you'd expect with 64% OH% that the second tick must have OH'd and the third tick would be useless, but the snipe rate for that is only 2%, i.e. only 2% of his penance dots did nothing, rather than the 33% expected. So perhaps there was a lot of damage in between ticks; I'm not sure.
Perl implementation started from the above snippet. I broke this down per/healer and per/spell for completeness.
#!/usr/bin/perl
use warnings;
use strict;
my $char = '"Swampdonky"'; # Character to analyze (use quotes) #my$char = '"Baptistine"'; # Character to analyze (use quotes)
my %HEALS = ();
my ($heal,$overheal, $crit,$overcrit) = (0, 0, 0, 0);
my ($heal_count,$overheal_count, $crit_count,$overcrit_count) = (0, 0, 0, 0);
while (<STDIN>) {
next unless /SPELL_HEAL/;
my @event = split /,/;
#next unless $event[2] eq$char;
my $toon =$event[2];
my $spell =$event[ 8];
my $heal =$event[10];
my $over =$event[11];
my $is_crit = ($event[13] !~ /nil/);
$toon =~ s/"//g;$spell =~ s/"//g;
if ( not exists $HEALS{$toon}{$spell} ) {$HEALS{$toon}{$spell} = {
crit=>{ count=>0, size=>0, over_size=>0, over_count=>0, bad_size=>0, bad_count=>0 },
heal=>{ count=>0, size=>0, over_size=>0, over_count=>0, bad_size=>0, bad_count=>0 },
};
}
my $data; if ($is_crit ) { $data =$HEALS{$toon}{$spell}{'crit'} }
else { $data =$HEALS{$toon}{$spell}{'heal'} }
$data->{'count'}++;$data->{'size' } += $heal; if ($over > 0 ) {
$data->{'over_size' } +=$over;
$data->{'over_count'}++; } if ($heal == 0 ) {
$data->{'bad_size' } +=$over;
$data->{'bad_count'}++; } } printf "%-30s %-30s %10s %10s %10s | %10s %10s %10s | %10s %10s %10s | %10s %10s %10s\n", "Toon", "Spell", "Count", "RawHeals", "TotalHeals", "CritRate", "CritFrac", "SnipedRate", "ORate", "OHealRate", "OCritRate", "OFrac", "OHealFrac", "OCritFrac", ; foreach my$toon ( sort keys %HEALS ) {
next if scalar keys %{$HEALS{$toon}} < 4;
foreach my $spell ( sort keys %{$HEALS{$toon}} ) { my$heal = $HEALS{$toon}{$spell}{'heal'}; my$crit = $HEALS{$toon}{$spell}{'crit'}; my$total_count = $heal->{'count'} +$crit->{'count'};
my $total_heal_size =$crit->{'size'} + $heal->{'size'}; my$total_over_size = $crit->{'over_size'} +$heal->{'over_size'};
my $total_size =$total_heal_size + $total_over_size; printf "%-30s %-30s %10u %9uk %9uk | %9.0f%% %9.0f%% %9.0f%% | %9.0f%% %9.0f%% %9.0f%% | %9.0f%% %9.0f%% %9.0f%%\n",$toon,
substr($spell, 0, 30),$total_count,
$total_heal_size / 1000,$total_size / 1000,
$crit->{'count'} /$total_count * 100,
$crit->{'size' } /$total_size * 100,
($crit->{'bad_count'} +$heal->{'bad_count'}) / $total_count * 100, ($heal->{'over_count'} + $crit->{'over_count'}) /$total_count * 100,
$heal->{'over_count'} / ($heal->{'count'}||1) * 100,
$crit->{'over_count'} / ($crit->{'count'}||1) * 100,
($heal->{'over_size' } +$crit->{'over_size' }) / $total_size * 100,$heal->{'over_size' } / (($heal->{'size'}+$heal->{'over_size'})||1) * 100,
$crit->{'over_size' } / (($crit->{'size'}+\$crit->{'over_size'})||1) * 100,
;
}
}
Toon: Who cast it Spell: Spell cast Count: Number of times cast RawHeals: Real healing accomplished (i.e. excluding overheals) TotalHeals: Total healing, including overheals CritRate: % of casts resulting in crits CritFrac: % of real healing resulting from crits SnipedRate: % of casts resulting in 0 real healing (i.e. too slow or wrong target) ORate: % of casts resulting in overhealing OHealRate: % of non-crits resulting in overhealing OCritRate: % of crits resulting in overhealing OFrac: % of total that is overhealing OHealFrac: % of non-crit subtotal that is overhealing OCritFrac: % of crit subtotal that is overhealing
Toon Spell Count RawHeals TotalHeals | CritRate CritFrac SnipedRate | ORate OHealRate OCritRate | OFrac OHealFrac OCritFrac
Adrianromero Death Coil 9 60k 89k | 0% 0% 0% | 67% 67% 0% | 33% 33% 0%
Adrianromero Drain Life 21 31k 36k | 0% 0% 0% | 24% 24% 0% | 14% 14% 0%
Adrianromero Fel Synergy 891 436k 763k | 0% 0% 0% | 64% 64% 0% | 43% 43% 0%
Adrianromero Master Healthstone 2 8k 8k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Adrianromero Siphon Life 1785 3200k 5619k | 0% 0% 0% | 81% 81% 0% | 43% 43% 0%
Baptistine Binding Heal 18 136k 182k | 72% 59% 6% | 44% 40% 46% | 25% 28% 24%
Baptistine Circle of Healing 167 916k 1201k | 29% 31% 0% | 38% 32% 53% | 24% 21% 28%
Baptistine Desperate Prayer 8 83k 93k | 50% 55% 0% | 38% 25% 50% | 11% 20% 5%
Baptistine Divine Hymn 30 253k 327k | 40% 40% 0% | 67% 61% 75% | 23% 23% 22%
Baptistine Empowered Renew 110 375k 418k | 35% 43% 0% | 19% 19% 18% | 10% 11% 9%
Baptistine Flash Heal 2302 18804k 26917k | 44% 36% 1% | 58% 51% 67% | 30% 26% 34%
Baptistine Glyph of Power Word: Shield 1124 2369k 3315k | 43% 39% 0% | 44% 41% 48% | 29% 27% 30%
Baptistine Greater Heal 168 3705k 3886k | 32% 36% 0% | 8% 9% 7% | 5% 5% 5%
Baptistine Holy Nova 123 376k 720k | 41% 27% 0% | 93% 96% 90% | 48% 48% 47%
Baptistine Leeching Swarm 65 87k 87k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Baptistine Penance 782 3959k 6311k | 44% 34% 2% | 64% 62% 66% | 37% 37% 38%
Baptistine Prayer of Healing 205 1084k 1608k | 44% 37% 0% | 58% 54% 64% | 33% 30% 35%
Baptistine Prayer of Mending 892 5227k 6950k | 40% 39% 0% | 51% 46% 59% | 25% 21% 28%
Baptistine Spirit of Redemption 8 0k 0k | 25% 33% 0% | 0% 0% 0% | 0% 0% 0%
Cassharn Ardent Defender 4 50k 50k | 25% 31% 0% | 0% 0% 0% | 0% 0% 0%
Cassharn Holy Light 5 36k 38k | 0% 0% 0% | 40% 40% 0% | 4% 4% 0%
Cassharn Lay on Hands 3 187k 187k | 33% 27% 0% | 0% 0% 0% | 0% 0% 0%
Cassharn Leeching Swarm 65 96k 96k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Cassharn Master Healthstone 1 4k 4k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Cassharn Seal of Light 118 91k 139k | 0% 0% 0% | 54% 54% 0% | 34% 34% 0%
Darklawlz Ancestral Awakening 8 16k 22k | 0% 0% 0% | 38% 38% 0% | 23% 23% 0%
Darklawlz Chain Heal 116 691k 1107k | 34% 26% 0% | 68% 66% 72% | 38% 35% 41%
Darklawlz Earth Shield 26 122k 187k | 31% 30% 0% | 62% 67% 50% | 34% 38% 30%
Darklawlz Lesser Healing Wave 6 45k 76k | 50% 34% 0% | 83% 100% 67% | 41% 50% 31%
Darklawlz Riptide 13 56k 74k | 38% 36% 0% | 46% 38% 60% | 25% 24% 26%
Famous Beacon of Light 14 99k 180k | 0% 0% 0% | 86% 86% 0% | 45% 45% 0%
Famous Flash of Light 65 415k 612k | 43% 36% 0% | 58% 49% 71% | 32% 30% 34%
Famous Glyph of Holy Light 10 15k 22k | 0% 0% 0% | 50% 50% 0% | 33% 33% 0%
Famous Holy Light 5 84k 149k | 60% 38% 0% | 100% 100% 100% | 44% 48% 42%
Famous Holy Shock 26 196k 279k | 42% 37% 0% | 58% 60% 55% | 30% 30% 30%
Kahoo Death Coil 1 6k 6k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Kahoo Fel Synergy 681 631k 1226k | 0% 0% 0% | 95% 95% 0% | 49% 49% 0%
Kahoo Leeching Swarm 65 89k 89k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Kahoo Soul Leech 65 147k 272k | 0% 0% 0% | 89% 89% 0% | 46% 46% 0%
Khunanya Binding Heal 2 7k 7k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Khunanya Circle of Healing 432 1560k 2365k | 27% 24% 0% | 62% 60% 69% | 34% 33% 35%
Khunanya Desperate Prayer 1 6k 6k | 100% 100% 0% | 0% 0% 0% | 0% 0% 0%
Khunanya Devouring Plague 38 5k 10k | 0% 0% 0% | 82% 82% 0% | 45% 45% 0%
Khunanya Empowered Renew 235 488k 670k | 28% 27% 0% | 39% 36% 47% | 27% 25% 31%
Khunanya Flash Heal 59 410k 603k | 10% 10% 0% | 64% 60% 100% | 32% 29% 44%
Khunanya Greater Heal 16 235k 416k | 31% 24% 0% | 94% 91% 100% | 43% 45% 42%
Khunanya Holy Nova 8 23k 37k | 12% 11% 0% | 62% 57% 100% | 38% 34% 50%
Khunanya Leeching Swarm 65 88k 88k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Khunanya Prayer of Healing 5 27k 53k | 20% 14% 0% | 100% 100% 100% | 48% 47% 50%
Khunanya Prayer of Mending 111 784k 1064k | 31% 30% 0% | 53% 45% 71% | 26% 22% 32%
Ladsexington Lifebloom 47 227k 438k | 19% 13% 0% | 94% 92% 100% | 48% 48% 48%
Ladsexington Living Seed 2 5k 8k | 0% 0% 0% | 50% 50% 0% | 34% 34% 0%
Ladsexington Regrowth 19 130k 216k | 53% 37% 0% | 79% 78% 80% | 40% 40% 40%
Ladsexington Swiftmend 8 97k 133k | 38% 35% 0% | 75% 60% 100% | 27% 28% 25%
Lessa Leeching Swarm 65 72k 72k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Lessa Lifebloom 102 434k 816k | 15% 9% 0% | 95% 95% 93% | 47% 47% 45%
Lessa Living Seed 52 137k 178k | 0% 0% 0% | 48% 48% 0% | 23% 23% 0%
Lessa Nourish 159 1227k 2057k | 67% 44% 1% | 81% 79% 81% | 40% 38% 41%
Lessa Regrowth 73 492k 850k | 60% 40% 3% | 81% 76% 84% | 42% 43% 42%
Lessa Tranquility 34 220k 392k | 0% 0% 0% | 88% 88% 0% | 44% 44% 0%
Mirilyn Death Coil 1 2k 4k | 0% 0% 0% | 100% 100% 0% | 50% 50% 0%
Mirilyn Fel Synergy 497 305k 557k | 0% 0% 0% | 83% 83% 0% | 45% 45% 0%
Mirilyn Leeching Swarm 65 81k 81k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Mirilyn Master Healthstone 1 6k 12k | 100% 57% 0% | 100% 0% 100% | 43% 0% 43%
Mirilyn Soul Leech 137 236k 409k | 0% 0% 0% | 75% 75% 0% | 42% 42% 0%
Nassah Healing Touch 9 190k 222k | 22% 33% 0% | 33% 43% 0% | 15% 22% 0%
Nassah Lifebloom 245 1501k 2376k | 13% 9% 0% | 70% 65% 100% | 37% 34% 48%
Nassah Living Seed 200 688k 832k | 0% 0% 0% | 34% 34% 0% | 17% 17% 0%
Nassah Master Healthstone 1 4k 4k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Nassah Nourish 536 4963k 6119k | 40% 41% 0% | 33% 27% 41% | 19% 16% 22%
Nassah Regrowth 218 1492k 2355k | 42% 33% 0% | 66% 60% 73% | 37% 34% 39%
Nassah Swiftmend 139 1614k 1882k | 14% 15% 0% | 40% 40% 42% | 14% 15% 13%
Reanimated Ancestral Awakening 102 234k 318k | 0% 0% 0% | 42% 42% 0% | 26% 26% 0%
Reanimated Chain Heal 1243 8553k 13725k | 43% 33% 0% | 66% 61% 73% | 38% 35% 40%
Reanimated Earth Shield 123 535k 618k | 27% 35% 0% | 29% 29% 30% | 13% 17% 9%
Reanimated Healing Wave 3 32k 50k | 0% 0% 0% | 67% 67% 0% | 35% 35% 0%
Reanimated Leeching Swarm 65 79k 79k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Reanimated Lesser Healing Wave 100 706k 1160k | 60% 44% 10% | 70% 60% 77% | 39% 35% 41%
Reanimated Riptide 109 474k 743k | 42% 33% 0% | 71% 65% 78% | 36% 35% 38%
Semirhagge Death Pact 1 13k 13k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Semirhagge Death Strike 1 3k 3k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Semirhagge Leeching Swarm 61 83k 83k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Semirhagge Unholy Strength 353 397k 665k | 28% 22% 0% | 71% 70% 72% | 40% 40% 40%
Themanone Beacon of Light 228 1761k 3132k | 0% 0% 0% | 90% 90% 0% | 44% 44% 0%
Themanone Flash of Light 169 900k 1557k | 58% 39% 0% | 84% 73% 92% | 42% 39% 44%
Themanone Holy Light 47 717k 1309k | 51% 34% 0% | 94% 96% 92% | 45% 45% 45%
Themanone Holy Shock 22 150k 257k | 50% 35% 0% | 91% 91% 91% | 42% 40% 42%
Touchmyclam Healing Touch 1 10k 21k | 0% 0% 0% | 100% 100% 0% | 50% 50% 0%
Touchmyclam Living Seed 24 60k 80k | 0% 0% 0% | 46% 46% 0% | 25% 25% 0%
Touchmyclam Nourish 71 470k 700k | 51% 41% 0% | 58% 46% 69% | 33% 25% 37%
Touchmyclam Regrowth 38 233k 399k | 53% 37% 0% | 84% 72% 95% | 41% 41% 42%
Utholas Beacon of Light 324 3669k 6664k | 0% 0% 0% | 88% 88% 0% | 45% 45% 0%
Utholas Crazy Alchemist's Potion 3 13k 23k | 33% 24% 0% | 100% 100% 100% | 43% 37% 50%
Utholas Flash of Light 188 1163k 2072k | 40% 28% 0% | 83% 79% 88% | 44% 42% 46%
Utholas Glyph of Holy Light 459 792k 1431k | 0% 0% 0% | 79% 79% 0% | 45% 45% 0%
Utholas Holy Light 182 3027k 5562k | 47% 31% 0% | 96% 95% 96% | 46% 45% 46%
Utholas Leeching Swarm 52 73k 73k | 0% 0% 0% | 0% 0% 0% | 0% 0% 0%
Utholas Rejuvenation Potion 2 9k 12k | 50% 50% 0% | 50% 100% 0% | 25% 50% 0%
Utholas Seal of Light 24 21k 38k | 0% 0% 0% | 83% 83% 0% | 45% 45% 0%
Xaali Ancestral Awakening 284 1475k 1677k | 0% 0% 0% | 23% 23% 0% | 12% 12% 0%
Xaali Chain Heal 798 6010k 9060k | 34% 29% 0% | 68% 61% 80% | 34% 32% 36%
Xaali Earth Shield 332 1584k 1944k | 36% 41% 0% | 38% 34% 45% | 19% 17% 20%
Xaali Healing Wave 173 3993k 4291k | 33% 39% 0% | 17% 13% 26% | 7% 6% 8%
Xaali Lesser Healing Wave 294 2341k 3044k | 52% 48% 0% | 38% 30% 46% | 23% 19% 25%
Xaali Riptide 305 1655k 1999k | 27% 31% 0% | 31% 29% 39% | 17% 17% 18%
ÃÂndras Ancestral Awakening 18 36k 44k | 0% 0% 0% | 28% 28% 0% | 17% 17% 0%
ÃÂndras Chain Heal 271 1744k 2881k | 38% 27% 0% | 76% 76% 75% | 39% 39% 40%
ÃÂndras Earth Shield 4 14k 14k | 75% 82% 0% | 0% 0% 0% | 0% 0% 0%
ÃÂndras Lesser Healing Wave 4 28k 56k | 50% 31% 0% | 100% 100% 100% | 50% 50% 50%
ÃÂndras Riptide 35 177k 271k | 46% 37% 0% | 69% 74% 62% | 35% 38% 32%
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Now that heals have been nerfed in pvp wouldn't haste result in a higher HPS in PVP?
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Now that heals have been nerfed in pvp wouldn't haste result in a higher HPS in PVP?
That's one of the reasons why good Priests get the haste offset for their PVP gear. Hydra posted a full gear guide and stuff on http://hydramist.net if you're interested in looking into it.
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Good priests don't get crit because so many of their defensive spells can't crit. And their offensive spells that can crit only crit at 1.5x damage before resilience.
Priest mana regen is peeling from team mates and drinking.
So spellpower and haste are left as a viable stacking stat. Some priests go spell power, some go haste. If you're part of a good team you'll play less of a healing role and more of an offensive/utility role. Haste is probably better for this.
Stacking haste has very little to do with the healing nerf in arena, as haste stacking happened long before that change was implemented.
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https://bookdown.org/valerie_hase/TextasData_HS2021/tutorial-3-objects-structures-in-r.html | # 3 Tutorial 3: Objects & structures in R
After working through Tutorial 3, you’ll…
• know about types of data in R (numbers, text, factors, dates, logical/other operators)
• know about types of objects in R (scalars, vectors, matrices, data frames, lists)
## 3.1 Types of data
Objects in R can contain a variety of different types of data. We’ll get to know a selection of these, namely
To understand and work with these types of data, we’ll now import a data set. The data set is called “data_tutorial3.csv” via OLAT/Materials/Data for R.
The data set consists of data that is completely made up - a survey with 20 fictional students in a fictional seminar. We only use this data here so you can understand differences between types of data and types of objects.
The datafile data_tutorial3.csv is structured as follows:
• Each row contains the answer for a single student.
• Each column contains all values given by students for a single variable.
The variables included here are:
• name: the name of each student
• age: the age of each student (in years)
• date: the date on which each student was surveyed (in YYYY-MM-DD)
• outlet: the type of media outlet each student’s mainly uses to get information
• outlet_use: the time each student uses this media outlet daily (in hours)
• outlet_trust: how much each student trusts this media outlet (from 1 = not at all to 5 = very much)
We’ll read in the file with read.csv(). Here, we specify where to find the data file with the argument x as well as that the first row contains variable names with the argument header = TRUE2.
survey <- read.csv2("data_tutorial3.csv", header = TRUE)
survey
## name age date outlet outlet_use outlet_trust
## 1 Alexandra 20 2021-09-09 TV 2 5
## 2 Alex 25 2021-09-08 Online 3 5
## 3 Maximilian 29 2021-09-09 Zeitung 4 1
## 4 Moritz 22 2021-09-06 TV 2 2
## 5 Vanessa 25 2021-09-07 Online 1 3
## 6 Andrea 26 2021-09-09 Online 3 4
## 7 Fabienne 26 2021-09-09 TV 3 2
## 8 Fabio 27 2021-09-09 Online 0 1
## 9 Magdalena 8 2021-09-08 Online 1 4
## 10 Tim 26 2021-09-07 TV NA 2
## 11 Alex 27 2021-09-09 Online NA 2
## 12 Tobias 26 2021-09-07 Online 2 2
## 13 Michael 25 2021-09-09 Online 3 2
## 14 Sabrina 27 2021-09-08 Online 1 2
## 15 Valentin 29 2021-09-09 TV 1 5
## 16 Tristan 26 2021-09-09 TV 2 5
## 17 Martin 21 2021-09-09 Online 1 2
## 18 Anna 23 2021-09-08 TV 3 3
## 19 Andreas 24 2021-09-09 TV 2 5
## 20 Florian 26 2021-09-09 Online 1 5
### 3.1.1 Numbers
A typical type of data you may come is numeric data. In our data set, the age of respondents is saved as numbers.
In R, this type of data is called numeric. To check if age is really saved as numbers, you can use mode(). The function tells you an object’s type.
mode(survey$age) ## [1] "numeric" ### 3.1.2 Text In this seminar, we will often work with letters, words, or whole texts. Such text is saved in the character format. In our data set, we know that the variable name is likely to contain textual content. We’ll check this with the same command mode(): #find out which type of data the object word consists of mode(survey$name)
## [1] "character"
You may have already spotted something else: If you want to save something in character format, you have to set quotation marks "" before and after the respective values. For instance, if you save the number 1 with quotation marks, R will use the character format:
word <- "1"
mode(word)
## [1] "character"
However, if you save the number 1 without quotation marks, R will use the numeric format:
word <- 1
mode(word)
## [1] "numeric"
Why is this important? In R, many functions for conducting statistical analyses, for example, can only be executed with objects containing numeric data, not character data.
For instance: Let’s say you want R to calculate the mean of some numbers:
numbers <- c("1", "8", "9", "4")
mean(numbers)
## Warning in mean.default(numbers): Argument ist weder numerisch noch boolesch: gebe NA zurück
## [1] NA
R will throw you an error message. Why? Because you saved these numbers as character values, not as numeric values:
mode(numbers)
## [1] "character"
If we save these numbers in a numeric format, i.e., as numbers not text, the command works just fine:
numbers <- c(1,8,9,4)
mean(numbers)
## [1] 5.5
### 3.1.3 Factors
Factors constitute an additional type of data. Factors can include both numeric data and characters. What is important, however, is that they treat any types of data as categorical.
We can, for instance, convert the variable name - which now contains each student’s name in character format - to a factor using as.factor():
survey$name <- as.factor(survey$name)
As you can see, the variable age now contains each name as an independent level. You can get each unique level with the levels() command:
levels(survey$name) ## [1] "Alex" "Alexandra" "Andrea" "Andreas" "Anna" "Fabienne" "Fabio" "Florian" ## [9] "Magdalena" "Martin" "Maximilian" "Michael" "Moritz" "Sabrina" "Tim" "Tobias" ## [17] "Tristan" "Valentin" "Vanessa" You may have noted that the data set survey contains 20 observations but that the variable age only included 19 levels. Why? If you inspect the data set, you can see that two students have the same name (here, Alex in row 2 and 11) - which is why we have 20 observations, but only 19 unique levels for the variable name: survey ## name age date outlet outlet_use outlet_trust ## 1 Alexandra 20 2021-09-09 TV 2 5 ## 2 Alex 25 2021-09-08 Online 3 5 ## 3 Maximilian 29 2021-09-09 Zeitung 4 1 ## 4 Moritz 22 2021-09-06 TV 2 2 ## 5 Vanessa 25 2021-09-07 Online 1 3 ## 6 Andrea 26 2021-09-09 Online 3 4 ## 7 Fabienne 26 2021-09-09 TV 3 2 ## 8 Fabio 27 2021-09-09 Online 0 1 ## 9 Magdalena 8 2021-09-08 Online 1 4 ## 10 Tim 26 2021-09-07 TV NA 2 ## 11 Alex 27 2021-09-09 Online NA 2 ## 12 Tobias 26 2021-09-07 Online 2 2 ## 13 Michael 25 2021-09-09 Online 3 2 ## 14 Sabrina 27 2021-09-08 Online 1 2 ## 15 Valentin 29 2021-09-09 TV 1 5 ## 16 Tristan 26 2021-09-09 TV 2 5 ## 17 Martin 21 2021-09-09 Online 1 2 ## 18 Anna 23 2021-09-08 TV 3 3 ## 19 Andreas 24 2021-09-09 TV 2 5 ## 20 Florian 26 2021-09-09 Online 1 5 ### 3.1.4 Dates Since we’ll be working with textual data and are often interested in when these texts are published, we’ll need to know about dates. The date format can include any type of information on time. Let’s check out an example using the variable date in our data set, which specifies the day on which students were surveyed. survey$date
## [1] "2021-09-09" "2021-09-08" "2021-09-09" "2021-09-06" "2021-09-07" "2021-09-09" "2021-09-09" "2021-09-09"
## [9] "2021-09-08" "2021-09-07" "2021-09-09" "2021-09-07" "2021-09-09" "2021-09-08" "2021-09-09" "2021-09-09"
## [17] "2021-09-09" "2021-09-08" "2021-09-09" "2021-09-09"
When inspecting our data set, we can see that the variable is saved in the following format: “YYYY-MM-DD” (i.e., the year followed by the month and day on which a student was surveyed). However, as of yet this date is saved in character format.
For dates and times, that format is not too helpful. If we want to work with dates in a more flexible way, we should save them in the date format using the as.Date() format:
survey$date <- as.Date(survey$date)
### 3.1.5 Missing data/NAs
Something we will often encounter is missing data, also called NAs. In our data set, the variable outlet_use which describes how many hours a day a student uses a specific outlet contains missing data:
survey$outlet_use ## [1] 2 3 4 2 1 3 3 0 1 NA NA 2 3 1 1 2 1 3 2 1 When working with data, you should always check how much of it may be missing (and of course, why). To do so, you can use the is.na() command: It tells you which values in a specific objects are missing (TRUE) or not missing (FALSE): is.na(survey$outlet_use)
## [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [20] FALSE
For instance, the first 9 observations do not contain any missing values for the variable age, but the 10th observation does.
With larger data sets, you obviously would not want to check this manually for every single observation. We can use a little trick here: Since the value FALSE counts as 0 and the value TRUE counts as a 1…
as.numeric(FALSE)
## [1] 0
as.numeric(TRUE)
## [1] 1
… we can simply calculate the sum of these values for survey$outlet_use to see whether how many observations are missing: sum(is.na(survey$outlet_use))
## [1] 2
### 3.1.6 Logical/other operators
Lastly, you should know about logical (or other) operators. These are often used to check whether certain statements are true/false, whether some values take on a certain value or not, or whether some values are higher/lower than others. Since we won’t need this right away,I won’t go into details here - but we will need these later:
Table 3.1: Logical/other operators
Operator Meaning
TRUE indicates that a certain statement applies, i.e., is true
FALSE indicates that a certain statement does not apply, i.e., is not true
& connects two statements which should both be true
| connects two statements of which at least one should be true
== indicates that a certain value should equal another one
!= indicates that a certain value should not equal another one
> indicates that a certain value should be larger than another one
< indicates that a certain value should be smaller than another one
>= indicates that a certain value should be larger than or equal to another one
<= indicates that a certain value should be smaller than or equal to another one
You may wonder: Why should I use this?
Glad you asked: We’ll need these operators, for instance, to filter data sets by specific values or apply functions if some conditions (but not others) are true.
For instance, you could use a logical operator to only keep those students in the data set that are older than 21 (17 out of 20 students). We filter the data set using the > operator. We can get basic information about this data set using the str() command:
survey_subset <- survey[survey$age>21,] str(survey_subset) ## 'data.frame': 17 obs. of 6 variables: ##$ name : Factor w/ 19 levels "Alex","Alexandra",..: 1 11 13 19 3 6 7 15 1 16 ...
## $age : int 25 29 22 25 26 26 27 26 27 26 ... ##$ date : Date, format: "2021-09-08" "2021-09-09" "2021-09-06" "2021-09-07" ...
## $outlet : chr "Online " "Zeitung" "TV" "Online " ... ##$ outlet_use : int 3 4 2 1 3 3 0 NA NA 2 ...
## \$ outlet_trust: int 5 1 2 3 4 2 1 2 2 2 ...
## 3.2 Types of objects
Great, now you know all about types of data. As promised, this tutorial will also teach you about different types of objects:
### 3.2.1 Scalars
The “smallest” type of data you will encounter are scalars.
Scalars are objects consisting of a single value - for example, a letter, a word, a sentence, a number, etc.
You’ve already seen what a scalar looks like - remember the very first time you were Writing Code? Here, we defined an objects word which only consisted of the word “hello”.
word <- "hello"
word
## [1] "hello"
We could also save a single number this way:
number <- 1
number
## [1] 1
The same applies to a single sentence:
sentence <- "I would like to be saved as a scalar"
sentence
## [1] "I would like to be saved as a scalar"
### 3.2.2 Vectors
The next type of data you should know are vectors:
Vectors are objects that consist of several values of the same type of data.
For instance, if we want to save several words or numbers separately, we can save them as vectors. This means that the first element of our vector will be our first word, the second the second word and so forth. Importantly, a vector can only contain data of the same type - for instance, you cannot save data in numeric and character format in the same format.
In principle, you can often (but not always) compare vectors with variables in data sets: They contain values for all observations in your data set (with all of these values being of the same data type).
An example would be the numbers from 1 to 20 that we worked with before. Now, it becomes apparent what the c() stands for - it specifies the vector format.
We define the object numbers to consist of the a vector c() which contains the values 1 to 20. Here, we ask R to include all values from 1 to 20 by inserting a colon between both numbers:
numbers <- c(1:20)
numbers
## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
If you want the vector numbers to only contain selected numbers from 1 to 20, you can specify these numbers and separate them using a comma:
numbers <- c(1,5,8,19,20)
numbers
## [1] 1 5 8 19 20
We can do the same thing for data in character format:
fruits <- c("Apple", "Banana", "Orange", "Lemon")
fruits
## [1] "Apple" "Banana" "Orange" "Lemon"
In practice, a vector is nothing else than scalars of the same typed chained together - in this case, the words “Apple”, “Banana”, “Orange”, and “Lemon”.
If you love to write inefficient code, you could also define separate objects (here: scalars) consisting of every element of the vector you want to define and then chain them together (10/10 wouldn’t recommend because this is unnecessarily complicated):
apple <- "Apple"
banana <- "Banana"
orange <- "Orange"
lemon <- "Lemon"
fruits <- c(apple, banana, orange, lemon)
fruits
## [1] "Apple" "Banana" "Orange" "Lemon"
This would lead to very similar results, but you would have to write 4 more lines of code compared to the previous example.
### 3.2.3 Data frames & matrices
An even “bigger” type of data, in this sense, are data frames and matrices:
Data frames & matrices consist of several vectors of the same length.
Matrices and data frames come closest to how you may understand data sets (for instance, in SPSS). Our survey data set is one example for a data frame:
survey
## name age date outlet outlet_use outlet_trust
## 1 Alexandra 20 2021-09-09 TV 2 5
## 2 Alex 25 2021-09-08 Online 3 5
## 3 Maximilian 29 2021-09-09 Zeitung 4 1
## 4 Moritz 22 2021-09-06 TV 2 2
## 5 Vanessa 25 2021-09-07 Online 1 3
## 6 Andrea 26 2021-09-09 Online 3 4
## 7 Fabienne 26 2021-09-09 TV 3 2
## 8 Fabio 27 2021-09-09 Online 0 1
## 9 Magdalena 8 2021-09-08 Online 1 4
## 10 Tim 26 2021-09-07 TV NA 2
## 11 Alex 27 2021-09-09 Online NA 2
## 12 Tobias 26 2021-09-07 Online 2 2
## 13 Michael 25 2021-09-09 Online 3 2
## 14 Sabrina 27 2021-09-08 Online 1 2
## 15 Valentin 29 2021-09-09 TV 1 5
## 16 Tristan 26 2021-09-09 TV 2 5
## 17 Martin 21 2021-09-09 Online 1 2
## 18 Anna 23 2021-09-08 TV 3 3
## 19 Andreas 24 2021-09-09 TV 2 5
## 20 Florian 26 2021-09-09 Online 1 5
Let’s say we don’t have a data set yet but want to create one ourselves. For instance, we have three vectors of the same length that contain the following data:
• a vector called fruits consisting of the names of four different fruits (character)
• a vector called price consisting of the amount of money each fruit costs (numeric)
• a vector called color consisting of the color of each fruit (character)
fruits <- c("Apple", "Banana", "Orange", "Lemon")
price <- c(0.8,0.6,0.78,0.9)
color <- c("red", "yellow", "orange", "yellow")
Since these vectors describe the same data, we want them to be saved as one data set. To create the data set data_fruits, we use the command data.frame(), which contains:
• an vector called fruits as its first column
• a vector called price as its second column
• a vector color as its third column
Here, columns describe different variables and rows observations.
data_fruits <- data.frame("fruits"= fruits,
"price" = price,
"color" = color)
data_fruits
## fruits price color
## 1 Apple 0.80 red
## 2 Banana 0.60 yellow
## 3 Orange 0.78 orange
## 4 Lemon 0.90 yellow
In difference to data frames that can encompass different types of data, matrices consist of vectors of exactly the same type of data - for instance, only character vectors or numeric vectors.
This is also why we will mostly be working with data frames in this seminar - they are just more flexible and more suitable for the data we will work with.
For a nice example for comparing scalars, vectors, data frames, and matrices, check out this example here.
### 3.2.4 Lists
Lastly, you should know about lists.
Lists can be described as an array of (different) objects.
As discussed before, data frames can encompass vectors consisting of different types of data (for instance, character and numeric vectors). However, these all have to be vectors (and of the same length). In some case, you may want to save very different types of objects - scalars, vectors, data frames, and objects -, which will differ in length3.
In this case, lists offer the most flexible way of saving very different objects within one object (i.e., the list). Let’s say that we want to save our data frame data_fruits as well the data frame survey in a single object. We do so by specifying a list which includes data_fruits as its first and survey as its second element:
list <- list(data_fruits, survey)
View(list)
Image: List
As you see, the object list consists of two elements:
• the first element [[1]] encompasses the data frame data_fruits
• the second element [[2]] encompasses the data frame survey
### 3.2.5 Other types of objects
Since we’ll be learning how to conduct automated content analysis in this tutorial, we’ll encounter a bunch of other types of data such as the corpus or the document-feature-matrix (dfm). You don’t have to know anything about these data types for now - just be aware that other types of objects exist.
## 3.3 Take Aways
• Types of data: In R, you can work with different types of data in numeric, character, factor, or data format. Important commands: mode(), str(), as.character(), as.numeric(), as.factor(), levels(), as.Date() & is.na()
• Types of objects: In R, you can work with different types of objects, for instance scalars, vectors, data frames, matrices, or lists. Important commands: c(), data.frame(), list()
## 3.4 More tutorials on this
You still have questions? The following tutorials & papers can help you with that:
You’ve worked through all the material of Tutorial 3? Let’s see it - the following tasks will test your knowledge.
Create a data frame called data. The data frame should contain the following variables (in this order):
• a vector called food. It should contain 5 elements, namely the names of your five favourite dishes.
• a vector called description. For every dish mentioned in food, please describe the dish in a single sentence (for instance, if the first food you describe is “pizza”, you could write: “This is an Italian dish, which I prefer with a lot of cheese.”)
• a vector called rating. Rate every dish mentioned in food with 1-5 (using every number only once), i.e., by rating your absolute favorite dish out of all five with a 1 and your least favorite dish out of all five with a 5.
Hint: For me, the data frame would look something like this:
## food description Rating
## 1 pizza Italian dish, I actually prefer mine with little cheese 3
## 2 pasta Another Italian dish 1
## 3 ice cream The perfect snack in summer 2
## 4 crisps Potatoes and oil - a luxurious combination 4
## 5 passion fruit A fruit that makes me think about vacation 5
Can you sort the data in your data set by rating - with your favorite dish (i.e., the one rated “1”) on top of the list and your least favourite dish (i.e., the one rated “5”) on the bottom?
Important: You do not yet know this command - you’ll have to google for the right solution. Please do and note down the exact search terms you used for googling, so we can discuss them next week.
Hint: For me, the data frame would look something like this:
## food description Rating
## 1 pasta Another Italian dish 1
## 2 ice cream The perfect snack in summer 2
## 3 pizza Italian dish, I actually prefer mine with little cheese 3
## 4 crisps Potatoes and oil - a luxurious combination 4
## 5 passion fruit A fruit that makes me think about vacation 5
This is where you’ll find solutions for tutorial 3.
Let’s keep going: Tutorial 4: Reading data in/out
1. Don’t worry, we’ll discuss the read.csv2() command in a bit. Ignore it for now.↩︎
2. Length means that these objects may contain different numbers of elements (in the case of scalars and vectors, for example) or different numbers of rows/columns (in the case of data frames and matrices, for example)↩︎ | 2022-06-28 20:53:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2805405557155609, "perplexity": 2576.141507253436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00386.warc.gz"} |
http://mathhelpforum.com/number-theory/74184-solve-congruences-print.html | # solve congruences
• February 17th 2009, 07:42 PM
mathisthebestpuzzle
solve congruences
3x^5=1 (mod 29)
and
3^x=2 (mod 29)
find all solutions for both congruences.
• February 17th 2009, 10:32 PM
ThePerfectHacker
Quote:
Originally Posted by mathisthebestpuzzle
3x^5=1 (mod 29)
Write, $x^5 \equiv 10(\bmod 29)$ (since $10$ is an inverse of $3$ mod $29$).
Now find the index of $10$ for a given primitive root of $29$.
Can you finish up the problem?
• February 17th 2009, 11:00 PM
mathisthebestpuzzle
hacker i thank you so much, but i had already solved that one since i posted it.
i am stuck on the second one though...any help would be wonderful! | 2015-11-29 18:22:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.963234543800354, "perplexity": 1870.6489725597644}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398459214.39/warc/CC-MAIN-20151124205419-00171-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://warosu.org/sci/thread/9271013#p9272130 | [ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports / report a bug ] [ 4plebs / archived.moe / rbt ]
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# /sci/ - Science & Math
[ Toggle deleted replies ]
File: 94 KB, 996x1170, DNny-tgVoAA9xE6.jpg [View same] [iqdb] [saucenao] [google]
Can /sci solve this?
>> Anonymous Thu Nov 2 08:28:56 2017 No.9271015 No they can't.
>> Anonymous Thu Nov 2 08:57:00 2017 No.9271053 Ill-formed question. Won't bother.
>> Anonymous Thu Nov 2 09:52:46 2017 No.9271143 >>9271013~31,6%
>> Anonymous Thu Nov 2 10:04:55 2017 No.9271165 >>9271013((1/16)^m)*16
>> Anonymous Thu Nov 2 10:08:29 2017 No.9271171 >>9271165oh I missed this((1/m)^n)*m)
>> Anonymous Thu Nov 2 10:18:18 2017 No.9271186 >>9271165>>9271171read correctly will you?
>> Anonymous Thu Nov 2 10:27:03 2017 No.9271200 So, exactly one instance of exactly three darts in the same square? That's a lot of cases.
>> Anonymous Thu Nov 2 10:30:21 2017 No.9271207
>> Anonymous Thu Nov 2 10:41:58 2017 No.9271220 >>9271013Does more than 2 darts in a square count? Is it ok if we have multiple squares with 3 or more darts?We need more information here.
>> Anonymous Thu Nov 2 10:57:03 2017 No.9271243 ${n \choose m} {\frac{1}{d}}^{m-1}$
>> Anonymous Thu Nov 2 11:02:45 2017 No.9271259 >>9271013How can you throw the darts both uniformly and randomly?
>> Anonymous Thu Nov 2 11:15:34 2017 No.9271286 >>9271259The (X, Y) of where each dart lands has a Uniform([0,1]x[0,1]) distribution, brainlet.
>> Anonymous Thu Nov 2 11:28:14 2017 No.9271320 >>9271013Surely it's just a binomial thing?
