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https://jax.readthedocs.io/en/stable/_autosummary/jax.scipy.linalg.lu_factor.html
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# jax.scipy.linalg.lu_factor¶
jax.scipy.linalg.lu_factor(a, overwrite_a=False, check_finite=True)[source]¶
Compute pivoted LU decomposition of a matrix.
LAX-backend implementation of lu_factor(). Original docstring below.
The decomposition is:
A = P L U
where P is a permutation matrix, L lower triangular with unit diagonal elements, and U upper triangular.
Parameters
• a ((M, M) array_like) – Matrix to decompose
• overwrite_a (bool, optional) – Whether to overwrite data in A (may increase performance)
• check_finite (bool, optional) – Whether to check that the input matrix contains only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs.
Returns
• lu ((N, N) ndarray) – Matrix containing U in its upper triangle, and L in its lower triangle. The unit diagonal elements of L are not stored.
• piv ((N,) ndarray) – Pivot indices representing the permutation matrix P: row i of matrix was interchanged with row piv[i].
lu_solve()
solve an equation system using the LU factorization of a matrix
Notes
This is a wrapper to the *GETRF routines from LAPACK.
Examples
>>> from scipy.linalg import lu_factor
>>> A = np.array([[2, 5, 8, 7], [5, 2, 2, 8], [7, 5, 6, 6], [5, 4, 4, 8]])
>>> lu, piv = lu_factor(A)
>>> piv
array([2, 2, 3, 3], dtype=int32)
Convert LAPACK’s piv array to NumPy index and test the permutation
>>> piv_py = [2, 0, 3, 1]
>>> L, U = np.tril(lu, k=-1) + np.eye(4), np.triu(lu)
>>> np.allclose(A[piv_py] - L @ U, np.zeros((4, 4)))
True
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https://docs.geotools.org/latest/javadocs/org/opengis/metadata/quality/package-summary.html
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Data quality and positional accuracy. The following is adapted from OpenGIS® Spatial Referencing by Coordinates (Topic 2) specification.
The parameters that define a coordinate reference system are chosen rather than measured to satisfy the degrees-of-freedom problem in the changeover from observation to coordinate quantities. Coordinate reference systems are therefore by definition error-free (i.e., non-stochastic). A coordinate reference system is realised through a network of control points. The coordinates of those control points, derived from surface and/or from satellite observations, are stochastic. Their accuracy can be expressed in a covariance matrix, which, due to the degrees-of-freedom problem, will have a rank deficiency, described in geodetic literature.
Coordinate transformations between coordinate reference systems usually have parameter values derived from two sets of point coordinates, one set in system 1, the other set in system 2. As these coordinates are stochastic (i.e., have random-error characteristics) the derived transformation parameter values will also be stochastic. Their covariance matrix can be calculated.
Coordinates that have not been "naturally" determined in coordinate reference system 2, but have been determined in coordinate system 1 and then transformed to system 2, have the random error effects of the transformation superimposed on their original error characteristics. It may be possible in well-controlled cases to calculate the covariance matrices of the point coordinates before and after the transformation, and thus isolate the effect of the transformation, but in practice a user will only be interested in the accuracy of the final transformed coordinates.
Nevertheless the option is offered to specify the covariance matrix of point coordinates resulting exclusively from the transformation. It is outside the scope of this specification to describe how that covariance matrix should be used. Because a covariance matrix is symmetrical, only the upper or lower diagonal part (including the main diagonal) needs to be specified.
For some transformations, this accuracy information is compacted in some assessment of an average impact on horizontal position and vertical position, allowing specification of average absolute accuracy and, when relevant and available, average relative accuracy. Hence separate quality measures may be specified for horizontal and for vertical position in those objects.
Since:
GeoAPI 2.0
• Interface Summary
Interface Description
AbsoluteExternalPositionalAccuracy
Closeness of reported coordinate values to values accepted as or being true.
AccuracyOfATimeMeasurement
Correctness of the temporal references of an item (reporting of error in time measurement).
Completeness
Presence and absence of features, their attributes and their relationships.
CompletenessCommission
Excess data present in the dataset, as described by the scope.
CompletenessOmission
Data absent from the dataset, as described by the scope.
ConceptualConsistency
Adherence to rules of the conceptual schema.
ConformanceResult
Information about the outcome of evaluating the obtained value (or set of values) against a specified acceptable conformance quality level.
DataQuality
Quality information for the data specified by a data quality scope.
DomainConsistency
Adherence of values to the value domains.
Element
Type of test applied to the data specified by a data quality scope.
FormatConsistency
Degree to which data is stored in accordance with the physical structure of the dataset, as described by the scope.
GriddedDataPositionalAccuracy
Closeness of gridded data position values to values accepted as or being true.
LogicalConsistency
Degree of adherence to logical rules of data structure, attribution and relationships (data structure can be conceptual, logical or physical).
NonQuantitativeAttributeAccuracy
Accuracy of non-quantitative attributes.
PositionalAccuracy
Accuracy of the position of features.
QuantitativeAttributeAccuracy
Accuracy of quantitative attributes.
QuantitativeResult
Information about the value (or set of values) obtained from applying a data quality measure.
RelativeInternalPositionalAccuracy
Closeness of the relative positions of features in the scope to their respective relative positions accepted as or being true.
Result
Base interface of more specific result classes.
Scope
Description of the data specified by the scope.
TemporalAccuracy
Accuracy of the temporal attributes and temporal relationships of features.
TemporalConsistency
Correctness of ordered events or sequences, if reported.
TemporalValidity
Validity of data specified by the scope with respect to time.
ThematicAccuracy
Accuracy of quantitative attributes and the correctness of non-quantitative attributes and of the classifications of features and their relationships.
ThematicClassificationCorrectness
Comparison of the classes assigned to features or their attributes to a universe of discourse.
TopologicalConsistency
Correctness of the explicitly encoded topological characteristics of the dataset as described by the scope.
• Class Summary
Class Description
EvaluationMethodType
Type of method for evaluating an identified data quality measure.
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# R/example.R In varycoef: Modeling Spatially Varying Coefficients
#### Documented in sample_SVCdata
```#' Sample Function for GP-based SVC Model for Given Locations
#'
#' @description Samples SVC data at given locations. The SVCs parameters and the
#' covariance function have to be provided. The sampled model matrix can be
#' provided or it is sampled. The SVCs are sampled according to their given parametrization and at
#' respective observation locations. The error vector is sampled from a nugget
#' effect. Finally, the response vector is computed. Please note that the
#' function is not optimized for sampling large data sets.
#'
#' @param df.pars (\code{data.frame(p, 3)}) \cr
#' Contains the mean and covariance parameters of SVCs. The three columns
#' must have the names \code{"mean"}, \code{"var"}, and \code{"scale"}.
#' @param nugget.sd (\code{numeric(1)}) \cr
#' Standard deviation of the nugget / error term.
#' @param cov.name (\code{character}(1)) \cr
#' Character defining the covariance function, c.f. \code{\link{SVC_mle_control}}.
#' @param locs (\code{numeric(n)} or \code{matrix(n, d)}) \cr
#' The numeric vector or matrix contains the observation locations and
#' therefore defines the number of observations to be \code{n}. For a vector,
#' we assume locations on the real line, i.e., \eqn{d=1}.
#' @param X (\code{NULL} or \code{matrix(n, p)}) \cr
#' If \code{NULL}, the covariates are sampled, where the first column contains
#' only ones to model an intercept and further columns are sampled from a
#' standard normal. If it is provided as a \code{matrix}, then the dimensions
#' must match the number of locations in \code{locs} (\code{n}) and the number of SVCs
#' defined by the number of rows in \code{df.pars} (\code{p}).
#'
#' @return \code{list} \cr
#' Returns a list with the response \code{y}, model matrix
#' \code{X}, a matrix \code{beta} containing the sampled SVC at given
#' locations, a vector \code{eps} containing the error, and a matrix
#' \code{locs} containing the original locations. The \code{true_pars}
#' contains the data frame of covariance parameters that were used to
#' sample the GP-based SVCs. The nugget variance has been added to the
#' original argument of the function with its respective variance, but
#' \code{NA} for \code{"mean"} and \code{"scale"}.
#'
#' @details The parameters of the model can be chosen such that we obtain data
#' from a not full model, i.e., not all covariates are associated with a
#' fixed and a random effect. Using \code{var = 0} for instance yields a
#' constant beta coefficient for respective covariate. Note that in that
#' case the \code{scale} value is neglected.
#'
#' @examples
#' set.seed(123)
#' # SVC parameters
#' (df.pars <- data.frame(
#' var = c(2, 1),
#' scale = c(3, 1),
#' mean = c(1, 2)))
#' # nugget standard deviation
#' tau <- 0.5
#'
#' # sample locations
#' s <- sort(runif(500, min = 0, max = 10))
#' SVCdata <- sample_SVCdata(
#' df.pars = df.pars, nugget.sd = tau, locs = s, cov.name = "mat32"
#' )
#' @importFrom spam rmvnorm cov.exp cov.mat cov.sph cov.wend1 cov.wend2
#' @importFrom stats rnorm
#' @export
sample_SVCdata <- function(
df.pars, nugget.sd, locs,
cov.name = c("exp", "sph", "mat32", "mat52", "wend1", "wend2"),
X = NULL
) {
# transform to matrix for further computations
if (is.vector(locs)) {
locs <- matrix(locs, ncol = 1)
}
# check covariance parameters and locations
stopifnot(
is.data.frame(df.pars),
all(df.pars\$var >= 0),
all(df.pars\$scale > 0),
nugget.sd > 0,
is.matrix(locs)
)
# dimensions
d <- dim(locs)[2]
n <- dim(locs)[1]
p <- nrow(df.pars)
## build SVC models depending on covariance function, i.e., Sigma_y
D <- as.matrix(dist(locs, diag = TRUE, upper = TRUE))
## covariance functions
cov_fun <- function(theta) {
do.call(
what = MLE.cov.func(cov.name),
args = list(h = D, theta = theta)
)
}
## sample SVCs (including mean effect)
beta <- apply(df.pars, 1, function(x) {
if (x["var"] == 0) {
rep(x["mean"], n)
} else {
spam::rmvnorm(
n = 1,
mu = rep(x["mean"], n),
Sigma = cov_fun(theta = x[c("scale", "var")])
)
}
})
# nugget
eps <- rnorm(n, sd = nugget.sd)
# data
if (is.null(X)) {
X <- cbind(1, matrix(rnorm(n*(p-1)), ncol = p-1))
} else {
stopifnot(
is.matrix(X),
dim(X)[1] == n,
dim(X)[2] == p
)
}
y <- apply(beta*X, 1, sum) + eps
list(
y = y, X = X, beta = beta, eps = eps, locs = locs,
true_pars = rbind(
df.pars,
data.frame(
var = nugget.sd^2,
scale = NA, mean = NA
)
)
)
}
```
## Try the varycoef package in your browser
Any scripts or data that you put into this service are public.
varycoef documentation built on Sept. 18, 2022, 1:07 a.m.
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# Axiom of Choice
(redirected from Choice axiom)
Also found in: Dictionary.
## axiom of choice
[¦ak·sē·əm əv ′chȯis]
(mathematics)
The axiom that for any family A of sets there is a function that assigns to each set S of the family A a member of S.
## Axiom of Choice
(mathematics)
(AC, or "Choice") An axiom of set theory:
If X is a set of sets, and S is the union of all the elements of X, then there exists a function f:X -> S such that for all non-empty x in X, f(x) is an element of x.
In other words, we can always choose an element from each set in a set of sets, simultaneously.
Function f is a "choice function" for X - for each x in X, it chooses an element of x.
Most people's reaction to AC is: "But of course that's true! From each set, just take the element that's biggest, stupidest, closest to the North Pole, or whatever". Indeed, for any finite set of sets, we can simply consider each set in turn and pick an arbitrary element in some such way. We can also construct a choice function for most simple infinite sets of sets if they are generated in some regular way. However, there are some infinite sets for which the construction or specification of such a choice function would never end because we would have to consider an infinite number of separate cases.
For example, if we express the real number line R as the union of many "copies" of the rational numbers, Q, namely Q, Q+a, Q+b, and infinitely (in fact uncountably) many more, where a, b, etc. are irrational numbers no two of which differ by a rational, and
Q+a == q+a : q in Q
we cannot pick an element of each of these "copies" without AC.
An example of the use of AC is the theorem which states that the countable union of countable sets is countable. I.e. if X is countable and every element of X is countable (including the possibility that they're finite), then the sumset of X is countable. AC is required for this to be true in general.
Even if one accepts the axiom, it doesn't tell you how to construct a choice function, only that one exists. Most mathematicians are quite happy to use AC if they need it, but those who are careful will, at least, draw attention to the fact that they have used it. There is something a little odd about Choice, and it has some alarming consequences, so results which actually "need" it are somehow a bit suspicious, e.g. the Banach-Tarski paradox. On the other side, consider Russell's Attic.
AC is not a theorem of Zermelo Fr?nkel set theory (ZF). G?del and Paul Cohen proved that AC is independent of ZF, i.e. if ZF is consistent, then so are ZFC (ZF with AC) and ZF(~C) (ZF with the negation of AC). This means that we cannot use ZF to prove or disprove AC.
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## LEETCODE 1564. Put Boxes Into the Warehouse I 解题思路分析
Given two arrays of positive integers `boxes` and `warehouse` representing the heights of some boxes of unit width, and the heights of `n` rooms in a warehouse, respectively. The warehouse’s rooms are labeled from `0` to `n - 1` from left to right where `warehouse[i]` (0-indexed) is the height of the `ith` room.
Boxes are put into the warehouse by the following rules:
• Boxes can’t be piled up.
• You can rearrange the order of the boxes.
• Boxes can only be pushed into the warehouse from left to right only.
• If the height of some room in the warehouse is less than the height of a box, then the box will be stopped before that room, so are the boxes behind it.
Return the maximum number of boxes you can put into the warehouse.
## LEETCODE 1561. Maximum Number of Coins You Can Get 解题思路分析
• 每一轮中，你将会选出 任意 3 堆硬币（不一定连续）。
• Alice 将会取走硬币数量最多的那一堆。
• 你将会取走硬币数量第二多的那一堆。
• Bob 将会取走最后一堆。
• 重复这个过程，直到没有更多硬币。
## LEETCODE 1553. Minimum Number of Days to Eat N Oranges 解题思路分析
• 吃掉一个橘子。
• 如果剩余橘子数 `n` 能被 2 整除，那么你可以吃掉 n/2 个橘子。
• 如果剩余橘子数 `n` 能被 3 整除，那么你可以吃掉 2*(n/3) 个橘子。
## LEETCODE 1545. Find Kth Bit in Nth Binary String 解题思路分析
• S1 = “0”
• 当 i > 1 时，Si = Si-1 + “1” + reverse(invert(Si-1))
• `S1 = "0"`
• `S2 = "011"`
• `S3 = "0111001"`
• `S4 = "011100110110001"`
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# Numbering Measures
February 11, 2020, 9:43 PM · Is there anywhere a nice set of guidelines for numbering measures in orchestra parts? Questions include but are not limited to:
(1) Do you start numbering again at 1 for each movement of a concerto or other multi-movement work?