>> Anonymous Thu Nov 2 11:48:01 2017 No.9271366 >>9271320Depends on >>9271200If it means exactly one instance, then I don't think it's binomial. If it doesn't, then it seems like it'd be P(X = 2), X ~ Binomial(7, 1/16) since the first one goes wherever it goes and the other two must match it.
>> Anonymous Thu Nov 2 11:50:13 2017 No.9271372 >>9271013counting problems are kind of boring
>> Anonymous Thu Nov 2 12:00:28 2017 No.9271389 >>92710130.00024%
>> Anonymous Thu Nov 2 16:56:17 2017 No.9271855 File: 58 KB, 931x777, Screenshot from 2017-11-02 16-55-59.png [View same] [iqdb] [saucenao] [google] Bumping with simulations.
>> Anonymous Thu Nov 2 18:10:26 2017 No.9271960 >>9271855I don't understand this.
>> Anonymous Thu Nov 2 18:20:54 2017 No.9271978 >>9271960brainlet
>> Anonymous Thu Nov 2 18:26:53 2017 No.9271995 File: 64 KB, 889x848, Screenshot from 2017-11-02 18-25-31.png [View same] [iqdb] [saucenao] [google] >>9271960Do you mean the code or the results. Here's a different version with some expository examples.
>> Anonymous Thu Nov 2 18:50:52 2017 No.9272042 >>9271013d choices for the square.$\binom{n}{m}$ choices for which darts you'll pick so that you can put in the square.$d \binom{n}{m}$ acceptable configurations in total. $\binom{n+d-1}{d-1}$ possible configurations in total (just put the squares in an array and think of it as the classic problem with the cells and the balls.Therefore, the probability is $d \binom{n}{m} / \binom{n+d-1}{d-1}$ .
>> Anonymous Thu Nov 2 18:56:58 2017 No.9272050 18.274004% for the first one
>> Anonymous Thu Nov 2 19:06:07 2017 No.9272063 >>9272042Oh shit forgot to put the rest n-m darts in other squares .Multiply by $\binom{(n-m)+(d-1)-1}{(d-1)-1}$ if you want exactly m balls and by $\binom{(n-m)+d-1}{d-1}$ if you want at least m balls.
>> Anonymous Thu Nov 2 19:08:59 2017 No.9272068 >>9272063actually nvm, ignore that, it give probabilities greater than 1
>> Anonymous Thu Nov 2 19:50:05 2017 No.9272130 >>9272042>>9272063>>9272068Oh I figured it out. I cosidered the darts being different. Remove $\binom{n}{m}$ and everything is fine.You choose a square (d ways). Put m darts in it. Ignore them. Put the rest of the darts in (if you want exactly m, you ignore the square you put the m darts in, otherwise you include it) .Probability of Exactly m dots in one square is $d \binom{n-m+d-2}{d-2} / \binom{n+d-1}{d-1}$ .Probability of m or More dots in one square is $d \binom{n-m+d-1}{d-1} / \binom{n+d-1}{d-1}$ .For d=16, n=8 , m=3 the probabilities are 96/253 ~ 38% and 128/253 ~ 50.6 %
>> Anonymous Thu Nov 2 21:53:38 2017 No.9272315 >>9272130Plug 16 darts into the first one and you get something greater than 1.Seems like you're just pulling stuff out of your ass.
>> Anonymous Thu Nov 2 23:15:49 2017 No.9272444 >>9272315Yeah I noticed that too, but I can't see where is the damn flaw. It's driving me mad.
>> Anonymous Fri Nov 3 02:19:44 2017 No.9272744 >>9271013100%
>> /psy/mon|0-X's|/sci/mon Fri Nov 3 02:28:01 2017 No.9272750 >>9272315Any attempt to isolate any SINGULAR variable will always result in the following constant: 0.716531310573789250425604096925379667453112059821479157140...Derived by this formula: $\frac{1}{e^{1/3}}$Because you are allow for 'square' then the probability will be 100%, if you never allow one variable to submit to the above constant (to your desired level of machine precision).
>> Anonymous Fri Nov 3 11:07:23 2017 No.9273292 bump
>> Anonymous Fri Nov 3 11:10:10 2017 No.9273296 100%
>> Anonymous Fri Nov 3 13:22:43 2017 No.9273568 >>9271013dCm*(1/d)^m*(1-(1/d))^(n-m)Probability function
>> Anonymous Sat Nov 4 02:14:57 2017 No.9274696 bumping
>> Anonymous Sat Nov 4 02:37:20 2017 No.9274731 >>92710131)16 * {Probability of 3 darts in the first square} - 120 * {Probability of 3 darts in the first and second square}2)d * {Probability of m darts in the first square} - {d choose 2} * {Probability of m darts in the first and second square} + {d choose 3} * {Probability of m darts in the first, second and third square} - ...
>> Anonymous Sat Nov 4 02:41:05 2017 No.9274736 >thread is up for 2 days>still no correct solution/sci/ is really filled with brainlets
>> Anonymous Sat Nov 4 11:37:04 2017 No.9275248 >>9274736Nigga it's x^n where x is the chance of a specific square with one try and n is the amount of times you throw(1/16)^3 is the solution
>> Anonymous Sat Nov 4 12:45:59 2017 No.9275333 >>9275248Wrong brainlet.
>> Anonymous Sat Nov 4 12:48:34 2017 No.9275336 >>9271013The real problem is whether or not to do what you asked. I have chosen not to do it. Problem solved. Tohohohoho, hohoho, hohohohoho.
>> Anonymous Sat Nov 4 12:58:17 2017 No.9275352 >>9275333Wrong brainlet, right answer?
>> Anonymous Sat Nov 4 13:04:21 2017 No.9275362 >>9271013At least correct the formulation of the problem as "at least 3 darts" t. Taleb
>> Anonymous Sat Nov 4 13:06:10 2017 No.9275368 How about the old $\frac{1}{16}^3 \times \frac{15}{16}^5$ (if they're talking about exactly 3 hits in one square)
>> Anonymous Sat Nov 4 13:08:02 2017 No.9275370 >>9275333Ok nigger.
>> Anonymous Sat Nov 4 13:10:44 2017 No.9275372 >>9275368and with this(1/d)^m * (d-1/d)^(m-n)for the second problem i see is the same way
>> Anonymous Sat Nov 4 13:14:24 2017 No.9275377 >>9275372obviously n-m
>> Anonymous Sat Nov 4 13:22:16 2017 No.9275389 >>9275333>>9275352Sorry I shouldn't have teased you. I actually normally don't on 4chan, my apologies.
>> Anonymous Sat Nov 4 14:02:01 2017 No.9275417 Pick a square, and pick m darts then the probability those m darts land in that square is (1/d)^m.There are d squares and $n \choose m$ sets of m darts.So the probability there is at least one square with exactly m darts in it is:${n \choose m} d ({\frac{1}{d}})^{m}={n \choose m} ({\frac{1}{d}})^{m-1}$To get exactly one square we need to look at the remaining d-1 squares and the remaining n-m darts. The probability that there are no squares with m darts is 1-P(at least one square with m darts). We already know the formula for that probability since it's the same one as previously, but now with d-1 squares and n-m darts.The final probability is therefore:${n \choose m} ({\frac{1}{d}})^{m-1}(1-{n-m \choose m} ({\frac{1}{d-1}})^{m-1})$
>> Anonymous Sat Nov 4 14:25:57 2017 No.9275440 >>9275417For m = 1 this gives negative probabilities.
>> Anonymous Sat Nov 4 14:30:08 2017 No.9275446 >>9275440Think about what m=1 would mean in terms of there being exactly 1 square with m darts.
>> Anonymous Sat Nov 4 14:33:50 2017 No.9275452 >>9275446It means there's one square with one dart, a perfectly reasonable scenario, yet for e.g. n = 3, m = 1, d = 3 you get a "probability" of -3.
>> Anonymous Sat Nov 4 14:37:06 2017 No.9275457 >>9271013Someone follows nn taleb
>> Anonymous Sat Nov 4 14:39:50 2017 No.9275460 >>9275452Hmm. Good point. Where's the flaw in my argument?
>> Anonymous Sat Nov 4 14:41:48 2017 No.9275464 File: 127 KB, 1190x808, IMG_0053.jpg [View same] [iqdb] [saucenao] [google] >>9271013Solution
>> Anonymous Sat Nov 4 15:00:21 2017 No.9275490 File: 75 KB, 980x812, Screenshot from 2017-11-04 14-58-01.png [View same] [iqdb] [saucenao] [google] >>9275464Agrees with simulations but a little stupid that his solution is to brute force the problem with a method that won't scale to large n, m, d. What's the point of posting something like this online?
>> Anonymous Sat Nov 4 15:24:12 2017 No.9275519 File: 52 KB, 935x656, Screenshot from 2017-11-04 15-23-28.png [View same] [iqdb] [saucenao] [google] Actually I think these are the predicates you'd want to use.
>> Anonymous Sat Nov 4 15:28:40 2017 No.9275521 >>9271259autist detected
>> Anonymous Sun Nov 5 00:30:00 2017 No.9276476 >>9275464I don't understand it.
>> Anonymous Sun Nov 5 00:36:21 2017 No.9276482 File: 4 KB, 117x125, 1433132237110s.jpg [View same] [iqdb] [saucenao] [google] Who gives a shit, how does this help me figure out how to pay for my rent AND get the brakes fixed on my car?If there was less maths bullshit, there would be more people working as mechanics and it would cost less to get my brakes fixed. That's real math in the real word.
>> Anonymous Sun Nov 5 02:05:02 2017 No.9276644 File: 2 KB, 464x134, Solved.png [View same] [iqdb] [saucenao] [google] >>9271013An exact symbolic answer is difficult.I just do 1 minus the probability of getting less than m in each box.Generating functions ftw.1) 371759562752/2199023255552 ~ 0.16905667632) Pic
>> Anonymous Sun Nov 5 02:11:35 2017 No.9276651 >>9276644This is for at least 1 box with at least m btw.
>> Anonymous Sun Nov 5 03:40:46 2017 No.9276730 >>9275464>>9276476nvm I do now
>> Anonymous Sun Nov 5 05:56:35 2017 No.9276867 >>9276482xd fuck you nigger
>>
Delete posts | 2022-06-28 19:33:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5720359086990356, "perplexity": 12372.573707204321}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00125.warc.gz"} |
http://statistica.it/gianluca/post/2016-06-03-large-enough-probability/ | # Large enough probability?
I’ve deliberately stayed away from the Brexit referendum (that is in terms of spending time doing some modelling to see what the predictions may be) $-$ mostly because I was otherwise busy and didn’t really have much time.
However, I was talking earlier to my “polls guy” Roberto (yes: I have a polls guy) and he’s suggested I took a look at Stephen Fisher’s blog, who is predicting a 72% chance that the outcome of the referendum will be “remain” (in the EU). I think I’ll take that as a good sign.
In the past couple of days, people have put forward the argument that the UK should take their economy back and “do like New Zealand in the 1980s $-$ that was very good for their economy, so surely it’ll work for us too”. _[start rather annoyed irony]Surely the global economy and trading system were exactly what they are today, back in the 1980s, so surely this is super relevant evidence and something we should be basing our judgement on, today…[end rather annoyed irony]_ | 2020-01-20 00:07:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39722856879234314, "perplexity": 1922.6267141363703}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250595787.7/warc/CC-MAIN-20200119234426-20200120022426-00406.warc.gz"} |
https://advanceddataanalytics.net/page/2/ | # Document worth reading: “Interpreting Blackbox Models via Model Extraction”
Interpretability has become an important issue as machine learning is increasingly used to inform consequential decisions. We propose an approach for interpreting a blackbox model by extracting a decision tree that approximates the model. Our model extraction algorithm avoids overfitting by leveraging blackbox model access to actively sample new training points. We prove that as the number of samples goes to infinity, the decision tree learned using our algorithm converges to the exact greedy decision tree. In our evaluation, we use our algorithm to interpret random forests and neural nets trained on several datasets from the UCI Machine Learning Repository, as well as control policies learned for three classical reinforcement learning problems. We show that our algorithm improves over a baseline based on CART on every problem instance. Furthermore, we show how an interpretation generated by our approach can be used to understand and debug these models. Interpreting Blackbox Models via Model Extraction
# Book Memo: “Statistical Shape Analysis”
With Applications in R A thoroughly revised and updated edition of this introduction to modern statistical methods for shape analysis. Shape analysis is an important tool in the many disciplines where objects are compared using geometrical features. Examples include comparing brain shape in schizophrenia; investigating protein molecules in bioinformatics; and describing growth of organisms in biology. This book is a significant update of the highly-regarded `Statistical Shape Analysis’ by the same authors. The new edition lays the foundations of landmark shape analysis, including geometrical concepts and statistical techniques, and extends to include analysis of curves, surfaces, images and other types of object data. Key definitions and concepts are discussed throughout, and the relative merits of different approaches are presented. The authors have included substantial new material on recent statistical developments and offer numerous examples throughout the text. Concepts are introduced in an accessible manner, while retaining sufficient detail for more specialist statisticians to appreciate the challenges and opportunities of this new field. Computer code has been included for instructional use, along with exercises to enable readers to implement the applications themselves in R and to follow the key ideas by hands-on analysis. Statistical Shape Analysis: with Applications in R will offer a valuable introduction to this fast-moving research area for statisticians and other applied scientists working in diverse areas, including archaeology, bioinformatics, biology, chemistry, computer science, medicine, morphometics and image analysis
# Magister Dixit
“Today, you are much less likely to face a scenario in which you cannot query data and get a response back in a brief period of time. Analytical processes that used to require month, days, or hours have been reduced to minutes, seconds, and fractions of seconds. But shorter processing times have led to higher expectations. Two years ago, many data analysts thought that generating a result from a query in less than 40 minutes was nothing short of miraculous. Today, they expect to see results in under a minute. That’s practically the speed of thought – you think of a query, you get a result, and you begin your experiment. “It’s about moving with greater speed toward previously unknown questions, defining new insights, and reducing the time between when an event happens somewhere in the world and someone responds or reacts to that event,” says Erickson. A rapidly emerging universe of newer technologies has dramatically reduced data processing cycle time, making it possible to explore and experiment with data in ways that would not have been practical or even possible a few years ago. Despite the availability of new tools and systems for handling massive amounts of data at incredible speeds, however, the real promise of advanced data analytics lies beyond the realm of pure technology. “Real-time big data isn’t just a process for storing petabytes or exabytes of data in a data warehouse,” says Michael Minelli, co-author of Big Data, Big Analytics. “It’s about the ability to make better decisions and take meaningful actions at the right time. It’s about detecting fraud while someone is swiping a credit card, or triggering an offer while a shopper is standing on a checkout line, or placing an ad on a website while someone is reading a specific article. It’s about combining and analyzing data so you can take the right action, at the right time, and at the right place.” For some, real-time big data analytics (RTBDA) is a ticket to improved sales, higher profits and lower marketing costs. To others, it signals the dawn of a new era in which machines begin to think and respond more like humans.” Mike Barlow ( 2013 )
# If you did not already know
Deep Rotation Equivariant Network (DREN)
Recently, learning equivariant representations has attracted considerable research attention. Dieleman et al. introduce four operations which can be inserted to CNN to learn deep representations equivariant to rotation. However, feature maps should be copied and rotated four times in each layer in their approach, which causes much running time and memory overhead. In order to address this problem, we propose Deep Rotation Equivariant Network(DREN) consisting of cycle layers, isotonic layers and decycle layers.Our proposed layers apply rotation transformation on filters rather than feature maps, achieving a speed up of more than 2 times with even less memory overhead. We evaluate DRENs on Rotated MNIST and CIFAR-10 datasets and demonstrate that it can improve the performance of state-of-the-art architectures. Our codes are released on GitHub. …
Semantic Matching
Semantic matching is a technique used in computer science to identify information which is semantically related. Given any two graph-like structures, e.g. classifications, taxonomies database or XML schemas and ontologies, matching is an operator which identifies those nodes in the two structures which semantically correspond to one another. For example, applied to file systems it can identify that a folder labeled “car” is semantically equivalent to another folder “automobile” because they are synonyms in English. This information can be taken from a linguistic resource like WordNet. In the recent years many of them have been offered. S-Match is an example of a semantic matching operator. It works on lightweight ontologies, namely graph structures where each node is labeled by a natural language sentence, for example in English. These sentences are translated into a formal logical formula (according to an artificial unambiguous language) codifying the meaning of the node taking into account its position in the graph. For example, in case the folder “car” is under another folder “red” we can say that the meaning of the folder “car” is “red car” in this case. This is translated into the logical formula “red AND car”. The output of S-Match is a set of semantic correspondences called mappings attached with one of the following semantic relations: disjointness (⊥), equivalence (≡), more specific (⊑) and less specific (⊒). In our example the algorithm will return a mapping between ”car” and ”automobile” attached with an equivalence relation. Information semantically matched can also be used as a measure of relevance through a mapping of near-term relationships. Such use of S-Match technology is prevalent in the career space where it is used to gauge depth of skills through relational mapping of information found in applicant resumes. Semantic matching represents a fundamental technique in many applications in areas such as resource discovery, data integration, data migration, query translation, peer to peer networks, agent communication, schema and ontology merging. It using is also being investigated in other areas such as event processing. In fact, it has been proposed as a valid solution to the semantic heterogeneity problem, namely managing the diversity in knowledge. Interoperability among people of different cultures and languages, having different viewpoints and using different terminology has always been a huge problem. Especially with the advent of the Web and the consequential information explosion, the problem seems to be emphasized. People face the concrete problem to retrieve, disambiguate and integrate information coming from a wide variety of sources. …
Waterfall Plot
A waterfall plot is a three-dimensional plot in which multiple curves of data, typically spectra, are displayed simultaneously. Typically the curves are staggered both across the screen and vertically, with ‘nearer’ curves masking the ones behind. The result is a series of ‘mountain’ shapes that appear to be side by side. The waterfall plot is often used to show how two-dimensional information changes over time or some other variable such as rpm. The term ‘waterfall plot’ is sometimes used interchangeably with ‘spectrogram’ or ‘Cumulative Spectral Decay’ (CSD) plot. …
# Whats new on arXiv
This paper presents two unsupervised learning layers (UL layers) for label-free video analysis: one for fully connected layers, and the other for convolutional ones. The proposed UL layers can play two roles: they can be the cost function layer for providing global training signal; meanwhile they can be added to any regular neural network layers for providing local training signals and combined with the training signals backpropagated from upper layers for extracting both slow and fast changing features at layers of different depths. Therefore, the UL layers can be used in either pure unsupervised or semi-supervised settings. Both a closed-form solution and an online learning algorithm for two UL layers are provided. Experiments with unlabeled synthetic and real-world videos demonstrated that the neural networks equipped with UL layers and trained with the proposed online learning algorithm can extract shape and motion information from video sequences of moving objects. The experiments demonstrated the potential applications of UL layers and online learning algorithm to head orientation estimation and moving object localization.
Variational inference is a powerful approach for approximate posterior inference. However, it is sensitive to initialization and can be subject to poor local optima. In this paper, we develop proximity variational inference (PVI). PVI is a new method for optimizing the variational objective that constrains subsequent iterates of the variational parameters to robustify the optimization path. Consequently, PVI is less sensitive to initialization and optimization quirks and finds better local optima. We demonstrate our method on three proximity statistics. We study PVI on a Bernoulli factor model and sigmoid belief network with both real and synthetic data and compare to deterministic annealing (Katahira et al., 2008). We highlight the flexibility of PVI by designing a proximity statistic for Bayesian deep learning models such as the variational autoencoder (Kingma and Welling, 2014; Rezende et al., 2014). Empirically, we show that PVI consistently finds better local optima and gives better predictive performance.
Generative adversarial networks (GAN) approximate a target data distribution by jointly optimizing an objective function through a ‘two-player game’ between a generator and a discriminator. Despite their empirical success, however, two very basic questions on how well they can approximate the target distribution remain unanswered. First, it is not known how restricting the discriminator family affects the approximation quality. Second, while a number of different objective functions have been proposed, we do not understand when convergence to the global minima of the objective function leads to convergence to the target distribution under various notions of distributional convergence. In this paper, we address these questions in a broad and unified setting by defining a notion of adversarial divergences that includes a number of recently proposed objective functions. We show that if the objective function is an adversarial divergence with some additional conditions, then using a restricted discriminator family has a moment-matching effect. Additionally, we show that for objective functions that are strict adversarial divergences, convergence in the objective function implies weak convergence, thus generalizing previous results.
We present a neural network technique for the analysis and extrapolation of time-series data called Neural Decomposition (ND). Units with a sinusoidal activation function are used to perform a Fourier-like decomposition of training samples into a sum of sinusoids, augmented by units with nonperiodic activation functions to capture linear trends and other nonperiodic components. We show how careful weight initialization can be combined with regularization to form a simple model that generalizes well. Our method generalizes effectively on the Mackey-Glass series, a dataset of unemployment rates as reported by the U.S. Department of Labor Statistics, a time-series of monthly international airline passengers, the monthly ozone concentration in downtown Los Angeles, and an unevenly sampled time-series of oxygen isotope measurements from a cave in north India. We find that ND outperforms popular time-series forecasting techniques including LSTM, echo state networks, ARIMA, SARIMA, SVR with a radial basis function, and Gashler and Ashmore’s model.
This work presents a rigorous statistical analysis of adversarial training for generative models, advancing recent work by Arjovsky and Bottou [2]. A key element is the distinction between the objective function with respect to the (unknown) data distribution, and its empirical counterpart. This yields a straight-forward explanation for common pathologies in practical adversarial training such as vanishing gradients. To overcome such issues, we pursue the idea of smoothing the Jensen-Shannon Divergence (JSD) by incorporating noise in the formulation of the discriminator. As we show, this effectively leads to an empirical version of the JSD in which the true and the generator densities are replaced by kernel density estimates. We analyze statistical consistency of this objective, and demonstrate its practical effectiveness.
Recurrent neural networks have achieved remarkable success at generating sequences with complex structures, thanks to advances that include richer embeddings of input and cures for vanishing gradients. Trained only on sequences from a known grammar, though, they can still struggle to learn rules and constraints of the grammar. Neural Attribute Machines (NAMs) are equipped with a logical machine that represents the underlying grammar, which is used to teach the constraints to the neural machine by (i) augmenting the input sequence, and (ii) optimizing a custom loss function. Unlike traditional RNNs, NAMs are exposed to the grammar, as well as samples from the language of the grammar. During generation, NAMs make significantly fewer violations of the constraints of the underlying grammar than RNNs trained only on samples from the language of the grammar.
Machine learning and data analysis now finds both scientific and industrial application in biology, chemistry, geology, medicine, and physics. These applications rely on large quantities of data gathered from automated sensors and user input. Furthermore, the dimensionality of many datasets is extreme: more details are being gathered about single user interactions or sensor readings. All of these applications encounter problems with a common theme: use observed data to make inferences about the world. Our work obtains the first provably efficient algorithms for Independent Component Analysis (ICA) in the presence of heavy-tailed data. The main tool in this result is the centroid body (a well-known topic in convex geometry), along with optimization and random walks for sampling from a convex body. This is the first algorithmic use of the centroid body and it is of independent theoretical interest, since it effectively replaces the estimation of covariance from samples, and is more generally accessible. This reduction relies on a non-linear transformation of samples from such an intersection of halfspaces (i.e. a simplex) to samples which are approximately from a linearly transformed product distribution. Through this transformation of samples, which can be done efficiently, one can then use an ICA algorithm to recover the vertices of the intersection of halfspaces. Finally, we again use ICA as an algorithmic primitive to construct an efficient solution to the widely-studied problem of learning the parameters of a Gaussian mixture model. Our algorithm again transforms samples from a Gaussian mixture model into samples which fit into the ICA model and, when processed by an ICA algorithm, result in recovery of the mixture parameters. Our algorithm is effective even when the number of Gaussians in the mixture grows polynomially with the ambient dimension
Content popularity prediction has been extensively studied due to its importance and interest for both users and hosts of social media sites like Facebook, Instagram, Twitter, and Pinterest. However, existing work mainly focuses on modeling popularity using a single metric such as the total number of likes or shares. In this work, we propose Diffusion-LSTM, a memory-based deep recurrent network that learns to recursively predict the entire diffusion path of an image through a social network. By combining user social features and image features, and encoding the diffusion path taken thus far with an explicit memory cell, our model predicts the diffusion path of an image more accurately compared to alternate baselines that either encode only image or social features, or lack memory. By mapping individual users to user prototypes, our model can generalize to new users not seen during training. Finally, we demonstrate our model’s capability of generating diffusion trees, and show that the generated trees closely resemble ground-truth trees.
We study implicit regularization when optimizing an underdetermined quadratic objective over a matrix $X$ with gradient descent on a factorization of $X$. We conjecture and provide empirical and theoretical evidence that with small enough step sizes and initialization close enough to the origin, gradient descent on a full dimensional factorization converges to the minimum nuclear norm solution.
# Document worth reading: “Living Together: Mind and Machine Intelligence”
In this paper we consider the nature of the machine intelligences we have created in the context of our human intelligence. We suggest that the fundamental difference between human and machine intelligence comes down to \emph{embodiment factors}. We define embodiment factors as the ratio between an entity’s ability to communicate information vs compute information. We speculate on the role of embodiment factors in driving our own intelligence and consciousness. We briefly review dual process models of cognition and cast machine intelligence within that framework, characterising it as a dominant System Zero, which can drive behaviour through interfacing with us subconsciously. Driven by concerns about the consequence of such a system we suggest prophylactic courses of action that could be considered. Our main conclusion is that it is \emph{not} sentient intelligence we should fear but \emph{non-sentient} intelligence. Living Together: Mind and Machine Intelligence
# Book Memo: “Health 4.0”
How Virtualization and Big Data are Revolutionizing Healthcare This book describes how the creation of new digital services—through vertical and horizontal integration of data coming from sensors on top of existing legacy systems—that has already had a major impact on industry is now extending to healthcare. The book describes the fourth industrial revolution (i.e. Health 4.0), which is based on virtualization and service aggregation. It shows how sensors, embedded systems, and cyber-physical systems are fundamentally changing the way industrial processes work, their business models, and how we consume, while also affecting the health and care domains. Chapters describe the technology behind the shift of point of care to point of need and away from hospitals and institutions; how care will be delivered virtually outside hospitals; that services will be tailored to individuals rather than being designed as statistical averages; that data analytics will be used to help patients to manage their chronic conditions with help of smart devices; and that pharmaceuticals will be interactive to help prevent adverse reactions. The topics presented will have an impact on a variety of healthcare stakeholders in a continuously global and hyper-connected world.
# Distilled News
One of the most common question we get on Analytics Vidhya is: ‘How much maths do I need to learn to be a data scientist?’ Even though the question sounds simple, there is no simple answer to the the question. Usually, we say that you need to know basic descriptive and inferential statistics to start. That is good to start. But, once you have covered the basic concepts in machine learning, you will need to learn some more math. You need it to understand how these algorithms work. What are their limitations and in case they make any underlying assumptions. Now, there could be a lot of areas to study including algebra, calculus, statistics, 3-D geometry etc. If you get confused (like I did) and ask experts what should you learn at this stage, most of them would suggest / agree that you go ahead with Linear Algebra. But, the problem does not stop there. The next challenge is to figure out how to learn Linear Algebra. You can get lost in the detailed mathematics and derivation and learning them would not help as much! I went through that journey myself and hence decided to write this comprehensive guide. If you have faced this question about how to learn & what to learn in Linear Algebra – you are at the right place. Just follow this guide.
Our Two Sigma Financial Modeling Challenge ran from December 2016 to March 2017 this year. Asked to search for signal in financial markets data with limited hardware and computational time, this competition attracted over 2000 competitors. In this winners’ interview, 2nd place winners’ Nima and Chahhou describe how paying close attention to unreliable engineered features was important to building a successful model.
It’s been a while since my last post on some TB WHO data. A lot has happened since then, including the opportunity to attend the Open Data Science Conference (ODSC) East held in Boston, MA. Over a two day period I had the opportunity to listen to a number of leaders in various industries and fields. It was inspiring to learn about the wide variety of data science applications ranging from finance and marketing to genomics and even the refugee crisis. One of the workshops at ODSC was text analytics, which includes basic text processing, dendrograms, natural language processing and sentiment analysis. This gave me the thought of applying some text analytics to visualize some data I was working on last summer. In this post I’m going to walk through how I used regular expression to label classification codes in a large dataset (NHAMCS) representing emergency department visits in the United States and eventually visualize the data.
As also described in Cormen, et al (2009) p. 65, in algorithm design, divide-and-conquer paradigm incorporates a recursive approach in which the main problem is:
• Divided into smaller sub-problems (divide),
• The sub-problems are solved (conquer),
• And the solutions to sub-problems are combined to solve the original and “bigger” problem (combine).
Instead of constructing indefinite number of nested loops destroying the readability of the code and the performance of execution, the “recursive” way utilizes just one block of code which calls itself (hence the term “recursive”) for the smaller problem. The main point is to define a “stop” rule, so that the function does not sink into an infinite recursion depth. While nested loops modify the same object (or address space in the low level sense), recursion moves the “stack pointer”, so each recursion depth uses a different part of the stack (a copy of the objects will be created for each recursion). This illustrates a well-known trade-off in algorithm design: Memory versus performance; recursion enhances performance at the expense of using more memory.
This is a keynote highlight from the Strata Data Conference in London 2017.
In this episode of the Data Show, I spoke with Jeremy Stanley, VP of data science at Instacart, a popular grocery delivery service that is expanding rapidly. As Stanley describes it, Instacart operates a four-sided marketplace comprised of retail stores, products within the stores, shoppers assigned to the stores, and customers who order from Instacart. The objective is to get fresh groceries from popular retailers delivered to customers in a timely fashion. Instacart’s goals land them in the center of the many opportunities and challenges involved in building high-impact data products.
This is a highlight from Ted Malaska’s Introduction to Apache Spark for Java and Scala developers.
DataScience.com new Python library, Skater, uses a combination of model interpretation algorithms to identify how models leverage data to make predictions.
Data science and machine learning are iterative processes. It is never possible to successfully complete a data science project in a single pass. A data scientist constantly tries new ideas and changes steps of his pipeline:
1. extract new features and accidentally find noise in the data
2. clean up the noise, find one more promising feature
3. extract the new feature
4. rebuild and validate the model, realize that the learning algorithm parameters are not perfect for the new feature set
5. change machine learning algorithm parameters and retrain the model
6. find the ineffective feature subset and remove it from the feature set
7. try a few more new features
8. try another ML algorithm. And then a data format change is required.
This is only a small episode in a data scientist’s daily life and it is what makes our job different from a regular engineering job.
The machine learning revolution leaves no stone unturned. Natural language processing is yet another field that underwent a small revolution thanks to the second coming of artificial neural networks. Let’s just briefly discuss two advances in the natural language processing toolbox made thanks to artificial neural networks and deep learning techniques.
Data science is an interdisciplinary field where scientific techniques from statistics, mathematics, and computer science are used to analyze data and solve problems more accurately and effectively. It is no wonder, then, that languages such as R and Python, with their extensive packages and libraries that support statistical methods and machine learning algorithms are cornerstones of the data science revolution. Often times, beginners find it hard to decide which language to learn first. This guide will help you make that decision.
The simplest solutions are usually the most powerful ones, and Naive Bayes is a good proof of that. In spite of the great advances of the Machine Learning in the last years, it has proven to not only be simple but also fast, accurate and reliable. It has been successfully used for many purposes, but it works particularly well with natural language processing (NLP) problems. Naive Bayes is a family of probabilistic algorithms that take advantage of probability theory and Bayes’ Theorem to predict the category of a sample (like a piece of news or a customer review). They are probabilistic, which means that they calculate the probability of each category for a given sample, and then output the category with the highest one. The way they get these probabilities is by using Bayes’ Theorem, which describes the probability of a feature, based on prior knowledge of conditions that might be related to that feature. We’re going to be working with an algorithm called Multinomial Naive Bayes. We’ll walk through the algorithm applied to NLP with an example, so by the end not only will you know how this method works, but also why it works. Then, we’ll lay out a few advanced techniques that can make Naive Bayes competitive with more complex Machine Learning algorithms, such as SVM and neural networks.
Welcome to Part 2 of our tour through modern machine learning algorithms. In this part, we’ll cover methods for Dimensionality Reduction, further broken into Feature Selection and Feature Extraction. In general, these tasks are rarely performed in isolation. Instead, they’re often preprocessing steps to support other tasks.
Brian Hopkins of Forrester Research recently penned an excellent blog post about why companies are getting disrupted and why they realize it so late. The post draws from Ray Kurzweil’s Law Of Accelerating Returns and speaks to the fact that the human brain doesn’t do well with exponential growth.
This post outlines an entire 6-part tutorial series on the MXNet deep learning library and its Python API. In-depth and descriptive, this is a great guide for anyone looking to start leveraging this powerful neural network library.
If you build a model and never update it you’re missing a trick. Behaviours change so your model will tend to perform worse over time. You’ve got to regularly refresh it, whether that’s adjusting the existing model to fit the latest data (recalibration) or building a whole new model (retraining), but this means you’ve got new versions of your model that you have to handle. You need to think about your methodology for versioning R model objects, ideally before you lose any versions. You could store models with ye olde YYYYMMDD style of versioning but that means regularly changing your code to use the latest model version. I’m too lazy for that! If we’re storing our R model objects in SQL Server then we can utilise another SQL Server capability, temporal tables, to take the pain out of versioning and make it super simple. Temporal tables will track changes automatically so you would overwrite the previous model with the new one and it would keep a copy of the old one automagically in a history table. You get to always use the latest version via the main table but you can then write temporal queries to extract any version of the model that’s ever been implemented. Super neat! For some of you, if you’re not interested in the technical details you can drop off now with the knowledge that you can store your models in a non-destructive but easy to use way in SQL Server if you need to. If you want to see how it’s done, read on!
In this special guest feature, Irshad Raihan, Product Marketing Manager at Red Hat Storage, discusses how organizations can save money and realize greater flexibility by moving data with lower business value to a more affordable storage solution. Irshad Raihan is a product manager at Red Hat Storage, responsible for product strategy, messaging, and go to market activities. Previously, he held senior product marketing and product management positions at HP and IBM responsible for big data and data management products. Irshad holds a Masters in Computer Science from Clemson University, and an MBA from Carnegie Mellon University.
This is a short blog post to introduce the concept of an ontology for those who are unfamiliar with the term, or who have previously encountered explanations that make little or no sense, as I have. I’m aiming to “democratise knowledge of this topic” as one of my colleagues put it.
# R Packages worth a look
Constructing an Epistemic Model for the Games with Two Players (EpistemicGameTheory)
Constructing an epistemic model such that, for every player i and for every choice c(i) which is optimal, there is one type that expresses common belief in rationality.
Bayesian Network Belief Propagation (BayesNetBP)
Belief propagation methods in Bayesian Networks to propagate evidence through the network. The implementation of these methods are based on the article: Cowell, RG (2005). Local Propagation in Conditional Gaussian Bayesian Networks <http://…/>.
Quick Generalized Full Matching (quickmatch)
Provides functions for constructing near-optimal generalized full matching. Generalized full matching is an extension of the original full matching method to situations with more intricate study designs. The package is made with large data sets in mind and derives matches more than an order of magnitude quicker than other methods.
Convolution of Gamma Distributions (coga)
Convolution of gamma distributions in R. The convolution of gamma distributions is the sum of series of gamma distributions and all gamma distributions here can have different parameters. This package can calculate density, distribution function and do simulation work.
Graphical User Interface for Generalized Multistate Simulation Model (GUIgems)
A graphical user interface for the R package Gems. Apart from the functionality of Gems package in the Graphical User interface, GUIgems allows adding states to a defined model, merging states for the analysis and plotting progression paths between states based on the simulated cohort. There is also a module in the GUIgems which allows to compare costs and QALYs between different cohorts.
# R Packages worth a look
A Faster Implementation of the Poisson-Binomial Distribution (poisbinom)
Provides the probability, distribution, and quantile functions and random number generator for the Poisson-Binomial distribution. This package relies on FFTW to implement the discrete Fourier transform, so that it is much faster than the existing implementation of the same algorithm in R.
Evolutionary Monte Carlo (EMC) Methods for Clustering (EMCC)
Evolutionary Monte Carlo methods for clustering, temperature ladder construction and placement. This package implements methods introduced in Goswami, Liu and Wong (2007) <doi:10.1198/106186007X255072>. The paper above introduced probabilistic genetic-algorithm-style crossover moves for clustering. The paper applied the algorithm to several clustering problems including Bernoulli clustering, biological sequence motif clustering, BIC based variable selection, mixture of Normals clustering, and showed that the proposed algorithm performed better both as a sampler and as a stochastic optimizer than the existing tools, namely, Gibbs sampling, “split-merge” Metropolis-Hastings algorithm, K-means clustering, and the MCLUST algorithm (in the package ‘mclust’).
JAR Files of the Apache Commons Mathematics Library (commonsMath)
Java JAR files for the Apache Commons Mathematics Library for use by users and other packages.
Interactive Graphs with R (RJSplot)
Creates interactive graphs with ‘R’. It joins the data analysis power of R and the visualization libraries of JavaScript in one package.
Simulation Education (simEd)
Contains various functions to be used for simulation education, including queueing simulation functions, variate generation functions capable of producing independent streams and antithetic variates, functions for illustrating random variate generation for various discrete and continuous distributions, and functions to compute time-persistent statistics. Also contains two queueing data sets (one fabricated, one real-world) to facilitate input modeling. | 2017-05-29 22:47:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27763286232948303, "perplexity": 1097.8734781286378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463613135.2/warc/CC-MAIN-20170529223110-20170530003110-00455.warc.gz"} |
https://matheducators.stackexchange.com/questions/3987/what-is-the-best-way-to-intuitively-explain-understand-and-approach-p-vs-np/4009 | # What is the best way to intuitively explain, understand and approach P vs NP
P vs NP is a million dollar millenium problem.
Essentially it boils down to, If it is easy to check that a solution to a problem is correct, is it also easy to solve the problem? This is the essence of the P vs NP question. Typical of the NP problems is that of the Hamiltonian Path Problem: given N cities to visit, how can one do this without visiting a city twice? If you give me a solution, I can easily check that it is correct. But I cannot so easily find a solution.
Per a comment, it became clear that I should add some extra clarity! :)
In what ways can we intuitively teach HS-college students how to understand and attack a problem like this. Not P vs NP in particular but rather, a truly difficult, if not unsolvable problem. I think finding ways to to reconcile these types of problems, is important to developing really fine tuned critical thinking and analysis skills. With that, I do want them to have an intuition around P vs NP in this particular case, to set the stage for a very difficult problem. And I do want them to try out ways to approach difficult problems
• To be clear: What exactly is the goal? The title says "to intuitively explain," but the body of the post includes "...and attack." I think it may be very difficult to help high school students unfamiliar with P/NP to mount a direct attack on it; maybe they can gain an understanding of it (intuitive or otherwise) that motivates them to pursue the problem later on, but I strongly believe encouraging them to solve a problem without a strong foundation will lead to discouragement and, in some cases, crankery. So: Is it just the (intuitive) explanation, or do you really want them to attack it? – Benjamin Dickman Jul 14 '14 at 11:44
• @BenjaminDickman Thank you for the comment. I intended the "approach" component of the title to refer to the "attack" portion of the body. I asked the best ways to attack a problem like this. Not P Vs NP itself, but a truly difficult, if not unsolvable problem. I think finding ways to to reconcile these types of problems, is important to developing really fine tuned critical thinking and analysis skills. With that, I do want them to have an intuition around P vs NP in particular, to set the stage for a very difficult problem. And I do want them to try out ways to approach difficult problems. – David BasedMathematician Coven Jul 14 '14 at 16:22
• I wouldn't encourage high school students to attack P vs NP, or even to develop intuition about it. I'd rather encourage them to attack the Riemann hypothesis or to find an $O(n^2)$ algorithm for matrix multiplication, which are simpler. – user173 Jul 15 '14 at 2:27
• "I asked the best ways to attack a problem like this... a truly difficult if not unsolvable problem. I think finding ways to reconcile these types of problems is important to developing really fine tuned critical thinking and analysis skills." What is the last sentence based upon? A more common approach would be to give students problems that are within their grasp or just beyond it (cf. Vygotsky's ZPD) rather than unsolvable millennium problems! Moreover, there are much more easily stated intractable problems; e.g. the Twin Prime Conjecture or even math.stackexchange.com/q/776447/37122 – Benjamin Dickman Jul 17 '14 at 1:10 | 2019-09-19 17:51:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6386834979057312, "perplexity": 436.2440753584562}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573561.45/warc/CC-MAIN-20190919163337-20190919185337-00528.warc.gz"} |
http://www.turkmath.org/beta/seminer.php?id_seminer=1896 | #### Bilkent University Topology Seminars
An algebraic approach to Mislin's theorem
Hatice Mutlu
Bilkent University, Turkey
Özet : Mislin shows that an isomorphism of mod-p cohomology groups implies control of p-fusion between corresponding compact Lie groups, in particular, between finite groups. I will give a talk on the alternative approach to Mislin's theorem given by Peter Symonds based on global cohomological Mackey functors.