And other questions that I'm guessing will arise organically in the ensuing discussion thread ...
## Replies (22)
Edited: February 11, 2020, 10:12 PM · As I understand it, it's generally printed measures excluding any partial measure at the beginning. (This means a second ending gets the measure number after the first ending.) Always start over at 1 for every movement, unless something different is printed in the part.
At least this would apply if everyone has repeats and first and second endings in the same places. It gets confusing in editions where different parts are printed differently.
February 11, 2020, 11:45 PM · Andrew is correct. Beware of partial measures at the ends of lines.
February 12, 2020, 2:41 AM · I agree with the above. Just makes more natural sense that way
February 12, 2020, 3:50 AM · First and second endings in different places in different parts may save ink and, occasionally, paper, but it wreaks havoc with bar numbering. You'd have to solve it by putting two sets of numbers above the affected bars, or even numbering something like "58a" and "58b" in the parts where the 1st and 2nd endings are the same and not printed separately.
February 12, 2020, 5:00 AM · If this is for music in an ensemble, you should ask your conductor how he/she wants the endings numbered. There is no universally accepted system for doing this, although I tell the members of my ensembles to number straight through so the first measure of a 2nd ending gets the number immediately after the number of the last measure of the 1st ending, as others have said.
Also in my ensembles if I have asked everybody to number their measures I go through the movements/pieces to tell them my numbers and be certain that everybody has the same numbers in the same places.
The only thing worse than movements where some people have 1st/2nd endings while others simply have a repeat sign is when people in an ensemble don't all number their measures the same way.
For such pieces I will often also add rehearsal letters in key places so that it's even easier to get everybody to start in the same place.
Also to be considered is whether you should write a small number under/over each individual measure (some school music directors ask for that) or whether you can get by simply by putting the correct measure number at the start of each line, in the margin.
Depending on the music, it can get quite crowded and harder to read if a measure number is over or under every single measure.
February 12, 2020, 7:34 AM · Thanks so far ... I'm kind of taking charge here because our director is not super communicative.
During a recent rehearsal we discovered that his score is a different edition that has a DC al fine, whereas the parts all have the recap written out. He said "let's play from B to the end" and we're all wondering why he's going back so far...
Obviously it's imperative to see what is going on in the score before numbering the parts.
And yes -- partial measures, etc., I know about that stuff, but I appreciate the advice anyway. Keep it coming.
February 12, 2020, 8:31 AM · I agree with Paul about numbering parts from a previously numbered score. I have played chamber music in which one of the parts is written out where others have repeats - BEWARE of those.
February 12, 2020, 10:31 AM · I think Paul is waiting for the conflicting advice to start coming in as usually happens. Mine would be to use your common sense and, if in doubt, your fingers
Edited: February 12, 2020, 12:41 PM · Haha Steve I might be an evil troll but hopefully not bad enough to hope that my own thread will turn into a dumpster fire.
I think most parts I have seen just have the first measure of every line numbered, unless that measure is a half-measure. Then the new measure number appears at the start of the first full bar. But even this number can be written in the margin and then it's just understood that it applies to the measure after the first barline.
Also with starting the numbering at "1" for each movement, my guess is that there are exceptions such as "Minuet and Trio" which is really a single movement. Is that right? (This is important because one of the pieces my community orchestra needs to number is Mozart 15, which, in case anyone cares, is the worst drivel Mozart ever wrote).
February 12, 2020, 1:10 PM · Minuet and Trio can be numbered either way successfully -- Calling out "Measure 17 in the Trio" works fine, just as does numbering from the start of the minuet continuously through the trio.
February 12, 2020, 2:03 PM · I think trios are usually numbered separately but it doesn't make any difference as long as everyone does it the same way.
Edited: February 12, 2020, 5:21 PM · I prefer rehearsal numbers to rehearsal letters because there are conductors who do not speak clearly enough when enunciating rehearsal letters B, C, D, E, G, and other letters ending in the "ee" sound. This causes confusion to those of us of the "Silent Generation", and even its successor Generation X, whose ears can easily misinterpret the subtle high-frequency start of the consonant in the letter name, and hear "B" instead of "D", for example.
There are some conductors who are aware of this possibility of confusion and will say "C for Charlie", "D for David" etc, but they seem to be in the minority. I once played under a conductor who had a sense of fun coupled with a knowledge of English Literature. When he wanted to refer to rehearsal letter "O" he would say "Oh for a muse of fire" (that's from Shakespeare's Henry V, btw).
A trap to be aware of, which I fell into a few weeks ago, was when the conductor was using a score fitted out with rehearsal numbers and we in the orchestra were using a German edition with rehearsal letters. The conductor would call out "rehearsal number 12", so we count through to the 12th rehearsal letter on our fingers - and get it wrong. In an English edition that would be rehearsal letter "L". However, the German edition didn't use rehearsal letter "I" but jumped from "H" to "J". Consequently, not noticing the missing "I", we would finger count up to 12, ending up at rehearsal letter "M", the conductor's rehearsal number "13".
February 12, 2020, 5:22 PM · As has been said, the thing that matters most is matching to the conductor's score. Whatever choices have been made for numbering in the score should be made for the parts.
When we don't have numbered parts, we generally send out a list of the rehearsal letters in the piece, and what measure numbers they correspond to in the conductor's score, so that people numbering their parts can make sure they are doing the right thing.
February 12, 2020, 6:30 PM · Lydia, yes, that's what we did in the case of our contretemps. We replaced the German rehearsal letters in our parts with numbers corresponding to those in the conductor's score.
February 13, 2020, 2:48 AM · Somewhat like Trevor's experience, it always annoys me when a conductor calls out "14 after C"; half the orchestra then starts in the fourteenth bar and the rest at the fourteenth barline, not including the barline of C. From my strategic position underneath his left nostril I'd ask "Do you mean "the fourteenth bar of C?"". This of course would annoy the conductor who would then find something trivial in our playing to whinge about.
But have you ever noticed that conductors count bars much more quickly than you (or most of the orchestra) do? This is because they've discovered that it's quicker to count the music than the barlines.
February 13, 2020, 8:14 AM · The correct way to call this--or so we were instructed by one of our conductors--is like this: the conductor would call out: "From C (or "before C" as the case may be): one, two, three...(counting loud)., fourteen". The musicians were expected to count right along as the conductor counted the measures (silently, with a finger pointing at the counted measures) so that at the end everybody would know simultaneously where to start. There were of course always a few who only began counting measures when the conductor stopped speaking, causing the conductor to whinge about wasting time.
Measure numbers facilitate this sort of proceedings and I prefer them vastly over rehearsal marks.
February 13, 2020, 11:26 AM · "From C count one two three ... fourteen" ... that works just fine unless that "fourteen" lands you somewhere in the midst of an 8-bar rest and there are meter changes (Respighi!). If the counting is deliberate enough then it's not that hard to count out a portion of the multi-measure rest, but it's also easy to be wrong.
February 13, 2020, 11:54 AM · Surely you don't mean Respighi planted meter changes in the middle of a rest block without marking them? The only composer I've encountered who did that was Emile Jaques-Dalcroze. He also invented "Eurhythmics" but we don't play that Euro-stuff any more on this side of the Channel
Edited: February 13, 2020, 2:47 PM · Steve -- no, all the meter changes are properly marked. Respighi was likely smoking something but not enough to become uncivilized.
PS I just had an email in which a student, in the process of asking for a professional reference, started a sentence with the word "So." I'm not making this up.
February 13, 2020, 3:18 PM ·
PS I just had an email in which a student, in the process of asking for a professional reference, started a sentence with the word "So." I'm not making this up.