Tarih : 26.11.2018 Saat : 13:40 Yer : Mathematics Seminar Room, SA - 141. Dil : English | 2018-12-18 13:53:38 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9075348973274231, "perplexity": 7063.113946409846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376829399.59/warc/CC-MAIN-20181218123521-20181218145521-00416.warc.gz"} |
https://wikimili.com/en/Computer_algebra | # Computer algebra
Last updated
In mathematics and computer science, [1] computer algebra, also called symbolic computation or algebraic computation, is a scientific area that refers to the study and development of algorithms and software for manipulating mathematical expressions and other mathematical objects. Although computer algebra could be considered a subfield of scientific computing, they are generally considered as distinct fields because scientific computing is usually based on numerical computation with approximate floating point numbers, while symbolic computation emphasizes exact computation with expressions containing variables that have no given value and are manipulated as symbols.
## Contents
Software applications that perform symbolic calculations are called computer algebra systems , with the term system alluding to the complexity of the main applications that include, at least, a method to represent mathematical data in a computer, a user programming language (usually different from the language used for the implementation), a dedicated memory manager, a user interface for the input/output of mathematical expressions, a large set of routines to perform usual operations, like simplification of expressions, differentiation using chain rule, polynomial factorization, indefinite integration, etc.
Computer algebra is widely used to experiment in mathematics and to design the formulas that are used in numerical programs. It is also used for complete scientific computations, when purely numerical methods fail, as in public key cryptography, or for some non-linear problems.
## Terminology
Some authors distinguish computer algebra from symbolic computation using the latter name to refer to kinds of symbolic computation other than the computation with mathematical formulas. Some authors use symbolic computation for the computer science aspect of the subject and "computer algebra" for the mathematical aspect. [2] In some languages the name of the field is not a direct translation of its English name. Typically, it is called calcul formel in French, which means "formal computation". This name reflects the ties this field has with formal methods.
Symbolic computation has also been referred to, in the past, as symbolic manipulation, algebraic manipulation, symbolic processing, symbolic mathematics, or symbolic algebra, but these terms, which also refer to non-computational manipulation, are no longer used in reference to computer algebra.
## Scientific community
There is no learned society that is specific to computer algebra, but this function is assumed by the special interest group of the Association for Computing Machinery named SIGSAM (Special Interest Group on Symbolic and Algebraic Manipulation). [3]
There are several annual conferences on computer algebra, the premier being ISSAC (International Symposium on Symbolic and Algebraic Computation), which is regularly sponsored by SIGSAM. [4]
There are several journals specializing in computer algebra, the top one being Journal of Symbolic Computation founded in 1985 by Bruno Buchberger. [5] There are also several other journals that regularly publish articles in computer algebra. [6]
## Computer science aspects
### Data representation
As numerical software is highly efficient for approximate numerical computation, it is common, in computer algebra, to emphasize exact computation with exactly represented data. Such an exact representation implies that, even when the size of the output is small, the intermediate data generated during a computation may grow in an unpredictable way. This behavior is called expression swell. To obviate this problem, various methods are used in the representation of the data, as well as in the algorithms that manipulate them.
#### Numbers
The usual numbers systems used in numerical computation are floating point numbers and integers of a fixed bounded size. None of these is convenient for computer algebra, due to expression swell.[ citation needed ]
Therefore, the basic numbers used in computer algebra are the integers of the mathematicians, commonly represented by an unbounded signed sequence of digits in some base of numeration, usually the largest base allowed by the machine word. These integers allow to define the rational numbers, which are irreducible fractions of two integers.
Programming an efficient implementation of the arithmetic operations is a hard task. Therefore, most free computer algebra systems and some commercial ones such as Mathematica and Maple (software), use the GMP library, which is thus a de facto standard.
#### Expressions
Except for numbers and variables, every mathematical expression may be viewed as the symbol of an operator followed by a sequence of operands. In computer algebra software, the expressions are usually represented in this way. This representation is very flexible, and many things that seem not to be mathematical expressions at first glance, may be represented and manipulated as such. For example, an equation is an expression with “=” as an operator, a matrix may be represented as an expression with “matrix” as an operator and its rows as operands.
Even programs may be considered and represented as expressions with operator “procedure” and, at least, two operands, the list of parameters and the body, which is itself an expression with “body” as an operator and a sequence of instructions as operands. Conversely, any mathematical expression may be viewed as a program. For example, the expression a + b may be viewed as a program for the addition, with a and b as parameters. Executing this program consists in evaluating the expression for given values of a and b; if they do not have any value—that is they are indeterminates—, the result of the evaluation is simply its input.
This process of delayed evaluation is fundamental in computer algebra. For example, the operator “=” of the equations is also, in most computer algebra systems, the name of the program of the equality test: normally, the evaluation of an equation results in an equation, but, when an equality test is needed,—either explicitly asked by the user through an “evaluation to a Boolean” command, or automatically started by the system in the case of a test inside a program—then the evaluation to a boolean 0 or 1 is executed.
As the size of the operands of an expression is unpredictable and may change during a working session, the sequence of the operands is usually represented as a sequence of either pointers (like in Macsyma) or entries in a hash table (like in Maple).
### Simplification
The raw application of the basic rules of differentiation with respect to x on the expression ${\displaystyle a^{x}}$ gives the result
${\displaystyle x\cdot a^{x-1}\cdot 0+a^{x}\cdot \left(1\cdot \log a+x\cdot {\frac {0}{a}}\right).}$
Such a complicated expression is clearly not acceptable, and a procedure of simplification is needed as soon as one works with general expressions.
This simplification is normally done through rewriting rules. [8] There are several classes of rewriting rules that have to be considered. The simplest consists in the rewriting rules that always reduce the size of the expression, like EE → 0 or sin(0) → 0. They are systematically applied in computer algebra systems.
The first difficulty occurs with associative operations like addition and multiplication. The standard way to deal with associativity is to consider that addition and multiplication have an arbitrary number of operands, that is that a + b + c is represented as "+"(a, b, c). Thus a + (b + c) and (a + b) + c are both simplified to "+"(a, b, c), which is displayed a + b + c. What about ab + c? To deal with this problem, the simplest way is to rewrite systematically E, EF, E/F as, respectively, (1)⋅E, E + (1)⋅F, EF1. In other words, in the internal representation of the expressions, there is no subtraction nor division nor unary minus, outside the representation of the numbers.
A second difficulty occurs with the commutativity of addition and multiplication. The problem is to recognize quickly the like terms in order to combine or cancel them. In fact, the method for finding like terms, consisting of testing every pair of terms, is too costly for being practicable with very long sums and products. For solving this problem, Macsyma sorts the operands of sums and products with a function of comparison that is designed in order that like terms are in consecutive places, and thus easily detected. In Maple, the hash function is designed for generating collisions when like terms are entered, allowing to combine them as soon as they are introduced. This design of the hash function allows also to recognize immediately the expressions or subexpressions that appear several times in a computation and to store them only once. This allows not only to save some memory space but also to speed up computation, by avoiding repetition of the same operations on several identical expressions.
Some rewriting rules sometimes increase and sometimes decrease the size of the expressions to which they are applied. This is the case of distributivity or trigonometric identities. For example, the distributivity law allows rewriting ${\displaystyle (x+1)^{4}\rightarrow x^{4}+4x^{3}+6x^{2}+4x+1}$ and ${\displaystyle (x-1)(x^{4}+x^{3}+x^{2}+x+1)\rightarrow x^{5}-1.}$ As there is no way to make a good general choice of applying or not such a rewriting rule, such rewritings are done only when explicitly asked for by the user. For the distributivity, the computer function that applies this rewriting rule is generally called "expand". The reverse rewriting rule, called "factor", requires a non-trivial algorithm, which is thus a key function in computer algebra systems (see Polynomial factorization).
## Mathematical aspects
In this section we consider some fundamental mathematical questions that arise as soon as one wants to manipulate mathematical expressions in a computer. We consider mainly the case of the multivariate rational fractions. This is not a real restriction, because, as soon as the irrational functions appearing in an expression are simplified, they are usually considered as new indeterminates. For example,
${\displaystyle (\sin(x+y)^{2}+\log(z^{2}-5))^{3}}$
is viewed as a polynomial in ${\displaystyle \sin(x+y)}$ and ${\displaystyle \log(z^{2}-5)}$
### Equality
There are two notions of equality for mathematical expressions. The syntactic equality is the equality of the expressions which means that they are written (or represented in a computer) in the same way. Being trivial, the syntactic equality is rarely considered by mathematicians, although it is the only equality that is easy to test with a program. The semantic equality is when two expressions represent the same mathematical object, like in
${\displaystyle (x+y)^{2}=x^{2}+2xy+y^{2}.}$
It is known from Richardson's theorem that there may not exist an algorithm that decides if two expressions representing numbers are semantically equal, if exponentials and logarithms are allowed in the expressions. Therefore, (semantical) equality may be tested only on some classes of expressions such as the polynomials and rational fractions.
To test the equality of two expressions, instead of designing specific algorithms, it is usual to put expressions in some canonical form or to put their difference in a normal form, and to test the syntactic equality of the result.
Unlike in usual mathematics, "canonical form" and "normal form" are not synonymous in computer algebra. [9] A canonical form is such that two expressions in canonical form are semantically equal if and only if they are syntactically equal, while a normal form is such that an expression in normal form is semantically zero only if it is syntactically zero. In other words, zero has a unique representation by expressions in normal form.
Normal forms are usually preferred in computer algebra for several reasons. Firstly, canonical forms may be more costly to compute than normal forms. For example, to put a polynomial in canonical form, one has to expand by distributivity every product, while it is not necessary with a normal form (see below). Secondly, it may be the case, like for expressions involving radicals, that a canonical form, if it exists, depends on some arbitrary choices and that these choices may be different for two expressions that have been computed independently. This may make impracticable the use of a canonical form.
## History
At the beginning of computer algebra, circa 1970, when the long-known algorithms were first put on computers, they turned out to be highly inefficient. [10] Therefore, a large part of the work of the researchers in the field consisted in revisiting classical algebra in order to make it effective and to discover efficient algorithms to implement this effectiveness. A typical example of this kind of work is the computation of polynomial greatest common divisors, which is required to simplify fractions. Surprisingly, the classical Euclid's algorithm turned out to be inefficient for polynomials over infinite fields, and thus new algorithms needed to be developed. The same was also true for the classical algorithms from linear algebra.
## Related Research Articles
Algebraic geometry is a branch of mathematics, classically studying zeros of multivariate polynomials. Modern algebraic geometry is based on the use of abstract algebraic techniques, mainly from commutative algebra, for solving geometrical problems about these sets of zeros.
In mathematics, an equation is a statement that asserts the equality of two expressions, which are connected by the equals sign "=". The word equation and its cognates in other languages may have subtly different meanings; for example, in French an équation is defined as containing one or more variables, while in English, any equality is an equation.
In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. An example of a polynomial of a single indeterminate x is x2 − 4x + 7. An example in three variables is x3 + 2xyz2yz + 1.
A computer algebra system (CAS) or symbolic algebra system (SAS) is any mathematical software with the ability to manipulate mathematical expressions in a way similar to the traditional manual computations of mathematicians and scientists. The development of the computer algebra systems in the second half of the 20th century is part of the discipline of "computer algebra" or "symbolic computation", which has spurred work in algorithms over mathematical objects such as polynomials.
In logic and computer science, unification is an algorithmic process of solving equations between symbolic expressions.
Maple is a symbolic and numeric computing environment as well as a multi-paradigm programming language. It covers several areas of technical computing, such as symbolic mathematics, numerical analysis, data processing, visualization, and others. A toolbox, MapleSim, adds functionality for multidomain physical modeling and code generation.
In symbolic computation, the Risch algorithm is an algorithm for indefinite integration. It is used in some computer algebra systems to find antiderivatives. It is named after the American mathematician Robert Henry Risch, a specialist in computer algebra who developed it in 1968.
In mathematics, and more specifically in computer algebra, computational algebraic geometry, and computational commutative algebra, a Gröbner basis is a particular kind of generating set of an ideal in a polynomial ring K[x1, ..,xn] over a field K. A Gröbner basis allows many important properties of the ideal and the associated algebraic variety to be deduced easily, such as the dimension and the number of zeros when it is finite. Gröbner basis computation is one of the main practical tools for solving systems of polynomial equations and computing the images of algebraic varieties under projections or rational maps.
In mathematics and computer science, a canonical, normal, or standardform of a mathematical object is a standard way of presenting that object as a mathematical expression. Often, it is one which provides the simplest representation of an object and which allows it to be identified in a unique way. The distinction between "canonical" and "normal" forms varies from subfield to subfield. In most fields, a canonical form specifies a unique representation for every object, while a normal form simply specifies its form, without the requirement of uniqueness.
The Knuth–Bendix completion algorithm is a semi-decision algorithm for transforming a set of equations into a confluent term rewriting system. When the algorithm succeeds, it effectively solves the word problem for the specified algebra.
In computational algebraic geometry and computational commutative algebra, Buchberger's algorithm is a method of transforming a given set of generators for a polynomial ideal into a Gröbner basis with respect to some monomial order. It was invented by Austrian mathematician Bruno Buchberger. One can view it as a generalization of the Euclidean algorithm for univariate GCD computation and of Gaussian elimination for linear systems.
In calculus, symbolic integration is the problem of finding a formula for the antiderivative, or indefinite integral, of a given function f(x), i.e. to find a differentiable function F(x) such that
Affine arithmetic (AA) is a model for self-validated numerical analysis. In AA, the quantities of interest are represented as affine combinations of certain primitive variables, which stand for sources of uncertainty in the data or approximations made during the computation.
In mathematics and computer algebra, factorization of polynomials or polynomial factorization expresses a polynomial with coefficients in a given field or in the integers as the product of irreducible factors with coefficients in the same domain. Polynomial factorization is one of the fundamental components of computer algebra systems.
In mathematics and computer science, a word problem for a set S with respect to a system of finite encodings of its elements is the algorithmic problem of deciding whether two given representatives represent the same element of the set. The problem is commonly encountered in abstract algebra, where given a presentation of an algebraic structure by generators and relators, the problem is to determine if two expressions represent the same element; a prototypical example is the word problem for groups. Less formally, the word problem in an algebra is: given a set of identities E, and two expressions x and y, is it possible to transform x into y using the identities in E as rewriting rules in both directions? While answering this question may not seem hard, the remarkable result that emerges, in many important cases, is that the problem is undecidable.
In computer algebra, the Faugère F4 algorithm, by Jean-Charles Faugère, computes the Gröbner basis of an ideal of a multivariate polynomial ring. The algorithm uses the same mathematical principles as the Buchberger algorithm, but computes many normal forms in one go by forming a generally sparse matrix and using fast linear algebra to do the reductions in parallel.
Numerical linear algebra, sometimes called applied linear algebra, is the study of how matrix operations can be used to create computer algorithms which efficiently and accurately provide approximate answers to questions in continuous mathematics. It is a subfield of numerical analysis, and a type of linear algebra. Computers use floating-point arithmetic and cannot exactly represent irrational data, so when a computer algorithm is applied to a matrix of data, it can sometimes increase the difference between a number stored in the computer and the true number that it is an approximation of. Numerical linear algebra uses properties of vectors and matrices to develop computer algorithms that minimize the error introduced by the computer, and is also concerned with ensuring that the algorithm is as efficient as possible.
A system of polynomial equations is a set of simultaneous equations f1 = 0, ..., fh = 0 where the fi are polynomials in several variables, say x1, ..., xn, over some field k.
In computer algebra, a triangular decomposition of a polynomial system S is a set of simpler polynomial systems S1, ..., Se such that a point is a solution of S if and only if it is a solution of one of the systems S1, ..., Se.
ISSAC, the International Symposium on Symbolic and Algebraic Computation, is an academic conference in the field of computer algebra. ISSAC has been organized annually since 1988, typically in July. The conference is regularly sponsored by the Association for Computing Machinery special interest group SIGSAM, and the proceedings since 1989 have been published by ACM. ISSAC is considered as being one of the most influential conferences for the publication of scientific computing research.
## References
1. "ACM Association in computer algebra".
2. Watt, Stephen M. (2006). Making Computer Algebra More Symbolic (Invited) (PDF). Proc. Transgressive Computing 2006: A conference in honor of Jean Della Dora, (TC 2006). pp. 43–49.
3. "SIGSAM list of conferences". Archived from the original on 2013-08-08. Retrieved 2012-11-15.
4. Cohen, Joel S. (2003). . AK Peters, Ltd. p. 14. ISBN 978-1-56881-159-8.
5. Kevin G. Cassidy (Dec 1985). The Feasibility of Automatic Storage Reclamation with Concurrent Program Execution in a LISP Environment (PDF) (Master's thesis). Naval Postgraduate School, Monterey/CA. Here: p.15
6. Buchberger, Bruno, and Rüdiger Loos. "Algebraic simplification." Computer algebra. Springer, Vienna, 1982. 11-43.
7. Davenport, J. H., Siret, Y., & Tournier, É. (1988). Computer algebra. London: Academic.
8. Kaltofen, Erich (1982), "Factorization of polynomials", in Buchberger, B.; Loos, R.; Collins, G. (eds.), Computer Algebra, Springer Verlag, CiteSeerX
For a detailed definition of the subject:
For textbooks devoted to the subject: | 2021-09-18 15:48:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8100892901420593, "perplexity": 488.3390320575177}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00263.warc.gz"} |
https://html-shark.com/HTML/Equations.htm | # Math formulas in HTML
## The problem
Formulas written in HTML is one of the areas where development has been soemwhat sparse. Not in HTML 5 either. For web sites you therefore see various more or less elegant workarounds. For formulas you have the following options:
• Pictures of formulas, done with a formula/equation editor, e.g. in Open Office or Office 365. It is an easy solution in regards to insertion using the IMG tag. Editing can be a bit difficult, because you need to go through several programs. Visually is rarely looks nice.
• Tables. By being a bit creative in construction tables, possibly along with some SVG, you can obtain a visually beautiful result. The methos is a bit time consuming, until you have a library of formulas to use from. On the other han, the editing can be very easy, and most importantly, there are no limitations to what you can do.
• MathML. As a part of HTML 5, you have the tag MATH, with which you can construct formulas. The basic ideas is good, and the result looks nice, on the browsers that can use the tag. Unfortunately, not all browsers can use the tag. Right now, only Firefox and Opera handles the tag correctly. Safari is handling some of it right. For all intents and purposes, this solution is currently not usable.
• MathJax. To get around the limitations in formulas, an open-source project has been started, that can create formulas, based on JavaScript. There are some limitations to what you can show, and you are limited to the font types that comes with the system. Also you have to be aware, that opening the page involves an API call. API calls can cause som delays in showing pages, especially since MathJax isn't loading asynchronous (i.e. the other things to be loaded have to wait for the call to MathJax to be in place).
## Using tables
Building formulas using tables is a matter of thinking in cells that are merged or not, and whether the table is inline or not.
If we start with inline, that could be something like having two numbers on top of each other in a sentence, like the isotope
2 1
H, or the fraction
1 x
. Here you use the style display:inline-table, and from here it is just a table, like any other.
If we look at the two examples, the respective codes are
and
<TABLE STYLE="display:inline-table; font-size:65%; border-collapse:collapse; vertical-align:middle"><TR><TD STYLE="padding:0px; border-bottom: solid 1px; width:8px; text-align:center">1</TD></TR> <TR><TD STYLE="padding:0px; text-align:center">x</TD></TR></TABLE>
The way the codes are written here, they are quite long and look a bit complicated. Normally you would define some classes instead, making it easier to work with.
Bigger formulas requre bigger tables, and you don't have them inline, but basically you do it the same way. If we look at the acid constant for acetic acid, which looks like this:
Ka = [H+] · [CH3COO−] [CH3COOH]
It is only a matter of an adapted table with text. If you show the borders, it becomes clear:
Ka = [H+] · [CH3COO−] [CH3COOH]
The advantage in using tables, compared to images or MathJax, is that the formulas scale nicely, and look nice and readable on both screen and print-outs, and, as seen, you get to keep the font from the text in the formulas, thus getting a nice, seamless visual experience when reading the page.
## MathJax
MathJax is an API (application programming interface), i.e. an external program/tool, you can call like other scripts. To make it work, you need the following two lines in your HEAD tag:
<SCRIPT SRC='https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML'></SCRIPT>
<SCRIPT>MathJax.Hub.Config({ tex2jax: {inlineMath: [['$','$'], ['\$','\$']]}})</SCRIPT>
From here, you just have to write the right code for the formulas, and MathJax will convert it to readable formulas when the page is opened. Instead of a start and stop tag, you use $$before and after. So, writing the solution to a second degree equations, the code looks like this:$$ x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$On the screen, it looks like this:$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.
On MathJax' own web site and on various tutorials you can find instructions on the names of the commands, and how to use them, to get the various characters and figures, if you want to use this solution. If you don't mind the possible delays from connecting to MathJax' server, and having another typography for you formulas than the rest of the text (or take it the other way around and use a font that looks like the one used for the formulas), MathJax is definitely a fine solution for formulas. | 2019-01-19 08:03:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6401420831680298, "perplexity": 1358.7571339070703}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583662863.53/warc/CC-MAIN-20190119074836-20190119100836-00287.warc.gz"} |
https://blog.csdn.net/u010185894/article/details/71104387 | # 深度学习的Xavier初始化方法
variance_scaling_initializer(
factor=2.0,
mode='FAN_IN',
uniform=False,
seed=None,
dtype=tf.float32
)
Returns an initializer that generates tensors without scaling variance.
When initializing a deep network, it is in principle advantageous to keep the scale of the input variance constant, so it does not explode or diminish by reaching the final layer. This initializer use the following formula:
if mode='FAN_IN': # Count only number of input connections.
n = fan_in
elif mode='FAN_OUT': # Count only number of output connections.
n = fan_out
elif mode='FAN_AVG': # Average number of inputs and output connections.
n = (fan_in + fan_out)/2.0
truncated_normal(shape, 0.0, stddev=sqrt(factor / n)) | 2018-12-15 04:23:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3907623589038849, "perplexity": 5030.35221442729}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00547.warc.gz"} |
http://latkin.org/blog/ | # Muller's Recurrence - roundoff gone wrong
A while back I came upon a seemingly not-too-difficult programming exercise:
Define a recurrence $x_n$ by
$f(y, z) = 108 - \frac{815 - 1500/z}{y}$
$x_0 = 4$
$x_1 = 4.25$
$x_i = f(x_{i-1}, x_{i-2})$
Compute $x_{30}$.
This isn't too hard to code up, using perhaps a recursive function to represent $x_i$. With normal double-precision floats, as $i$ increases, the result converges neatly toward 100. Super!
Unfortunately, 100 is not even close to the right answer. This recurrence actually converges to 5. Continue reading
# A simple benchmark of various math operations
How computationally expensive are various fundamental floating point mathematical operations? Here's a quick and dirty benchmark, which, although surely quite naive, seems to capture the rough relative cost of a few operations.
# Response to "Little Performance Explorations: F#"
I recently saw a tweet from Ryan Riley linking to an article exploring F# performance and floating point processing:
I poked around the code and tested it a bit myself, and figured I would take up the author's call for feedback. Comments on the original blog are locked down, so I've written up my results here, instead. Continue reading
# An elliptical pool
@StevenStrogatz posed a fun question this evening:
People quickly realized the answer is "no," but how to show it? Continue reading
# F# extension methods in Roslyn
If you scan the source code for the Roslyn project, the platform on which the next-gen C# and VB compilers are based, you might stumble across an interesting special behavior that was added in for the sole purpose of preserving backward-compatibility with F#.
From [roslyn]\Src\Compilers\CSharp\Source\Symbols\Metadata\PE\PEAssemblySymbol.cs (as of commit dc3171c8a878):
public override bool MightContainExtensionMethods
{
get
{
if (!this.lazyContainsExtensionMethods.HasValue())
{
var moduleSymbol = this.PrimaryModule;
var module = moduleSymbol.Module;
// The F# compiler may not emit an assembly-level ExtensionAttribute, and previous versions of C# never checked for it.
// In order to avoid a breaking change (while preserving the perceived performance benefits of not looking for extension
// methods in assemblies that don't contain them), we'll also look for FSharpInterfaceDataVersionAttribute.
var mightContainExtensionMethods = module.HasExtensionAttribute(this.assembly.Handle, ignoreCase: false) ||
module.HasFSharpInterfaceDataVersionAttribute(this.assembly.Handle);
this.lazyContainsExtensionMethods = mightContainExtensionMethods.ToThreeState();
}
return this.lazyContainsExtensionMethods.Value();
}
}
The comment does a nice job of summing up the issue, but allow me to provide some additional context. Continue reading
# Project Euler visualizations - problem 144
Here's another nifty Project Euler visualization, one I put together while solving problem 144 a while ago. Continue reading
# Project Euler visualizations - problem 431
I recently solved Project Euler problem 431, which (through a bizarre and unnecessarily confusing back-story) requires one to calculate the volume trapped above a cone, inside a cylinder. The point of the cone is taken to be level with the top of the cylinder, but the cone angle and its horizontal offset compared to the cylinder are variable.
I solved this one with Mathematica, and along the way I cobbled together a little visualization that allows one to easily see the curve of intersection where the cone and the cylinder join together. Why not take this opportunity to share the visualization via an embedded CDF? I've been wanting to try this for a while. Continue reading
# Selecting a random element from a linked list - 3 approaches in F#
You have a linked list (i.e. no random access) with an unknown number of elements and a standard random number generator. Write an algorithm that selects a random element from the list, going for best O(N) performance and memory usage. Continue reading
# A Twitter search client in 10 lines of code with F# and the JSON type provider
Twitter's basic search API is pretty simple to use, and returns JSON back to you with tweets matching your search parameters.
JSON, eh? "There's an app a type provider for that." I wonder if I could cook something up... Continue reading
# Tracking a new year's resolution with F# and FSharpChart
This year, I have a goal of running 500 miles. That's not a crazy-ambitious goal, but between work, school, hobbies, friends, and (occasional) downtime, I think it's plenty for me. In contrast, the CEO of RunKeeper is planning to run 1,500 miles this year! That's an admirable goal, and I hope he succeeds (though I would prefer that he focus on releasing a Windows Phone app, instead. Ah well...).
In order to stay motivated (and because it's cool) I have decided to track my runs and chart my progress throughout the year. Excel works just fine for this, but I want to try something a little different. Why not use this as an opportunity to use FSharpChart? Continue reading | 2014-11-23 00:39:44 | {"extraction_info": {"found_math": true, "script_math_tex": 8, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31481656432151794, "perplexity": 2608.2325969905505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416400378862.11/warc/CC-MAIN-20141119123258-00078-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://nrich.maths.org/public/leg.php?code=-68&cl=3&cldcmpid=4903 | # Search by Topic
#### Resources tagged with Visualising similar to Product Sudoku:
Filter by: Content type:
Age range:
Challenge level:
### There are 180 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Visualising
### Cuboids
##### Age 11 to 14 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Triangles to Tetrahedra
##### Age 11 to 14 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### Isosceles Triangles
##### Age 11 to 14 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Square Coordinates
##### Age 11 to 14 Challenge Level:
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
### Tetrahedra Tester
##### Age 11 to 14 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
### Picturing Triangular Numbers
##### Age 11 to 14 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Frogs
##### Age 11 to 14 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Semi-regular Tessellations
##### Age 11 to 16 Challenge Level:
Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?
### Marbles in a Box
##### Age 11 to 16 Challenge Level:
How many winning lines can you make in a three-dimensional version of noughts and crosses?
### Shady Symmetry
##### Age 11 to 14 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Christmas Chocolates
##### Age 11 to 14 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Hidden Rectangles
##### Age 11 to 14 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Reflecting Squarely
##### Age 11 to 14 Challenge Level:
In how many ways can you fit all three pieces together to make shapes with line symmetry?
### Square It
##### Age 11 to 16 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### On the Edge
##### Age 11 to 14 Challenge Level:
If you move the tiles around, can you make squares with different coloured edges?
### Cubes Within Cubes
##### Age 7 to 14 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Cubes Within Cubes Revisited
##### Age 11 to 14 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### The Spider and the Fly
##### Age 14 to 16 Challenge Level:
A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly?
### Tourism
##### Age 11 to 14 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Travelling Salesman
##### Age 11 to 14 Challenge Level:
A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs?
### Wari
##### Age 14 to 16 Challenge Level:
This is a simple version of an ancient game played all over the world. It is also called Mancala. What tactics will increase your chances of winning?
### Diagonal Dodge
##### Age 7 to 14 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Dice, Routes and Pathways
##### Age 5 to 14
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Rati-o
##### Age 11 to 14 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Picturing Square Numbers
##### Age 11 to 14 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### An Unusual Shape
##### Age 11 to 14 Challenge Level:
Can you maximise the area available to a grazing goat?
### Inside Out
##### Age 14 to 16 Challenge Level:
There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you. . . .
### Trice
##### Age 11 to 14 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### Framed
##### Age 11 to 14 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Tetra Square
##### Age 11 to 14 Challenge Level:
ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square.
### Take Ten
##### Age 11 to 14 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube so that the surface area of the remaining solid is the same as the surface area of the original?
### Clocked
##### Age 11 to 14 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Icosian Game
##### Age 11 to 14 Challenge Level:
This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at.
### Pattern Power
##### Age 5 to 14
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
### Convex Polygons
##### Age 11 to 14 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
### Coloured Edges
##### Age 11 to 14 Challenge Level:
The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set?
### Linkage
##### Age 11 to 14 Challenge Level:
Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?
### Khun Phaen Escapes to Freedom
##### Age 11 to 14 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Sliced
##### Age 14 to 16 Challenge Level:
An irregular tetrahedron has two opposite sides the same length a and the line joining their midpoints is perpendicular to these two edges and is of length b. What is the volume of the tetrahedron?
### Cuboid Challenge
##### Age 11 to 16 Challenge Level:
What's the largest volume of box you can make from a square of paper?
### Double Trouble
##### Age 14 to 16 Challenge Level:
Simple additions can lead to intriguing results...
### Route to Infinity
##### Age 11 to 14 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### 3D Stacks
##### Age 7 to 14 Challenge Level:
Can you find a way of representing these arrangements of balls?
### Auditorium Steps
##### Age 7 to 14 Challenge Level:
What is the shape of wrapping paper that you would need to completely wrap this model?
### Polygon Rings
##### Age 11 to 14 Challenge Level:
Join pentagons together edge to edge. Will they form a ring?
### Troublesome Dice
##### Age 11 to 14 Challenge Level:
When dice land edge-up, we usually roll again. But what if we didn't...?
### Triangles in the Middle
##### Age 11 to 18 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
### Partly Painted Cube
##### Age 14 to 16 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Mystic Rose
##### Age 14 to 16 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Speeding Boats
##### Age 14 to 16 Challenge Level:
Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling? | 2019-06-18 02:50:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38124656677246094, "perplexity": 1326.2582546652543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998605.33/warc/CC-MAIN-20190618023245-20190618045245-00340.warc.gz"} |
https://www.dreamwings.cn/poj2442/4492.html | # POJ 2442 Sequence （堆）
Sequence
Time Limit: 6000MS Memory Limit: 65536K Total Submissions: 9316 Accepted: 3113
Description
Given m sequences, each contains n non-negative integer. Now we may select one number from each sequence to form a sequence with m integers. It’s clear that we may get n ^ m this kind of sequences. Then we can calculate the sum of numbers in each sequence, and get n ^ m values. What we need is the smallest n sums. Could you help us?
Input
The first line is an integer T, which shows the number of test cases, and then T test cases follow. The first line of each case contains two integers m, n (0 < m <= 100, 0 < n <= 2000). The following m lines indicate the m sequence respectively. No integer in the sequence is greater than 10000.
Output
For each test case, print a line with the smallest n sums in increasing order, which is separated by a space.
Sample Input
1
2 3
1 2 3
2 2 3
Sample Output
3 3 4
### AC 代码
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<map>
typedef __int64 LL;
int num[2001];
priority_queue<int,vector<int>,greater<int> >d; //小顶堆
priority_queue<int>q; //大顶堆
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int m,n,x;
scanf("%d%d",&m,&n);
for(int i=0; i<n; i++)
{
scanf("%d",&x);
d.push(x); //首先当只有一组的时候
}
for(int i=1; i<m; i++) //添加剩下的组
{
for(int j=0; j<n; j++)
scanf("%d",num+j);
while(!d.empty())
{
x=d.top();
d.pop();
for(int j=0; j<n; j++) //二重循环计算
{
if((int)q.size()<n)
q.push(x+num[j]);
else if(x+num[j]<q.top())
{
q.pop();
q.push(x+num[j]);
}
}
}
while(!q.empty())
{
d.push(q.top());
q.pop();
}
}
printf("%d",d.top());
d.pop();
while(!d.empty())
{
printf(" %d",d.top());
d.pop();
}
printf("\n");
}
return 0;
} | 2021-09-23 02:58:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47902849316596985, "perplexity": 2197.0353965722475}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057416.67/warc/CC-MAIN-20210923013955-20210923043955-00506.warc.gz"} |
https://chat.stackexchange.com/transcript/2165/2014/3/30 | 9:03 AM
I wanted to vote to close this question
1
$$\sum_{k=0}^{i+1} \binom {i+1} k$$ I can't find an identity for this summation :( To clarify I'm trying to prove using induction that this sum is equal to $2^{i+1}$, I have my basis and inductive hypothesis done, this is just my inductive step
as a duplicate of
0
This is a question I came across in an old midterm and I'm not sure how to do it. Any help is appreciated. $$2^n = C(n,0) + C(n,1) + \cdots + C(n,n).$$ Prove this statement is true for all $n \ge 0$ by induction.
I got the error message: This closure would result in the 'duplicate of' navigation only leading in a circle.
Probably the circle is cause by this question?
5
I want to know if I can get some help with this proof. I tried, but I just cannot seem to get $2^{k}$. It states that, For $k \in \mathbb{Z}_{\ge 0}$, $$\sum^{k}_{m=0}\binom{k}{m} = 2^k$$ Thank You.
I voted to close a duplicate of:
11
I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a series of algebraic manipulations that can lead from $\sum\limits_{i=0}^n \binom{n}{i}$ to $2^n$.
I still think that the first choice would be a little bit better, since they are both specifically about inductive proofs.
3 hours later…
11:44 AM
@MartinSleziak As you surely noticed, 177405 and 27539 were somehow closed as duplicates of each other. 27539 was also closed as a duplicate of 18690. I changed it so that 18690 is the only stated duplicate of 27539.
12:08 PM
@ArthurFischer Thanks for that!
5 hours later…
4:41 PM
I saw one proof using this formula: $$\sum_{i=0}^{z} {n_1 \choose i}{n_2 \choose z-i} = {n_1+n_2 \choose z}$$ Can anyone help explain it, thank you! | 2019-07-19 00:36:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7760564088821411, "perplexity": 287.3393114027292}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525863.49/warc/CC-MAIN-20190718231656-20190719013656-00056.warc.gz"} |
https://physics.stackexchange.com/questions/448649/temperature-heat-transfer-through-multiple-layers-of-insulation-materials | # Temperature / heat transfer through multiple layers of insulation materials
how can I determine the resultant temperature on one side of a wall after a specified time when the wall has a series of different insulation layers and when the temperature on the other side is known, and is constant?
• What exactly are the boundary conditions on the two sides of the wall, and what is the initial condition? – Chet Miller Dec 16 '18 at 12:57
• The answer depends on the accuracy you're seeking. Strictly speaking, temperature is discontinuous across an interface, although usually, but not always, this fact is completely neglected. – AccidentalBismuthTransform Dec 16 '18 at 14:17
• cecs.wright.edu/~sthomas/htchapter03.pdf – Gert Dec 16 '18 at 16:06
• The standard numerical approach is the finite difference method, discussed in a heat transfer context by Incropera and DeWitt. An alternate method appears here ("1-D Heat Transfer in Multilayer Materials Using a Finite Volume Approach"). Özışık discusses the analytical solution in Chap. 8 of Heat Conduction here. The search for more convenient analytical solutions is an area of active research. – Chemomechanics Dec 17 '18 at 20:18
• If this was for example a home it would depend on the furnace setting, no furnace and the house eventually be the same temperature as the outside. An interesting question to as his what is the rate of heat loss so that you can figure out how often your furnace needs to run. – PhysicsDave Dec 18 '18 at 0:47
The relevant partial differential equation is the heat equation:
$$\frac{\partial T}{\partial t}=\alpha(\vec{x})\nabla^2 T$$
where $$\alpha$$ is the thermal diffusivity of whatever material a layer is made of, and so will depend on position for a wall made of different materials. The spatial variation in temperature along the wall is likely negligible, so we can treat this as a 1-dimensional problem:
$$\frac{\partial T}{\partial t}=\alpha(x)\frac{\partial^2T}{\partial x^2}$$
where the $$x$$-direction points perpendicular to the wall. It is likely that this must be solved numerically at this point, given the variation of $$\alpha$$ within the wall.
• "so will depend on position for a wall made of different materials." What needs to be calculated is the overall thermal resistance of the composite wall. – Gert Dec 16 '18 at 16:08
• Also, because this is transient problem, heat capacities must come into play at some point. – Gert Dec 16 '18 at 16:11
• @Gert Heat capacity is included in the definition of thermal diffusivity, which is equivalent to thermal conductivity divided by density and specific heat capacity. – probably_someone Dec 16 '18 at 16:17
• Yep, momentarily forgot that. But I'm sure this problem does not need a numerical solution, not with this simple geometry. – Gert Dec 16 '18 at 16:20
To show just how tedious this problem is, mathematically, let's take the simpler case of a homogeneous wall (no layers):
Because heat only flows in the $$x$$ direction, we can treat it as a quasi-one dimensional problem.
As @probably_someone stated, the relevant partial differential equation (PDE) is Fourier's Equation:
$$\frac{\partial T}{\partial t}=\alpha\nabla^2 T$$
Or in partial derivatives shorthand: $$T_t=\alpha T_{xx}$$
The problem states that for all times, $$T(t,0)=T_0$$ and for that reason I make a small substitution:
$$U(t,x)=T(t,x)-T_0$$
So that:
$$U(t,0)=T(t,0)-T_0=0$$
This is our first boundary condition.