I feel your pain, Paul. Unfortunately, the practice is becoming widespread. Listen to any radio interview; it's amazing (and depressing) how many people start off a response with either "So" or "Yeah", spoken in a way that suggests they were busy playing with their smartphones and have to drag their attention back to the interview now that the interviewer has stopped talking. It sounds very disrespectful.
But we digress...
Re: "Oh for a muse of fire" - sounds like the conductor is edging toward Cockney rhyming slang. Our conductor uses the "A for apple" approach to rehearsal letters - being a pilot and a nitpicker, it irritates me that he doesn't use the ICAO phonetic alphabet. But I found it amusing that when we were rehearsing Dance of the Hours he would call out "E for elephant", which put us at exactly the point where the elephants enter in the rendition in Fantasia.
As for calling out something like "8 bars before C", I figure that since our conductor doesn't bring a printed score but works entirely from memory, keeping up with him on my printed score is the least I can do.
Oh all right... while we're drifting off topic, why do they call it a "trio"? There are usually a lot more than three instruments playing...
Edited: February 13, 2020, 4:57 PM · Traditionally, there were only three instruments playing. (Most typically two oboes and a bassoon.) Haydn moved away from that.
Edited: February 13, 2020, 5:09 PM · Our conductor uses modern composers. "S for Stockhausen."
I've had one conductor with a penchant for using non-phonetic words... "G for gnocchi", "J for jalapeƱo", and "P for psychology" come to mind.
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• PPS is the current price per share ($/share) • #S is the number of outstanding shares To calculate the market price, multiply the current price per share by the number of outstanding shares. ## How to Calculate Market Price? The following example problems outline how to calculate Market Price. Example Problem #1: 1. First, determine the current price per share ($/share).
1. The current price per share ($/share) is given as: 10.00. 2. Next, determine the number of outstanding shares. 1. The number of outstanding shares is provided as: 1,000,000. 3. Finally, calculate the Market Price using the equation above: MP = PPS * #S The values given above are inserted into the equation below: MP = 10.00 * 1,000,000 = 10,000,000.00 ($)
Example Problem #2:
For this problem, the variables needed are provided below:
current price per share (\$/share) = 50
number of outstanding shares = 150105
This example problem is a test of your knowledge on the subject. Use the calculator above to check your answer.
MP = PPS * #S = ?
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# The perfect tip
### 6 thoughts on “The perfect tip”
1. Yoav – it’s still 3.14159…
Since none of the digits need to be greater than 26.86 nothing changes. If, however, Pi were, say, 31.4159, in base 26.86 it would be 14.###something
1. banachtarskibanachtarski
First of all, you can’t have a non-integer base. You can have negative integer bases, you can even have Gaussian integer bases (look it up if you’re interested), but you can’t have a non-integer real base like 26.86. As such, let’s assume that we’re dealing with base 27 (rounding the value to the nearest integer). The representation of pi actually does change. Think about it: (3.14159265…)10 means 3*100+1*10-1+4*10-2… . 100=270, so the units place would remain at 3, but for all exponents less than 0, the base representation should change. To be exact, pi base 27 is 3.3M5Q3M…, using the convention where A=10, B=11, C=12, and so on.
1. banachtarskibanachtarski
Sorry, that should have been (pi) subscript 10, as in base 10, and then 3*10^0+1*10^-1+4*10^-2… and 10^0=27^0. Something must have been wrong with my HTML.
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# Jordan form of a matrix
Let $$A = \left( {\matrix{ 0 & 1 & 0 & 0 \cr 0 & 0 & 2 & 0 \cr 0 & 0 & 0 & 3 \cr 0 & 0 & 0 & 0 \cr } } \right)$$
The characteristic polynomial is $f_A(x)=x^4$.
Questions:
1. How do I conclude that $m_A=x^4$? Do I have to evaulate $A^3$ and figure that $A^3\ne 0$?
2. The jordanian form is: $A = \left( {\matrix{ 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 \cr } } \right)$. Why?
• The characteristic polynomial is $f_A(x)=x^4$. – user35603 Dec 20 '14 at 13:01
• @user35603, I've edited the question. – AlonAlon Dec 20 '14 at 13:17
• have you computed $ker(A) = \{x: Ax = 0 \}?$ – abel Dec 20 '14 at 13:20
• @abel, can you please explain the use of $\ker(A)$? – AlonAlon Dec 20 '14 at 13:20
• the number of jordan blocks corresponding to the eigenvalue $0$ is the dimension of $ker(A).$ – abel Dec 20 '14 at 13:22
i will use $e_1, e_2, e_3$ and $e_4$ to stand for the standard basis. with this convention, we have $$Ae_1 = 0, Ae_2 = e_1, Ae_3= 2e_2 \mbox{ and }Ae_4 = 3e_3.$$ so with respect to the basis $\{e_1, e_2, \frac{1}{2}e_3, \frac{1}{6}e_4 \} = \{f_1,f_2,f_3, f_4\}$ in the new $f$ basis, the linear transformation is $$Tf_1 = 0, Tf_2 = f_1, Tf_3 = f_2, Tf_4 = f_3$$ so $T$ has the representation $\pmatrix{0&1&0&0\cr0&0&1&0\cr0&0&0&1\cr0&0&0&0},$ called the jordan form, similar to the matrix in question.
if we permute the basis,e.g,$\{f_4,f_3,f_2,f_1\}$ the same transformation is now represented by $$\pmatrix{0&0&0&0\cr1&0&0&0\cr0&1&0&0\cr0&0&1&0}$$
• We've learned in class that the $1$-s are below the main diagonal. Can you explain why at this example the $1$-s are above? – AlonAlon Dec 20 '14 at 13:50
(2) As you observe, the characteristic polynomial of $A$ is $x^4$, so it has sole eigenvalue zero (of multiplicity $4$). Now, if we denote by $J_k$ the $k \times k$ Jordan block with eigenvalue $0$, the possible Jordan forms of such a matrix are $$J_4, \quad J_3 \oplus J_1, \quad J_2 \oplus J_2, \quad J_2 \oplus J_1 \oplus J_1, \quad 0.$$
Now, $$\ker A = \langle e_4 \rangle,$$ and in particular $\dim \ker A = 1$, but the only Jordan-form matrix above with $1$-dimensional kernel is $J_4$, and the dimension of the kernel of two similar matrices agrees, so $J_4$ must be the Jordan form of $A$. In fact, as abel observes, the number of Jordan blocks in the Jordan form a matrix with all zero eigenvalues (a nilpotent matrix) is exactly the dimension of its kernel.
(1) Jordan blocks $J_k$ of eigenvalue zero have the property that $J_k^r \neq 0$ for $r < k$ and $J_k^r = 0$ for $r \geq k$. So, $J_4^3 \neq 0$ and hence $A^3 \neq 0$, that is, the minimal polynomial of $A$ must be $x^3$. More generally, the minimal polynomial of a matrix with all zero eigenvalues is $x^r$, where $r$ is the size of the largest Jordan block in its Jordan form.
• You're welcome, I hope you found it useful. – Travis Dec 20 '14 at 19:58
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# Exponential Moving Average. EMA Forex, Forex EMA
Exponential Moving Average (EMA) is applied in Forex and is one of the sorts of simple moving averageExponential Moving Average represents one of the cases of weighted moving average, because not only price values, but weight values as well as are used in its measurement. The difference between them is that price for entire period of observation are accounted, whereas weight is exponentially decreased and is never equal to 0, thereby assigning more weight to new prices. Upon that, exponential smoothing formula is used rather than linear arithmetical or another progression. This exponential smoothing is applied in forecasting number series.
Forex EMA is calculated as per the formula:
,
where:
EMAt – EMA in the point corresponding to a certain moment of time;
EMAt-1 – EMA in the point preceding preset moment of time;
pt source value of price on chart corresponding to a certain moment of time;
α – smoothing constant – designates speed of reducing weight, by default it takes 0<α<1 .
In order words, to calculate Exponential Moving Average in Forex, a previous price value is taken multiplied into smoothing constant and added to previous price value. This way, the more α value is, the less impact on current EMA value is asserted by the previous figure. As a rule, the constant equal to 2/3 is taken.
As with other kinds of MA, various prices can be taken into account: Close, Open, High, Low, Median Price, Typical Price, Weighted Close.
When it comes to EMA of arbitrary Forex order, as a rule, two special cases are distinguished: double exponential moving average and triple exponential moving average. The order sets smoothing extent of price: the higher order is, the stronger smoothing gets. Upon that, EMA value is smoothed rather that initial price pt.
In this case, the formula looks as below:
## Application of EMA in Forex
Exponential Moving Average is applied in the same way as any type of moving average: for defining entrance and exit signals. However, EMA represents the line, which is rougher to initial chart and allows to get more precise signals quicker that, for example, SMA – simple moving average (watch picture 1) It has a special importance, at the moments of publication of important economic news, large interventions and other similar cases, when prices promptly change.
It is better to apply EMA in short-term trading, because it responses to Forex changes with a maximum speed and is not as demonstrative on high time-frames. You can find out that on pic 1 (daily chart) EMA line is more smoothed than on pic. 2 (5-min time-frame)
Picture 1 Simple and Exponential Moving Average with period 21
Picture 2. Exponential Moving Average with period 21, 5-min time-frame
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# How does the dressed Klein-Gordon propagator look in position space?
The free Klein-Gordon propagator in momentum space $\sim (p^2-m^2+i\epsilon)^{-1}$ has just a single pole at $p^2=m^2$. The passage to Fourier space is difficult but possible. The result is very illuminating in terms of how disturbances of a free scalar field propagate.
In an interacting theory, with a dressed Klein-Gordon propagator in momentum space,
$$G\sim\frac{1}{p^2-m^2-\Sigma(p^2)+i\epsilon}$$
is there a picture of what disturbances look like in position space?
just use the Kallen-Lehmann spectral representation: once Fourier Transformed in real space you have an integral representation of the (time-ordered) 2-point functions $$\langle T\Phi(x_1)\Phi(x_2)\rangle=\int_0^\infty d\mu^2 \rho(\mu^2)\Delta(x_1-x_2;\mu^2)$$ where $\Delta(x;\mu^2)$ is the free propagator for a scalar of mass (squared) $\mu^2$, and $\rho(\mu^2)$ is positive distribution that sum up to $1=\int_0^\infty d\mu^2 \rho(\mu^2)$ and has a delta function at $\mu^2=m^2$ if $\Phi$ is associated to some particle state of mass $m$.
$\Delta_{F}(t,\vec{x})=\frac{im}{4\pi^{2}\sqrt{|\vec{x}|^{2}-t^{2}}}K_{1}\Big(m\sqrt{|\vec{x}|^{2}-t^{2}} \Big)$
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# Force and Energy Question
#### Jacob Huskisson
1. Homework Statement
"You are investigating an accident where a 1500 kg car rolls down a hill with a slope of 8 degrees, a distance of 150 m (measured along the road) and hit a parked van. How fast was the car moving when it hit the van, assuming little friction?"
2. Homework Equations
If I knew all of the equations, I would have most likely figured out the problem.
Force= mass x acceleration
K-work=velocity x impulse
K-work= Force x distance
3. The Attempt at a Solution
Now what I did so far was calculated the force of gravity pushing down on the car which is mass x gravity acceleration constant which I ended up getting 14,700 N. I then figured out using the 8 degrees and the force I just calculated, the force of the hypotenuse of the right angle I created. When I figured that all out I got some huge acceleration. Can someone please help me with this? I feel like it's so simple.
Related Introductory Physics Homework Help News on Phys.org
#### Doc Al
Mentor
Now what I did so far was calculated the force of gravity pushing down on the car which is mass x gravity acceleration constant which I ended up getting 14,700 N.
Good.
I then figured out using the 8 degrees and the force I just calculated, the force of the hypotenuse of the right angle I created. When I figured that all out I got some huge acceleration.
Show exactly what you did.
If the car fell off a cliff, the most its acceleration could be is equal to g. When it rolls down an incline, the acceleration will be less. So if you got a huge acceleration, you must have made a mistake somewhere.
"Force and Energy Question"
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# Shifting frequency peaks such that w[i] moves to w[i/n] using STFT and IFFT
I am trying to do a kind of frequency compression on say 20 most energetic frequencies in a particular slice of a STFT. If w[i] is the frequency peak for a slice of STFT at a particular time, then I want to move w[i] to w[i/n] where n is arbitrary. To make sure invertability is still maintained I'm using a 2 sided STFT and just switch w[i] with w[i/n] and making sure that conjugate symmetry is still maintained. Here the n factor im shifting down by is 4
import numpy as np
from scipy import signal
import matplotlib.pyplot as plt
rng = np.random.default_rng()
fs = 10e3
N = 10e4
amp = 2 * np.sqrt(2)
noise_power = 0.01 * fs / 2
time = np.arange(N) / float(fs)
mod = 500*np.cos(2*np.pi*0.25*time)
carrier = amp * np.sin(2*np.pi*3e3*time + mod)
noise = rng.normal(scale=np.sqrt(noise_power),size=time.shape)
noise *= np.exp(-time/5)
x = carrier + noise
f, t, Zxx = signal.stft(x, fs, nperseg=1000,return_onesided=False)
Spectra= Zxx
Zxx= abs(Zxx)
plt.plot(f,Zxx[:,100])
plt.show()
freqdic = {}
Spectra=Spectra[:,100] #I'm considering a slice at the 100th time element of STFT
list = sorted(range(len(Zxx[:,100])), key=lambda i: Zxx[:,100][i])[-20:] #20 most energetic peaks
for i in list :
print (Zxx[:,100][i])
peaklist =list[::2]
for i in peaklist:
lowered = int(i/4) #shiftdown factor
if lowered not in freqdic:
freqdic[lowered] = 1
Spectra[i], Spectra[lowered] = Spectra[lowered], Spectra[i]
Spectra[len(Spectra)-i], Spectra[len(Spectra)-lowered] = Spectra[len(Spectra)-lowered], Spectra[len(Spectra)-i]
plt.plot(f,abs(Spectra))
plt.show()
That's the code, I'm using an example from SciPy's doc on STFT. This is what the STFT looks like
This is what the 100th time element slice looks like
This is what the same slice looks like after running it through my code
Something has clearly moved, but the I am not sure why w[i] is still strong and hasn't been swapped, even turning the thing to 0 seems to do nothing, I am not sure whats going on. I'm getting more unexpected and random results by varying the number of spectral peaks considered
Further, what I want to do after this is, hopefully after getting rid of the spectral peak where it used to be, is to take IFFT of the STFT slices and stitch them together and thus in effect move the frequency peaks down by an arbitrary n factor.