The PDE becomes:
$$U_t=\alpha U_{xx}$$
For a second boundary condition we'll assume that no heat flows from the $$x=L$$ boundary because air is a good insulator. This means:
$$\dot{Q}(t,L)=0 \Rightarrow T_t(t,L)=U_t(t,L)=0$$
Finally we need an initial condition ($$t=0$$) and we'll assume $$T(0,x)=T_1$$, so that:
$$U(0,x)=T_1-T_0$$
For PDEs of this type we use the Ansatz:
$$U(t,x)=\Theta (t)X(x)$$
where $$\Theta (t)$$ is a function dependent on $$t$$ only and $$X(x)$$ a function dependent on $$x$$ only. Now insert $$U(t,x)$$ into the PDE:
$$X(x)\Theta'(t)=\alpha \Theta(t)X''(x)$$
Now divide by $$\Theta X$$ (abbreviated):
$$\frac{\Theta'}{\Theta}=\alpha \frac{X''}{X}$$
Introducing a separation constant (a Real number):
$$\frac{1}{\alpha}\frac{\Theta'}{\Theta}= \frac{X''}{X}=-m^2$$
The original PDE can now be separated into two ODEs:
$$\frac{X''}{X}=-m^2$$
Or: $$X''+m^2X=0$$ and: $$\frac{\Theta'}{\Theta}=-\alpha m^2$$
Starting with the first, it has a general solution of: $$X(x)=A\sin mx +B\cos mx$$
Using our first boundary condition: $$U(t,0)=0 \Rightarrow X(0)=0$$
$$0=A\sin 0 +B\cos 0 \Rightarrow B=0$$
So that $$X(x)=A \sin mx$$
$$U_t(t,L)=0 \Rightarrow X'(L)=0$$
Or: $$A\cos mL=0 \Rightarrow \cos mL=0$$
Cosines become zero for $$\pi/2, 3\pi/2, 5\pi/2, etc$$
So:
$$mL=\frac{n\pi}{2}\:\mathrm{for}\:\ n=1,3,5,7,...$$
So we have: $$m=\frac{n\pi}{2L}$$
Finally: $$X(x)=A\sin \Big(\frac{n\pi x}{2L}\Big)$$
The eigenvalues $$m$$ can now be used in the first ODE:
$$\frac{\Theta'}{\Theta}=-\alpha \Big(\frac{n\pi}{2L}\Big)^2$$
Solved we get:
$$\Theta(t)=Ce^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}$$
with $$C$$ an integration constant.
Going back to the Ansatz: $$U=\Theta X$$, then:
$$U_n(t,x)=D_ne^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}\sin \Big(\frac{n\pi x}{2L}\Big)$$
For $$n=1,3,5,7,...$$ The integration constants $$A$$ and $$C$$ have been lumped together.
Each $$n$$ presents a solution but the Superposition Principle demands we take the sum of all these solution to have the complete solution:
$$U(t,x)=\displaystyle\sum_{n=1}^{+\infty} U_n(t,x)$$ $$U(t,x)=\displaystyle\sum_{n=1}^{+\infty}D_ne^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}\sin \Big(\frac{n\pi x}{2L}\Big)$$
We now need to determine the coefficients $$D_n$$, using Fourier. The initial condition was set as:
$$U(0,x)=T_1-T_0$$
Inserting this into our solution, the exponential terms all become $$1$$, so:
$$T_1-T_0=\displaystyle\sum_{n=1}^{+\infty}D_n\sin \Big(\frac{n\pi x}{2L}\Big)$$
$$D_n=\frac{2}{L}\int_0^{L}(T_1-T_0)\sin\Big(\frac{n\pi x}{2L}\Big)dx$$
$$D_n=\frac{2}{L}(T_1-T_0)\Big(\frac{2L}{n\pi }\Big)\Big[1-\cos\Big(\frac{n\pi }{2}\Big)\Big]$$
$$D_n=4(T_1-T_0)\Big(\frac{1}{n\pi }\Big)$$
Finally we need to add $$T_0$$ to $$U(t,x)$$ to obtain $$T(t,x)$$. Then put Humpty Dumpty back together and presto, done!
From $$T(t,x)$$ can then be determined the time evolution at the boundary $$L$$:
$$\boxed{T(t,L)=T_0+\frac{4}{\pi}(T_1-T_0)\displaystyle\sum_{n=1}^{+\infty}\Big[\frac{1}{n} e^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}\sin \Big(\frac{n\pi }{2}\Big)\Big]}$$ For $$n=1,3,5,7,...$$
Although this is a complete solution to the problem for a simple homogeneous wall, its very user unfriendly.
Furthermore, in this 'simple' case $$\alpha$$ is a constant but in the case of a composite wall, $$\alpha=f(x)$$, complicating matters enormously. | 2021-04-22 17:36:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 68, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8207423686981201, "perplexity": 460.7293427695896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00367.warc.gz"} |
https://freshergate.com/mechanical-engineering/engineering-mechanics/discussion/1453 | Home / Mechanical Engineering / Engineering Mechanics :: Discussion
### Discussion :: Engineering Mechanics
1. When a rigid body is suspended vertically, and it oscillates with a small amplitude under the action of the force of gravity, the body is known as
2. A. simple pendulum B. compound pendulum C. torsional pendulum D. second's pendulum | 2022-01-26 00:05:17 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.850208580493927, "perplexity": 2536.4474143117745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00058.warc.gz"} |
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## nuhcolee 2 years ago Simplify: (5x^2 + 3x + 4) + (2x^2 − 6x + 3) 7x2 + 9x + 7 7x2 + 3x + 7 7x2 − 3x + 1 7x2 − 3x + 7
• This Question is Closed
1. Champs
Factorize?
2. jim_thompson5910
5x^2 + 2x^2 = ???
3. eliassaab
$\left(2 x^2-6 x+3\right)+\left(5 x^2+3 x+4\right)=7 x^2-3 x+7$
4. eliassaab
Just add them up
5. dpflan
I imagine that your goal now is make it into binomial?
6. dpflan
like $(a+b)^2$
7. eliassaab
Just add them up. Thant is all what you have to do.
8. nuhcolee
thanks !
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Privacy Policy | 2015-03-31 18:07:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5600951313972473, "perplexity": 6655.2200362443255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131300905.36/warc/CC-MAIN-20150323172140-00002-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/222698/correlations-between-test-question-scores-with-discrete-and-small-number-of-valu | # Correlations between test question scores with discrete and small number of values
I am a newbie at statistical analysis. I have been using the Pearson correlation coefficient for pairs of test question scores.
The test had questions some of which carried 1, some 2 and a few carried 3 marks. The questions were graded with a least count of 0.5 marks - so in a 3-marks question, a student can get either of the following scores: 0, 0.5, 1, 1.5, 2, 2.5,3.
I am not sure if using the Pearson correlation coefficient for such data that takes less number of possible values is still valid. I suspect that the values of $r$ that I get may not reflect the real correlations.
What is a better way to calculate correlations in this situation?
• Welcome to our site! There's no need to add "thanks" at the end of your question - the best way to thank the person who provides the answer that helps you the most is to "accept" it (green tick) – Silverfish Jul 8 '16 at 6:21 | 2020-08-09 03:42:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6557660698890686, "perplexity": 502.85612342270184}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738380.22/warc/CC-MAIN-20200809013812-20200809043812-00000.warc.gz"} |
https://www.esaral.com/q/the-given-plots-represents-the-variation-of-the-concentration-54366 | # The given plots represents the variation of the concentration
Question:
The given plots represents the variation of the concentration of a reactant $R$ with time for two different reactions (i) and (ii). The respective orders of the reactions are:
1. 1,0
2. 1,1
3. 0,1
4. 0,2
Correct Option: 1
Solution:
In graph (i), In [Reactant vs time is linear with positive intercept and negative slope. Hence it is $1^{\text {st }}$ order In graph (ii), [Reactant] vs time is linear with positive intercept and negative slope. Hence, it is zero order. | 2023-03-21 14:30:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6817712187767029, "perplexity": 1948.860617918278}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00074.warc.gz"} |
https://bytepawn.com/optimal-coverage-for-wordle-with-monte-carlo-methods-part-iii.html | # Optimal coverage for Wordle with Monte Carlo methods - Part III
Marton Trencseni - Sat 22 January 2022 - Machine Learning
## Introduction
In the previous article, we saw that a relatively simple Monte Carlo method can find a lot of 24-letter-unique wordlists, about 1 per second running at 16x parallelism. Here I use a crucial insights from the english language to make the search more efficient, and find a 25-letter-unique Worlde wordlist. The ipython notebook is up on Github.
## Vowels
The key insight came from my friend and collegaue Rameez Kakodker, who introduced me to Wordle. We were discussing the methods for solving this problems, when he said that vowels are likely to duplicate. This led me to the following idea: to get a 25-unique-solution, we need 5 words with 25 unqiue letters. But if a word has two (or more) vowels, like BRAVE, it is very unlikely that this word would be in such a wordlist, because it contains both A and E. But there are only 5+1 vowels in english: AIEUO+Y, so ignoring Y, if a word uses up 2 vowels, like BRAVE, then we won't have enough vowels left for all the 5 words.
So the idea is:
1. Like before, prune the vowel wordlist, and keep only words which have 5 unique letters.
2. Further prune the wordlist, and keep only words that have 1 vowel. So eg. BRAVE and ALIEN is pruned, but STICK is kept.
3. When searching for the 5 words, always pick 5 words (from the above pruned words) which contain one of the 5 vowels AIEUO. In other words, treat the 5 word list like 5 slots, and always pick an A-word for the first slot, always pick an I-word for the second slot, and so on, eg. ['bawty', 'crimp', 'fjeld', 'vughy', 'zonks']
Step 1 and 2 in code:
vowels = sorted(['a', 'e', 'i', 'o', 'u'])
ltw = defaultdict(set) # letter-to-word index
for word in words:
if len(set(list(word))) == len(word): # make sure all 5 letters are unique
if len(set(vowels) & set(list(word))) == 1: # make sure there is exactly 1 vowel in the word
for letter in word:
len(words), sum([len(v) for _, v in ltw.items()])
For step 3, we will use the same method as before: randomly generate wordlists, and if we find one that is promising (22 or more unique letters), then stick to that and try a few variations:
def solve_wordle(num_tests=1000, num_improve_attempts=1000, random_seed=0):
random.seed(random_seed * 131071)
solutions = []
for _ in range(num_tests):
wordlist = [random.sample(ltw[vowel], 1)[0] for vowel in vowels]
letters = set(list(''.join(wordlist)))
if len(letters) >= 22:
for _ in range(num_improve_attempts):
v = random.randint(0, len(vowels) - 1) # pick out one of the vowels randomly
word = random.sample(ltw[vowels[v]], 1)[0] # pick a new word which contains this vowel
new_wordlist = deepcopy(wordlist)
new_wordlist[v] = word
new_letters = set(list(''.join(new_wordlist)))
if len(new_letters) > len(letters):
wordlist, letters = new_wordlist, new_letters
if len(letters) >= 24:
print(len(letters), wordlist)
solutions.append(wordlist)
return solutions
The rest of the code is as before, using joblib to run in parallel:
def flatten_list(li):
return [item for sublist in li for item in sublist]
def clean_solutions(solutions):
return [t.split(',') for t in sorted({','.join(sorted(c)) for c in solutions})]
# solve_wordle(num_tests=10*1000*1000, num_improve_attempts=1*1000*1000)
n_jobs = 16
num_tests = 10*1000*1000
num_improve_attempts = 100*1000
solutions = Parallel(n_jobs=n_jobs)(delayed(solve_wordle)
(num_tests=num_tests,
num_improve_attempts=num_improve_attempts,
random_seed=i*random.randint(0, 100*1000))
for i in range(n_jobs))
solutions = clean_solutions(flatten_list(solutions))
for wordlist in solutions:
num_unique_letters = len(set(list(''.join(wordlist))))
print(f'{num_unique_letters}-unique-letter solution: {wordlist}')
if num_unique_letters == 25:
print('*** JACKPOT! ***')
print(f'Found {len(solutions)} solutions...')
On my 12-core AMD system, after running it for 13 minutes, I got:
24-unique-letter solution: ['backs', 'flint', 'grews', 'jumpy', 'vozhd']
24-unique-letter solution: ['balky', 'cinqs', 'grump', 'vozhd', 'wheft']
...
24-unique-letter solution: ['brick', 'flews', 'jumpy', 'thanx', 'vozhd']
25-unique-letter solution: ['brick', 'glent', 'jumpy', 'vozhd', 'waqfs']
*** JACKPOT! ***
24-unique-letter solution: ['brick', 'gulpy', 'meynt', 'vozhd', 'waqfs']
24-unique-letter solution: ['bring', 'chomp', 'junky', 'veldt', 'waqfs']
...
24-unique-letter solution: ['jumby', 'knelt', 'pyric', 'vozhd', 'waqfs']
24-unique-letter solution: ['jumby', 'prong', 'veldt', 'wakfs', 'zilch']
Found 249 solutions...
## Jackpot!
The following wordlist has 25 unique letters: ['brick', 'glent', 'jumpy', 'vozhd', 'waqfs']. The english alphabet has 26 letters, the one missing letter is x. Playing these 5 words in Wordle will usually reveral a lot of letters from the solutions, and probably a few positions. For example, for today's Wordle:
So, without any thinking, we already know the 5 letters (ICENW), and 2 positions (W**C*). The word is WINCE:
## Words
3 of the 5 words are unknown to me, 2 of them are not really english:
1. glent: to move quickly especially in an oblique direction
2. vozhd: a Russian leader (the word is russian, and means "to lead")
3. waqfs: plural of waqf (probably from arabic), a Muslim religious or charitable foundation created by an endowed trust fund
## Conclusion
I find it pleasing that hard problems can be solved on a modern desktop computer by brute-forcing with some minimal insights and minor optimizations. | 2022-09-28 22:57:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37341639399528503, "perplexity": 8557.889682601093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335286.15/warc/CC-MAIN-20220928212030-20220929002030-00010.warc.gz"} |
https://math.stackexchange.com/questions/2967684/is-a-y-in-bbbr-y-lim-n-to-inftyfx-n-textfor-some-sequence-x-n | # Is $A=\{y \in \Bbb{R}: y=\lim_{n\to \infty}f(x_n), \text{for some sequence}~ x_n \to +\infty\}$ compact for a continuous $f$?
Question. Let $$f:\Bbb{R} \to \Bbb{R}$$ be a continuous function and $$A\subset \Bbb{R}$$ be defined by $$A=\{y \in \Bbb{R}: y=\lim_{n\to \infty}f(x_n), \text{for some sequence}~ x_n \to +\infty\}$$ Then the set $$A$$ is necessarily
A. A connected set
B. A compact set
C. A singleton set
D. None of the above
My attempt. I don't have enough progress in this problem.
First, I want to understand the definition of $$A$$ properly. Since $$f$$ is given continuous, then for a sequence $$(x_n)$$ with $$x_n \to +\infty$$, $$f(x_n)$$ may converge or diverge to $$+\infty$$ or may be oscilatory. So, for a $$(x_n)$$ with $$x_n \to +\infty$$ if it happens that $$f(x_n)$$ is convergent then we collect that limit, $$\lim f(x_n)$$, in the set $$A$$.
Now if we take $$f(x)=\cos\pi x$$ and $$x_n=2n+1, y_n=2n$$ then both $$x_n,y_n\to +\infty$$ but $$\lim f(x_n)=-1$$ and $$\lim f(y_n)=1$$. So, $$\{1,-1\} \subset A$$. Therefore Option C is False.
Now I try to prove Option A:
To prove $$A$$ to be connected I have to show that any continuos function $$g: A\to \{\pm 1\}$$ is constant. But I cannot proceed with this further.
So I try to prove: $$A=cl(B)$$ for some connected set $$B \subset \Bbb{R}$$. Now looking on the description of the set $$A$$ I try to find out $$B$$. It seems $$B$$ should contains all $$f(x_n)$$'s for some $$x_n \to +\infty$$. Since, every such $$(x_n)$$ can be made $$>1$$ after all but finitely many terms, so $$B=\{f(x):x>1\}=f((1,+\infty))$$. Since $$f$$ is continuous so $$B$$ is connected being continuous image of a connected set and so is $$A=cl(B)$$. Therefore, Option A is True.
But I cannot conclude Option B. Please help me in B. Is this solution correct? Please tell if there is any mistake in my solution.
• Try $e^x \sin(x)$. – JonathanZ Oct 23 '18 at 14:25
• @JonathanZ ...But how to compute $A$ for $e^x \sin(x)$? OR, How to prove it to be non compact? – Indrajit Ghosh Oct 23 '18 at 14:29
• If you try to picture the function you will see that for $x \rightarrow + \infty$ it forms periodic waves of increasing amplitude. Thus for any $y \in \mathbb{R}$ there will be a wave that touches it, moreover all the following waves will reach that value. so it is easy to see that you can find an increasing sequence whose limit of the image is that $y$. Therefore $A= \mathbb{R}$. – N.B. Oct 23 '18 at 14:34
• @IndrajitGhosh (+1) Very interesting problem. If you don't mind can I ask you what is the source of this problem. I mean from where you got this. – StammeringMathematician Oct 23 '18 at 15:11
• @StammeringMathematician...It was given in the TIFR PhD entrance examination in 2015. – Indrajit Ghosh Oct 23 '18 at 15:12
I think the understanding of the question is hampering your progress here.
The meaning of $$x_n \to +\infty$$ is that for all $$r > 0$$ there exists $$N$$ such that $$n > N$$ implies $$x_n > r$$. In other words, $$x_n$$ is eventually larger than every real number.
Now, we are asking if $$f(x_n)$$ has a limit for any such sequence $$x_n \to + \infty$$. $$A$$ is the set of all such limits.
To see the intuition for $$A$$, try to think of points that would obviously belong to $$A$$. For example, if there is a sequence $$x_n \to +\infty$$ such that $$f(x_n)$$ are all equal. Then, this number will be in $$A$$.
This allows us to construct $$\sin x$$ as a counterexample to $$A$$ being a singleton : it is periodic, so $$A$$ is the range of this function, which is not a singleton.
For connectedness, you seem to have made a mistake in interpreting $$A$$. $$A$$ does not contain $$f(x_n)$$ for any sequence $$x_n \to +\infty$$, but rather any limit point that may arise from a choice of $$x_n$$ for which this is convergent. Points of a sequence are not its limit points, therefore we must be more careful in this aspect.
To show that $$A$$ is actually connected, we note that $$A$$ is a subset of the real numbers, and here we have the description that $$A$$ is connected if and only if it is an interval. This is what must be used : suppose that $$a < c < b$$ is there with $$a,b \in A$$, then show that $$c \in A$$ as well. This is the definition of an interval.
If $$a,b \in A$$, then there exist two sequences $$x_n , w_n \to \infty$$ such that $$f(x_n) \to a$$ and $$f(w_n) \to b$$. We want to construct $$z_n$$ such that $$z_n \to \infty$$ and $$f(z_n)$$ goes to $$c$$.
Let us perform a trimming : Let $$y_n$$ be a subsequence of $$w_n$$ defined as follows : $$y_n = w_{l_n}$$, where $$l_n = min\{ k > l_{n-1} : w_{k} > x_n\}$$, with $$l_0 = 0$$. Now, $$y_n \to +\infty$$ (why?) and $$y_n > x_n$$ (why?). Furthermore, a subsequence of a convergent sequence is convergent, so $$f(y_n) \to b$$.
First, denote $$\delta = \frac{\min(c-a,b-c)}{2}$$. Then, by definition of $$f(y_n) \to b$$ we have $$N$$ such that $$f(y_n) > b-\delta > c$$ for $$n > N$$. Similarly, we have $$f(x_n) < a+\delta < c$$ for some $$n > M$$. Taking the maximum, we have for $$L = \max \{N,M\}$$ that $$n > L$$ implies $$f(x_n) < c < f(y_n)$$.
Now, by the intermediate value theorem (applied infinitely many times) for each $$x_n < y_n$$, after $$n > L$$ there exists a $$z_n$$ in between such that $$f(z_n) = c$$, since $$c$$ lies between $$f(x_n)$$ and $$f(y_n)$$. Reindex $$z_n$$(it is now defined only for $$n> L$$) to start from $$1$$ by shifting. This $$z_n$$ goes to $$+\infty$$ (why?) , however we even get that $$f(z_n) \to c$$ , because it is the constant sequence $$c,c,...$$!
Hence, the connectedness follows.
As for compactness, the idea is that a compact set is bounded. A continuous function can possible have very wild oscillaing behaviour (if you have seen $$\sin \frac 1x$$ near zero, like that) near infinity. This oscillating behaviour, like $$\sin x$$, leads to the creation of many elements of $$A$$. If this oscillation can be made unbounded, then $$A$$ can possibly be made unbounded!
That is the thinking behind the counterexample function $$x \sin x$$.
If you look at this function, it will have increasing oscillation with increasing $$x$$. This increasing oscillation means, that if you draw a horizontal line at any height on the grid, it will intersect the graph of this function at infinitely many points, starting from some point onwards, and will have a subsequence increasing to infinity.
To prove this a little more rigorously, we will show that $$A = \mathbb R$$ in this case, with a few steps skipped.
The easy way to see this is to see that $$f(n\pi + \frac \pi 2) = n\pi + \frac \pi 2$$, and $$f(n \pi - \frac{\pi}{2}) = n \pi - \frac \pi 2$$ for every $$n \geq 1$$, so you have two subsequences going to infinity : now use the intermediate value theorem like we did for connectedness between these points to conclude that every point has in its preimage a sequence going to infinity. | 2019-05-27 13:22:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 114, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9490915536880493, "perplexity": 107.45080466227698}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232262600.90/warc/CC-MAIN-20190527125825-20190527151825-00185.warc.gz"} |
http://hmassignmentwhsl.firdaus.info/censored-data-truncated-data.html | # Censored data truncated data
1 censored and truncated models data on durations are often censored, either to the right (common) or to the / no truncation / reg y x . The analysis of censored data is a major issue in survival studies censored data are together with truncated data, missing data, current status data, and others, among the complex data structures in which only partial information on the variable(s) of interest is available. When performing estimations with truncated or censored data, we need to use tools that account for that type of incomplete data for truncated linear regression, we can use the truncreg command, and for censored linear regression, we can use the intreg or tobit command.
Data in many settings are censored: we know only an inequality about some values - for example, that a particular value weighs less than the scale permits i. The left truncated right censored competing risks data under a fairly general set of priors, and that is the major contribution of this paper although, we have assumed that the competing. Statistically speaking, when we see a continuous outcome variable we often worry about outliers and how these extreme observations can impact our model but have you ever had an outcome variable with no outliers because there was a boundary value at which accurate measurements couldn’t be or weren . Journal of modern applied statistical methods volume 13|issue 2 article 22 11-2014 double bootstrap confidence interval estimates with censored and truncated data.
In my case i don't have left-censored data since i know when all individuals became at risk (the year they turned 18) so individuals who turned before 2003 (which is the year where my data set starts at) are left truncated but are included in my data. Survival analysis: techniques for censored and truncated data (statistics for biology and health) by moeschberger, melvin l,klein, john p and a great selection of similar used, new and collectible books available now at abebookscom. Censored and truncated data censored data point are those whose measured properties are not known precisely, but are known to lie above or below some limiting sensitivity truncated data points are those which are missing from the sample altogether due to sensitivity limits.
Truncated data this problem appears at mackay’s book, at the beginning of chapter 3: unstable particles are emitted from a source and decay at a distance x, a real number that has an exponential probability distribution with characteristic length $$\lambda$$. Regression analysis of doubly truncated data censored and truncated data doubly truncated data, besides efron and petrosian (1999)’s work, bilker and wang . Both truncated-only data and censored and truncated data were generated in the simulation study the proposed point and variance estimators showed good performance in various simulated settings the bone marrow transplant registry data were analyzed as the illustrative example.
## Censored data truncated data
Useful data will be excluded when data is censored but not accounted for, and biases can be introduced when data is truncated since censoring and truncation are often confused, a brief. Left censoring for survival data in r you can think of left censored data as going from -infinity until survival analysis on left truncated data with . Survival analysis for bivariate truncated data provides readers with a comprehensive review on the existing works on survival analysis for truncated data, mainly focusing on the estimation of univariate and bivariate survival function the most distinguishing feature of survival data is known as censoring, which occurs when the survival time .
Therefore, insurance loss data is left-truncated because the insurance company doesn't know if there are values below the deductible d because policyholders won't make a claim the insurance loss is also right-censored if the loss is greater than u because u is the most the insurance company will pay. Stat 798l, spring 2008 eric slud left-truncated right-censored data maximum likelihood & nelson-aalen estimators as we discussed in class, left-truncated and right-censored survival data.
The practical di erence between censored and truncated data is that the number of censored values is known, but the number of truncated values is not observed and censored environmental data are much more common than truncated data. So what exactly are censored data regression terminology censoring share | cite it's important to distinguish censored versus truncated as well as missing data. The left truncated right censored competing risks data under a fairly general set of priors, and that is the major contribution of this paper although, we have . Modeling left-truncated and right-censored survival data with longitudinal covariates1 by yu-ru su and jane-ling wang university of california, davis and national .
Censored data truncated data
Rated 5/5 based on 18 review | 2018-10-24 00:32:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40272998809814453, "perplexity": 1555.541488094002}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583517628.91/warc/CC-MAIN-20181024001232-20181024022732-00053.warc.gz"} |
http://math.stackexchange.com/questions/193482/find-minimum-planks-for-constructing-surface | # Find minimum planks for constructing surface
I have pieces of wooden planks 1.22m by 2.44m. I want to find the minimum number of planks required to construct a wooden surface of sides L length and W width.
The first intuition is that I can arrange the planks "horizontally" or "vertically" so I would calculate total number of planks N = (L/1.22 x W/2.44) or (L/2.44 x W/1.22) where the divisions needs to get rounded up to nearest integer. Since this is a simple case, I just calculate the two cases and get the minimum option.
For fun and interest, tried to express this in calculus but have failed. I don't know how to handle the discrete maths involved here. Anyone can advise or link to study material? Thanks.
-
Calculate $n=\lceil \frac L{1.22}\rceil$ and $m=\lceil \frac W{1.22}\rceil$, whete $\lceil\ \rceil$ denotes rounding up. if at least one of $n,m$ is even, you need $\frac{nm}2$ planks. Otherwise you need $\frac{nm+\min\{n,m\}}2$ planks. If you are allowed to mix horizontal and vertical arrangement, then $\lceil \frac{nm}2\rceil$ planks are enough.
You need $\lceil \frac{LW}{(1.22)(2.44)} \rceil$ tiles.
How do you plan to tile a $1.22\sqrt{2} \times 1.22\sqrt{2}$ square using one tile? – Erick Wong Sep 10 '12 at 13:16
I just need enough tiles to cover the area. Each tile has area $1.22 \cdot 2.44$. The total area is $LW$. The problem was to construct a surface of side $L$ and $W$. Presumably the tiles can be cut. – copper.hat Sep 10 '12 at 14:31 | 2016-02-12 23:09:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7282200455665588, "perplexity": 547.4056271155349}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701165378.58/warc/CC-MAIN-20160205193925-00104-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-are-the-charges-of-two-ions-of-copper | What are the charges of two ions of copper?
Copper (I) ions have a ${1}^{+}$ charge. This happens when copper atoms lose one electron. Its formula is $\text{Cu"^+}$.
Copper (II) ions have a ${2}^{+}$ charge. This happpens when copper atoms lose two electrons. Its formula is $\text{Cu"^(2+)}$. | 2022-01-17 01:54:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4711473286151886, "perplexity": 2147.797203460405}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00187.warc.gz"} |
https://gmatclub.com/forum/henry-purchased-3-items-during-a-sale-he-received-a-126613.html | It is currently 21 Apr 2018, 08:37
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# Henry purchased 3 items during a sale. He received a 20
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Henry purchased 3 items during a sale. He received a 20 [#permalink]
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Updated on: 08 Mar 2015, 13:29
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Henry purchased 3 items during a sale. He received a 20 percent discount off the regular price of the most expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total amount of the 3 discounts greater than 15 percent of the sum of the regular prices of the 3 items?
(1) The regular price of the most expensive item was $50, and the regular price of the next most expensive item was$20
(2) The regular price of the least expensive item was $15 [Reveal] Spoiler: Attachment: Henry discount on items.GIF [ 14.71 KiB | Viewed 15608 times ] [Reveal] Spoiler: OA Originally posted by docabuzar on 27 Jan 2012, 06:24. Last edited by Bunuel on 08 Mar 2015, 13:29, edited 3 times in total. Added the OA Math Expert Joined: 02 Sep 2009 Posts: 44598 Re: Henry purchased 3 items during a sale. He received a 20 [#permalink] ### Show Tags 27 Jan 2012, 06:32 13 This post received KUDOS Expert's post 25 This post was BOOKMARKED docabuzar wrote: Henry purchased 3 items during a sale. He received a 20 percent discount off the regular price of the most expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total discount of these three items greater than 15 percent of the sum of the regular prices of the 3 items? (1) The regular price of the most expensive item was$50, and the regular price of the next most expensive item was $20 (2) The regular price of the least expensive item was$15
Let the prices be a, b, and c, so that a > b > c.
Basically the questions: is $$0.2a+0.1b+0.1c>0.15(a+b+c)$$? --> is $$a>b+c$$?
(1) The regular price of the most expensive item was $50 and the regular price of the next most expensive item was$20 --> $$a=50$$, $$b=20$$, $$c\leq{20}$$ (as the second most expensive item was $20 then the least expansive item, the third one, must be less than or equal to 20). So the question becomes: is $$50>20+c$$ --> is $$c<30$$? As we got that $$c\leq{20}$$, hence the above is always true. Sufficient. (2) The regular price of the least expensive item was$15. Clearly insufficient.
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink]
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27 Jan 2012, 07:21
Then it makes sense. Thanks
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27 Jan 2012, 07:25
THe Q asks for the which of discounts will be greater 20%, 10%, 10% on A, B, C resp or 15% on (A+B+C). How can we say its same as 50> 20+C ? Now I m totally lost.
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27 Jan 2012, 07:30
docabuzar wrote:
THe Q asks for the which of discounts will be greater 20%, 10%, 10% on A, B, C resp or 15% on (A+B+C). How can we say its same as 50> 20+C ? Now I m totally lost.
The questions asks: is $$0.2a+0.1b+0.1c>0.15(a+b+c)$$? After simplifying we'll get that the question becomes: is $$a>b+c$$?
(1) says $$a=50$$ and $$b=20$$. Also we concluded that: $$c\leq{20}$$. Now, if we substitute the values of a and b we'll get that the question boils down to: is $$50>20+c$$? or is $$c<30$$?
Hope it's clear.
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink]
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27 Jan 2012, 07:34
Thanks. Its clear now.
Just how do you know to keep A on one side & B & C on other? This simplied the Ans a lot.
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink]
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27 Jan 2012, 07:38
docabuzar wrote:
Thanks. Its clear now.
Just how do you know to keep A on one side & B & C on other? This simplied the Ans a lot.
It really doesn't matter how you write it: $$a>b+c$$, $$a-b-c>0$$, $$a-b>c$$, ... You still will get the same result. For me $$a>b+c$$ was the most "attractive" form, so I chose this one.
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink]
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22 Aug 2014, 20:30
Bunuel: if the first item is 50, the next 2 are 20. If I take 20% of 50 I get 10 and 10% of 20 I get 2. Total discount is 14 and total price is 90. 14/90>.15. What's am I doing wrong?
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23 Aug 2014, 12:21
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bankerboy30 wrote:
Henry purchased 3 items during a sale. He received a 20 percent discount off the regular price of the most expensive item and a 10 percent discount off the regular price of each of the other 2 items. Was the total discount of these three items greater than 15 percent of the sum of the regular prices of the 3 items?
(1) The regular price of the most expensive item was $50, and the regular price of the next most expensive item was$20
(2) The regular price of the least expensive item was $15 Bunuel: if the first item is 50, the next 2 are 20. If I take 20% of 50 I get 10 and 10% of 20 I get 2. Total discount is 14 and total price is 90. 14/90>.15. What's am I doing wrong? Total discount =$14.
The sum of the regular prices of the 3 items = $90. 15% of$90 = $13.5.$14 > \$13.5.
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink]
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15 Jul 2017, 11:24
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I believe this question is a perfect example of an "Easy C" trap. Just by realizing that A&B would be sufficient and make the problem way too easy, we can eliminate C,D and E. This is a simple rule that could improve one's chances on 700+ problems
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink]
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19 Dec 2017, 00:35
Fortunately or Unfortunately, the OA is C for this one. I'm staring at the answer on GMATPREP software :/ Attaching screenshot
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Henry Purchased 3 items.docx [823.94 KiB]
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink]
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19 Dec 2017, 00:47
vishal205 wrote:
Fortunately or Unfortunately, the OA is C for this one. I'm staring at the answer on GMATPREP software :/ Attaching screenshot
The OA for the question discussed above is A, NOT C.
The question you attached is not the one discussed above, it's another one discussed here: https://gmatclub.com/forum/henry-purcha ... 94280.html
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Re: Henry purchased 3 items during a sale. He received a 20 [#permalink] 19 Dec 2017, 00:47
Display posts from previous: Sort by | 2018-04-21 15:37:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4945172071456909, "perplexity": 4544.608120091684}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945232.48/warc/CC-MAIN-20180421145218-20180421165218-00175.warc.gz"} |
https://www.transtutors.com/questions/problem-3-23a-comprehensive-cvp-analysis-lo-3-3-3-4-3-5-rosenthal-company-makes-and--2566002.htm | # Problem 3-23A Comprehensive CVP analysis LO 3-3, 3-4, 3-5 Rosenthal Company makes and sells produ...
Need help with part g
Problem 3-23A Comprehensive CVP analysis LO 3-3, 3-4, 3-5 Rosenthal Company makes and sells products with variable costs of $67 each. Rosenthal incurs annual fixed costs of$30,800. The current sales price is $81. Required: The following requirements are interdependent. For example, the$4,200 desired profit introduced in Requirement c also applies to subsequent requirements. Likewise, the \$77 sales price introduced in Requirement d applies to the subsequent requirements. a Determine the contribution margin per unit. 14 V Contribution margin per unit b-1. Determine the break-even point in units and in dollars 2,200 Break-even point in units Break-even point in dollars 178,200 b-2. Prepare an income statement using the contribution margin format. ROSENTHAL COMPANY Income Statement 178,200 Sales 147, 400V Variable cost 30,800V Contribution margin 30,800 Fixed cost Net income | 2018-11-13 17:54:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28056201338768005, "perplexity": 9358.832851750707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195927-00498.warc.gz"} |
http://www.juliabloggers.com/tag/programming/page/2/ | # Video Blog: Developing and Editing Julia Packages
Google Summer of Code is starting up, so I thought it would be a good time to share my workflow for developing my own Julia packages, as well as my workflow for contributing to other Julia packages. This does not assume familiarity with commandline Git, and instead shows you how to use a GUI (GitKraken) to make branches and PRs, as well as reviewing and merging code. You can think of it as an update to my old blog post on package development in Julia. However, this is not only updated but also improved since I am now able to walk through the “non-code” parts of package developing (such as setting up AppVeyor and code coverage).
Enjoy! (I quite like this video blog format: it was a lot less work)
The post Video Blog: Developing and Editing Julia Packages appeared first on Stochastic Lifestyle.
# Using pipes while running external programs in Julia
Re-posted from: http://perfectionatic.org/?p=340
Recently I was using Julia to run ffprobe to get the length of a video file. The trouble was the ffprobe was dumping its output to stderr and I wanted to take that output and run it through grep. From a bash shell one would typically run:
ffprobe somefile.mkv 2>&1 |grep Duration
This would result in an output like
Duration: 00:04:44.94, start: 0.000000, bitrate: 128 kb/s
This works because we used 2>&1 to redirect stderr to stdout which would in be piped to grep.
If you were try to run this in Julia
julia> run(ffprobe somefile.mkv 2>&1 |grep Duration)
you will get errors. Julia does not like pipes | inside the backticks command (for very sensible reasons). Instead you should be using Julia’s pipeline command. Also the redirection 2>&1 will not work. So instead, the best thing to use is and instance of Pipe. This was not in the manual. I stumbled upon it in an issue discussion on GitHub. So a good why to do what I am after is to run.
julia> p=Pipe()
Pipe(uninit => uninit, 0 bytes waiting)
julia> run(pipeline(ffprobe -i somefile.mkv,stderr=p))
This would create a pipe object p that is then used to capture stderr after the execution of the command. Next we need to close the input end of the pipe.
julia> close(p.in)
Finally we can use the pipe with grep to filter the output.
julia> readstring(pipeline(p,grep Duration))
" Duration: 00:04:44.94, start: 0.000000, bitrate: 128 kb/s\n"
We can then do a little regex magic to get the duration we are after.
julia> matchall(r"(\d{2}:\d{2}:\d{2}.\d{2})",ans)[1]
"00:04:44.94"
# HDF5 in Julia
Re-posted from: http://albi3ro.github.io/M4/programming/HDF5.html
So, last summer, my program was producing three dimensional data, and I needed a way to export and save that data from my C++ program. Simple ASCII files, my default method, no longer covered my needs. Of course, I wasn’t the first person to encounter this problem, so I discovered the HDF5 standard.
Instead of storing data in a human readable format like ASCII, the Hierarchical Data Format, HDF, stores data in binary format. This preserves the shape of the data in the computer and keeps it at its minimum size. WOHOO!!
Sadly, the syntax for HDF5 in C++ and Fortran is just as bad as FFTW or OpenBLAS. But happily, just like FFTW and OpenBLAS, HDF5 has wonderful syntax in Julia, Python, and julia, among others.
So how does it work?
We don’t just print a single variable. Each HDF5 file is like its own file system. In my home directory, I have my documents folder, my programming folder, my pictures, configuration files,… and inside each folder I can have subfolders or files.
The same is true for an HDF5 file. We have the root, and then we have groups and subgroups. A group is like a folder. Then we can have datasets. Datasets are objects that hold data (files).
## Installing the Package
While running Pkg.add("HDF5"); should hopefully add the HDF5 library, additional steps may be required. I remember having a horrible time with the HDF installation when using C++ a year ago. If at all possible, just use a package manager, and do not try and install it from source! See the HDF5.jl or HDFGroup pages for details.
# Hello World
Firstly, lets open a file and then write some data to it.
We can open a file in three ways:
Symbol Meaning
“w” Write. Will overwrite anything already there.
If we open with this syntax, we have to always remember to close it with close()
Now lets see if we were successful by reading. Instead of reading the dataset, we are going to checkout the structure of the file first.
names(fid) tells us what is inside the location fid.
dump(fid) is much more in depth, exploring everything below fid. If we had a bunch of subdirectories, it would go down each one to see what was there.
names
Union{ASCIIString,UTF8String}["string"]
dump
HDF5.HDF5File len 1
string: HDF5Dataset () : Hello World
nothing
Now when we are reading data, we need to know the difference between dataset and the data the dataset contains.
Look at the below example
the dataset: HDF5.HDF5Dataset
the string: ASCIIString Hello World
read another way: ASCIIString Hello World
A dataset is like the filename “fairytale.txt”, so we then need to read the file to get “Once upon a time …”.
### Groups
I’ve talked about groups, but we haven’t done anything with them yet. Let’s make some!
Here we use g_create to create two groups, one inside the other. For the subgroup, it’s parent is g, so we have to create it at location g. Just like in a filesystem, it’s name/ path is nested within its parent’s path.
HDF5.HDF5File len 1
mygroup: HDF5.HDF5Group len 1
mysubgroup: HDF5.HDF5Group len 0
nothing
path of h: /mygroup/mysubgroup
### Attributes
Say in a file I want to include the information that I ran the simulation with 100 sites, at 1 Kelvin, for 100,000 timesteps. Instead of creating new datasets for each of these individual numbers, I can create attributes and tie them to either a group or a dataset.
typeof attrs: HDF5.HDF5Attributes
Temp: 1
N Sites: 100
### Final Tips
Before diving in to learn how to use this, think about whether you need it or not. How large and complex is your data? Is it worth the time to learn? While the syntax might be relatively simple in Julia, ASCII files are still much easier to deal with.
If you are going to play around or use this format, I recommend getting an HDF viewer, like HDFViewer. While you can have much more control via code, sometimes it is just that much simpler to check everything is working with a GUI.