• Moving peaks around in the spectrum requires resampling the spectrum, which is also sorta expensive. My recommendation is to use the STFT and phase vocoder to time-stretch or time-compress the original recording, And then use resampling techniques to scrunch or stretch this back to the original length. When you connect the individual frequency components, be sure you continued to adjust the phases for each component in successive frames so that the splice of that sinusoid to the corresponding sinusoid of the previous frame is seamless. Commented Feb 24 at 19:20
• @robertbristow-johnson, what do you mean by resampling the spectrum? I don't really need to even move peaks around, I don't even mind finding a few peaks and writing it to a new and blank spectrum and then taking the IFFT, do you feel that'll be less expensive? Coming to phase vocoder, my understanding was it does something quite similar so I don't see how it'll be less expensive tbh. Wrt Phase transients, I'm not too concerned Commented Feb 25 at 1:51
• If you want to compress everything towards DC by a factor of $\frac{1}{n}$, you can resample the original signal to a new sample rate of $f_{s}n$. However, this will leave you with a $\approx nN$ length signal, so I believe he’s giving you a way to keep the data length roughly the same. If you aren’t concerned about this, a polyphase resampler will do the trick. If you need different $n$ for different frequencies, mix with $e^{\pm j\omega_{0}t}$ where $\omega_{0} = \omega_{p}-\frac{\omega_{p}}{n}$ and then have bandpass filters centered at $\pm \omega_{0}$ Commented Feb 25 at 7:57
• @Baddioes, I understand, I'm probably going to end up using a phase vocoder, it works pretty well and is really fast. but now what I really want to know is whether I can do it this way because as I see it, able to mess around with signals in STFT leads to a wide range of applications. I ended up solving this by writing complex peaks on blank frequency domain representation, I ended up with another problem however : dsp.stackexchange.com/questions/93094/… Commented Feb 25 at 9:22
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https://documen.tv/the-ratio-of-boys-to-girls-at-a-soccer-camp-is-3-5-if-there-are-51-boys-at-the-camp-then-how-man-28944954-31/
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Question
The ratio of boys to girls at a soccer camp is 3:5. If there are 51 boys at the camp, then how many of the campers are girls? HINT: Solve using a proportion.
:} thank uuu
85 girls
Step-by-step explanation:
A proportion of 3:5 means that there are 5 girls at the camp for every 3 boys.
You can find the number of girls by taking the amount of boys and multiplying by the proportion.
51 * 5:3
51 * 5/3
85 girls
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# Search by Topic
#### Resources tagged with Odd and even numbers similar to Sorting Numbers:
Filter by: Content type:
Stage:
Challenge level:
### There are 61 results
Broad Topics > Numbers and the Number System > Odd and even numbers
### Sorting Numbers
##### Stage: 1 Challenge Level:
Use the interactivity to sort these numbers into sets. Can you give each set a name?
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Odd Tic Tac
##### Stage: 1 Challenge Level:
An odd version of tic tac toe
### More Numbers in the Ring
##### Stage: 1 Challenge Level:
If there are 3 squares in the ring, can you place three different numbers in them so that their differences are odd? Try with different numbers of squares around the ring. What do you notice?
### Share Bears
##### Stage: 1 Challenge Level:
Yasmin and Zach have some bears to share. Which numbers of bears can they share so that there are none left over?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Venn Diagrams
##### Stage: 1 and 2 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Light the Lights
##### Stage: 1 Challenge Level:
Investigate which numbers make these lights come on. What is the smallest number you can find that lights up all the lights?
### I Like ...
##### Stage: 1 Challenge Level:
Mr Gilderdale is playing a game with his class. What rule might he have chosen? How would you test your idea?
### Odd Squares
##### Stage: 2 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Take One Example
##### Stage: 1 and 2
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Arrangements
##### Stage: 2 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Ring a Ring of Numbers
##### Stage: 1 Challenge Level:
Choose four of the numbers from 1 to 9 to put in the squares so that the differences between joined squares are odd.
### Various Venns
##### Stage: 2 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### How Odd
##### Stage: 1 Challenge Level:
This problem challenges you to find out how many odd numbers there are between pairs of numbers. Can you find a pair of numbers that has four odds between them?
### Carroll Diagrams
##### Stage: 1 Challenge Level:
Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers?
### Lots of Biscuits!
##### Stage: 1 Challenge Level:
Help share out the biscuits the children have made.
### Cube Bricks and Daisy Chains
##### Stage: 1 Challenge Level:
Daisy and Akram were making number patterns. Daisy was using beads that looked like flowers and Akram was using cube bricks. First they were counting in twos.
### Diagonal Trace
##### Stage: 2 Challenge Level:
You can trace over all of the diagonals of a pentagon without lifting your pencil and without going over any more than once. Can the same thing be done with a hexagon or with a heptagon?
### Pairs of Legs
##### Stage: 1 Challenge Level:
How many legs do each of these creatures have? How many pairs is that?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Odd Times Even
##### Stage: 1 Challenge Level:
This problem looks at how one example of your choice can show something about the general structure of multiplication.
### Domino Sorting
##### Stage: 1 Challenge Level:
Try grouping the dominoes in the ways described. Are there any left over each time? Can you explain why?
### Largest Even
##### Stage: 1 Challenge Level:
How would you create the largest possible two-digit even number from the digit I've given you and one of your choice?
### Two Numbers Under the Microscope
##### Stage: 1 Challenge Level:
This investigates one particular property of number by looking closely at an example of adding two odd numbers together.
### Pairs of Numbers
##### Stage: 1 Challenge Level:
If you have ten counters numbered 1 to 10, how many can you put into pairs that add to 10? Which ones do you have to leave out? Why?
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### What Number?
##### Stage: 1 Short Challenge Level:
I am less than 25. My ones digit is twice my tens digit. My digits add up to an even number.
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### The Set of Numbers
##### Stage: 1 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### The Thousands Game
##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### Numbers as Shapes
##### Stage: 1 Challenge Level:
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
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https://calculationcalculator.com/5-10-as-a-decimal
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# 5/10 as a Decimal
## What is 5/10 as a Decimal?
How to write 5/10 as a Decimal? Convert a fraction value to a decimal format. A fraction belongs to numerator divided by denominator. So enter the numerator and denominator value in given input box, then press calculate button, the system will automatically calculate the decimal value.
To conversion Fraction to Decimal mainly there are two methods available, Those are-
1. Calculator Method: If you have a calculator, then there is a simple tricks to instantly transform any fraction to a decimal. Just divide the numerator value by the denominator value. Like this -
numerator ÷ denominator
2. Long Division Method: If you haven't a calculator, Then you have to use the long division or old school method to convert fraction value to decimal value. Like this -
denominator numerator
For calculation, here's how to convert 5/10 as a Decimal using the formula above, step by step instructions are given below
Calculator Method:
5 (numerator)/10 (denominator)
= 5 ÷ 10; Which is = 0.5;
Long Division Method:
Fraction 5/10 is equal to 0.5 as a Decimal.
Fraction 5/10 is 50% as a Percentage.
Fraction Decimal Percentage
5/2 2.5 250%
5/3 1.66667 166.667%
5/4 1.25 125%
5/6 0.833333 83.3333%
5/7 0.714286 71.4286%
5/8 0.625 62.5%
5/9 0.555556 55.5556%
5/10 0.5 50%
5/11 0.454545 45.4545%
5/12 0.416667 41.6667%
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https://www.businessmanagementideas.com/mutual-funds/return/determining-the-variability-of-return-on-mutual-funds-financial-management/16597
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The following points highlight the top eight methods used for determining the variability of return on mutual funds. The methods are: 1. Standard Deviation 2. Sharpe Ratio 3. Treynor Ratio 4. Sortino Ratio 5. Differential Return 6. Jensen’s Alpha 7. Appraisal Ratio 8. M2 Measure.
#### Method # 1. Standard Deviation:
Standard deviation looks at funds volatility in terms of rise and fall in its returns. Maximum volatility in a security is the riskiest, considering the unevenness it brings about in its performance. Standard deviation of a fund measures this risk by measuring the degree to which the fund fluctuates in relation to its mean return. That is the average return of a fund over a period of time.
The fund with higher standard deviation is more riskier because it fluctuates widely between negative and positive returns with in a short period. To determine how well a fund is maximizing its returns received for its volatility, you can compare the fund to another with a similar investment strategy and similar returns.