For more information, checkout the Package page at HDF5.jl or the HDFGroup page at HDFGroup
I’ve shown some of the basic functionality in simple test cases. If you want more control, you might just have to work a bit for it. | 2017-07-25 18:44:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.272738516330719, "perplexity": 2486.326600477178}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549425352.73/warc/CC-MAIN-20170725182354-20170725202354-00152.warc.gz"} |
https://www.semanticscholar.org/paper/First-Order-Primal%E2%80%93Dual-Methods-for-Nonsmooth-Valkonen/5e1b5dc2dccc38c9655ef8dbee4206f099b0c807 | # First-Order Primal–Dual Methods for Nonsmooth Non-convex Optimisation
@article{Valkonen2019FirstOrderPM,
title={First-Order Primal–Dual Methods for Nonsmooth Non-convex Optimisation},
author={Tuomo Valkonen},
journal={arXiv: Optimization and Control},
year={2019}
}
• T. Valkonen
• Published 30 September 2019
• Mathematics
• arXiv: Optimization and Control
We provide an overview of primal-dual algorithms for nonsmooth and non-convex-concave saddle-point problems. This flows around a new analysis of such methods, using Bregman divergences to formulate simplified conditions for convergence.
4 Citations
Multikernel Regression with Sparsity Constraint | SIAM Journal on Mathematics of Data Science | Vol. 3, No. 1 | Society for Industrial and Applied Mathematics
• Computer Science, Mathematics
• 2021
A Banach-space formulation of supervised learning with generalized totalvariation (gTV) regularization is provided, and it is shown that the solution admits a multikernel expansion with adaptive positions.
Multikernel Regression with Sparsity Constraint
• Mathematics, Computer Science
SIAM J. Math. Data Sci.
• 2021
This paper identifies the class of kernel functions that are admissible in this Banach-space formulation of supervised learning with generalized total-variation (gTV) regularization and proposes a variation of supervisedLearning in a continuous-domain hybrid search space with gTV regularization.
Regularisation, optimisation, subregularity
Regularisation theory in Banach spaces, and non-norm-squared regularisation even in finite dimensions, generally relies upon Bregman divergences to replace norm convergence. This is comparable to the
Primal-dual block-proximal splitting for a class of non-convex problems
• Mathematics
ArXiv
• 2019
We develop block structure adapted primal-dual algorithms for non-convex non-smooth optimisation problems whose objectives can be written as compositions $G(x)+F(K(x))$ of non-smooth block-separable
## References
SHOWING 1-10 OF 91 REFERENCES
and T
• Valkonen, PrimalâĂŞdual proximal splitting and generalized conjugation in non-smooth non-convex optimization,
• 2019
Primal-dual proximal splitting and generalized conjugation in nonsmooth nonconvex optimization, Applied Mathematics and Optimization (2020), doi:10.1007/s00245-020-09676-1, arXiv:1901.02746
• 1901
Primal-dual block-proximal splitting for a class of non-convex problems
• Mathematics
ArXiv
• 2019
We develop block structure adapted primal-dual algorithms for non-convex non-smooth optimisation problems whose objectives can be written as compositions $G(x)+F(K(x))$ of non-smooth block-separable
Testing and Non-linear Preconditioning of the Proximal Point Method
• T. Valkonen
• Computer Science
Applied Mathematics & Optimization
• 2018
This work formalises common arguments in convergence rate and convergence proofs of optimisation methods to the verification of a simple iteration-wise inequality and demonstrates the effectiveness of the general approach on several classical algorithms, as well as their stochastic variants.
and T
• Valkonen, Primal–dual block-proximal splitting for a class of non-convex problems,
• 2019
Acceleration and Global Convergence of a First-Order Primal-Dual Method for Nonconvex Problems
• Mathematics
SIAM J. Optim.
• 2019
The primal-dual hybrid gradient method, modified (PDHGM, also known as the Chambolle--Pock method), has proved very successful for convex optimization problems involving linear operators arising in...
A primal-dual hybrid gradient method for non-linear operators with applications to MRI
We study the solution of minimax problems $\min_x \max_y G(x) + \langle K(x),y\rangle - F^*(y)$ in finite-dimensional Hilbert spaces. The functionals $G$ and $F^*$ we assume to be convex, but the
and a at
• Chemistry
The William Makepeace Thackeray Library
• 2018
The xishacorene natural products are structurally unique apolar diterpenoids that feature a bicyclo[3.3.1] framework. These secondary metabolites likely arise from the well-studied, structurally
Relaxed Gauss-Newton methods with applications to electrical impedance tomography
• Mathematics
SIAM J. Imaging Sci.
• 2020
It is proved that the Gauss--Newton-type method with inexact relaxed steps converges to a set of disjoint critical points given that the linearisation of the forward operator for the inverse problem is sufficiently precise.
Fenchel Duality Theory and a Primal-Dual Algorithm on Riemannian Manifolds
• Mathematics
Found. Comput. Math.
• 2021
This paper introduces a new notion of a Fenchel conjugate, which generalizes the classical Fenchel conjugation to functions defined on Riemannian manifolds. We investigate its properties, e.g., the | 2022-07-07 08:46:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7358174324035645, "perplexity": 3609.592933978605}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00271.warc.gz"} |
https://proofwiki.org/wiki/Algorithm_to_determine_whether_Polynomial_Diophantine_Equation_has_Integer_Solution/Historical_Note | Algorithm to determine whether Polynomial Diophantine Equation has Integer Solution/Historical Note
Historical Note on Algorithm to determine whether Polynomial Diophantine Equation has Integer Solution
Hilbert's $10$th problem was the question:
Given a Diophantine equation with any number of unknown quantities and with rational integer numerical coefficients: To devise a process according to which it can be determined in a finite number of operations whether the equation is solvable in rational integers.
The MRDP Theorem has shown that recursively enumerable sets are equivalent to Diophantine sets.
Matiyasevich's Theorem shows that solutions to Diophantine equations may grow exponentially.
Yuri Vladimirovich Matiyasevich proved this in $1970$ by using a method involving Fibonacci numbers, which grow exponentially. | 2019-09-18 22:33:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.645950198173523, "perplexity": 466.5047191896566}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573368.43/warc/CC-MAIN-20190918213931-20190918235931-00235.warc.gz"} |
https://math.stackexchange.com/questions/632372/minimal-counterexamples-of-the-isomorphism-problem-for-integral-group-rings | # Minimal counterexamples of the isomorphism problem for integral group rings
The isomorphism problem for integral group rings asks if two finite groups $G,H$ are isomorphic when their integral group rings $\mathbb{Z}[G]$, $\mathbb{Z}[H]$ are isomorphic. Quite a lot has been done towards solving this and related problems. See for example Chapter 9 in Milies, Sehgal, "An Introduction to Group Rings". Now it seems that the first (?!) counterexample has been found by Martin Hertweck in his 2001 paper "A Counterexample to the Isomorphism Problem for Integral Group Rings". He has constructed two counterexamples, the one group has order $2^{25} \cdot 97^2$ and the other group has $2^{21} \cdot 97^{28}$. Do we really have to consider such huge groups? The thesis by Geoffrey Janssens discusses Hertweck's construction in detail, and claims that this is the only known counterexample. Is this still correct?
Question. What is known about the minimal counterexamples of the isomorphism problem for integral group rings?
I think, it's correct that the smallest known counterexample is still the Hertweck's pair of non-isomorphic groups of order $2^{21} \cdot 97^{28}$.
Please note that the group of order $2^{25} \cdot 97^2$ is used in the construction, but there is no known pair of groups of that order that provide a counterexample.
Of course, it may happen that a smaller counterexample exists. Remember, for example, the situation with the 2nd Zassenhaus conjecture, where by know we know a variety of counterexamples and the minimal one is 30 times smaller than the first discovered. To give a more detailed account, known counterexamples to the 2nd Zassenhaus conjecture are:
• of order 2880 – Roggenkamp & Scott, 1988
• of orders 2880 and 6720 – Klingler, 1991
• of order 1140 (metabelian) – Hertweck, 2002
• Hertweck, 2003 – of orders: 180 (metabelian), 360 (supersolvable), 72600 (Frobenius)
• of order 96 (three groups) – Blanchard, 2001 (who also verified using GAP 3.4.4 that these are counterexamples of minimal possible order)
• Thank you. Does one know that the isomorphism problem holds for groups of order $\leq n$ for some specific large values of $n$? Jan 9 '14 at 13:35
• @MartinBrandenburg: to my best knowledge, no. There were results formulated for classes of groups, but not for groups of consecutive orders smaller than some bound. The Modular Isomorphism Problem stands a bit better, e.g. for $p=2$ it is checked that it holds for all 2-groups of order up to $2^8$ (theoretical proof for up to $2^6$, the rest using computers). Jan 9 '14 at 15:51 | 2021-11-27 09:37:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8917868733406067, "perplexity": 512.3613604276221}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00612.warc.gz"} |
https://electronicsreference.com/module2/dc-ohms-law/ | # Ohm’s Law
Ohm’s Law is considered one of the most important formulas in the fields of physics and electricity. Ohm’s Law describes the relationship between voltage, current, and resistance.
Ohm’s Law states that the current through two points of a conductor is linearly proportional to the voltage, and defines electrical resistance as the ratio between the voltage and current. It is expressed by the formula:
### V = IR
Which means ‘voltage (V) is equal to the current (I) times the resistance (R)’.
By exploring Ohm’s Law, we gain valuable insight into the relationship between three important concepts that are critical to every electrical circuit and electronic device.
Pro Tip: Check out our Ohm’s Law Calculator for a handy reference!
## Ohm’s Law: Voltage, Current, and Resistance
Up to this point, we have explored the concepts of voltage, current, and resistance.
### Voltage
Voltage is the electrical potential difference between two points in any circuit. Voltage is the energetic ‘push’ that causes electrons to move, creating electrical current. Voltage is measured in units of volts, which is a ratio of energy to charge: 1 volt equals one joule of energy per coulomb of charge.
There are technically two quantities that are measured in volts:
(1) Electromotive force (EMF) is the energy per unit of electric charge (electric potential) supplied to a circuit by a power source like a generator or battery.
(2) Voltage, which describes the equivalent loss in electric potential (called a voltage ‘drop‘) as electric charge travels through a circuit.
A good way to think about voltage is ‘energy per unit of electric charge’, or ‘how much energy is given to and dissipated by each electron’.
Ohm’s law tells us that the voltage ‘drop’ across any part of a circuit is equal to the current through the circuit times the resistance between those two points.
### Current
Current is the flow of electric charge (i.e. electrons) through any given point in a circuit.
When we apply a suitable voltage to a closed circuit, electrons will flow through the circuit. Current is the measure of that flow. The unit of measurement for current is amperes, or coulombs (electrical charge) per second.
Ohm’s Law tells us that the current through a path in a circuit is equal to the voltage divided by the resistance.
### Resistance
Resistance is the opposition to current flow from one point to another in a circuit.
Resistance is intentionally contributed by resistors, but everything in a circuit adds some resistance. This includes conducting wires, interconnections, and components other than resistors.
Most materials do not have free electrons, so they have high electrical resistance. Some materials have free electrons available for conduction, and the internal structure determines how good they are at conducting electricity. The better a material is at conducting, the lower its resistance. Resistance is measured in the unit Ohm, which is defined as 1 volt (potential difference or ‘drop’) per 1 ampere applied to the material. This relation actually describes Ohm’s Law in terms of units.
Ohm’s law tells us that resistance is the ratio of voltage to current. A very low resistance value will therefore result in a very high current, and a high resistance value will result in a low current (if the applied voltage is kept constant).
## Ohm’s Law: The Relationship Between Voltage, Current, and Resistance
The significance of Ohm’s Law is that it describes the relationship between voltage, current, and resistance.
Although Ohm’s Law seems almost obvious today, it actually took many years of scientific experimentation and theory before George Ohm published his famous Law. It wasn’t easily accepted, either; scientists around the world denounced Ohm and his law. It took almost 30 years before his work was widely accepted by the scientific community.
## Ohm’s Law Triangle
Ohm’s Law is commonly represented in a triangular format. This format can be helpful in remembering or determining the formulations of Ohm’s Law.
There are three possible terms to solve for: voltage, current, and resistance.
The triangle allows us to solve for any of the three terms without having to memorize the equation (or in case we forget).
To use the triangle to solve for each one, you simply ignore the term being solved for.
For example, to solve for voltage (V) you ignore the top of the triangle (voltage) and are left with IR. Therefore, V = IR.
To solve for current, you ignore the lower left and are left with V over R; voltage over resistance.
To solve for resistance, you ignore the resistance block on the lower right and are left with V over I; voltage over current.
## What is Ohm’s Law?
Ohm’s Law states that the voltage (V) is equal to the current (I) times the resistance (R):
V = IR
Written as ‘V = IR’, Ohm’s Law tells us that the voltage, or potential difference (also called voltage ‘drop’) is equal to the current times the resistance of the circuit between the two points. If the current is increased, the voltage drop will also increase. If the resistance of the circuit between the points being measured is increased, the voltage drop will also increase.
You can re-write this to calculate the current:
I = \frac{V}{R}
This tells us that the current increases when the voltage is increased, and decreases as the resistance is increased.
If the voltage across that section is increased (by adding a battery, for instance), the current (charge flow) in the circuit will also increase. If the resistance across the section is increased (by adding a resistor), the current will decrease. Voltage is proportional to the current but resistance is inversely proportional. Decrease the resistance, and current will increase.
If the resistance is extremely low, for example, in a circuit without a resistor, the current will be extremely high, which explains why it’s so important to always have a resistor (or resistive element like a lightbulb) in every current carrying path of a circuit.
We can also solve for resistance as the ratio of voltage to current:
R = \frac{V}{I}
Thinking about resistance as the ratio of voltage to current can help us better understand the nature of these principles. A 1 ohm resistor will yield 1 amp of current for every 1 volt (1 joule per coulomb of charge) supplied to the circuit by a battery.
If you increase the voltage, you increase the resistance. If you increase the current, you decrease the resistance. One important thing to note however is that resistance is defined by the material, not the applied voltage or resulting current. So the correct way to think about this form of Ohm’s Law, is that resistance is proportional to the the voltage drop of the circuit between the points being measured. We can manipulate the voltage drop across that section by using a different resistor or combinations of resistors, or by adding or removing voltage sources.
## Ohm’s Law Examples
### Example 1
Let’s look at a simple circuit consisting of a battery and resistor:
On the left we have a typical DC Voltage source; a 1.5 Volt Battery. We connect our battery terminals with a 100 Ohm resistor between, forming a circuit.
Reminder: It is always important to place at least one resistor in a circuit to prevent accidentally damaging the circuit and causing an unsafe condition.
Current in this circuit flows from the positive terminal of the battery clockwise through the resistor and into the negative terminal of the battery.
Let’s use Ohm’s Law to calculate the current:
I = \frac{V}{R} = \frac{1.5V}{100\Omega} = .015A
So, the current in this circuit is .015 amps.
### Example 2
What happens if we replace the 100 Ohm resistor with a 5 Ohm resistor?
In this case:
I = \frac{V}{R} = \frac{1.5V}{5\Omega}=.3A
With a much lower resistance than in the first example, the current in the circuit is much higher. It is now .3 amps.
### Example 3
We want to connect a new device to a 1.5 volt battery but it has a maximum current rating of .25 A. Determine the minimum value of resistor needed to protect the device.
Identify the value of the resistor needed:
R = \frac{V}{I} = \frac{1.5V}{.25A} = 6\Omega
To protect our device, we will need to use a 6 Ohm resistor. In reality, we wouldn’t want the current in the circuit to approach the maximum current rating, so we would want to use a resistor with a higher value.
We will explore Ohm’s Law in much greater detail as we use it for examples and conceptual understanding in Module 3.
Before we tackle more complex circuits, let’s cover one more important concept in Lesson 7: Electric Power.
Module 2 – Fundamental Concepts:
Lesson 0: Module 2 Introduction
Lesson 1: Electric Circuits
Lesson 2: Electric Power Sources
Lesson 3: Electric Current
Lesson 4: Voltage
Lesson 5: Electrical Resistance
Lesson 6: Ohm’s Law
Lesson 7: Electric Power | 2022-05-24 02:38:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6181702017784119, "perplexity": 612.7330782344362}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00270.warc.gz"} |
https://www.physicsforums.com/threads/logarithms-problem.778193/ | # Logarithms problem
1. Oct 25, 2014
### Rectifier
Hey!
I have a question regarding a statement in my physics book. I dont see how
$$|H(f)|_{dB} = - 10log (1+( \frac{f}{f_B})^2)$$
approaches this equation below for big values on f.
$$|H(f)|_{dB} = - 20log ( \frac{f}{f_B})$$
EDIT: I am sorry if this is posted to a wrong sub-forum.
Last edited: Oct 25, 2014
2. Oct 25, 2014
### ShayanJ
When f is large, you can ignore 1 and so there only remains $-10 \log{(\frac{f}{f_B})^2}$. Now you can use the property $\log a^n=n\log a$ to get what you want.
3. Oct 25, 2014
### Rectifier
Oh gosh. Thank you for your help :) | 2018-07-22 17:11:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5468260645866394, "perplexity": 1359.3306914104278}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676593378.85/warc/CC-MAIN-20180722155052-20180722175052-00574.warc.gz"} |
https://fr.maplesoft.com/support/help/errors/view.aspx?path=DocumentTools/Components/Plot&L=F | Plot Component - Maple Help
DocumentTools[Components]
Plot
generate XML for a Plot Component
Calling Sequence Plot( p, opts )
Parameters
p - (optional) ; a 2D or 3D plot, or the result from InlinePlot opts - (optional) ; one or more keyword options as described below
Options
• captionalignment : identical(center,left,right):=center ; horizontal alignment of the component caption.
• captionposition : identical(bottom,top):=bottom ; position of the component caption.
• clickaction : string:=""; A string which parses to one or more valid statements in 1-D Maple Notation. These statements form the Click Action Component Code that executes when the Plot is clicked. The default is the empty string (no such action code).
• clickdefault : truefalse:=false; Indicates whether the manipulator on a pointer click is to execute action code. The default value is false.
• clickx : realcons; The initial x value of the click selector (as if the cursor were clicked).
• clicky : realcons; The initial y value of the click selector (as if the cursor were clicked).
• continuous : truefalse:=false ; Indicates whether the animation plays continuously or stops after displaying all the frames once.
• delay : posint:=100 ; Delay in milliseconds between animation frames. The default value is 100.
• dragendaction : string:=""; A string which parses to one or more valid statements in 1-D Maple Notation. These statements form the Drag End Action Component Code that executes when dragging the the pointer over the plot ends. The default is the empty string (no such action code).
• drageventaction : string:=""; A string which parses to one or more valid statements in 1-D Maple Notation. These statements form the Dragged Action Component Code that executes when the pointer is dragged over the plot. The default is the empty string (no such action code).
• drawcaption : truefalse:=false ; Indicates whether the component caption is shown. The caption on a PlotComponent is distinct from any caption in the contained plot, and is automatically generated by the graphical interface.
• enabled : truefalse; Indicates whether the component is enabled. The default is true. If enabled is false then the inserted component is grayed out and interaction with it cannot be initiated.
• endx : realcons:=0.0 ; The initial x value of the drag end selector, (as if a drag had ended).
• endy : realcons:=0.0 ; The initial y value of the drag end selector, (as if a drag had ended).
• pixelheight : posint; The height in pixels of the component.
• fillcolor : {list(nonnegint),symbol,string}:=[255,255,255] ; Specifies the background color of the component. The passed value can be either a named color or a list of three integers each between 0 and 255. A list of nonnegative integers is interpreted as RGB values in a 24bit 3-channel color space. The default value is [255,255,255] which corresponds to white.
• hoveraction : string:=""; A string which parses to one or more valid statements in 1-D Maple Notation. These statements form the Hover Action Component Code that executes when the pointer hovers over the plot when in focus, if the component's Manipulator is set to Click and Drag. The default is the empty string (no such action code).
• hoverx : realcons; The initial x value of the hover selector (as if the cursor were hovered).
• hovery : realcons; The initial y value of the hover selector (as if the cursor were hovered).
• identity : {name,string}; The reference name of the component.
• legendvisibility : truefalse ; Indicates whether the legend is visible.
• showborders : truefalse; Indicates whether component borders are shown
• startx : realcons:=0.0 ; The initial x value of the drag start selector, (as if a drag had occurred).
• starty : realcons:=0.0 ; The initial y value of the drag start selector, (as if a drag had occurred).
• tooltip : string; The text that appears when the mouse pointer hovers over the component. The default is the empty string (no tooltip).
• visible : truefalse; Indicates whether the component is visible. The default is true.
• pixelwidth : posint; The width in pixels of the component.
Description
• The Plot command in the Component Constructors package returns an XML function call which represents a Plot Component.
• The generated XML can be used with the results of commands in the Layout Constructors package to create an entire Worksheet or Document in XML form. Such a representation of a Worksheet or Document can be inserted into the current document using the InsertContent command.
Examples
> $\mathrm{with}\left(\mathrm{DocumentTools}\right):$
> $\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Layout}\right):$
> $\mathrm{with}\left(\mathrm{DocumentTools}:-\mathrm{Components}\right):$
Executing the Plot command produces a function call.
> $P≔\mathrm{Plot}\left(\mathrm{identity}="Plot0"\right)$
{P}{≔}{\mathrm{_XML_EC-Plot}}{}\left({"id"}{=}{"Plot0"}{,}{"captionalignment"}{=}{"1"}{,}{"captionposition"}{=}{"1"}{,}{"clickdefault"}{=}{"false"}{,}{"clickx"}{=}{"0."}{,}{"clicky"}{=}{"0."}{,}{"continuous"}{=}{"false"}{,}{"delay"}{=}{"100"}{,}{"drawcaption"}{=}{"false"}{,}{"enabled"}{=}{"true"}{,}{"endx"}{=}{"0."}{,}{"endy"}{=}{"0."}{,}{"hoverx"}{=}{"0."}{,}{"hovery"}{=}{"0."}{,}{"pixel-height"}{=}{"500"}{,}{"pixel-width"}{=}{"500"}{,}{"show-border"}{=}{"true"}{,}{"startx"}{=}{"0."}{,}{"starty"}{=}{"0."}{,}{"fillcolor"}{=}{"\left[255,255,255\right]"}{,}{"visible"}{=}{"true"}\right) (1)
By using commands from the Layout Constructors package a nested function call can be produced which represents a worksheet.
> $\mathrm{xml}≔\mathrm{Worksheet}\left(\mathrm{Group}\left(\mathrm{Input}\left(\mathrm{Textfield}\left(P\right)\right)\right)\right):$
That XML representation of a worksheet can be inserted directly.
> $\mathrm{InsertContent}\left(\mathrm{xml}\right):$
> $P≔\mathrm{Plot}\left(\mathrm{plot}\left(\mathrm{sin}\left(x\right),x=-\mathrm{\pi }..\mathrm{\pi }\right),\mathrm{identity}="Plot0",\mathrm{tooltip}="My example plot",\mathrm{pixelheight}=200\right):$
> $\mathrm{xml}≔\mathrm{Worksheet}\left(\mathrm{Group}\left(\mathrm{Input}\left(\mathrm{Textfield}\left(P\right)\right)\right)\right):$
The previous example's call to the InsertContent command inserted a component with identity "Plot0", which still exists in this worksheet. Inserting additional content whose input contains another component with that same identity "Plot0" incurs a substitution of the input identity in order to avoid a conflict with the identity of the existing component.
The return value of the following call to InsertContent is a table which can be used to reference the substituted identity of the inserted component.
> $\mathrm{lookup}≔\mathrm{InsertContent}\left(\mathrm{xml},\mathrm{output}=\mathrm{table}\right)$
${\mathrm{lookup}}{≔}{table}{}\left(\left[{"Plot0"}{=}{"Plot1"}\right]\right)$ (2)
> $\mathrm{lookup}\left["Plot0"\right]$
${"Plot1"}$ (3)
> $\mathrm{GetProperty}\left(\mathrm{lookup}\left["Plot0"\right],\mathrm{visible}\right)$
${"true"}$ (4)
The next example makes use of action code for a Plot and a Button component.
> $P≔\mathrm{Plot}\left(\mathrm{plot}\left(\left[\left[0,0\right]\right],\mathrm{color}=\mathrm{white},\mathrm{axes}=\mathrm{normal},\mathrm{view}=\left[0..10,0..10\right]\right),\mathrm{identity}="Plot0",\mathrm{clickdefault},\mathrm{clickaction}="update\left(%Plot0\right);"\right):$
> $B≔\mathrm{Button}\left("Clear",\mathrm{identity}="Button0",\mathrm{action}="unassign\left(\text{'}:-L\text{'}\right); clear\left(%Plot0\right);"\right):$
The action code passed with the clickaction option of the Plot component constructor calls a procedure update, and the action code of the Button component calls a procedure clear. These procedures are defined below. The update procedure uses the clickx and clicky values taken on when the mouse pointer is left clicked in the plot.
> update := proc(comp) global L; local cx, cy, v; uses DocumentTools, plots, CurveFitting; (cx,cy) := Do(comp(clickx)), Do(comp(clicky)); if type(L,listlist) then L := [L[], [cx, cy]]; else L := [[cx, cy]]; end if; Do(comp = display(plot(L, style=point, symbol=solidcircle, symbolsize=15, color=blue, axes=normal), plot(LeastSquares(L, v), v=0..10), view=[0..10, 0..10])); end proc:
> clear := proc(comp) DocumentTools:-Do(comp = plot([[0,0]], color=white, axes=normal, view=[0..10,0..10])); end proc:
Now we insert the components. Following insertion every single left click in the plot augments L with a new data point, and the plot is updated with a point plot and a least squares linear fit of the points in L.
> $\mathrm{InsertContent}\left(\mathrm{Worksheet}\left(\mathrm{Group}\left(\mathrm{Input}\left(\mathrm{Textfield}\left(B,P\right)\right)\right)\right)\right):$
The Plot component in the previous example is constructed by supplying identity="Plot0" as an option. But the first example on this page had already inserted content with a component with that identity. Hence the identity of the Plot component in the last example is replaced at insertion time. Note that literal instances of "Plot0" and Plot0 are also replaced in all component action code of the inserted content. The consistent replacement allows the action code to function as intended. Such replacement can be confirmed selecting the inserted Plot and Button components and viewing the component names in the Context Panel and then clicking the Edit Code buttons to view the Code Editor for these components.
The fillcolor can be specified to match that of a parent Table Cell.
> $P≔\mathrm{Plot}\left(\mathrm{identity}="Plot0",\mathrm{plot3d}\left({y}^{2}\mathrm{sin}\left(x\right),x=-\mathrm{\pi }..\mathrm{\pi },y=-2..2,\mathrm{axis}=\left[\mathrm{color}="DarkBlue"\right]\right),\mathrm{fillcolor}="LightGray",\mathrm{showborders}=\mathrm{false}\right):$\mathrm{xml}≔\mathrm{Worksheet}\left(\mathrm{Table}\left(\mathrm{alignment}=\mathrm{center},\mathrm{width}=380,\mathrm{widthmode}=\mathrm{pixels},\mathrm{Column}\left(\right),\mathrm{Row}\left(\mathrm{Cell}\left(\mathrm{Textfield}\left(P\right),\mathrm{fillcolor}="LightGray"\right)\right)\right)\right):
> $\mathrm{InsertContent}\left(\mathrm{xml}\right):$
The following example programmatically inserts a PlotComponent with a scaled 3D animation which can be played using SetProperty.
> $A≔\mathrm{plots}:-\mathrm{animate}\left(\mathrm{plot3d},\left[\mathrm{sin}\left(a{x}^{2}\right){y}^{2},x=-\mathrm{\pi }..\mathrm{\pi },y=-1..1,\mathrm{grid}=\left[200,50\right],\mathrm{labels}=\left["","",""\right],\mathrm{orientation}=\left[85,75,0\right],\mathrm{scaling}=\mathrm{constrained}\right],a=1.0..3.0,\mathrm{frames}=20\right):$
> $P≔\mathrm{Plot}\left(\mathrm{identity}="MyAnim",\mathrm{showborders}=\mathrm{false},\mathrm{pixelwidth}=800,\mathrm{pixelheight}=400,\mathrm{InlinePlot}\left(A,\mathrm{scale}=2.7,\mathrm{width}=800,\mathrm{height}=400\right)\right):$\mathrm{xml}≔\mathrm{Worksheet}\left(\mathrm{Table}\left(\mathrm{alignment}=\mathrm{center},\mathrm{widthmode}=\mathrm{pixels},\mathrm{width}=720,\mathrm{Column}\left(\right),\mathrm{Row}\left(\mathrm{Cell}\left(\mathrm{Textfield}\left(P\right)\right)\right)\right)\right):
> $\mathrm{tb}≔\mathrm{InsertContent}\left(\mathrm{xml},':-\mathrm{output}'=\mathrm{table}\right):$
> $\mathrm{compnm}≔\mathrm{tb}\left["MyAnim"\right]$
${\mathrm{compnm}}{≔}{"MyAnim"}$ (5)
> $\mathrm{DocumentTools}:-\mathrm{SetProperty}\left(\mathrm{compnm},\mathrm{play},\mathrm{true}\right)$
Compatibility
• The DocumentTools:-Components:-Plot command was introduced in Maple 2015. | 2023-03-30 07:35:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 28, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6544414162635803, "perplexity": 3997.8134032546086}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00163.warc.gz"} |
http://journal.svmo.ru/en/archive/article?id=1244 | ### On classification of gradient-like diffeomorphisms on surfaces by means automorphisms of three-color graphs.
#### V. Z. Grines1, S. H. Kapkaeva2
Annotation This article is a continuation of the paper \cite{kapkaeva6}, in which the conditions of topological conjugacy of gradient-like diffeomorphisms are found, under suggestion that wandering set consists of only fixed points. In this paper we consider the class of orientation preserving gradient-like diffeomorphisms whose nonwandering set admits an existence of periodic orbits of period greater than one. To each diffeomorphism we appreciate three-color graph equipped by an automorphism given on the set of vertices of the graph. It is stated that all vertices of the graph have the same period under action of the automorphism. It is proved that the three-color graph equipped with the automorphism, is a complete topological invariant in the considered class of diffeomorphisms Morse-Smale diffeomorphisms, gradient-like diffeomorphisms, topological conjugate diffeomorphisms, three-color graph.
1Professor, Lobachevsky State University of Nizhni Novgorod, Nizhni Novgorod; vgrines@yandex.ru.
2Student, Mordovian State University after N.P. Ogarev, Saransk; kapkaevasvetlana@yandex.ru.
Citation: V. Z. Grines, S. H. Kapkaeva, "[On classification of gradient-like diffeomorphisms on surfaces by means automorphisms of three-color graphs.]", Zhurnal Srednevolzhskogo matematicheskogo obshchestva,15:2 (2013) 12–22 (In Russian) | 2022-09-26 05:15:36 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8410549163818359, "perplexity": 2291.4513136799346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334802.16/warc/CC-MAIN-20220926051040-20220926081040-00146.warc.gz"} |
https://www.jobilize.com/course/section/proof-the-cross-product-by-openstax?qcr=www.quizover.com | # 2.4 The cross product (Page 2/16)
Page 2 / 16
Notice what this means for the direction of $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}.$ If we apply the right-hand rule to $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u},$ we start with our fingers pointed in the direction of $\text{v},$ then curl our fingers toward the vector $\text{u}.$ In this case, the thumb points in the opposite direction of $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}.$ (Try it!)
## Anticommutativity of the cross product
Let $\text{u}=⟨0,2,1⟩$ and $\text{v}=⟨3,-1,0⟩.$ Calculate $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ and $\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}$ and graph them.
We have
$\begin{array}{ccc}\hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & ⟨\left(0+1\right),\text{−}\left(0-3\right),\left(0-6\right)⟩=⟨1,3,-6⟩\hfill \\ \hfill \text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}& =\hfill & ⟨\left(-1-0\right),\text{−}\left(3-0\right),\left(6-0\right)⟩=⟨-1,-3,6⟩.\hfill \end{array}$
We see that, in this case, $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)$ ( [link] ). We prove this in general later in this section.
Suppose vectors $\text{u}$ and $\text{v}$ lie in the xy -plane (the z -component of each vector is zero). Now suppose the x - and y -components of $\text{u}$ and the y -component of $\text{v}$ are all positive, whereas the x -component of $\text{v}$ is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}$ point?
Up (the positive z -direction)
The cross products of the standard unit vectors $\text{i},\text{j},$ and $\text{k}$ can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that
$\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}=\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}=\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}=0.$
(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar $0.\right)$ It’s up to you to verify the calculations on your own.
Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of $\text{i}$ and $\text{j}$ is parallel to $\text{k}.$ Similarly, the vector product of $\text{i}$ and $\text{k}$ is parallel to $\text{j},$ and the vector product of $\text{j}$ and $\text{k}$ is parallel to $\text{i}.$ We can use the right-hand rule to determine the direction of each product. Then we have
$\begin{array}{cccccccc}\hfill \text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =\hfill & \text{k}\hfill & & & \hfill \text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}& =\hfill & \text{−}\text{k}\hfill \\ \hfill \text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}& =\hfill & \text{i}\hfill & & & \hfill \text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}& =\hfill & \text{−}\text{i}\hfill \\ \hfill \text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}& =\hfill & \text{j}\hfill & & & \hfill \text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}& =\hfill & \text{−}\text{j}.\hfill \end{array}$
These formulas come in handy later.
## Cross product of standard unit vectors
Find $\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right).$
We know that $\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}=\text{i}.$ Therefore, $\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{k}\right)=\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}=0.$
Find $\left(\text{i}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{k}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{i}\right).$
$\text{−}\text{i}$
As we have seen, the dot product is often called the scalar product because it results in a scalar. The cross product results in a vector, so it is sometimes called the vector product . These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.
## Properties of the cross product
Let $\text{u},\text{v},$ and $\text{w}$ be vectors in space, and let $c$ be a scalar.
$\begin{array}{cccccccc}\text{i.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & \text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right)\hfill & & \text{Anticommutative property}\hfill \\ \text{ii.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{v}+\text{w}\right)& =\hfill & \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}+\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\hfill & & \text{Distributive property}\hfill \\ \text{iii.}\hfill & & & \hfill c\left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)& =\hfill & \left(c\text{u}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(c\text{v}\right)\hfill & & \text{Multiplication by a constant}\hfill \\ \text{iv.}\hfill & & & \hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0& =\hfill & 0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}=0\hfill & & \text{Cross product of the zero vector}\hfill \\ \text{v.}\hfill & & & \hfill \text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =\hfill & 0\hfill & & \text{Cross product of a vector with itself}\hfill \\ \text{vi.}\hfill & & & \hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =\hfill & \left(\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}\right)·\text{w}\hfill & & \text{Scalar triple product}\hfill \end{array}$
## Proof
For property $\text{i}.,$ we want to show $\text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}=\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right).$ We have
$\begin{array}{cc}\hfill \text{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{v}& =⟨{u}_{1},{u}_{2},{u}_{3}⟩\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}⟨{v}_{1},{v}_{2},{v}_{3}⟩\hfill \\ & =⟨{u}_{2}{v}_{3}-{u}_{3}{v}_{2},\text{−}{u}_{1}{v}_{3}+{u}_{3}{v}_{1},{u}_{1}{v}_{2}-{u}_{2}{v}_{1}⟩\hfill \\ & =\text{−}⟨{u}_{3}{v}_{2}-{u}_{2}{v}_{3},\text{−}{u}_{3}{v}_{1}+{u}_{1}{v}_{3},{u}_{2}{v}_{1}-{u}_{1}{v}_{2}⟩\hfill \\ & =\text{−}⟨{v}_{1},{v}_{2},{v}_{3}⟩\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}⟨{u}_{1},{u}_{2},{u}_{3}⟩\hfill \\ & =\text{−}\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u}\right).\hfill \end{array}$
Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
how did you get the value of 2000N.What calculations are needed to arrive at it
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https://forum.allaboutcircuits.com/threads/do-i-have-to-tin-the-tip-of-my-soldering-iron-every-single-time-other-questions.117323/page-3 | # Do I have to tin the tip of my soldering iron every single time? + other questions
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#### GopherT
Joined Nov 23, 2012
8,012
After being so concerned with a solder tip, I'm interested to see how timid you will be about applying power to a chip or LED.
#### #12
Joined Nov 30, 2010
18,217
After being so concerned with a solder tip, I'm interested to see how timid you will be about applying power to a chip or LED.
Excellent point. A good quality soldering iron can last for years, even if treated badly. An IC can smoke in an instant if you get even one connection wrong.
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
Ha-ha-ha. I want to read info and ask questions about my new tool, I think it's not that hard to get, my Mohawk friend.
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
Yeah, he's joking. He is a famous Canadian YouTuber that teach electrical stuff in a very funny way. He's awesome. Still not seen any possible way of hurting your eyes while soldering.
#### shortbus
Joined Sep 30, 2009
8,443
Ever heard of the infamous, "Murphy's law"? So worried about a ~$10 iron tip, but not about priceless eyes. #### dl324 Joined Mar 30, 2015 12,684 Still not seen any possible way of hurting your eyes while soldering. Here's an OT thread about solder/soldering injuries. At least one poster mentions having molten solder splashed on their person... I have a solder pot and I was extremely cautious the first time I used it. Thread Starter #### rambomhtri Joined Nov 9, 2015 346 What's up with some of you guys? What the heck is wrong about asking questions about how to take care of the soldering iron, which practically refers to the tip? That's the name of the post, for god's sake. This is a forum, I asked some questions about the tip cause I had questions. That's it. Some of you need to get over it. My god... If some of you think it does not take any kind of care, then that's fine, just say: -"Hi, IMO, the tip is not that important, it really does not matter if you use a sponge or wool, just clean it when you see it's dirty, and that's it, don't think about it too much". Others will say that it's crucial to take care of your tip, as I've already watched on YouTube and many other sites. I'll just read the comments, think about them and their logic, and try to create the best answer for my question. But if you come and say: -"Lol, so much care about a$5 tip and you don't care about your eyes"
It's irritating and has no sense. Tell me:
Do you wear glasses while cracking walnuts?
I'm sure you don't, you would even laugh at someone that uses glasses for that purpose.
Do you wear glasses while cutting your nails? And your cat's nails?
What about while cracking hazelnuts?
And while it's raining? You know some raining is acidic, and that can be very bad to your eyes? And while it's snowing?
No, no, no and no.
Do you want me to tell you:
-Lol, wears glasses while soldering cause there's 0.0005% chances of getting hurt by a remotely possible accident, but breakfasts everyday 4 walnuts and smashes them with a hammer without any kind of protections for his eyes, and a piece could hit his eye and make it bleed... Dumb...
OK, thank you dl324, as you've provided at least 2 personal experiences from 2 different users that has suffer an accident injuring his eyes. I still think hurting your eyes while soldering, with a fan pointing in the opposite direction, is highly unlikely.
Another question here:
The melting point of a 60/40 wire solder is 180-190ºC. Is it enough if I set the temperature knob to 200ºC? Or it works faster and better if I set it to 230ºC?
PD: Thank you all for the responses, but some users are starting to annoy me with their unhelpful responses.
#### kubeek
Joined Sep 20, 2005
5,746
300 is the usual temperature for me.
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
300 is the usual temperature for me.
Wow, I did not expect that... Why so far away from the melting point?
I know setting it to 190ºC would be inefficient due to thermal dissipation and other facts... but almost doubling the temperature of melting is kind of overkill, isn't it?
I'm talking about ºC, not ºF.
#### dl324
Joined Mar 30, 2015
12,684
I use a 700°F (370°C) tip for most of my soldering. Tips for my iron only come in 600, 700, or 800°F.
I usually set my hot air tool to around 280-3oo°C.
#### GopherT
Joined Nov 23, 2012
8,012
What's up with some of you guys? What the heck is wrong about asking questions about how to take care of the soldering iron, which practically refers to the tip? That's the name of the post, for god's sake.
This is a forum, I asked some questions about the tip cause I had questions. That's it. Some of you need to get over it. My god...
If some of you think it does not take any kind of care, then that's fine, just say:
-"Hi, IMO, the tip is not that important, it really does not matter if you use a sponge or wool, just clean it when you see it's dirty, and that's it, don't think about it too much".
Others will say that it's crucial to take care of your tip, as I've already watched on YouTube and many other sites. I'll just read the comments, think about them and their logic, and try to create the best answer for my question. But if you come and say:
-"Lol, so much care about a $5 tip and you don't care about your eyes" It's irritating and has no sense. Tell me: Do you wear glasses while cracking walnuts? I'm sure you don't, you would even laugh at someone that uses glasses for that purpose. Do you wear glasses while cutting your nails? And your cat's nails? What about while cracking hazelnuts? And while it's raining? You know some raining is acidic, and that can be very bad to your eyes? And while it's snowing? No, no, no and no. Do you want me to tell you: -Lol, wears glasses while soldering cause there's 0.0005% chances of getting hurt by a remotely possible accident, but breakfasts everyday 4 walnuts and smashes them with a hammer without any kind of protections for his eyes, and a piece could hit his eye and make it bleed... Dumb... OK, thank you dl324, as you've provided at least 2 personal experiences from 2 different users that has suffer an accident injuring his eyes. I still think hurting your eyes while soldering, with a fan pointing in the opposite direction, is highly unlikely. Another question here: The melting point of a 60/40 wire solder is 180-190ºC. Is it enough if I set the temperature knob to 200ºC? Or it works faster and better if I set it to 230ºC? PD: Thank you all for the responses, but some users are starting to annoy me with their unhelpful responses. Wow, I did not expect that... Why so far away from the melting point? I know setting it to 190ºC would be inefficient due to thermal dissipation and other facts... but almost doubling the temperature of melting is kind of overkill, isn't it? I'm talking about ºC, not ºF. Just like in cooking, the higher (and closer) the heat, the more rare the meat. That is, you heat the surface so fast that the inside has no time to get hot. Same goes for soldering. If you hold heat on the solder so long that it finally melts and flows, #### atferrari Joined Jan 6, 2004 4,248 but it makes a mess and if something gets in the way, you can knock the end off your iron. Add me to the list... Thread Starter #### rambomhtri Joined Nov 9, 2015 346 Another question right here: When you're done soldering, job finished, it's time to save your iron soldering and stuff. Nevertheless, the sponge is wet, so you can't put the soldering station inside the box cause the sponge will not dry. So... what do you do? Do you live the soldering station just like that, in your work area? #### #12 Joined Nov 30, 2010 18,217 Do you live the soldering station just like that, in your work area? Yes. Either that or take the sponge out and store it later. #### wayneh Joined Sep 9, 2010 17,061 Ever heard of the infamous, "Murphy's law"? So worried about a ~$10 iron tip, but not about priceless eyes.
Excellent point! Not that it matters, but the tips are only about $1, not$10.
That said, I have to admit I don't wear safety glasses like I should. I always feel like I can't see quite as well. I need one of those magnifying things you wear on your head. Could help solve both problems.
#### dl324
Joined Mar 30, 2015
12,684
When you're done soldering, job finished, it's time to save your iron soldering and stuff. Nevertheless, the sponge is wet, so you can't put the soldering station inside the box cause the sponge will not dry. So... what do you do?
Do you live the soldering station just like that, in your work area?
My soldering iron and hot air tool have dedicated space on my workbench.
The sponge should be damp, not wet when you use it. It reduces heat loss and thermal shock to the tip.
#### dl324
Joined Mar 30, 2015
12,684
That said, I have to admit I don't wear safety glasses like I should. I always feel like I can't see quite as well.
I always were safety glasses. I've lost count the number of times I've scraped flux off the lenses after a soldering session...
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https://nla-group.org/blog/page/3/ | ## Numerical Behaviour of Tensor Cores
For many years, the arithmetic operations available on most hardware were $+, -, *, /, \sqrt{}$. More recently ($\sim 20$ years ago) the FMA (fused multiply-add) operation also became prevalent in general purpose devices such as CPUs and GPUs. Software builds on top of these operations, for example compilers use a series of these and other hardware instructions to implement more complex functions and algorithms such as the exponential function, matrix multiply, or the algorithms for pseudo random number generation. These arithmetic operations have been standardized by the IEEE 754 floating-point standard since 1985 and most current systems are compliant with it.
Recently, because of the increasing adoption of machine learning, general purpose devices started to include inner product or matrix multiply-accumulate (MMA) operations in hardware. This is a generalization of a scalar FMA to vectors and matrices. Since it is performed in hardware, the expected speedup is achieved due to parallelism — instead of using a few FMA hardware units sequentially to multiply matrices as is the case in software, all of the elements of a matrix of some size are computed in parallel. Using the inner product and matrix multiply-accumulate operations in hardware, compute-bound applications that have high-intensity usage of them are sped up significantly.
NVIDIA GPUs are widely used for machine learning and other applications. In the latest TOP500 supercomputer list published in June 2020 [1], 112 computers are equipped with NVIDIA graphics cards. The main feature of the recent NVIDIA GPUs is hundreds of arithmetic units for performing the MMA operation — NVIDIA calls these tensor cores. Various applications outside machine learning are being explored in order to utilize very high arithmetic throughput that can be achieved when using MMAs in hardware [2]. Table 1 lists three recent NVIDIA architectures as well as other hardware devices with an MMA operation on chip. The NVIDIA V100 has the first version and the T4 has the second version of tensor cores; in these devices tensor cores work on matrices of size $4 \times 4$ in mixed-precision of fp16 and fp32, as shown in Figure 1. The most recent revision of tensor cores in the A100 updated both the precisions available and the dimensions of input/output matrices. See the NVIDIA V100 and NVIDIA A100 whitepapers for more details.
The MMA operation is not standardized by the IEEE 754, therefore various numerical features of tensor cores are not clear. Knowing such features as the support for subnormal numbers, order of operations, and rounding modes and normalization of significands in various parts of the MMA computation can be important, for example when doing error analysis of algorithms that utilize MMAs in order to derive error bounds [3]. As the application space utilizing tensor cores is growing beyond machine learning, understanding of the numerical behaviour of tensor cores will become increasingly useful. Even before such hardware is put to use in some applications, one might want to simulate the behaviour of tensor cores in order to develop numerical codes targeted to tensor cores, on a conventional hardware. Furthermore, there is also a question of differences in numerical behaviour between tensor cores and software MMA using the standard FMA hardware calls, for example differences that would appear when transitioning the existent software to use tensor cores.
Motivated by this, in our recent preprint [4] we investigate an experimental testing methodology of the tensor cores. The method consists of the following steps.
1. Identify a numerical feature that needs to be tested, for example, the rounding mode in the 5-operand adders that accumulate products in MMA.
2. Identify possible implementations, for example round to nearest or round toward zero.
3. Find floating-point inputs that would result in different outputs for each possible hardware behaviour in 2.
4. Observe outputs and make a conclusion based on expected outputs for each possible behaviour in 2.
Using this approach, we identified the following list of numerical features of the NVIDIA tensor cores.
• Subnormal floating-point numbers are fully supported, both on the inputs and outputs.
• 5-operand adders accumulate 5 addends (4 products from $AB$ and a value from $C$) starting from the largest in magnitude.
• Round toward zero, rather than round to nearest (default in IEEE 754-compliant arithmetic), in the 5-operand adders is implemented.
• Different normalization behaviour from the MMA implemented in software (tensor cores normalize the answer of the whole 5-element dot product at the end rather than after each addition of products).
• Inner products without intermediate normalization are shown to be non-monotonic in rare cases (this result is more general than tensor cores, since to the best of our knowledge, most hardware implementations do not normalize on every addition due to lower hardware costs).
Our conclusion is that in the current version, the tensor cores on V100 and T4 (the A100 is not yet available to us) do not replicate the behaviour of the MMA implemented with IEEE 754 compliant FMA hardware operations. These numerical behaviours are expected in a hardware MMA optimized for reducing hardware cost and most likely is motivated by a fact that machine learning applications usually are claimed to not need all the IEEE 754 features and high precision in general. These results provide the parameters that can be used in rounding error analysis of tensor cores [3] which can be useful when developing numerical software.
Our CUDA code to test numerical features of tensor cores is available here.
### References
[2] A. Abdelfattah et al. A Survey of Numerical Methods Utilizing Mixed Precision Arithmetic. July 2020. Published online.
[3] P. Blanchard, N. J. Higham , F. Lopez, T. Mary, and S. Pranesh. Mixed Precision Block Fused Multiply-Add: Error Analysis and Application to GPU Tensor Cores. SIAM J. Sci. Comput. May 2020.
[4] M. Fasi, N. J. Higham, M. Mikaitis, and S. Pranesh. Numerical Behavior of NVIDIA Tensor Cores. July 2020; revised October 2020. MIMS Eprint, published online.
## Françoise Tisseur Awarded the LMS Fröhlich Prize
Professor Françoise Tisseur has been awarded the London Mathematics Society’s Fröhlich Prize for “her important and highly innovative contributions to the analysis, perturbation theory, and numerical solution of nonlinear eigenvalue problems”.
The Fröhlich Prize is awarded even two years in memory of Albrecht Fröhlich, FRS. The prize is awarded for original and extremely innovative work in any branch of mathematics.
The full prize citation is available here.
## Nick Higham Interviewed in The Actuary Magazine
Professor Nick Higham is featured in the June 2020 issue of The Actuary Magazine, the magazine of the Institute and Faculty of Actuaries.
In an interview with the editor, Dan Georgescu, Nick talks about his development in 2002 of his nearest correlation matrix algorithm, which is widely used in finance and data science.
The magazine is freely readable on the Issuu platform.
More details about the nearest correlation matrix problem and algorithms for solving it are available on Nick’s blog.
## Stochastic Rounding Has Unconditional Probabilistic Error Bounds
In the IEEE standard 754 for binary floating-point arithmetic, four rounding modes are defined.
• Round to nearest.
• Round towards $0$.
• Round towards $+\infty$.
• Round towards $-\infty$.
Recently stochastic rounding, an old idea that dates back to the beginning of the digital computer era, has gained popularity, most notably in deep learning. While the rounding modes defined in the IEEE standard are deterministic, stochastic rounding is inherently random. We can define two modes of stochastic rounding. Consider the figure below, where we have a real number $x$ and adjacent floating-point numbers $a$ and $b$. In what we call mode 1 stochastic rounding, we round $x$ to either $a$ or $b$ with equal probability. In mode 2 stochastic rounding, we round to $a$ (or $b$) with probability proportional to $1$ minus the distance of $x$ from $a$ (or $b$). So in the example shown, for mode 2 we are more likely to round to $b$ than to $a$.
In our recent EPrint Stochastic Rounding and its Probabilistic Backward Error Analysis, Nick Higham, Theo Mary and I generalized previous probabilistic error analysis results [SIAM J. Sci. Comput., 41 (2019), pp. A2815–A2835] to stochastic rounding.
We show that stochastic rounding (specifically mode 2) has the property of producing rounding errors $\delta_k$ that are random variables satisfying
• mean zero: $\mathbb{E}(\delta_k) = 0$,
• mean independent: $\mathbb{E}(\delta_k \mid \delta_{k-1}, \dots, \delta_1) = \mathbb{E}(\delta_k)$.
Here, $\mathbb{E}$ denotes the expectation. A key consequence of these results is that we can replace the worst case error bound proportional to $\gamma_n = nu + O(u^2)$, ubiquitous in backward error analyses, by a more informative probabilistic bound proportional to $\widetilde{\gamma}_n = \sqrt{n}u + O(u^2)$. What is a rule of thumb for round to nearest becomes a rule for stochastic rounding: it is proved that our rounding errors satisfy the above properties.
In the figure below, we compute the inner product $s = x^Ty$ of vectors sampled uniformly from $[0,1]$ in both round to nearest and stochastic rounding in IEEE half precision arithmetic. Shown is the backward error for each value of $n$ and the bounds $\gamma_n$ and $\widetilde{\gamma}_n$.
For increasing problem size $n$, with round to nearest the error no longer satisfies the $\sqrt{n}u$ bound. This is due to the phenomenon we call stagnation, and low precisions are particularly susceptible to it. As we compute recursively $s_i = s_{i-1} + x_iy_i$, eventually the partial sum $s_{i-1}$ can grow so large that the update $x_iy_i$ is less than half of the spacing of floating-point numbers around $s_{i-1}$ and under round to nearest the partial sum does not increase. This means we produce negative rounding errors, which are of course not mean zero. Stochastic rounding avoids this issue by rounding randomly. In fact we can prove that $\mathbb{E}(\widehat{s}) = s$, that is, the expected value of the computed result under stochastic rounding is the exact result.
## Online Seminar Series on Numerical Linear Algebra
Dr Stefan Güttel is one of the organizers of a new online series of seminars on Numerical Linear Algebra. The talks are broadcast live on Zoom and YouTube every Wednesday at 4pm Central European Summer Time (CEST). The programme is available on the seminar website. At the time of writing, 942 people from 79 countries have signed up to the seminar mailing list.
The other organizers are Melina Freitag (University of Potsdam), Daniel Kressner (EPF Lausanne), Jörg Liesen (TU Berlin), Valeria Simoncini (University of Bologna), and Bart Vandereycken (University of Geneva).
The first talk in the series was given by Professor Nick Higham, titled Are Numerical Linear Algebra Algorithms Accurate at Extreme Scale and at Low Precisions? The slides for the talk are available at this link and the video is below.
## Extreme-Scale Test Matrices with Specified 2-Norm Condition Number
The supercomputers in the TOP500 list are ranked using the High-Performance Linpack (HPL) Benchmark, which gauges the performance of a computer by solving a large dense linear system by Gaussian elimination with partial pivoting. The size of the coefficient matrix depends on the computational power of the machine being assessed, because more powerful systems require larger benchmark problems in order to reach their peak performance.
The test matrices used by HPL have random entries uniformly distributed on the interval $(-1/2,1/2]$. The 2-norm condition number of such matrices depends on their size, and can potentially be very large for the matrices required by today’s most powerful computers: the largest linear systems solved on Summit, the machines that leads the November 2019 TOP500 ranking, have order $10^7$, and even larger systems will be needed to benchmark the coming generations of exascale systems.
An $n \times n$ matrix with specified 2-norm condition number can be generated as $A := U \varSigma V^T$, where $U$ and $V$ are random real orthogonal $n \times n$ matrices from the Haar distribution (a natural uniform probability distribution on the orthogonal matrices) and $\varSigma$ is diagonal with nonnegative entries $\sigma_1\ge \cdots \ge \sigma_n \ge 0$. It is well known that $A$ has 2-norm condition number $\kappa_2(A) = \sigma_1/\sigma_n$ if $\sigma_n \neq 0$ and $\kappa(A) = \infty$ otherwise. This technique, which is used by the gallery('randsvd', ...) function, requires $2n^3$ floating-point operations to generate a test matrix of order $n$, and would thus be impractical in an extreme-scale setting.
In our recent EPrint Generating extreme-scale matrices with specified singular values or condition numbers, Nick Higham and I present four methods that, by giving up the requirement that the matrices $U$ and $V$ be from the Haar distribution, reduce the cost of the approach above from cubic to quadratic. The matrices generated retain a number of desirable properties that make them a suitable choice for testing linear solvers at scale.
These cheaper algorithms are particularly well suited to distributed-memory environments, since all communication between the nodes can be avoided at the price of a negligible increase in the overall computational cost. They are appealing to MATLAB users too, as the following example demostrates.
n = 10000; kappa = 1e6; mode = 2; rng(1)
% gallery('randsvd',...)
fprintf('gallery(''randsvd'',...): %5.2f seconds elapsed.\n',...
timeit(@()gallery('randsvd',n,kappa,mode,[],[],1)));
% Algorithm 3.1 in our EPrint.
method = 1; matrix = 0;
fprintf('Alg. 3.1 with Haar U: %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
matrix = 2;
fprintf('Alg. 3.1 with U=gallery(''orthog'',n,2): %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
% Algorithm 4.1 in our EPrint.
method = 3; matrix = 0;
fprintf('Alg. 4.1 with Haar U: %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
matrix = 2;
fprintf('Alg. 4.1 with U=gallery(''orthog'',n,2): %5.2f seconds elapsed.\n',...
timeit(@()randsvdfast(n,kappa,mode,method,matrix)));
In the listing above, randsvdfast is an implementation of our algorithms available on the MATLAB Central File Exchange. Setting the argument matrix to 0 tells our function to pick $U$ from the Haar distribution, whereas setting it to 2 causes $U$ to be chosen in a non-random way. Running this code in MATLAB 2019b on a machine equipped with an Intel processor I5-6500 running at 3.20GHz the script produces the output
gallery('randsvd',...): 79.52 seconds elapsed.
Alg. 3.1 with Haar U: 19.70 seconds elapsed.
Alg. 3.1 with U=gallery('orthog',n,2): 1.90 seconds elapsed.
Alg. 4.1 with Haar U: 19.28 seconds elapsed.
Alg. 4.1 with U=gallery('orthog',n,2): 1.43 seconds elapsed.
Therefore, for $n = 10{,}000$ randsvdfast is up to 56 times faster than gallery('randsvd',...).
## 2020 NLA group photo
This year’s group photo was taken on March 5, 2020 after the NLA group meeting. Most group members are in the photo; those missing include Jack Dongarra, Stefan Güttel, Ramaseshan Kannan and Marcus Webb.
The banner on this website has also been updated with the new group photo. A high resolution version of the photo is available here.
By row from the back: Craig Lucas, Nick Higham, Xinye Chen, Steven Elsworth, Xiaobo (Bob) Liu, Michael Connolly, Mantas Mikaitis, Len Freeman, Massimiliano Fasi, Pierre Blanchard, Sven Hammarling, Asad Raza Aitor Mehasi Mehasi, Stephanie Lai, Gian Maria Negri Porzio, Thomas McSweeney Mawussi Zounon, Françoise Tisseur, Srikara Pranesh, Yuqing (Mila) Zhang, Eleni Vlachopoulou.
## NLA Group at the SIAM Conference on Parallel Processing for Scientific Computing
Several members of the group attended the SIAM Conference on Parallel Processing for Scientific Computing held in Seattle on February 12-15, 2020.
The presentations given are as follows:
Nick Higham and Srikara Pranesh also organised a two part mini-symposium (Advances in Algorithms Exploiting Low Precision Floating-Point Arithmetic, MS10 and MS21) at the conference.
Max Fasi, Mantas Mikatis, Mawussi Zounon, Sri Pranesh, Theo Mary at SIAM Conference on Parallel Processing for Scientific Computing, Seattle, Washington, February 12-15, 2020.
## Conference Celebrating the 70th Birthday of Jack Dongarra
by Sven Hammarling, Nick Higham, and Françoise Tisseur
Jack Dongarra
July 18, 2020 is the 70th birthday of Professor Jack Dongarra, who holds appointments at the University of Tennessee, Oak Ridge National Laboratory, and the University of Manchester.
Jack has made seminal contributions to algorithms for numerical linear algebra and the design and development of high performance mathematical software for machines ranging from workstations to the largest parallel computers. His recent honours include election as a Foreign Member of the Royal Society and receipt of the
SIAM/ACM Prize in Computational Science and Engineering (2019)and the IEEE Computer Society Computer Pioneer Award (2020).
To celebrate Jack’s birthday we are organizing a conference New Directions in Numerical Linear Algebra and High Performance Computing: Celebrating the 70th Birthday of Jack Dongarra at The University of Manchester, July 17, 2020. Registration is now open and we welcome submission of posters.
## Sharper Probabilistic Backward Error Bounds
Most backward error bounds for numerical linear algebra algorithms are of the form $nu$, for a machine precision $u$ and a problem size $n$. The dependence on $n$ of these bounds is known to be pessimistic: together with Nick Higham, our recent probabilistic analysis [SIAM J. Sci. Comput., 41 (2019), pp. A2815–A2835], which assumes rounding errors to be independent random variables of mean zero, proves that $n$ can be replaced by a small multiple of $\sqrt{n}$ with high probability. However, even these smaller bounds can still be pessimistic, as the figure below illustrates.
The figure plots the backward error for summation (in single precision) of $n$ floating-point numbers randomly sampled from a uniform distribution. For numbers in the $[0,1]$ distribution, the bound $\sqrt{n}u$ is almost sharp and accurately predicts the error growth. However, for the $[-1,1]$ distribution, the error is much smaller, seemingly not growing with $n$. This strong dependence of the backward error on the data cannot be explained by the existing bounds, which do not depend on the values of the data.
In our recent preprint, we perform a new probabilistic analysis that combines a probabilistic model of the rounding errors with a second probabilistic model of the data. Our analysis reveals a strong dependence of the backward error on the mean of the data $\mu$: indeed, our new backward error bounds are proportional to $\mu\sqrt{n}u + u$. Therefore, for data with small or zero mean, these new bounds are much sharper as they bound the backward error by a small multiple of the machine precision independent of the problem size $n$.
Motivated by this observation, we also propose new algorithms that transform the data to have zero mean, so as to benefit from these more favorable bounds. We implement this idea for matrix multiplication and show that our new algorithm can produce significantly more accurate results than standard matrix multiplication. | 2021-11-29 20:00:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 82, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5951012372970581, "perplexity": 1169.8859679485834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00451.warc.gz"} |
https://undergroundmathematics.org/circles/conic-sections-in-real-life | ### Circles
Bigger picture
Euclid and Archimedes are just two of the ancient Greek mathematicians to have studied conic sections—the shapes created by slicing through a double cone with a flat plane. If the plane is perpendicular to the axis of the double cone, the intersection is a circle, and if the plane is angled parallel to the side of the cone the intersection is a parabola. If the plane cuts through both cones the intersection is a hyperbola. The only remaining case gives an ellipse.
As well as having a description in terms of slicing through a cone, these shapes also have a description in terms of loci (locations) of points. A circle is the locus of points that are the same distance from the centre (the focus of the circle). An ellipse is the locus of points whose distances from two foci sum to a constant $c$ (that is, $x+y = c$ in the image below). A hyperbola is the locus of points that have the same difference $c$ between distances from two foci (that is, $|x-y| = c$). A parabola is the locus of points equidistant from a focus and a straight line called the directrix (that is, $x=y$).
Skip forward a couple of millennia and the mathematical properties of the conic sections have a very modern application. Suppose that you are lost in the jungle, but that luckily you are equipped with a device that can transmit and receive information, a clock, and a map. Your device picks up a signal from my device, which tells you my exact location and the exact time I sent my signal.
Assuming you know how fast the signal travels, how can you use the information you have to determine how far away you are from me?
Suppose you have worked out that you are at distance $d$ from me. Which conic section describes all your possible locations?
Suppose you have picked up two signals, one from me and one from another friend, Anna, telling you where on your map each of us is and at what time we sent the signal. How does that narrow down your possible locations?
Many smartphones are now able to use information sent out by GPS satellites (of which there are about nine overhead at any one time) in a similar way. Since the satellites are moving around in space, and therefore can’t be located on a two-dimensional map of the Earth, we now need to consider three-dimensional geometry. Can you work out how many satellites are needed to be sure of the location of a smartphone?
This method for finding a target’s location is called trilateration. Now let’s explore another method, multireceiver radar, which uses another conic section but requires fewer transmissions.
Suppose again that you are lost in the jungle but that this time you have no map. My friend Anna and I need to pinpoint your location to come and rescue you. I send you a signal, you immediately reply and Anna picks up the reply. Anna and I can communicate to establish the time that passed between me sending the signal and Anna receiving your reply.
What distance can Anna and I work out using that information?
Which conic section describes all your possible locations?
How many more people are needed to pinpoint your exact location? How many transmissions are needed?
One downside of this method is that it assumes the target will immediately reply to any signal we send. What if we can’t guarantee they will respond? Perhaps they don’t want to be found? Another conic section provides the answer.
Suppose now that you are hiding out in the jungle and that you are transmitting signals to your conspirators elsewhere. So as to not give away your location too easily your signals do not include information as to when they were sent. Anna and I can both pick up a signal sent by you and communicate to each other the time at which we received it. This means that we can work out the difference between the time it took the signal to reach me and the time it took to reach Anna.
What distance can Anna and I work out using that information?
Which conic section describes all your possible locations?
To pinpoint your exact location I need more colleagues, in addition to Anna, who can also pick up your signal and communicate with me. How many more colleagues do I need?
This covert technique, called multilateration, was used in WWI to locate enemy artillery ranges by listening to the sound of their gunfire. Mathematical ideas that are over 2,000 years old are used to find lost souls and uncover hidden enemies! | 2020-04-07 22:46:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6309939026832581, "perplexity": 438.70432629842384}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00387.warc.gz"} |
https://tech.io/playgrounds/40701/contribute---help/writing-the-statement | ### Open Source Your Knowledge, Become a Contributor
The statement should be written in English. Make it as clear and concise as possible.
It should contain:
1. A quick description of the goal: 1 or 2 sentences maximum.
2. A list of rules to explain how the puzzle/game works.
3. The Winning/Losing conditions (for games created with the SDK only).
4. The protocol
• input parameters and their types
• expected output
• reasonable constraints
## Guidelines
The statement must be clear and non-ambiguous.
• Keep it clear and concise
A long statement can scare players away.
• Avoid flavour text
For instance, don’t say “The unit hits the other unit on the nose, causing 1 point of excruciating damage”, do say “The unit deals 1 damage”.
For instance don’t say “This next part looks complicated but you’ll get it, don’t worry!”.
• Avoid controversial topics like politics.
• Don't overlook the protocol
It must be clear, precise, with reasonable constraints...
## Puzzle Statement
(In/Out puzzle and CoC only)
### Syntax
<<Bold>> [[Variable]] {{Constant}} renders as:
Monospace renders as: Monospace
### Example
You must output the closest pair of points (euclidean distance).
Input:
<<Line 1:>> An integer [[N]] for the number of points to compare.
<<Next [[N]] lines:>> Two space separated integers [[x]] and [[y]] for the coordinates of a point.
Output:
<<Line 1 :>> The coordinates of the point from the closest pair of points in the same order as given in input.
<<Line 2 :>> The coordinates of the other point from the closest pair of points.
Constraints:
2 ≤ [[N]] ≤ 100
-1000 < [[x]],[[y]] < 1000
## Game Statement
(For all sdk games)
For all games created with the sdk, the statement is an html document. You can find a default one here
### Leagues
There is one statement file per league. They're all generated from one general statement file (located here: /config/statement_en.html.tpl) when building the project.
### Conditional logic
To show a piece of statement only in league x, write
<!-- BEGIN levelx -->
Piece of statement.
<!-- END -->
To highlight a new rule, use the same as above with the following div style:
<!-- BEGIN levelx -->
<div style="color: #7cc576;
background-color: rgba(124, 197, 118,.1); | 2020-10-30 05:12:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22438107430934906, "perplexity": 8923.845567056784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107907213.64/warc/CC-MAIN-20201030033658-20201030063658-00249.warc.gz"} |
https://physics.com.hk/2015/11/04/ground-states-and-annihilation-operators/ | # Ground states and Annihilation operators
1. The equation $a | 0 \rangle = 0$ means that the eigenvalue of $a$ on $| 0 \rangle$ is 0:
$a | 0 \rangle = 0 | 0 \rangle$
2. The length of the vector $a | 0 \rangle$ is 0:
$\langle 0 | a^\dagger a | 0 \rangle = 0$
3. The physical meaning is that the probability of the system being at state $a | 0 \rangle$ is 0.
In other words, there is no state with an eigen-energy lower than the ground state one.
4. For the equation $a | 0 \rangle = 0 | 0 \rangle$, the 0 at the right is a scalar.
5. For the equation $a | 0 \rangle = 0$, the 0 at the right is a zero vector – a state vector with length zero.
6. $| 0 \rangle$ is a state vector. However, it is NOT the zero vector.
Instead, it is the state vector of the ground state. Its length is 1 unit.
— Me@2015-11-03 03:26:58 PM | 2020-10-24 20:24:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9316912293434143, "perplexity": 207.4690305431968}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107884755.46/warc/CC-MAIN-20201024194049-20201024224049-00127.warc.gz"} |
https://www.physicsforums.com/threads/thermodynamics-idealgases-w-q-etc-much-help-needed-detailed-questions-in-descrip.663550/ | # Thermodynamics [IdealGases,W,Q, etc.] Much Help Needed-Detailed questions in descrip.
1. Jan 10, 2013
I'm sorry I cannot conform to the default format Physicsforums.com; it is because I do not even know the first step to solving these sorts of problems, I don't know which equations to use which is a major problem. Here are the types of questions I require understanding.
-
An ideal gas goes through three processes (A>B>C>[A]) (Triangular form) (PV Chart)
How would I figure out The Q, W, and ΔU (internal energy) for A to B, B to C, C to A?
-
On another graph using variables but this time with numerical values for P and V on the axis, how would I find the work done by a monatomic ideal gas as it expands from point A to point C along the path shown in the figure? Also, how much heat would be absorbed BY the gas during this process?
-
Finding the net work, heat, and ΔU in another PV Graph with data on the axis-es?
-
Calculating temperature, work, and/or internal energy in another PV Graph?
-
Whether or not W, Q, or ΔU is positive(gained) or negative(released) in an ideal gas system as well as how those three (Q,W,..U) are related?
-
I really wish to work on the problems myself, so I have only asked what procedures I should take.
Here is a list of equations I have scavenged.
ΔU = Won + Q
ΔU = (3/2)nRΔT
Won = -PΔV
P1V1 = P2V2
(P1V1)/T2 = (P2V2)/T2
PV = nRT
Is there any important equation that I am missing?
Last edited: Jan 10, 2013
2. Jan 12, 2013
### dikmikkel
Re: Thermodynamics [IdealGases,W,Q, etc.] Much Help Needed-Detailed questions in desc
I think that since you are working with idel gasses, you can dervie most of the stuff from 3-4 equations, namely:
The ideal gas law : pV = nRT
The Laws of thermodynamics.(2-3 laws is useful).