The fund with the lower standard deviation would be more optimal because it is more maximizing the return received for the amount of risk acquired. Standard deviation is a measure of dispersion in return. It quantifies the degree to which returns fluctuate around their average. A higher value of standard deviation means higher risk.
When used to measure the volatility of the performance of a security or a portfolio of securities, standard deviation is generally calculated for monthly returns over a specific time period- usually 36 months.
Annualized Monthly Deviation = Monthly Standard Deviation x Square Root of 12
Standard deviation allows portfolios with similar objectives to be used to compare over a particular time frame. It can also be used to gauge much more risk a fund in one category has, as compared to the other.
Problem 1:
PQR Fund has made a return of 13%, 15% and 8% respectively for 1st, 2nd and 3rd year. Calculate standard deviation of the return of the fund.
Solution:
Problem 2:
Mr. Kumar is a fund manager of an equity fund which is expected to provide risk premium of 10% and standard deviation of returns of 16%. Miss Akrita, a client of Mr. Kumar choose to invest Rs.70,000 in equity fund and Rs.30,000 in T-Bills. If T-Bills are trading at 7% p.a., what will be the expected return and standard deviation of return on the portfolio of Miss Akrita?
Solution:
Expected return on Equity fund =7+10 =17%
Expected return on portfolio of Miss Akrita = (0.70 x 17) + (0.30 x 7) = 14%
Expected standard deviation of the return on portfolio = (0.70 x 0.16) + (0.30 x 0) = 11.2%
#### Method # 2. Sharpe Ratio:
One approach is to calculate portfolio’s return in excess of the risk-free return and divide the excess return by the portfolio’s standard deviation. This risk adjusted return is called ‘Sharpe ratio’. This ratio named after William Sharpe, thus measures reward to variability. This ratio describes how much return is receiving for the extra volatility that an investor endures for holding a riskier asset. An investor always needs to be properly compensated for the additional risk he undertakes for not holding a risk-free asset.
Where, S = Sharpe ratio of the fund
Rp = Annualized average rate of return of the fund
Rf = Annualized risk-free rate of return
σp = Standard deviation of average return of the fund
This equation calls for three terms:
(a) Annualized return of the fund,
(b) Annualized risk-free return, and
(c) Annualized standard deviation.
This ratio is also known as ‘reward to variability ratio’. It measures the risk premium earned per unit to total risk. Risk premium is excess expected return over risk-free return and total risk involved in funds is standard deviation of expected return from the funds. When we compare position of two or more funds, a fund with higher Sharpe ratio gets a better rank. The Sharpe ratio is a measure of relative performance, in the sense that it enables the investor to compare two or more investment opportunities.
A fund with a higher Sharpe ratio in relation to another is preferable as it indicates that the fund has higher risk premium for every unit of standard deviation risk. Because Sharpe ratio adjusts return to the total portfolio risk, the implicit assumption of the Sharpe measure is that the portfolio will not be combined with any other risky portfolios. Thus the Sharpe measure is relevant for performance evaluation when we wish to evaluate several mutually exclusive portfolios.
Sharpe ratio is a risk-adjusted measure of return used to evaluate the performance of a portfolio. The ratio helps to make the performance of one portfolio comparable to that of another portfolio by making an adjustment for risk. A fund with a higher Sharpe ratio in relation to another is preferable as it indicates that the fund has higher risk premium for every unit of standard deviation risk.
A ratio of more than or equal to 1 is good, more than or equal to 2 is very good, and more than or equal to 3 is excellent. Sharpe ratio broken -down into three components, asset return, risk-free return, and standard deviation of return. After calculating the excess return, it is divided by the standard deviation of the risky asset to get its Sharpe ratio.
The objective of the Sharpe ratio is to observe how much additional return is receiving for accepting additional volatility of holding the risky asset over a risk-free asset – the higher the better.
Problem 3:
XYZ Ltd. has got 12.04% as annualized standard deviation. The annualized return for the same fund is 16.8% and the average yield on one-year Treasury paper is 6.8%.
Calculate Sharpe ratio of the fund.
Solution:
#### Method # 3. Treynor Ratio:
Treynor ratio developed by Jack Treynor, measures returns earned in excess of that which could have been earned on a risk less investment per each unit of market risk. The Treynor measure adjusts excess return for systematic risk. It is computed by dividing a portfolio’s excess return, by its beta as shown in equation.
Where, T = Treynor ratio
Rf = Risk-free return
Rp = Portfolio return
βp = Portfolio beta
As can be seen the numerator of the equation remains the same as in case of Sharpe ratio. The denominator standard deviation is replaced by Beta. As Treynor ratio indicates return per unit of systematic risk, it is a valid performance criterion when we wish to evaluate a portfolio in combination with the benchmark portfolio and other actively managed portfolios. Like Sharpe ratio it is a measure of relative performance.
Treynor ratio is a risk-adjusted measure of return based on systematic risk. It is similar to the Sharpe ratio with the difference being that the Treynor ratio uses beta as the measurement of volatility.
It is known as the ‘reward-to-volatility ratio’. For a completely diversified portfolio one without any unsystematic risk, the two measures give identical risk, because the total variance of a completely diversified portfolio is its systematic variance.
A poorly diversified portfolio could have a high ranking on the basis of Treynor ratio and a low ranking on the basis of Sharpe ratio. The difference in ranking is due to difference in diversification.
Problem 4:
For an ABC Fund, the portfolio return is 16,2% and risk-free return is 5.5%. The beta of the portfolio is 1.2. Calculate Treynor ratio of the fund.
Solution:
This indicates that the fund has generated 0.09 percentage point above the risk-free return for every unit of systematic risk.
#### Method # 4. Sortino Ratio:
Sortino ratio was developed by Frank A Sortino to differentiate between good and bad volatility in the Sharpe ratio. This differentiation of upwards and downwards volatility allows the calculation to provide a risk-adjusted measure of a security or fund’s performance in a clearer and comprehensive way.
It does not misses on the upward price changes and unlike standard deviation it does not discriminate between up and down volatility. The Sortino ratio is similar to the Sharpe ratio, except it uses downside deviation for the denominator instead of standard deviation.
#### Method # 5. Differential Return:
The second category of risk adjusted performance measure is referred to as differential return measure. The underlying objective of this category is to calculate the return that should be expected of the fund scheme given its realized risk and to compare that with the return actually realized over the period.
#### Method # 6. Jensen’s Alpha:
The most commonly used method of determining the return that should have been earned by the scheme at a given level of risk is by way of Alpha formulation:
∝ = (RP – RF) – βP(RB– RF)
Where, ∝ = the Jensen measure (alpha)
βp = portfolio beta
Rp = portfolio return
RB = benchmark return
RF = riskless return
Calculation of alpha is a fairly simple exercise. The intercept term in the regression equation is the Alpha. This number is usually very close to zero. A positive alpha means that return tends to be higher than expected given the beta statistic.
Conversely, a negative alpha indicates that the fund is an under performer. Alpha measures the value added of the portfolio given its level of systematic risk. The Jensen measure is also suitable for evaluating a portfolio’s performance in combination with other portfolios because it is based on systematic risk rather than total risk.
#### Method # 7. Appraisal Ratio:
If we wish to determine whether or not an observed alpha is due to skill or chance we can compute an appraisal ratio by dividing alpha by the standard error of the regression.
Where, A = the appraisal ratio
α =alpha
σC =the standard error of the regression (non-systematic risk)
To interpret this ratio, notice that the ‘a’ in the numerator represents the fund manager’s ability to use his skill and information to generate a portfolio return that differs from the benchmark against which his performance is being measured (e.g. BSE Sensitive Index or Nifty). The denominator measures the amount of residual (unsystematic) risk that the investor incurred in pursuit of those excess returns. Thus this ratio can be viewed as a benefit to cost ratio that assess the quality of fund manager’s skill.