If you want to find the Work done during a isothermal(T constant) step e.g.:
$W = -\int\limits_{V_1}^{V_2}\! p\,\text{d}V = -\int\limits_{V_1}^{V_2}\! \dfrac{nRT}{V}\,\text{d}V = -nRT\left(\ln(V_2)-\ln(V_1)\right) = -nRT\ln\left(\dfrac{V_2}{V_1}\right)$
I hope that I understood your question. | 2017-10-21 22:36:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5584793090820312, "perplexity": 1463.430627887772}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00785.warc.gz"} |
http://mymathforum.com/algebra/34778-advanced-analysis-quintics.html | My Math Forum Advanced Analysis of Quintics
Algebra Pre-Algebra and Basic Algebra Math Forum
March 22nd, 2013, 03:49 AM #1 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Advanced Analysis of Quintics Hello members of MMF, One of our senior members of MMF, agentredlum, requested me to create such a topic but withdrawn his request soon. This subject has stayed very foggy to majority of peoples like us and no one ever really shown the effort to describe this subject elegantly enough. I will give it a try to do this. Quintics are a mysterious subject. Unlike the other degrees < 5, quintics cannot be solved using elementary arithmetic operations and radicals. There are two proofs of this, one is by Abel and another is by Galois. Abel's theory is easier but it cannot specify which quintics are solvable and which are not, but Galois theory can be used to do the work. I will first describe the transformation of quintics to reduce it into several forms, then we will start the several methods to solve a quintic. NOTE : Any user is allowed to reply in this topic and ask for clarification. For all other comments, I ask that you post them in the auxiliary topic created just for this purpose. Thank you! Thanks from agentredlum and duc
March 22nd, 2013, 04:29 AM #2 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Advanced Analysis of Quintics Here, I will describe the transformation of a general quintic into a depressed quintic : $x^5 + a_1 x^4 + a_2 x^3 + a_3 x^2 + a_4 x + a_5= 0 \;\;\; (1)$ In our work throughout this topic, I will use the quintic above as your general quintic. We want to transform the quintic above into the depressed form : $y^5 + b_2 y^3 + b_3 y^2 + b_4 y + b_5= 0 \;\;\; (2)$ The first thing is to do is to write a linear relation between them : $y= x + k \;\;\; (3)$ Finding k by substituting (3) into (1) is rather tedious, so we'll avoid the arithmetic over polynomial ring n = 5 by noting that in (2), $\sum y_i= 0$ where yi for i = 1,2,3,4,5 are the five roots of (2). Now, since transformation (3) is showing a relation ship between the variables of quintic (2) and (1), it must also hold when we replace y in (3) by one of the roots of (2) and replace x by one of the roots of (1) corresponding to that of (2). Hence, we may write $y_i= x_i + k$ Now, sum up all of these relation for i = 1,2,3,4,5 : $\sum y_i= \sum x_i + 5 \cdot k$ Since the LHS is zero, and $\sum x_i= - a_1$ by NI, $k= -\frac{a_1}{5}$. So, we have the transformation and this way we can calculate the coefficients of the new depressed quintic which is much easier to handle than the original general form. The principal quintic form will be described in the second post. Balarka . Thanks from agentredlum and duc
March 23rd, 2013, 02:10 AM #3 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Advanced Analysis of Quintics OK, now for the principal quintic : In order to transform (1) into the principle form : $z^5 + 5a z^2 + 5bz + c= 0 \;\; (4)$ We need a quadratic Tschrinhausen transform : $z\,=\, x^2\, -\, ux\, +\, v \;\; (5)$ Now, similar to the method used before, we will sum the identity up for all the roots : $\sum z_k= \sum x_k ^2 - u \sum x_k + 5v$ Note that the LHS is zero. The sum $\sum x_k^2$ is $a_1^2 - 2a_2$ from elementary theory of Newton's identities. Hence, we reckon that $5v= -a_1 u - a_1 ^2 + 2a_2$. Now all we have to do is to derive the u in terms of the elementary symmetric polynomials. Let us square 5 and sum it up again : $\sum z_k^2= \sum (x_k ^2 - u x_k + v)^2$ LHS is again zero since equation (4) doesn't have the 'x^3' term. Expanding the RHS gives : $\sum x_k ^4 + u^2 \sum x_k ^2 + 5v^2 - 2u \sum x_k ^3 + 2v \sum x_k^2 - 2uv \sum x_k= 0$ Note that all the sums here are nothing but the Newton's symmetric formulas. After substituting the expressions obtained by NI, we get a quadratic $(2 a_1^2 - 5 a_2)u^2 - (4 a_1^3 - 13 a_1 a_2 + 15 a_3) u + (2 a_1^4 - 8 a_1^2 a_2 + 10 a_1 a_3 + 2 a_2^2 - 10 a_4)= 0$ After solving this quadratic, we will get u and substituting this into v will yield the full transform (5). In the next post, I will introduce how to avoid substituting (5) directly into (1) in order to calculate the coefficients of (4). Thanks from agentredlum and duc
March 24th, 2013, 11:58 AM #4 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Advanced Analysis of Quintics Now, consider the quintic $x^5 + b_0 x^4 + b_1 x^3 + b_2 x^2 + b_3 x + b_4= (x + u)^5 + a_0 (x + u)^4 + a_1 (x + u)^3 + a_2 (x + u)^2 + a_3 (x + u) + a_4 = 0$ We will write the principal quintic (4) with z replaced by v : $v^5 + 5av^2 + 5bv + c= \prod (v - z_k) \;\; (6)$ Now substitute (5) into the RHS of (6) to get : $v^5 + 5av^2 + 5bv + c= - \prod (x_k^2 - u x_k) = - \prod x_k \prod (x_k - u) = -a_4 \cdot b_4 \;\; (7)$ It can be easily seen that $b_4= u^5 + a_0 u^4 + a_1 u^3 + a_2 u^2 + a_3 u + a_4$ from the definition. Substituting the above into (7) and solving for c will give the result : $c= - a_4 (u^5 + u^5 + a_0 u^4 + a_1 u^3 + a_2 u^2 + a_3 u + a_4) - 5a v^2 - 5bv$ Similarly, $5b= a_3 (u^4 + a_0 u^3 + a_1 u^2 + a_2 u + a_3) - a_4 (5 u^3 + 4 a_0 u^2 + 3 a_1 u + a_2) - 5 v^4 - 10 a v$ $5a= - a_2 (u^3 + a_0 u^2 + a_1 u + a_2) + a_3 (4 u^2 + 3 a_0 u + 2 a_1) - a_4 (5u + 2a_0) - 10v^3$ We can obtain these by differentiating and double differentiating (6), respectively. The transformation of a principle quintic into a Brioschi Quintic will be explained in my next post. It would take a little knowledge in polyhedral symmetry, so be prepared! Thanks from agentredlum and duc
March 26th, 2013, 01:39 PM #5 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Advanced Analysis of Quintics We have been able to transform a general quintic into a principle quintic all the way to now. I will introduce how to transform the principle quintic (4) into the one-parameter Brioschi Quintic : $x^5 - 10 Z x^3 + 45 Z^2 x - Z^2= 0$ I said before that I would use Polyhedral Symmetry to geometrically describe the transformation but I and agentredlum decided together that we will save the Polyhedral Symmetry for Klein's method which will probably be described later and use the algebraic transformation instead. Consider the functions $g(x)= x^3 + ax^2 - \frac{1}{5} \sum x_k \;\;\; h(z) = x^4 + bx^2 - \frac{1}{5} \sum x_k$ We see that they have the property that $\sum g= 0$ and $\sum h= 0$ where the sum is taken over the roots of (4) and g, h are short notations for g(x) and h(x). Interestingly, they have to property $\sum x_k g= 0$ and $\sum x_k h= 0$ if and only if $\sum x_k^4 + a \sum x_k ^3= 0$ and $\sum x_k ^ 5 + b \sum x_k^3= 0$ and both of these can be satisfied with proper values of a & b. Now consider the function : $\psi(x)= \alpha g(x) + \beta h(x) \;\; (" /> This have the property $\psi^2= 0$ if $\alpha \sum g^2 + 2 \alpha \beta \sum g h + \beta^2 \sum h^2= 0$ The homogeneous equation above can be satisfied with proper non-zero choices of $$$\alpha, \beta$$$. So, the properties we have got so far are : $\sum x_k= 0 \;\, \sum x_k^2 = 0 \;\, \sum \psi = 0 \;\, \sum \psi^2 = 0 \;\, \sum x_k \psi = 0 \;\; (9)$ Now, from here, we can easily derive the Bring-Jerrard form but I'll save that for another day and proceed further to our current objective. It can be shown, that for a function $\psi$ defined as ( and follows the properties (9), there always exists values p, q, r, a, b, t such that $p \psi^2 + 2 q x \psi + r x^2 - a \psi - b x + t= 0 \pmod{f(x)} \;\; (10)$ Where $f$ is the principle quintic form (4). The proof is pretty elementary and I am not omitting it here; I'll surely post the proof if someone PMs for that reason. Now, if we sum up (10) for the roots of $f(x)$ and apply the properties (9), we see that t = 0 for obvious reasons. Now let $\chi$ and $\phi$ be linear functions(NOTE : Not [color=#FF0000]affine functions[/color] of...) of $x$ and $\psi$, respectively. Hence, by (9), $\sum \chi= 0 \;\, \sum \chi^2 = 0 \;\, \sum \phi = 0 \;\, \sum \phi^2 = 0 \;\, \sum \chi \phi = 0 \;\; (11)$ and from (10), $p \phi^2 + 2 q \chi \phi + r \chi^2 - a \phi - b \chi= 0 \pmod{f(x)} \;\; (12)$ We may need another identity so I am introducing it before the real calculations : $m(p \chi^2 + 2 q \phi \chi + r \phi^2)= (d \chi + e \phi )^2 - c (a \chi + b \phi)^2 \;\; (13)$ Where $m= pb^2 - 2qab + ra^2, \; c = q^2 - pr, \; d = bp - aq, \; e = bq - ar$ Now, let's apply a transformation to the principle quintic form : $y= \frac{d \chi + e \phi}{a \chi + b \phi}$ And let the transformed quintic be $y^5 + a_1 y^4 + a_2 y^3 + a_3 y^2 + a_4 y + a_5= 0 \;\; (14)$ All we gave to do is to determine its coefficients. Here it goes : $y + \sqrt{c}= \frac{d \chi + e \phi + \sqrt{c} (a \chi + b \phi)}{a \chi + b \phi} (15)$ Combining (12) and (13), we see that $m (a \chi + b \phi)= - (d \chi + e \phi )^2 + c (a \chi + b \phi)^2 \pmod{f(x)}$ Hence, from (15), $\frac{m}{y + \sqrt{c}}= - d \chi - e \phi + \sqrt{c} (a \chi + b \phi) \pmod{f(x)}$ Now from the properties (12), we see $\sum z= \sum z^2 = 0$ where $z= \frac{1}{y + \sqrt{c}} \;\; (16)$ If we substitute $y= \frac{1 - sqrt{c} z}{z}$ in the quintic (14), we have $(1 - \sqrt{c} z)^5 + a_1 z (1 - \sqrt{c} z)^4 + a_2 z^2 (1 - \sqrt{c} z)^3 + a_3 z^3 (1 - \sqrt{c} z)^2 + a_4 z^4 (1 - \sqrt{c} z) + a_5 z^5= 0$ and since we have another value, namely $- sqrt{c}$ for the square root of c, $(1 + \sqrt{c} z)^5 + a_1 z (1 + \sqrt{c} z)^4 + a_2 z^2 (1 + \sqrt{c} z)^3 + a_3 z^3 (1 + \sqrt{c} z)^2 + a_4 z^4 (1 + \sqrt{c} z) + a_5 z^5= 0$ From (16), we see that the coefficients of z^4 and z^3 is 0 and from subtracting and summing those coefficients, we get the following 4 equations : $5c^2 + 3 a_2 c + a_4= 0 \;\, 10c + 3a_2 = 0 \;\, 4a_1 c + 2 a_3 = 0 \;\, 6 a_1 c + a_3 = 0$ Hence, $a_1= a_3 = 0, \; a_2 = - 10 c/3$ and $a_4= 5 c^2$. The coefficient $a_5$ depends on the choice of $\chi$ and $\phi$. So, the quintic in (14) is actually $y^5 - \frac{10 c}{3} y^3 + 5 c^2 y + a_5= 0$ By means of a short Tschirnhausen transform and an elementary parameter Z, we can reduce this into $x^5 - 10 Z x^3 + 45 Z^2 x - Z^2= 0 \;\; (17)$ Thanks from agentredlum and duc
March 28th, 2013, 12:12 AM #6 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Advanced Analysis of Quintics In this post, I will describe how to transform a Brioschi quintic into a Jacobi sextic : Consider a quintic with roots of the form $r \zeta^k - \frac{1}{r \zeta^k}$ where $\zeta$ is the fifth root of unity. Multiplying out the factors, we see : $\prod $\chi - $$r \zeta^k - \frac{1}{r \zeta^k}$$$= \chi^5 + 5 \chi^3 + 5 \chi - $$r^5 - \frac{1}{r^5}$$$ [color=#FF0000]NOTE : The index is running through 0 to 4[/color]. If we replace $\chi$ by $\chi - 1$, $\prod $\chi - $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$$= \chi^5 - 5 \chi^4 + 15 \chi^3 - 25 \chi^2 + 25 \chi - 125 \sigma \;\; (1" /> Where $\sigma= \frac{1}{125} $$11 - \frac{1}{r^5} + r^5$$$ Now, if we replace $\chi$ by $- \chi$ in (1, multiply it by (1 and change the sign, we get : $$\prod \[ \chi - $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$$\]$\prod \[ \chi - $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$$ \]= \prod $\chi^2 - $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$^2$$ $= \chi^{10} + 5 \chi^8 + 25 \chi^6 + 125(1- 10 \sigma) \chi^4 + 625 (1 - 10 \sigma) \chi^2 - 125^2 \sigma^2 \;\; (19)$ Replacing $\chi$ by $5s/\rho$ in (19) and multiplying by $(\rho/5)^5 (s - \rho)$ gives $(s - \rho) \prod $s - \frac{\rho}{5} $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$^2$$ $s^6 - 10 \sigma \rho^3 s^3 - $$1 - 10 \sigma + 5 \sigma^2$$\rho^5 s + 5 \sigma^2 \rho^6 \;\; (20)$ We recognize the coefficients of s^3 and s as the Polyhedral polynomials $f$ and $H$ of that of the icosahedron, respectively. Hence, setting (20) as zero will introduce the Jacobi sextic : $s^6 - 10 f s^3 - H s + 5 f^2= 0 \;\; (21)$ The external root will be written as $s_5$. The other roots are of the form : $s_k= \frac{s_5}{5} $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$^2 \; 0 \le k \le 4$ All we have to do now is to relate the roots of this Jacobi sextic with that of the Brioschi quintic. I will describe O. Perron's results now : Define the variable $v_k$ as $v_k= (s_5 - s_k)(s_{k+2} - s_{k+3})(s_{k+4} - s_{k+1})$ Now we will calculate the product $\prod $v - $$\frac{v_k}{\sqrt{5}} - 4f$$$$. By using some tedious algebra or by a clever application of Newton's identities, one can derive $\prod $v - $$\frac{v_k}{\sqrt{5}} - 4f$$$= v^5 + 30 f^2 v^3 + 100 f^3 v^2 + 105 f^4 v + 36 f^5 + H^3$ Replacing $v$ by $\frac{w}{\sqrt{5}} - 4f$ and multiplying everything by $25 \sqrt{5}$ gives $\prod [w - v_k]= w^5 - 20 \sqrt{5} f w^4 + 950 f^2 w^3 - 4500 \sqrt{5} f^3 w^2 + 50625 f^4 w + 25 \sqrt{5} $$H^3 - 1728 f^5$$ = 0$ Now substituting $w= \sqrt{5} t^2$ and dividing by $25 \sqrt{5}$, we conclude that $t^{10} - 20f t^8 + 190 f^2 t^6 - 900 f^3 t^4 + 2025 f^4 t^2= - H^3 + 1728 f^5$ Looking at above carefully, we reckon that this is actually $$$t^5 - 10f t^3 + 45 f^2 t$$^2= - H^3 + 1728 f^5$ Applying the icosahedral identity to the RHS and taking square root of both sides gives $t^5 - 10 f t^3 + 45 f^2 t - T= 0 \;\; (22)$ Which is just the Brioschi normal form in two variables. We can obtain (17) by substituting $t= y T/f^2$ and introducing $Z= f^5/t^2$. Hence, the roots of Brioschi quintic and the Jacobi sextic are related as $t_k= \sqrt{\frac{v_k}{\sqrt{5}}}$ The next thing would be to solve the Jacobi sextic in terms of Weierstrass elliptic function which is analogues to that of the roots of cubic in terms of Trigonometric functions. The difference is that we perform period division by 3 on the periodic trigonometric function for a cubic, and we perform period division by 5 on the doubly periodic elliptic functions for a quintic. Thanks from agentredlum and duc
March 29th, 2013, 01:00 PM #7 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Advanced Analysis of Quintics In this post, I will describe the solution of Jacobi sextic in terms of Weierstrass elliptic function. But before that, we need to transform the coefficients of the Jacobi sextic (21), namely the polyhedral invariants, into the elliptic invariants : First, let us write the Brioschi normal form (22) like this $t^5 + \frac{10}{\Delta} + \frac{45}{\Delta^2} t - \frac{216 g_3}{\Delta^3}= 0 \;\; (23)$ Where $g_2, g_3$ are elliptic invariants and $\Delta= g_2^3 - 27 g_3^2$ is the elliptic discriminant. However, (23) implies $f= - \frac{1}{\Delta} \;\, H = \frac{12g_2}{\Delta^2} \;\, T = \frac{216 g_3}{\Delta^3}$ Hence, the Jacobi sextic in the elliptic form would be : $s^6 + \frac{10}{\Delta} s^3 - \frac{12 g_2}{\Delta^2} s - \frac{5}{\Delta^2}= 0 \;\; (24)$ Now we can begin our work. First let me introduce an interesting property of the Weierstrass elliptic function : $\frac{\sigma(5 u)}{\sigma^{(25)}(u)}= \frac{1}{82944} \left | \begin{matrix} \wp' &\wp''=&\wp'''=&\wp^{\text{IV}} \\ \wp''=&\wp'''=&\wp^{\text{IV}}=&\wp^{\text{V}} \\ \wp''=#39; & \wp^{\text{IV}} & \wp^{\text{V}} & \wp^{\text{VI}} \\ \wp^{\text{IV}} & \wp^{\text{V}} & \wp^{\text{VI}} & \wp^{\text{VII}} \end{matrix} \right | \;\; (25)$ ([color=#FF0000]NOTE : The sigma function is the Weierstrass sigma[/color]) Note that the determinant vanishes at the zeros of $\sigma$, explicitly, at $u_{m, n}= \frac{2m \omega + 2n \omega#39;}{5} \;\; (26)$ Where $\omega, \omega'$ are the two periods of $\wp$ and m,n are integers mod 5 not both zero. Expanding the determinant (25) and setting it to zero, we get the 12-th degree polynomial : $$$(\wp''^2 - 12 \wp (\wp'^2$$^3 - 16 (\wp'^4 \wp'' $$(\wp''^2 - 12 \wp (\wp'^2$$ - 64 (\wp'^8 = 0 \;\; (27)" /> The 24 roots of (26) relates to the 12 roots of (24) by the relation $\wp_{m, n}= \wp_{-m,-n}$ due to the mod 5 arithmetic of m,n. We will now set a variable defined as $y_{m, n}= \wp_{2m, 2n} - \wp_{m, n} = \wp$$\frac{4m \omega + 4n \omega}{5}$$ - \wp$$\frac{2m \omega + 2n \omega}{5}$$$ Using the formula $\wp(u) - \wp(2u)= \frac{\wp'(u) \wp'''(u) = \wp''(u)^2}{4 \wp#39;(u)^2}$ and some simplification on the derivatives we reckon, $y= y_{m, n} = \frac{\wp''_{m,n}^2 - 12 \wp_{m,n} \wp'_{m,n}^2}{4 \wp#39;_{m,n}} \;\; (2" /> Substituting this into (27) and dividing by $64 (\wp'^6" /> gives $y^3 - \wp'' y - (\wp'^2 = 0 \;\; (29)" /> Also, by solving for $(\wp''^2" /> in (2, $(\wp''^2 = (4y + 12 \wp)(\wp'^2 = (4y + 12 \wp)(y^3 - \wp'' y) \;\; (30)" /> By the closed form expression of the derivatives of $\wp$, we have two equations now $y^3 - (6 \wp^2 - g_2/2) y - (4 \wp^3 - g_2 \wp - g_3)= 0 \;\; (31) \\ (6 \wp^2 - g_2/2 )^2 = (4y + 12 \wp)(y^3 - 6 \wp^2 y + g_2 y /2) \;\; (32)$ Let us define an auxiliary variable z defiend as $z= 2 \wp + y$ so that the equations (30) and (31) becomes (respectively) : $z^3 - (3y^2 + g_2) z + 2 g_3= 0 \;\; (32) \\ 9z^4 - 6(5y^2 + g_2)z^2 + (5y^4 - 2 g_2 y^2 + g_2^2) = 0 \;\; (33)$ Using quadratic formula for (33), we get $3 z^2= 5 y^2 + g_2 + 2y \sqrt{5 y^2 + 3 g_2} \;\; (34)$ Now let us write (32) as $z(z^2 - 3y^2 - g_2)= 2g_2 \;\; (35)$ Squaring (35) and substituting (34) gives $$$5y^2 + g_2 + 2y \sqrt{5y^2 + 3g_2}$$$$-4y^2 - 2g_2 + 2y\sqrt{5y^2+3g_2}$$^2= 108 g_3^2$ Multiplying out the factors and dividing both sides by 4 gives the sextic $5 y^6 - 2 y^5 \sqrt{5y^2 + 3 g_2} + g_2^3= 27 g_2^3$ Using the definition of the elliptic discriminant, this is equivalent to $5 y^6 + \Delta= 2 y^5 \sqrt{5y^2 + 3 g_2}$ Squaring both sides, rearranging the equation by the powers of y gives $5y^{12} - 12 g_2 y^{10} + 10 \Delta y^6 + \Delta^2= 0$ If we let $y= \frac{1}{s^2}$ and multiply everything by $\frac{s^6}{\Delta^2}$, we get the Jacobi sextic (24) $s^6 + \frac{10}{\Delta} y^3 - \frac{12 g_2}{\Delta^2} s - \frac{5}{\Delta^2}= 0$ Hence, we reckon that the roots of (24) are actually of the form : $\sqrt{s_{m,n}}= \frac{1}{\wp$$\frac{4m\omega + 4n \omega'}{5}$$ - \wp$$\frac{2m\omega + 2n\omega#39;}{5}$$}$ After going through all the 12 roots, we see that the correct ones are : $\sqrt{s_5}= \frac{1}{\wp$$\frac{2\omega}{5}$$ - \wp$$\frac{4\omega}{5}$$}$ $\sqrt{s_k}= \frac{1}{\wp$$\frac{2\omega' + 48k\omega}{5}$$ - \wp$$\frac{4\omega#39; + 96 k \omega}{5}$$} \;\;\; 0 \le k \le 4$ Hence, we have the roots of the Jacobi sextic, undoing all the transformation will give the equivalent roots of general quintic equation, and this way, the Kiepert's algorithm ends. [1] R. B. King, Beyond the quartic equation [2] S. Bessels, One step beyond the solvable equation Thanks from agentredlum and duc
July 29th, 2019, 01:58 PM #8 Newbie Joined: Jul 2019 From: brisol,cct Posts: 1 Thanks: 0 Would be nice to see the equations.
July 29th, 2019, 02:16 PM #9
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Originally Posted by mathbalarka Quintics are a mysterious subject. Unlike the other degrees < 5, quintics cannot be solved using elementary arithmetic operations and radicals.
Yet the problem was solved by Galois, a political radical. Go figure.
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# Borrowing From Nature: Biopolymers and Biocomposites as Smart Wound Care Materials
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### Abstract
Wound repair is a complex and tightly regulated physiological process, involving the activation of various cell types throughout each subsequent step (homeostasis, inflammation, proliferation, and tissue remodeling). Any impairment within the correct sequence of the healing events could lead to chronic wounds, with potential effects on the patience quality of life, and consequent fallouts on the wound care management. Nature itself can be of inspiration for the development of fully biodegradable materials, presenting enhanced bioactive potentialities, and sustainability. Naturally-derived biopolymers are nowadays considered smart materials. They provide a versatile and tunable platform to design the appropriate extracellular matrix able to support tissue regeneration, while contrasting the onset of adverse events. In the past decades, fabrication of bioactive materials based on natural polymers, either of protein derivation or polysaccharide-based, has been extensively exploited to tackle wound-healing related problematics. However, in today's World the exclusive use of such materials is becoming an urgent challenge, to meet the demand of environmentally sustainable technologies to support our future needs, including applications in the fields of healthcare and wound management. In the following, we will briefly introduce the main physico-chemical and biological properties of some protein-based biopolymers and some naturally-derived polysaccharides. Moreover, we will present some of the recent technological processing and green fabrication approaches of novel composite materials based on these biopolymers, with particular attention on their applications in the skin tissue repair field. Lastly, we will highlight promising future perspectives for the development of a new generation of environmentally-friendly, naturally-derived, smart wound dressings.
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### Wound healing--aiming for perfect skin regeneration.
(1997)
The healing of an adult skin wound is a complex process requiring the collaborative efforts of many different tissues and cell lineages. The behavior of each of the contributing cell types during the phases of proliferation, migration, matrix synthesis, and contraction, as well as the growth factor and matrix signals present at a wound site, are now roughly understood. Details of how these signals control wound cell activities are beginning to emerge, and studies of healing in embryos have begun to show how the normal adult repair process might be readjusted to make it less like patching up and more like regeneration.
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### Factors affecting wound healing.
(2010)
Wound healing, as a normal biological process in the human body, is achieved through four precisely and highly programmed phases: hemostasis, inflammation, proliferation, and remodeling. For a wound to heal successfully, all four phases must occur in the proper sequence and time frame. Many factors can interfere with one or more phases of this process, thus causing improper or impaired wound healing. This article reviews the recent literature on the most significant factors that affect cutaneous wound healing and the potential cellular and/or molecular mechanisms involved. The factors discussed include oxygenation, infection, age and sex hormones, stress, diabetes, obesity, medications, alcoholism, smoking, and nutrition. A better understanding of the influence of these factors on repair may lead to therapeutics that improve wound healing and resolve impaired wounds.
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• Record: found
• Abstract: found
### Human skin wounds: a major and snowballing threat to public health and the economy.
(2009)
ABSTRACT In the United States, chronic wounds affect 6.5 million patients. An estimated excess of US$25 billion is spent annually on treatment of chronic wounds and the burden is rapidly growing due to increasing health care costs, an aging population and a sharp rise in the incidence of diabetes and obesity worldwide. The annual wound care products market is projected to reach$15.3 billion by 2010. Chronic wounds are rarely seen in individuals who are otherwise healthy. In fact, chronic wound patients frequently suffer from "highly branded" diseases such as diabetes and obesity. This seems to have overshadowed the significance of wounds per se as a major health problem. For example, NIH's Research Portfolio Online Reporting Tool (RePORT; http://report.nih.gov/), directed at providing access to estimates of funding for various disease conditions does list several rare diseases but does not list wounds. Forty million inpatient surgical procedures were performed in the United States in 2000, followed closely by 31.5 million outpatient surgeries. The need for post-surgical wound care is sharply on the rise. Emergency wound care in an acute setting has major significance not only in a war setting but also in homeland preparedness against natural disasters as well as against terrorism attacks. An additional burden of wound healing is the problem of skin scarring, a \$12 billion annual market. The immense economic and social impact of wounds in our society calls for allocation of a higher level of attention and resources to understand biological mechanisms underlying cutaneous wound complications.
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### Author and article information
###### Journal
Front Bioeng Biotechnol
Front Bioeng Biotechnol
Bioeng. Biotechnol.
Frontiers in Bioengineering and Biotechnology
Frontiers Media S.A.
2296-4185
02 October 2018
2018
: 6
###### Affiliations
1Smart Materials, Istituto Italiano di Tecnologia , Genoa, Italy
2In vivo Pharmacology Facility, Istituto Italiano di Tecnologia , Genoa, Italy
###### Author notes
Edited by: Gianni Ciofani, Politecnico di Torino, Italy
Reviewed by: Gozde Ozaydin Ince, Sabanci University, Turkey; Satoshi Arai, Waseda University, Japan; Filippo Rossi, Politecnico di Milano, Italy
*Correspondence: Giulia Suarato giulia.suarato@ 123456iit.it
This article was submitted to Nanobiotechnology, a section of the journal Frontiers in Bioengineering and Biotechnology
###### Article
10.3389/fbioe.2018.00137
6176001 | 2020-08-08 07:39:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21947018802165985, "perplexity": 7611.609929322086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737289.75/warc/CC-MAIN-20200808051116-20200808081116-00398.warc.gz"} |
https://math.stackexchange.com/questions/3248820/color-m-vertices-of-a-cyclic-graph-red-such-that-there-are-exactly-k-red-seg/3248859 | # Color $m$ vertices of a cyclic graph red such that there are exactly $k$ red segments.
Let $$n,m,k$$ be positive integers with $$n\ge m\ge k\ge2,\quad n\ge 2k$$ and consider the cyclic graph with $$n$$ vertices. How many different ways are there to color each vertex either blue or red such that
• A total of $$m$$ vertices is colored red
• There are $$k$$ red segments in total
If one coloring can be turned into another by rotating the graph, these two are considered different colorings (except of course when a rotation by $$0^\circ$$ suffices).
Of course counting graphs where just one of these conditions hold isn't hard.
• There are $${n\choose m}$$ different ways of coloring the graph such that exactly $$m$$ vertices are colorered red.
• There are $$2{n\choose 2k}$$ ways of coloring the graph such that there are exactly $$k$$ red segments.
• Shouldn't it be $2\binom{n+2k}{2k}$? I think this follows from the stars and bars method. – SmileyCraft Jun 2 at 19:31
• @SmileyCraft Well, for $n=4$ and $k=2$ there appear only to be $2$ options. Essentially, my thinking is that there are $2k$ verticies you label as 'switches', where, if you walk around the graph in a counter-clockwise direction while 'painting' each vertex, you switch colors at these vertices. – Mastrem Jun 2 at 19:38
• You are correct. I forgot to add the restriction that there needs to be a star to the left of each bar. Your method is more elegant anyways. – SmileyCraft Jun 2 at 20:06
Let us consider one extra restriction: vertex $$1$$ must be blue.
Then we have to distribute $$m$$ red vertices into $$k$$ non-empty distinguishable segments. By the stars and bars method we get $$k-1$$ bars and $$m$$ stars, but every bar needs a star to its left, and we need to end with a star. This gives $$\binom{m-1}{k-1}$$ options.
We then need to put the $$k$$ segments between the $$n-m-1$$ remaining blue vertices, and between any two segments there needs to be at least one blue vertex. By similar reasoning to the above we get $$\binom{n-m}{k}$$ options.
In total there are $$\binom{m-1}{k-1}\binom{n-m}{k}$$ options. By symmetry, the same holds if we pick any other vertex than $$1$$ to be blue. We can consider every vertex as possibly being blue, and then we have considered every valid colouring exactly $$n-m$$ times. This means there are $$\frac{n}{n-m}\binom{m-1}{k-1}\binom{n-m}{k}$$ valid colourings in total. | 2019-12-15 13:32:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6727811694145203, "perplexity": 235.88280384016213}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00236.warc.gz"} |
https://www.lmfdb.org/L/rational/8/63%5E8/1.1?Submit=sage&download=1&query=%7B'degree':%208,%20'conductor':%20248155780267521,%20'rational':%20True%7D | ## Results (15 matches)
Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim $\epsilon$ $r$ First zero Origin
8-63e8-1.1-c0e4-0-0 $1.40$ $15.3$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $$0.0, 0.0, 0.0, 0.0 0 1 0 0.164901 Modular form 3969.1.t.e 8-63e8-1.1-c0e4-0-1 1.40 15.3 8 3^{16} \cdot 7^{8} 1.1$$ $0.0, 0.0, 0.0, 0.0$ $0$ $1$ $0$ $0.217602$ Modular form 3969.1.j.b
8-63e8-1.1-c0e4-0-2 $1.40$ $15.3$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $$0.0, 0.0, 0.0, 0.0 0 1 0 0.322885 Modular form 3969.1.r.c 8-63e8-1.1-c0e4-0-3 1.40 15.3 8 3^{16} \cdot 7^{8} 1.1$$ $0.0, 0.0, 0.0, 0.0$ $0$ $1$ $0$ $0.508543$ Modular form 3969.1.k.e
8-63e8-1.1-c0e4-0-4 $1.40$ $15.3$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $$0.0, 0.0, 0.0, 0.0 0 1 0 0.587791 Modular form 3969.1.n.b 8-63e8-1.1-c0e4-0-5 1.40 15.3 8 3^{16} \cdot 7^{8} 1.1$$ $0.0, 0.0, 0.0, 0.0$ $0$ $1$ $0$ $0.622842$ Modular form 3969.1.m.c
8-63e8-1.1-c0e4-0-6 $1.40$ $15.3$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $$0.0, 0.0, 0.0, 0.0 0 1 0 0.654988 Modular form 3969.1.b.a 8-63e8-1.1-c1e4-0-0 5.62 1.00\times 10^{6} 8 3^{16} \cdot 7^{8} 1.1$$ $1.0, 1.0, 1.0, 1.0$ $1$ $1$ $0$ $0.440840$ Modular form 3969.2.a.t
8-63e8-1.1-c1e4-0-1 $5.62$ $1.00\times 10^{6}$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $$1.0, 1.0, 1.0, 1.0 1 1 0 0.533913 Modular form 3969.2.a.w 8-63e8-1.1-c1e4-0-2 5.62 1.00\times 10^{6} 8 3^{16} \cdot 7^{8} 1.1$$ $1.0, 1.0, 1.0, 1.0$ $1$ $1$ $4$ $1.05396$ Modular form 3969.2.a.s
8-63e8-1.1-c1e4-0-3 $5.62$ $1.00\times 10^{6}$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $$1.0, 1.0, 1.0, 1.0 1 1 4 1.05701 Modular form 3969.2.a.u 8-63e8-1.1-c1e4-0-4 5.62 1.00\times 10^{6} 8 3^{16} \cdot 7^{8} 1.1$$ $1.0, 1.0, 1.0, 1.0$ $1$ $1$ $4$ $1.06432$ Modular form 3969.2.a.y
8-63e8-1.1-c1e4-0-5 $5.62$ $1.00\times 10^{6}$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $$1.0, 1.0, 1.0, 1.0 1 1 4 1.15461 Modular form 3969.2.a.r 8-63e8-1.1-c1e4-0-6 5.62 1.00\times 10^{6} 8 3^{16} \cdot 7^{8} 1.1$$ $1.0, 1.0, 1.0, 1.0$ $1$ $1$ $4$ $1.23030$ Modular form 3969.2.a.v
8-63e8-1.1-c1e4-0-7 $5.62$ $1.00\times 10^{6}$ $8$ $3^{16} \cdot 7^{8}$ 1.1 $1.0, 1.0, 1.0, 1.0$ $1$ $1$ $4$ $1.35125$ Modular form 3969.2.a.x | 2021-11-28 21:42:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9675053358078003, "perplexity": 363.4414898215616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00131.warc.gz"} |
https://www.electropedia.org/iev/iev.nsf/17127c61f2426ed8c1257cb5003c9bec/2f17aa01cada1eb0c1257eec0049b108?OpenDocument | IEVref: 121-11-57 ID: Language: en Status: Standard Term: magnetic tension Synonym1: Synonym2: Synonym3: Symbol: Vm Definition: scalar quantity equal to the line integral of the magnetic field strength H along a specified path linking two points a and b: ${V}_{\text{m}\text{,ab}}=\underset{{r}_{\text{a}}}{\overset{{r}_{\text{b}}}{\int }}H\cdot \mathrm{d}r$ where ra and rb are the position vectors for a and b, respectively, and dr is the vector line element Publication date: 1998-08 Source: Replaces: Internal notes: 2017-06-02: Cleanup - Remove Attached Image 121-11-57.gif CO remarks: TC/SC remarks: VT remarks: Domain1: Domain2: Domain3: Domain4: Domain5: | 2021-05-17 21:17:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30487051606178284, "perplexity": 12660.852465591832}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991870.70/warc/CC-MAIN-20210517211550-20210518001550-00244.warc.gz"} |
http://jblevins.org/notes/heckman89identifiability | # Heckman and Honoré (1989)
The Identifiability of the Competing Risks Model
These notes are based on the following article:
Heckman, James J. and Bo E. Honoré (1989). The identifiability of the competing risks model. Biometrika, 76: 325–330.
## The Classical Competing Risks Model
• Suppose there are $J$ competing causes of death $\left\{1,2,\dots ,J\right\}$.
• Associated with each cause of death is a stochastic failure time ${T}_{j}$.
• We observe only the distribution of the identified minimum:
• The time of death $T={\mathrm{min}}_{j}{T}_{j}$.
• The cause of death $I=\mathrm{arg}{\mathrm{min}}_{j}{T}_{j}$.
• Goal: Identify the joint distribution of the latent failure times given that we only observe the distribution of the identified minimum.
• Note that we aren’t considering regressors yet.
## Cox and Tsiatis Nonidentification Theorem
• For any joint distribution of latent failure times, there exists another such distribution with independent failure times that yields the same distribution of the minimum (Cox, 1959, 1962; Tsiatis, 1975).
• That is, given r.v.’s $\left({T}_{1},{T}_{2},\dots ,{T}_{J}\right)$ there exist $\left({S}_{1},{S}_{2},\dots ,{S}_{J}\right)$ with ${S}_{i}⫫{S}_{j}$ for all $i\ne j$ such that $\left(T,{I}_{T}\right)$ and $\left(S,{I}_{S}\right)$ are observationally equivalent.
• In light of this result, any empirical work needed to proceed by placing some structure on the form of dependence across risks, for example, by assuming independence.
## Importance of Dependence
• We are concerned with conditional independence—independence of the risks ${T}_{1},\dots ,{T}_{J}$ conditional on $X$.
• Even conditional independence may not hold if, for example, we are studying an individual whose behavior may affect all of the risks.
• Yashin, Manton, and Stallard (1986): How do smoking, blood pressure, and body weight (regressors) affect time of death from cancer, heart disease, etc. (risks).
## Overview
• Establish an identification theorem for a general class of competing risks models with regressors.
• This class includes models with marginal distributions that follow:
• Proportional hazards.
• Mixed proportional hazards.
• Accelerated hazards.
• Results are presented for only two competing risks but generalize to any arbitrary finite number of risks.
## Proportional Hazards Model
• We want to model the time of death $T$ from a single risk conditional on some covariates $X$.
• Conditional on $X$, $T$ has cdf $F\left(t|x\right)$ and pdf $f\left(t|x\right)$.
• Hazard function: $\lambda \left(t|x\right)=\frac{f\left(t|x\right)}{1-F\left(t|x\right)}$.
• Integrated hazard: $\Lambda \left(t|x\right)={\int }_{0}^{t}\lambda \left(s|x\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{ds}$.
• If $\lambda \left(t|x\right)=z\left(t\right)\varphi \left(x\right)$ then $\Lambda \left(t|x\right)=Z\left(t\right)\varphi \left(x\right)$ with $Z\left(t\right)={\int }_{0}^{t}z\left(s\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{ds}$.
• Equivalently, we can work with the survivor function: $S\left(t|x\right)=\mathrm{Pr}\left(T>t|x\right)=\mathrm{exp}\left[-Z\left(t\right)\varphi \left(x\right)\right]$
• It is common in practice to use $\varphi \left(x\right)={e}^{x\beta }$.
• Suppose $F\left(t|x\right)=1-{e}^{-Z\left(t\right)\varphi \left(x\right)}$ where $Z\left(t\right)$ is the baseline integrated hazard and $\varphi \left(x\right)$ is a scaling term.
• If $Z$ is differentiable, then $Z\prime \left(t\right)$ is the baseline hazard.
## Proportional Hazards and Competing Risks
• Assuming for the moment that failure times are independent, we can easily generalize this to model competing risks.
• The distribution of each failure time has a proportional hazard specification.
• $Z\left(t\right)$ and $\varphi$ may differ across risks.
• The joint survivor function is $S\left({t}_{1},{t}_{2}|x\right)=1-\left(1-\mathrm{exp}\left[-{Z}_{1}\left({t}_{1}\right){\varphi }_{1}\left(x\right)\right]\right)\left(1-\mathrm{exp}\left[-{Z}_{2}\left({t}_{2}\right){\varphi }_{2}\left(x\right)\right]\right).$
## Introducing Dependence
• We could draw two independent failure times ${T}_{1}$ and ${T}_{2}$ by drawing (independently) ${U}_{j}\sim U\left(0,1\right)$ and solving for ${T}_{j}$:
${S}_{j}\left(t|x\right)=\mathrm{exp}\left\{-{Z}_{j}\left(t\right){\varphi }_{j}\left(x\right)\right\}.$
• If $K\left({u}_{1},{u}_{2}\right)={u}_{1}{u}_{2}$ is the CDF of ${U}_{1}$ and ${U}_{2}$, the joint survivor function is
$S\left({t}_{1},{t}_{2}|x\right)=K\left[\mathrm{exp}\left\{-{Z}_{1}\left(t\right){\varphi }_{1}\left(x\right)\right\},\mathrm{exp}\left\{-{Z}_{2}\left(t\right){\varphi }_{2}\left(x\right)\right\}\right].$
• We can introduce dependence in ${T}_{1}$ and ${T}_{2}$ by introducing dependence in ${U}_{1}$ and ${U}_{2}$ via $K$.
• Suppose $\left({U}_{1},{U}_{2}\right)\sim K\left(\cdot ,\cdot \right)$ on $\left[0,1{\right]}^{2}$ and assume that ${Z}_{1}\left(0\right)={Z}_{2}\left(0\right)=0$.
• Then the survivor function for $\left({T}_{1},{T}_{2}\right)$ is
(1)$S\left({t}_{1},{t}_{2}|x\right)=K\left(\mathrm{exp}\left[-{Z}_{1}\left({t}_{1}\right){\varphi }_{1}\left(x\right)\right],\mathrm{exp}\left[-{Z}_{2}\left({t}_{2}\right){\varphi }_{2}\left(x\right)\right]\right).$
## Generalization: Mixed Proportional Hazards
• Suppose that the competing risks are independent, ${\varphi }_{j}\left(x\right)={e}^{x{\beta }_{j}}$, and that one of the covariates, $\omega$, is not observed:
$S\left({t}_{1},{t}_{2}|x\right)={\int }_{\Omega }\mathrm{exp}\left[-{Z}_{1}\left({t}_{1}\right){e}^{x{\beta }_{1}+{c}_{1}\omega }\right]\mathrm{exp}\left[-{Z}_{2}\left({t}_{2}\right){e}^{x{\beta }_{2}+{c}_{2}\omega }\right]\phantom{\rule{thickmathspace}{0ex}}\mathrm{dG}\left(\omega \right).$
• We can arrive at this model by choosing $K$ such that:
$K\left({\eta }_{1},{\eta }_{2}\right)={\int }_{\Omega }{\eta }_{1}^{\mathrm{exp}\left({c}_{1}\omega \right)}{\eta }_{2}^{\mathrm{exp}\left({c}_{2}\omega \right)}\phantom{\rule{thickmathspace}{0ex}}\mathrm{dG}\left(\omega \right).$
## Generalization: Accelerated hazards
$S\left(t|x\right)=\mathrm{exp}\left[-Z\left\{t\varphi \left(x\right)\right\}\right]$
• Joint survivor with dependent competing risks:
$S\left({t}_{1},{t}_{2}|x\right)=K\left(\mathrm{exp}\left[-{Z}_{1}\left\{{t}_{1}{\varphi }_{1}\left(x\right)\right\}\right],\mathrm{exp}\left[-{Z}_{2}\left\{{t}_{2}{\varphi }_{2}\left(x\right)\right\}\right]\right).$
• For any $K$, the marginal distributions give rise to univariate accelerated hazard models.
## Identification Theorem
Assume that $\left({T}_{1},{T}_{2}\right)$ has joint distribution (1). Then ${Z}_{1}$, ${Z}_{2}$, ${\varphi }_{1}$, ${\varphi }_{2}$, and $K$ are identified from the minimum of $\left({T}_{1},{T}_{2}\right)$ under the following assumptions:
1. $K$ is continuously differentiable with partial derivatives ${K}_{1}$ and ${K}_{2}$ and for $i=1,2$, ${\mathrm{lim}}_{n\to \infty }{K}_{i}\left({\eta }_{1n},{\eta }_{2n}\right)$ is finite for all sequences ${\eta }_{1n}$, ${\eta }_{2n}$ for which ${\eta }_{1n}\to 1$ and ${\eta }_{2n}\to 1$ for $n\to \infty$. We also assume that $K$ is strictly increasing in each of its arguments.
2. ${Z}_{1}\left(1\right)={Z}_{2}\left(1\right)=1$ and ${\varphi }_{1}\left({x}_{0}\right)={\varphi }_{2}\left({x}_{0}\right)=1$ for some ${x}_{0}$.
3. The support of $\left\{{\varphi }_{1}\left(x\right),{\varphi }_{2}\left(x\right)\right\}$ is $\left(0,\infty \right)×\left(0,\infty \right)$.
4. ${Z}_{1}$ and ${Z}_{2}$ are nonnegative, differentiable, strictly increasing functions, except that we allow them to be infinite for finite $t$.
1. $K$ is already weakly increasing.
2. This is an innocuous normalization since ${\varphi }_{j}\left(x\right)$ and ${Z}_{j}\left(t\right)$ are not jointly identified to scale.
3. This is satisfied, for example, when ${\varphi }_{j}\left(x\right)=\mathrm{exp}\left(x{\beta }_{j}\right)$ and there is a common covariate with support $ℝ$ and different coefficients.
## Mapping Observables to Unobservables
Observed distributions:
${Q}_{1}\left(t|x\right)=\mathrm{Pr}\left({T}_{1}\ge t,{T}_{2}\ge {T}_{1}|x\right)\phantom{\rule{1em}{0ex}}{Q}_{2}\left(t|x\right)=\mathrm{Pr}\left({T}_{2}\ge t,{T}_{1}\ge {T}_{2}|x\right).$
Tsiatis (1975) establishes the following mappings:
$\frac{\partial {Q}_{1}}{\partial t}\left(t|x\right)={\left[\frac{\partial S}{\partial {t}_{1}}\right]}_{{t}_{1}={t}_{2}=t}\phantom{\rule{1em}{0ex}}\frac{\partial {Q}_{2}}{\partial t}\left(t|x\right)={\left[\frac{\partial S}{\partial {t}_{2}}\right]}_{{t}_{1}={t}_{2}=t}.$
We have $\frac{\partial {Q}_{1}}{\partial t}\left(t|x\right)=-{K}_{1}\left[\mathrm{exp}\left\{-{Z}_{1}\left(t\right){\varphi }_{1}\left(x\right)\right\},\mathrm{exp}\left\{-{Z}_{2}\left(t\right){\varphi }_{2}\left(x\right)\right\}\right]\mathrm{exp}\left\{-{Z}_{1}\left(t\right){\varphi }_{1}\left(x\right)\right\}Z{\prime }_{1}\left(t\right){\varphi }_{1}\left(x\right).$
## Identification of ${\varphi }_{j}$
Taking the ratio of $\frac{\partial {Q}_{1}\left(t|x\right)}{\partial t}$ at $x$ and ${x}_{0}$ yields
$\frac{{K}_{1}\left[\mathrm{exp}\left\{-{Z}_{1}\left(t\right){\varphi }_{1}\left(x\right)\right\},\mathrm{exp}\left\{-{Z}_{2}\left(t\right){\varphi }_{2}\left(x\right)\right\}\right]\mathrm{exp}\left\{-{Z}_{1}\left(t\right){\varphi }_{1}\left(x\right)\right\}Z{\prime }_{1}\left(t\right){\varphi }_{1}\left(x\right)}{{K}_{1}\left[\mathrm{exp}\left\{-{Z}_{1}\left(t\right){\varphi }_{1}\left({x}_{0}\right)\right\},\mathrm{exp}\left\{-{Z}_{2}\left(t\right){\varphi }_{2}\left({x}_{0}\right)\right\}\right]\mathrm{exp}\left\{-{Z}_{1}\left(t\right){\varphi }_{1}\left({x}_{0}\right)\right\}Z{\prime }_{1}\left(t\right){\varphi }_{1}\left({x}_{0}\right).}$
Taking $t\to 0$ and using the normalization yields ${\varphi }_{1}\left(x\right)$. Our choice of $x$ was arbitrary so ${\varphi }_{1}\left(x\right)$ is identified on the entire support of $X$. Similarly for ${\varphi }_{2}\left(x\right)$.
## Identification of $K$
We know $S\left(t,t|x\right)$ since $S\left(t,t|x\right)={Q}_{1}\left(t|x\right)+{Q}_{2}\left(t|x\right)$. Furthermore, $S\left(t,t|x\right)=K\left(\mathrm{exp}\left[-{Z}_{1}\left(t\right){\varphi }_{1}\left(x\right)\right],\mathrm{exp}\left[-{Z}_{2}\left(t\right){\varphi }_{2}\left(x\right)\right]\right).$
Setting $t=1$ gives $S\left(1,1|x\right)=K\left(\mathrm{exp}\left[-{\varphi }_{1}\left(x\right)\right],\mathrm{exp}\left[-{\varphi }_{2}\left(x\right)\right]\right).$ and letting ${\varphi }_{1}\left(x\right)$ and ${\varphi }_{2}\left(x\right)$ vary over $\left(0,\infty {\right)}^{2}$ (by Assumption 3) yields $K$.
## Identification of ${Z}_{j}$
$S\left(t,t|{x}_{n}\right)=K\left(\mathrm{exp}\left[-{Z}_{1}\left(t\right){\varphi }_{1}\left({x}_{n}\right)\right],\mathrm{exp}\left[-{Z}_{2}\left(t\right){\varphi }_{2}\left({x}_{n}\right)\right]\right)$
• Let ${\varphi }_{2}\left(x\right)\to 0$ while holding ${\varphi }_{1}\left(x\right)$ fixed.
• Then $S\left(t,t|x\right)\to K\left(\mathrm{exp}\left[-{Z}_{1}\left(t\right){\varphi }_{1}\left(x\right)\right],1\right).$
• Since $K$ and ${\varphi }_{1}$ are known and $K$ is strictly increasing in both arguments, we have ${Z}_{1}\left(1\right)=1$ for any $t$.
• Similarly for ${Z}_{2}\left(t\right)$.
## Conclusion
Identification argument:
• Given the distribution of $\left(T,I\right)$ and exploiting multiplicative separability gives us ${\varphi }_{j}\left(x\right)$ for $j=1,2$.
• Using the full range of ${\varphi }_{j}\left(x\right)$ on $\left(0,\infty \right)$ yields $K$.
• Using $K$, ${\varphi }_{j}$, and related properties gives us ${Z}_{j}\left(t\right)$.
Implications of Nonparametric Identification:
• Identification does not depend on parametric functional forms or assumed forms of risk dependence (modulo separability of the hazard).
• Highlights the role of regressors in identification in contrast to the Cox-Tsiatis nonidentification result.
• Suggests the possibility of a nonparametric estimator. | 2016-09-25 13:57:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 136, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9773763418197632, "perplexity": 738.0742405547386}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660214.50/warc/CC-MAIN-20160924173740-00113-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://homework.study.com/explanation/find-the-nth-term-of-a-sequence-whose-first-several-terms-are-given-1-5-9-13.html | # Find the nth term of a sequence whose first several terms are given. {eq}1, 5, 9, 13,... {/eq}
## Question:
Find the nth term of a sequence whose first several terms are given.
{eq}1, 5, 9, 13,... {/eq}
## Arithmetic Sequence:
An arithmetic sequence has a constant difference between successive terms.
If {eq}a{/eq} is the initial term and {eq}d{/eq} is the common difference then the nth term will be given by,
{eq}t_n=a+(n-1)d{/eq}
We can determine {eq}a{/eq} and {eq}d{/eq} for the given problem.
Then using the formula above we can find the nth term of given sequence.
Given sequence,
{eq}1, 5, 9, 13, ... {/eq}
First term of the sequence is {eq}a=1{/eq}
The difference between successive terms is constant and given by {eq}d=4{/eq}.
Hence given sequence is an arithmetic sequence with,
a=1, d=4
{eq}\begin{align} \text{nth term }&= a+(n-1)d \\ t_n &=1+(n-1)(4) \\ \color{blue}{t_n} & \color{blue}{=4n-3} \end{align} {/eq} | 2023-01-28 16:15:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9972511529922485, "perplexity": 2347.6113583011406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00526.warc.gz"} |
https://questions.examside.com/past-years/gate/gate-cse/database-management-system/relational-algebra | GATE CSE
Database Management System
Relational Algebra
Previous Years Questions
## Marks 1
Consider the following three relations in a relational database. Employee ( $$\underline {eld}$$ , Name), Brand ( $$\underline {bld}$$ , bName), Own...
What is the optimized version of the relation algebra expression $$\pi_{A1}(\pi_{A2}(\sigma_{F1}(\sigma_{F2}(r))))$$, where $$A1, A2$$ are sets of att...
Which of the following tuple relational calculus expression(s) is/are equivalent to $$\forall t \in r \left(P\left(t\right)\right)$$? I. $$\neg \exist... Consider the relations r1(P, Q, R) and r2(R, S, T) with primary keys P and R respectively. The relation r1 contains 2000 tuples and r2 contains 2500 t... Which of the following relational query languages have the same expressive power? I) Relational algebra II) Tuple relational calculus restricted to sa... Let r be a relation instance with schema R = (A, B, C, D). We define$${r_1} = {\pi _{A,B,C}}\left( r \right)$$and$${r_1} = {\pi _{A,D}}\left( r \ri...
Let R1 (A, B, C) and R2 (D, E) be two relation schema, where the primary keys are shown underlined, and let C be a foreign key in R1 referring to R2. ...
Consider the following SQL query: Select distinct a1, a2, ..., an From r1, r2, ..., rm Where P; For an arbitrary predicate P, this query is equi...
With regard to the expressive power of the formal relational query languages, which of the following statements is true?
Given the relations employee (name, salary, deptno), and department (deptno, deptname, address) Which of the following queries cannot be expressed u...
Consider the join of a relation R with a relation S. If R has m tuples and S has n tuples then the maximum and minimum sizes of the join respectively ...
The relational algebra expression equivalent to the following tuple calculus expression: $$\left\{ {t|t \in r \wedge \left( {t\left[ A \right] = 10 \w... An instance of a relational scheme R(A, B, C) has distinct values for attribute A. Can you conclude that A is a candidate key for R? Give a relational algebra expression using only the minimum number of operators from$$\left( { \cup ,\, - } \right)$$which is equivalent to$$R \cap...
## Marks 2
A relation r(A, B) in a relational database has 1200 tuples. The attribute A has integer values ranging from 6 to 20, and the attribute B has integer ...
Consider the following relations P(X,Y,Z), Q(X,Y,T) and R(Y,V). How many tuples will be returned by the following relational algebra query? ...
Consider the relations $$r(A, B)$$ and $$s(B, C),$$ where $$s.B$$ is a primary key and $$r.B$$ is a foreign key referencing $$s.B.$$ Consider the quer...
Consider two relations $${R_1}\left( {A,B} \right)$$ with the tuples $$(1,5), (3,7)$$ and $${R_2}\left( {A,C} \right) = \left( {1,7} \right),\left( {4... Consider a join (relation algebra) between relations r(R)and s(S) using the nested loop method. There are 3 buffers each of size equal to disk block s... Consider the relational schema given below, where eId of the relation dependent is a foreign key referring to empId of the relation employee. Assume t... Consider the following relations A, B and C: A .tg {border-collapse:collapse;border-spacing:0;} .tg td{font-family:Arial, sans-serif;font-size:14px;... Consider the following relations A, B and C: A .tg {border-collapse:collapse;border-spacing:0;} .tg td{font-family:Arial, sans-serif;font-size:14px;... Suppose R1 (A, B) and R2 (C, D) are two relation schemas. Let r1 and r2 be the corresponding relation instances. B is a foreign key that refers to C i... Consider a relational table r with sufficient number of records, having attributes A1, A2,....., An and let 1$$ \le $$p$$ \le $$n. Two queries Q1 ... The following functional dependencies hold for relations R(A, B, C) and S(B, D, E):$$\eqalign{ & B \to A \cr & A \to C \cr} The re...
Let R and S be two relations with the following schema R (P, Q, R1, R2, R3) S (P, Q, S1, S2) Where {P, Q} is the key for both schemas. Which of the fo...
Consider a selection of the form σA ≤ 100(r), where r is a relation with 1000 tuples. Assume that the attribute values for A among the tuples are unif...
Information about a collection of students is given by the relation studInfo(studId, name, sex). The relation enroll(studId, courseId) gives which stu...
Consider the relation employee(name, sex, supervisorName) with name as the key, supervisorName gives the name of the supervisor of the employee under ...
Consider the following relation schemas : b-Schema = (b-name, b-city, assets) a-Schema = (a-num, b-name, bal) d-Schema = (c-name, a-number) Let branch...
A table ‘student’ with schema (roll, name, hostel, marks), and another table ‘hobby’ with schema (roll, hobbyname) contains records as shown below: Ta...
Consider the relation Student (name, sex, marks), where the primary key is shown underlined, pertaining to students in a class that has at least one b...
Which of the following relational calculus expressions is not safe?
Which of the following query transformations (i.e. replacing the l.h.s. expression by the r.h.s. expression) is incorrect? R1 and R2 are relations, C1...
A library relational database system uses the following schema USERS (User #, User Name, Home Town) BOOKS (Books # Book Title, Author Name) ISSUED (Bo...
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NEET | 2023-03-28 06:17:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6480692028999329, "perplexity": 4611.4951691811375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00787.warc.gz"} |
https://ftp.aimsciences.org/article/doi/10.3934/ipi.2021054 | Article Contents
Article Contents
# A fuzzy edge detector driven telegraph total variation model for image despeckling
• * Corresponding author: Rajendra K. Ray
• Speckle noise suppression is a challenging and crucial pre-processing stage for higher-level image analysis. In this work, a new attempt has been made using telegraph total variation equation and fuzzy set theory for image despeckling. The intuitionistic fuzzy divergence function has been used to distinguish between edges and noise. To the best of the authors' knowledge, most of the studies on the multiplicative speckle noise removal process focus only on diffusion-based filters, and little attention has been paid to the study of fuzzy set theory. The proposed approach enjoys the benefits of both telegraph total variation equation and fuzzy edge detector, which is robust to noise and preserves image structural details. Moreover, we establish the existence and uniqueness of weak solutions of a regularized version of the present system using the Schauder fixed point theorem. With the proposed technique, despeckling is carried out on natural, real synthetic aperture radar, and real ultrasound images. The experimental results computed by the suggested method are reported, which are found better in terms of noise elimination and detail/edge preservation, concerning the existing approaches.
Mathematics Subject Classification: 35L70, 65M06, 68U10.
Citation:
• Figure 1. Set of sixteen $3 \times 3$ templates
Figure 2. Original images
Figure 3. Boat image ($512\times 512$). (a) Speckled image: $L = 3$. (b)-(f) Despeckled by various approaches. (g) Speckled image: $L = 10$. (h)-(l) Despeckled by various approaches
Figure 4. Brick image ($256\times 256$). (a) Speckled image: $L = 3$. (b)-(f) Despeckled by various approaches. (g) Speckled image: $L = 10$. (h)-(l) Despeckled by various approaches
Figure 5. (a) Ratio image for the original image 2b, (b)-(f) Ratio images for the despeckled images 4h-4l. (g) Indicate the one-dimensional slices. (h) Results for the Slice-1. (i) Results for the Slice-2. (j) Results for the Slice-3
Figure 6. Circle image ($299\times 299$). (a) Speckled image: $L = 3$. (b)-(f) Despeckled by various approaches. (g) Speckled image: $L = 10$. (h)-(l) Despeckled by various approaches
Figure 7. (a)-(f) 2D Contour map of the images 6g-6l. (g)-(l) 3D surface plot of the images 6g-6l
Figure 8. Results are plotted for the circle image when the image is degraded by $L = 10$. (a) Relative error vs. the iteration number. (b) Logarithmic Relative error vs. the iteration number. (c) PSNR value vs. the corresponding iteration number
Figure 9. (a) SAR Image-1: One look radar image [3]. (b)-(d) Restored by different models
Figure 10. (a) SAR Image-2: Image of KOMPSAT/Arirang-5 of a part of the Himalayan Arc [20]. (b)-(d) Restored by different models
Figure 11. A Ultrasound image of fetal foot and restored by different models
Figure 12. A Ultrasound image of liver cyst and restored by different models
Figure 13. Relative error vs. the iteration number for various models
Table 1. MSSIM, PSNR, and SI
Image $L$ AA [7] Dong[19] DDD[61] ZZDB[62] Proposed MSSIM PSNR SI MSSIM PSNR SI MSSIM PSNR SI MSSIM PSNR SI MSSIM PSNR SI Boat 1 0.5577 16.90 0.3695 0.5526 16.78 0.3368 0.5510 16.92 0.3417 0.5705 16.98 0.3289 0.5816 17.04 0.3162 3 0.6780 22.40 0.3759 0.6680 22.41 0.3712 0.6806 22.20 0.3720 0.6810 22.30 0.3569 0.6976 22.54 0.3472 5 0.7210 24.27 0.3783 0.7128 24.27 0.3755 0.7205 24.06 0.3662 0.7200 24.14 0.3637 0.7386 24.46 0.3558 10 0.7757 26.16 0.3796 0.7658 26.17 0.3782 0.7729 26.11 0.3794 0.7745 26.20 0.3709 0.7885 26.39 0.3658 Brick 1 0.2875 12.10 0.0816 0.2874 12.18 0.0805 0.2873 12.14 0.0779 0.2880 12.16 0.0728 0.2961 12.23 0.0719 3 0.3710 16.95 0.0933 0.3737 17.00 0.0901 0.3646 16.86 0.0879 0.3650 16.87 0.0865 0.3823 17.09 0.0854 5 0.4167 19.17 0.0998 0.4174 19.21 0.0978 0.4176 18.61 0.0955 0.4175 18.65 0.0923 0.4234 19.32 0.0908 10 0.4790 21.84 0.1063 0.4855 21.86 0.1051 0.4874 21.88 0.1043 0.4877 21.90 0.1005 0.4889 22.00 0.0996 Circle 1 0.9510 33.19 0.3106 0.9501 32.22 0.3165 0.9458 33.69 0.3219 0.9502 33.48 0.3098 0.9670 34.87 0.2982 3 0.9633 36.88 0.3270 0.9654 36.89 0.3245 0.9572 36.72 0.3271 0.9603 36.90 0.3215 0.9765 38.90 0.3163 5 0.9688 37.85 0.3285 0.9688 37.86 0.3271 0.9617 37.64 0.3289 0.9634 37.58 0.3249 0.9784 39.82 0.3198 10 0.9726 39.65 0.3291 0.9756 39.67 0.3279 0.9761 39.86 0.3290 0.9732 39.76 0.3253 0.9821 41.40 0.3241
Table 2. Comparison of SI and BRISQUE (BQ) values of despeckled images
Image AA [7] Dong[19] DDD[61] ZZDB[62] Proposed SI BQ SI BQ SI BQ SI BQ SI BQ SAR Image-1 0.5076 43.21 0.5034 43.99 0.5283 42.83 0.4806 42.56 0.4398 42.45 SAR Image-2 0.6985 45.26 0.6874 45.58 0.6845 43.96 0.6563 40.75 0.6270 38.38 Fetal foot 1.052 45.40 1.055 42.80 1.0642 40.17 1.0507 40.94 1.024 39.09 Liver cyst 0.8480 39.28 0.8484 41.15 0.8580 40.39 0.8252 45.95 0.8101 38.18
Figures(13)
Tables(2) | 2023-03-20 18:25:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.5395520925521851, "perplexity": 3193.050170069593}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943555.25/warc/CC-MAIN-20230320175948-20230320205948-00258.warc.gz"} |
http://semantic-domain.blogspot.fr/2012/11/in-this-post-ill-show-how-to-turn.html | ## Wednesday, November 28, 2012
### The Geometry of Interaction, as an OCaml program
In this post, I'll show how to turn the Geometry of Interaction construction --- a model of linear logic --- into a runnable Ocaml code. . The basic idea behind this paper is that we can view higher-order behavior, as a disciplined mode of use of a first-order language with recursion --- basically, if you fix a calling convention, then you can compile first-class functions directly and compositionally into a first-order program.
This is interesting because there are a lot of combinator libraries out there which can be seen as DSLs with a natural notion of "first-order program", but which lack a good notion of "higher-order program". (For example, most Haskell libraries based on arrows qualify.) As a programming model, this is very unsatisfying, since this restriction basically says that the author of the library can build interesting abstractions, but clients of the library can't. Wouldn't it be nice if there were a systematic way to turn any first-order language (with a bit of structure) into a higher-order one?
That's exactly what the G construction (also called the Int construction) does. It says that if you have a first-order language with sequential and parallel composition, plus feedback, then you've really got a higher-order language. Stated more formally, in categorical jargon, it says that there is a way of constructing a monoidal closed category from any symmetric monoidal category with a trace operator. (Monoidal products mean parallel composition, categories give you sequential composition, and trace is how category theorists pronounce recursion.)
Instead of giving the categorical derivation, I'll work very concretely, following the same example Abramsky gave in his classic paper Retracing Some Paths in Process Algebra.
First, let's get some Ocaml preliminaries out of the way. I'm just defining function composition and empty and sum types, and their operations. These should really be in the Ocaml Pervasives environment, but they aren't. The Haskell prelude is nicer in this regard.
let ($) f g x = f (g x) type ('a,'b) sum = Inl of 'a | Inr of 'b let inl x = Inl x let inr y = Inr y let case f g = function Inl x -> f x | Inr y -> g y let sum f g = case (inl$ f) (inr $g) let swap v = case inr inl v type zero = Void of zero let rec abort (Void v) = abort v Our running example will come from dataflow and visual programming languages like Simulink. These languages have a model of programming in which you have a bunch of stateful components (adders, multipliers, accumulators), which are connected together by means of wires. This is naturally a first-order formalism, in that it really only makes sense to send data over the wires --- you can't send components over the wire. One way of modeling these kinds of stateful components is by means of resumptions. A resumption from$I$(read "input") to$O$(read "output") is essentially an element of the solution to the domain equation$R = I \to (O \times R)$. You can think of a gadget of this type as something which takes an input, and then gives you an output, plus a new resumption to use on the next input. So it's a bit like a state machine, but formulated in such a way that you don't have to give an explicit state set. It's easy to turn this domain equation into a recursive ML type: module Resumption = struct type ('i,'o) r = R of ('i -> 'o * ('i,'o) r) The identity resumption just echoes its input as its output. (* val id : ('a, 'a) r *) let rec id = R(fun x -> (x, id)) To compose two resumptions$f$and$g$, we just feed an input into$f$, and take$f$'s output as the input into$g$. (* val ( >> ) : ('a, 'b) r -> ('b, 'c) r -> ('a, 'c) r *) let rec (>>) (R f) (R g) = R(fun x -> let (y, f') = f x in let (z, g') = g y in (z, f' >> g')) Note that with composition, we have a category$R$, the category of resumptions. The objects are types, a morphisms$f : I \to O$is a resumption taking$I$inputs and producing$O$outputs. This category is also "monoidal closed", with the tensor product$A \otimes C$getting defined as the sum type$A + C$. The intuition is that if we think of$A$and$C$as message types, we may want to consider how to react to$A$and$C$messages simultaneously. In particular, if we have two machines$f : A \to B$and$g : C \to D$, we can combine them by taking an$A+C$message and dispatching to either$f$or$g$respectively. (* val ( ** ) : ('a, 'b) r -> ('c, 'd) r -> (('a, 'c) sum, ('b, 'd) sum) r *) let rec ( ** ) (R f) (R g) = R(function | Inl x -> let (y, f') = f x in (inl y, f' ** (R g)) | Inr x -> let (y, g') = g x in (inr y, (R f) ** g')) We can also implement feedback --- if we have an$f : A \otimes B \to C \otimes B$, then we can construct a map$A \to C$by feeding$f$the$A$, and repeatedly sending the output back into the input until$f$coughs up a$C$rather than a$B$: (* val trace : (('a, 'b) sum, ('c, 'b) sum) r -> ('a, 'c) r *) let rec trace f = R(fun a -> let rec loop (R f) v = match f v with | (Inl c, f') -> (c, trace f') | (Inr b, f') -> loop f' (inr b) in loop f (inl a)) Now, we can also implement various associativity and commutativity properties. The names are not particularly great (they are just the standard names for the natural transformations associated with a tensor product in category theory), but all the real action is in the types. (* val sym : (('a, 'b) sum, ('b, 'a) sum) r *) let rec sym = R(fun v -> (swap v, sym)) (* val rho : (('a, zero) sum, 'a) r *) let rec rho : (('a,zero) sum, 'a) r = R(function | Inl a -> (a, rho) | Inr z -> abort z) (* val rho' : ('a, ('a, zero) sum) r *) let rec rho' : ('a, ('a, zero) sum) r = R(fun a -> (inl a, rho')) (* val lambda : ((zero, 'a) sum, 'a) r *) let rec lambda : ((zero,'a) sum, 'a) r = R(function | Inl z -> abort z | Inr a -> (a, lambda)) (* val lambda' : ('a, (zero, 'a) sum) r *) let rec lambda' : ('a, (zero, 'a) sum) r = R(fun a -> (inr a, lambda')) (* val alpha : ((('a, 'b) sum, 'c) sum, ('a, ('b, 'c) sum) sum) r *) let rec alpha : ((('a,'b) sum, 'c) sum, ('a, ('b,'c) sum) sum) r = R(function | Inl (Inl a) -> (inl a, alpha) | Inl (Inr b) -> (inr (inl b), alpha) | Inr c -> (inr (inr c), alpha)) (* val alpha' : (('a, ('b, 'c) sum) sum, (('a, 'b) sum, 'c) sum) r *) let rec alpha' : (('a, ('b,'c) sum) sum, (('a,'b) sum, 'c) sum) r = R(function | Inl a -> (inl (inl a), alpha') | Inr (Inl b) -> (inl (inr b), alpha') | Inr (Inr c) -> (inr c, alpha')) end Now, we give the G construction in terms of the resumption module we've just defined. In the G construction, we build a new category$G(R)$from the old one we've just defined. The basic idea here is to talk about bidirectional communication, and so for objects in our new category$G$will be pairs of objects from our old one. Objects of$G$will be pairs$(A^+, A^-)$, where you can think of$A^+$as a type of messages sent out, and the type$A^-$as the type of messages that are accepted. A morphisms of$G$,$f : (A^+,A^-) \to (B^+, B^-)$will be resumption of type$A^+ \otimes B^- \to A^- \otimes B^+$. Note that the polarities of$A$and$B$are flipped in this mapping -- we are transforming$A$messages out to$B$messages out, and we need to accept$B$messages in order to transform them into$A$messages in. module G = struct open Resumption Now, let's define the type of morphisms in$G(R)$. Since we don't have type-level pairing in ML, we end up having four type parameters. We use a dummy G constructor to get the type system to produce more useful inferred types. type ('aplus, 'aminus, 'bplus, 'bminus) map = G of (('aplus, 'bminus) sum, ('aminus, 'bplus) sum) r The identity map in$G(R)$is just the symmetry map in$R$. let gid = G sym Composition is trickier. If we have a map$f : (A^+,A^-) \to (B^+, B^-)$and$g : (B^+,B^-) \to (C^+,C^-)$, we can compose it by feeding$f$'s$B$-output into$g$'s input, and vice-versa, using the trace operator. It's best visualized with the following diagram: Actually doing so requires applying the associativity and commutativity properties of sum types$A + B \simeq B + A$and$(A + B) + C \simeq A + (B + C)$. let rec assoc : ((('aplus,'cminus) sum, ('bminus,'bplus) sum) sum, (('aplus, 'bminus) sum, ('bplus, 'cminus) sum) sum) r = R(function | Inl (Inl aplus) -> (inl (inl aplus), assoc) | Inl (Inr cminus) -> (inr (inr cminus), assoc) | Inr (Inl bminus) -> (inl (inr bminus), assoc) | Inr (Inr bplus) -> (inr (inl bplus), assoc)) let rec assoc2 : ((('aminus, 'bplus) sum, ('bminus, 'cplus) sum) sum, (('aminus, 'cplus) sum, ('bminus, 'bplus) sum) sum) r = R(function | Inl (Inl aminus) -> (inl (inl aminus), assoc2) | Inr (Inr cplus) -> (inl (inr cplus), assoc2) | Inl (Inr bplus) -> (inr (inr bplus), assoc2) | Inr (Inl bminus) -> (inr (inl bminus), assoc2)) (* val gcompose : ('a, 'b, 'c, 'd) map -> ('c, 'd, 'e, 'f) map -> ('a, 'b, 'e, 'f) map *) let gcompose (G f) (G g) = G (trace (assoc >> (f ** g) >> assoc2)) Once we have composition, we can define a tensor product on the$G(R)$category as well, so that$(A^+,A^-) \oplus (B^+,B^-) = (A^+ + B^+, A^- + B^-)$. Again, implementing the functorial action requires implementing the associativity maps. (* (A+, A-) * (X+, X-) = (A+ + X+, A- + A-) f : A+ + B- -> A- + B+ g : X+ + Y- -> X- + Y+ f ** g : (A+ + B-) + (X+ + Y-) -> (A- + B+) + (X- + Y+) r : (A+ + X+) + (B- + Y-) -> (A+ + B-) + (X+ + Y-) r' : (A- + B+) + (X- + Y+) -> (A- + X-) + (B+ + Y+) *) (* val ( + ) : ('a, 'b, 'c, 'd) map -> ('e, 'f, 'g, 'h) map -> (('a, 'e) sum, ('b, 'f) sum, ('c, 'g) sum, ('d, 'h) sum) map *) let (+) (G f) (G g) = let rec r = R(function | Inl (Inl aplus) -> (inl (inl aplus), r) | Inl (Inr xplus) -> (inr (inl xplus), r) | Inr (Inl bminus) -> (inl (inr bminus), r) | Inr (Inr yminus) -> (inr (inr yminus), r)) in let rec r' = R(function | Inl (Inl aminus) -> (inl (inl aminus), r') | Inl (Inr bplus) -> (inr (inl bplus), r') | Inr (Inl xminus) -> (inl (inr xminus), r') | Inr (Inr yplus) -> (inr (inr yplus), r')) in G (r >> f ** g >> r') One cool thing about this category (which leads to the coolest fact about it) is that it naturally supports a dualization, which sends$(A^+,A^-)^\ast \triangleq (A^-,A^+)$. This also has an action on morphisms, which flips around the direction, so that from$f : (A^+,A^-) \to (B^+,B^-)$we can get$f^\ast : (B^+,B^-)^\ast \to (A^+,A^-)^\ast$. (* val dualize : ('a, 'b, 'c, 'd) map -> ('d, 'c, 'b, 'a) map *) let dualize (G (R f)) = let rec dual f = R(fun v -> let (v', f') = f (swap v) in (swap v', dual f)) in G (dual f) Now, here's what I consider the neatest fact about the Int construction -- it turns a traced monoidal category into a monoidal closed category. So we can define the linear exponential$(A^+,A^-) \multimap (B^+,B^-) \triangleq (A^+,A^-)^\ast \otimes (B^+,B^-)$. (* Here's a bit of unrolling type definitions to make things easier to grok. (A+, A-) -o (B+, B-) = (A+, A-)* * (B+, B-) = (A-, A+) * (B+, B-) = (A- + B+, A+ + B-) Hom((A+,A-) * (B+,B-), (C+, C-)) = Hom((A+ + B+, A- + B-), (C+, C-)) = ((A+, B+) sum, (A- + B-) sum, C+, C-) map = (((A+, B+) sum, C+) sum, ((A- + B-) sum, C-) sum) r Hom((A+,A-), (B+,B-) -o (C+,C-)) = Hom((A+,A-), (B- + C+, B+ + C-)) = (A+, A-, (B-, C+) sum, (B+, C-) sum) map = ((A+, (B+, C-) sum) sum, (A-, (B-, C+) sum) sum) r *) (* val curry : (('a, 'b) sum, ('c, 'd) sum, 'e, 'f) map -> ('a, 'c, ('d, 'e) sum, ('b, 'f) sum) map *) let curry (G f) = let rec curry (R f) = R(fun v -> let (v', f') = f (case (inl$ inl) (case (inl $inr) inr) v) in let v'' = case (case inl (inr$ inl)) (inr $inr) v' in (v'', curry f')) in G(curry f) (* val uncurry : ('a, 'b, ('c, 'd) sum, ('e, 'f) sum) map -> (('a, 'e) sum, ('b, 'c) sum, 'd, 'f) map *) let uncurry (G f) = let rec uncurry (R f) = R(fun v -> let (v', f') = f (case (case inl (inr$ inl)) (inr $inr) v) in let v'' = case (inl$ inl) (case (inl \$ inr) inr) v' in
(v'', uncurry f'))
in
G (uncurry f)
end
The ML types for curry and uncurry somewhat obscure the fact that they implement the isomorphism $$\mathrm{Hom}((A^+,A^-) \otimes (B^+,B^-), (C^+,C^-)) \simeq \mathrm{Hom}((A^+,A^-), (B^+,B^-) \multimap (C^+,C^-))$$
However, now that we have this isomorphism, it's possible to compile arbitrary lambda-terms in multiplicative linear logic down into compositions of resumptions. But that's a story for another post!
In the meantime, if this intrigues you, you can play with the Verity compiler by Dan Ghica and his students, as well as the IntML compiler by Ulrich Schöpp and Ugo Dal Lago, both of which are compilers using the GoI interpretation. Verity is aimed at distributed computing, and IntML is a programming language whose type system allows you to write only (and all) the LOGSPACE programs. As you can see, there are a lot of potential applications...
1. Minor typo: "The intuition is that if we think of A and C as message types, we may want to consider how to react to A and B messages simultaneously." probably wants to be "how to react to A and C messages simultaneously."
1. Additionally, the type signature in "Composition is trickier. If we have a map f:(A+,A−)→(B+,B−) and g:(B+,B−)," appears to be truncated.
2. Thanks, Edward! I've fixed those typos.
3. An interesting way of looking at the GoI/Int construction (pointed out to me by Jeff Egger) is that it is the categorical analogue of the construction of an abelian group from a commutative monoid. If your original symmetry monoidal category was just the discrete category of the natural numbers with addition, then the GoI construction gives you the abelian group of integers (after quotienting out the isomorphism classes).
This connection was elaborated by Katis, Sabadini and Walters in their paper Feedback, trace and fixed-point semantics.
4. Hi Bob, I learned this same fact from Ulli Schoepp, who also pointed out that this makes "Int construction" a particularly well-chosen name. :)
One thing I don't understand about the Int construction, though, is what happens if the original category was already closed. For example, in the case of the natural numbers with addition, we have saturating subtraction as monoidal closed structure, and the closed structure on the integers is regular subtraction of integers. But is there an interesting relationship between these two?
5. I wonder is there a type level int construction that would allow encoding higher order generics in e.g c#? | 2014-09-24 00:53:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9541059732437134, "perplexity": 1689.131098102429}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657140890.97/warc/CC-MAIN-20140914011220-00116-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://crypto.stackexchange.com/tags/key-exchange/new | Tag Info
3
Yes, if you are using 3rd party key exchange, the 3rd party can read the messages. If that is not the security feature you want, use something else. There are many legitimate scenarios where users are fine with trusting the third party, however. For example, a system setup by my employer to allow encrypted chat between myself and our clients. My employer has ...
3
I have never heard of this reason, and I don't quite understand it. In general, the security of Diffie-Hellman key exchange is reduced to the DDH assumption. According to this assumption, the result of the key exchange is a group element that is computationally indistinguishable from a random/uniformly distributed element in the group. However, what is ...
0
To show that the protocol is secure under DDH, we need a reduction $R$ that takes a triple as input and outputs a transcript and key such that if the triple is a DDH triple, then the transcript and key are distributed identically to a real execution of the protocol if the triple is random, then the transcript and key are distributed as if you ran a real ...
Top 50 recent answers are included | 2015-09-05 03:58:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3818449079990387, "perplexity": 607.7523447288662}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645378542.93/warc/CC-MAIN-20150827031618-00143-ip-10-171-96-226.ec2.internal.warc.gz"} |
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power current | 2014-03-11 06:28:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2840583920478821, "perplexity": 2045.2443839417756}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011139063/warc/CC-MAIN-20140305091859-00028-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://www.ck12.org/arithmetic/Numbers-in-Expanded-Form/lesson/user:c3dpbGxpYW1zQHJ1c2hjaXR5LmsxMi5tbi51cw../Whole-and-Decimal-Numbers-in-Expanded-Form/r8/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Numbers in Expanded Form
## Expand given numbers by value of place up to the ten-thousandths.
Estimated2 minsto complete
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Practice Numbers in Expanded Form
Progress
Estimated2 minsto complete
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Whole and Decimal Numbers in Expanded Form
[Figure1]
### Introduction
Remember Kelsey and her decimal from the last Concept? She had the decimal .67 written in her notebook. In that Concept, you learned how to write identify the decimal digits according to place value.
Well, how could you write this decimal out the long way if you don't use words?
This is called expanded form, and it is the focus of this Concept. At the end of the Concept, you will know how to write any decimal in expanded form.
### Guided Learning
In the last Concept, you learned how to express decimals in words using a place value chart and in pictures using grids with tens and hundreds in them. We can also stretch out a decimal to really see how much value each digit of the decimal is worth.
This is called expanded form.
What is expanded form?
Expanded form is when a number is stretched out. Let’s look at a whole number first and then use this information with decimals.
265
If we read this number we can read it as two hundred and sixty-five. We can break this apart to say that we have two hundreds, six tens and five ones. HUH??? What does that mean? Let’s look at our place value chart to help us make sense of it.
Hundred Tens Ones Tenths Hundredths Thousandths
Ten
Thousandths
2 6 5 .
If you look at the chart you can see how we got those values for each digit. The two is in the hundreds place. The six is in the tens place and the five is in the ones place. Here it is in expanded form.
2 hundreds + 6 tens + 5 ones
This uses words, how can we write this as a number?
200 + 60 + 5
Think about this, two hundred is easy to understand. Six tens is sixty because six times 10 is sixty. Five ones are just that, five ones.
This is our number in expanded form.
How can we write decimals in expanded form?
We can work on decimals in expanded form in the same way. First, we look at a decimal and put it into a place value chart to learn the value of each digit.
.483\begin{align*}.483\end{align*}
Hundred Tens Ones Tenths Hundredths Thousandths
Ten
Thousandths
. 4 8 3
Now we can see the value of each digit.
4 = four tenths
8 = eight hundredths
3 = 3 thousandths
We have the values in words, now we need to write them as numbers.
Four tenths = .4
Eight hundredths = .08
Three thousandths = .003
What are the zeros doing in there when they aren’t in the original number?
The zeros are needed to help us mark each place. We are writing a number the long way, so we need the zeros to make sure that the digit has the correct value. If we didn’t put the zeros in there, then .8 would be 8 tenths rather than 8 hundredths. Now, we can write this out in expanded form.
.483\begin{align*}.483\end{align*}
.4 + .08 + .003 = .483
This is our answer in expanded form.
Now let's practice. Write each number in expanded form.
#### Example A
567\begin{align*}567\end{align*}
Solution:
#### Example B
.345\begin{align*}.345\end{align*}
Solution:
#### Example C
.99\begin{align*}.99\end{align*}
Solution:
Now let's apply this to the decimal that was in Kelsey's homework. Here is the original problem once again.
Well, in the last Concept, Kelsey had the decimal .67 written in her notebook. In that Concept, you learned how to write identify the decimal digits according to place value.
Well, how could you write this decimal out the long way if you don't use words?
Now let's write out .67 in expanded form. We have the tenths place and the hundredths place represented.
.6+.07=.67\begin{align*}.6 + .07 = .67\end{align*}
Here is one for you to try on your own.
Write the following decimal in expanded notation.
.4562\begin{align*}.4562\end{align*}
We have four places represented in this decimal. We have tenths, hundredths, thousandths and ten - thousandths represented in the decimal. We have to represent each of these places in the expanded form too.
### Video Review
Here is a video for review.
### Practice Set
Directions: Write each decimal out in expanded form.
1. 54
2. 173
3. 611
4. 5405
5. 62,310
6. 7.8
7. 34.5
8. .9
9. .23
10. .986
11. .3003
12. 2.821
13. 41.001
14. .8739
15. 10.9327
### Review
• Expanded form is when a number is stretched out.
• 2 hundreds + 6 tens + 5 ones = Word Form
• 200 + 60 + 5 = Expanded Form
• 265 = Standard Form
• We can work on decimals in expanded form in the same way.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes | 2016-10-23 08:53:46 | {"extraction_info": {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7331165075302124, "perplexity": 1751.2794135549198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719192.24/warc/CC-MAIN-20161020183839-00063-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-1-section-1-5-addition-of-real-numbers-exercise-set-page-63/70 | ## Introductory Algebra for College Students (7th Edition)
Published by Pearson
# Chapter 1 - Section 1.5 - Addition of Real Numbers - Exercise Set - Page 63: 70
#### Answer
$\displaystyle \frac{-15}{x}+\frac{4}{x}.$, simplifies to $\displaystyle \frac{-11}{x}$
#### Work Step by Step
the quotient of $-15$ and a number ... $\rightarrow$... $\displaystyle \frac{-15}{x}$ ...which is increased by ... $\rightarrow \displaystyle \frac{-15}{x}+...$ .... the quotient of $4$ and a number$\quad\rightarrow \displaystyle \frac{-15}{x}+\frac{4}{x}.$ This simplifies, because the demominator is common, to, $...=\displaystyle \frac{ -15+4}{x}$ ... then, adding different signed numbers, we keep the sign of $-15$, because it has the greater absolute value, and subtract their absolute values: $=\displaystyle \frac{-(15-4)}{x}$ $=\displaystyle \frac{-11}{x}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-11-20 18:23:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6896666884422302, "perplexity": 1431.13245243272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746528.84/warc/CC-MAIN-20181120171153-20181120193153-00497.warc.gz"} |
https://brilliant.org/practice/quant-finance-interview-expected-value/ | # Interview
Welcome to your expected value interview! In a quantitative finance interview, one of the most important things is to explain your thoughts. A good thought process is worth more than a correct answer with no ability to articulate your reasoning. This quiz is no different, as we’ll be breaking down a challenging problem into its logical steps.
## Interview
### Expected Value
# Interview
Here’s the problem we’ll be tackling:
An ant is at a vertex of a tetrahedron. Every second, it randomly chooses one of the other 3 vertices, and crawls to that vertex. What is the expected number of seconds until it has visited every vertex?
## Interview
### Expected Value
# Interview
First of all, to get a sense of the possible outcomes, what is the probability that the ant visits all 4 vertices in 3 seconds (that is, the minimum possible)?
## Interview
### Expected Value
# Interview
Let’s define some variables to help solve the original problem. Which do you think would be the most helpful?
• A: $$X_i,$$ the expected number of distinct vertices visited after $$i$$ seconds
• B: $$X_i,$$ the expected number of seconds until $$i$$ distinct vertices have been visited
## Interview
### Expected Value
# Interview
If $$X_i$$ is the expected number of seconds until $$i$$ distinct vertices have been visited. What is $$X_1?$$
Remember that the ant starts at a vertex (at time 0).
## Interview
### Expected Value
# Interview
If $$X_i$$ is the expected number of seconds until $$i$$ distinct vertices have been visited, what is $$X_2?$$
## Interview
### Expected Value
# Interview
Once 2 vertices have been visited, what is the probability of visiting a new vertex on the next step?
## Interview
### Expected Value
# Interview
What is the expected number of steps between visiting the second distinct vertex and visiting a third distinct vertex?
## Interview
### Expected Value
# Interview
Using the same logic, you can determine the expected number of seconds between visiting the third vertex and final vertex.
Using this and all prior knowledge, what is $$X_4,$$ the amount of total expected time in seconds from when the ant starts moving at 0 seconds?
## Interview
### Expected Value
# Interview
To wrap up, what we did in this interview was to determine that the time between reaching the $$k^\text{th}$$ and $$(k+1)^\text{th}$$ vertex had an expectation of $$\frac{3}{4-k}.$$ Thus, by linearity of expectation (the expected value to hit all 4 distinct vertices is the sum of the expected values to get from the 1st to the 2nd, 2nd to the 3rd, and 3rd to the 4th distinct vertex), the expected value was: $\frac{3}{3} + \frac{3}{2} + \frac{3}{1} = 5.5.$
## Interview
### Expected Value
× | 2018-04-23 21:29:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7881121635437012, "perplexity": 787.7135265993805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946199.72/warc/CC-MAIN-20180423203935-20180423223935-00424.warc.gz"} |
https://www.tcs.tifr.res.in/events/tifr-cafeteria-and-related-queues-game-arrivals | # TIFR Cafeteria and Related Queues - A Game of Arrivals
## Speaker:
Sandeep K. Juneja School of Technology and Computer Science Tata Institute of Fundamental Research Homi Bhabha Roa
## Time:
Wednesday, 11 November 2009 (All day)
## Venue:
• AG-69
In this talk, we introduce the cafeteria or the banquet hall queuing problem' strongly motivated by speaker's experience of waiting in the TIFR cafeteria queue: Fixed but a large number of users arrive into a queue which provides service starting at a fixed time, say, 12:15 pm. Users may (and some do) arrive before this time and queue up. Their cost is a function of their waiting time in the queue and time at which service is received. We analyze this system in an asymptotic regime and develop fluid limit for the resultant queuing system. The limiting system may be modeled as a non-atomic game for which we determine the Nash-Wardrop equilibrium arrival strategy under a variety of assumptions on the cost structure. Furthermore, we note that the price of anarchy' of this system equals 2 under linearity and homogeneity assumptions. We discuss some potential ways to limit this
anarchy. | 2023-02-02 17:33:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4030577838420868, "perplexity": 1627.5931796143707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00825.warc.gz"} |