#### Method # 8. M2 Measure:
Franco Modigliani and his grand-daughter Lea Modigliani in the year 1997 derived another risk adjusted performance measure by adjusting the risk of a particular portfolio so that it matches the risk of the market portfolio and then calculate the appropriate return for that portfolio. It operates on the concept that a scheme’s portfolio can be levered or de-levered to reflect a standard deviation that is identical with that of the market. The return that this adjusted portfolio earns is called M2.
m2 = (σmmf)x(RmrRf) + Rf
Where, σm = Standard deviation of the market
Rf = Risk-free return
σmf = Standard deviation of the scheme
Rmf = Return on the scheme
Since the standard deviation have been equalized, M2 can be directly compared with the return in the market. A high M2 indicates that the portfolio has outperformed and a low M2 indicates underperformed portfolio. The measures discussed above are extensively used in the mutual fund industry to comment on the performance of equity schemes.
The same measures can be used to evaluate the performance of debt securities. However, measures involving use of beta are considered theoretically unsound for debt schemes as beta is based on the capital assets pricing model, which is empirically tested for equities.
Home››Mutual Funds››Return››
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Search Results: 1 - 10 of 222 matches for " Sepideh Mirrahimi "
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Sepideh Mirrahimi Mathematics , 2012, DOI: 10.3934/dcdsb.2013.18.753 Abstract: A Hamilton-Jacobi formulation has been established previously for phenotypically structured population models where the solution concentrates as Dirac masses in the limit of small diffusion. Is it possible to extend this approach to spatial models? Are the limiting solutions still in the form of sums of Dirac masses? Does the presence of several habitats lead to polymorphic situations? We study the stationary solutions of a structured population model, while the population is structured by continuous phenotypical traits and discrete positions in space. The growth term varies from one habitable zone to another, for instance because of a change in the temperature. The individuals can migrate from one zone to another with a constant rate. The mathematical modeling of this problem, considering mutations between phenotypical traits and competitive interaction of individuals within each zone via a single resource, leads to a system of coupled parabolic integro-differential equations. We study the asymptotic behavior of the stationary solutions to this model in the limit of small mutations. The limit, which is a sum of Dirac masses, can be described with the help of an effective Hamiltonian. The presence of migration can modify the dominant traits and lead to polymorphic situations.
Mathematics , 2014, Abstract: Selection of a phenotypical trait can be described in mathematical terms by 'stage structured' equations which are usually written under the form of integral equations so as to express competition for resource between individuals whatever is their trait. The solutions exhibit a concentration effect (selection of the fittest); when a small parameter is introduced they converge to a Dirac mass. An additional space variable can be considered in order to take into account local environmental conditions. Here we assume this environment is a single nutrient which diffuses in the domain. In this framework, we prove that the solution converges to a Dirac mass in the physiological trait which depends on time and on the location in space with Lipschitz continuity. The main difficulties come from the lack of compactness in time and trait variables. Strong convergence can be recovered from uniqueness in the limiting constrained equation after Hopf-Cole change of unknown. Our analysis is motivated by a model of tumor growth introduced by Lorz et al. (preprint) in order to explain emergence of resistance to therapy.
Mathematics , 2013, Abstract: We study a non-local parabolic Lotka-Volterra type equation describing a population structured by a space variable x 2 Rd and a phenotypical trait 2 . Considering diffusion, mutations and space-local competition between the individuals, we analyze the asymptotic (long- time/long-range in the x variable) exponential behavior of the solutions. Using some kind of real phase WKB ansatz, we prove that the propagation of the population in space can be described by a Hamilton-Jacobi equation with obstacle which is independent of . The effective Hamiltonian is derived from an eigenvalue problem. The main difficulties are the lack of regularity estimates in the space variable, and the lack of comparison principle due to the non-local term.
Mathematics , 2011, Abstract: We consider populations structured by a phenotypic trait and a space variable, in a non-homogeneous environment. In the case of sex- ual populations, we are able to derive models close to existing mod- els in theoretical biology, from a structured population model. We then analyze the dynamics of the population using a simplified model, where the population either propagates through the whole space or it survives but remains confined in a limited range. For asexual pop- ulations, we show that the dynamics are simpler. In this case, the population cannot remain confined in a limited range, i.e. the popu- lation, if it does not get extinct, propagates through the whole space.
Mathematics , 2012, DOI: 10.1007/s00030-012-0156-3 Abstract: We consider the asymptotic behavior of an evolving weakly coupled Fokker-Planck system of two equations set in a periodic environment. The magnitudes of the diffusion and the coupling are respectively proportional and inversely proportional to the size of the period. We prove that, as the period tends to zero, the solutions of the system either propagate (concentrate) with a fixed constant velocity (determined by the data) or do not move at all. The system arises in the modeling of motor proteins which can take two different states. Our result implies that, in the limit, the molecules either move along a filament with a fixed direction and constant speed or remain immobile.
Mathematics , 2010, Abstract: Nonlocal Lotka-Volterra models have the property that solutions concentrate as Dirac masses in the limit of small diffusion. Is it possible to describe the dynamics of the limiting concentration points and of the weights of the Dirac masses? What is the long time asymptotics of these Dirac masses? Can several Dirac masses co-exist? We will explain how these questions relate to the so-called "constrained Hamilton-Jacobi equation" and how a form of canonical equation can be established. This equation has been established assuming smoothness. Here we build a framework where smooth solutions exist and thus the full theory can be developed rigorously. We also show that our form of canonical equation comes with a structure of gradient flow. Numerical simulations show that the trajectories can exhibit unexpected dynamics well explained by this equation. Our motivation comes from population adaptive evolution a branch of mathematical ecology which models darwinian evolution.
Mathematics , 2009, DOI: 10.4310/MAA.2009.v16.n3.a4 Abstract: We study two equations of Lotka-Volterra type that describe the Darwinian evolution of a population density. In the first model a Laplace term represents the mutations. In the second one we model the mutations by an integral kernel. In both cases, we use a nonlinear birth-death term that corresponds to the competition between the traits leading to selection. In the limit of rare or small mutations, we prove that the solution converges to a sum of moving Dirac masses. This limit is described by a constrained Hamilton-Jacobi equation. This was already proved by B. Perthame and G. Barles for the case with a Laplace term. Here we generalize the assumptions on the initial data and prove the same result for the integro-differential equation.
Mathematics , 2013, Abstract: We study the dynamics of phenotypically structured populations in environments with fluctuations. In particular, using novel arguments from the theories of Hamilton-Jacobi equations with constraints and homogenization, we obtain results about the evolution of populations in environments with time oscillations, the development of concentrations in the form of Dirac masses, the location of the dominant traits and their evolution in time. Such questions have already been studied in time homogeneous environments. More precisely we consider the dynamics of a phenotypically structured population in a changing environment under mutations and competition for a single resource. The mathematical model is a non-local parabolic equation with a periodic in time reaction term. We study the asymptotic behavior of the solutions in the limit of small diffusion and fast reaction. Under concavity assumptions on the reaction term, we prove that the solution converges to a Dirac mass whose evolution in time is driven by a Hamilton-Jacobi equation with constraint and an effective growth/death rate which is derived as a homogenization limit. We also prove that, after long-time, the population concentrates on a trait where the maximum of an effective growth rate is attained. Finally we provide an example showing that the time oscillations may lead to a strict increase of the asymptotic population size.
Mathematics , 2015, Abstract: We discuss a class of time-dependent Hamilton-Jacobi equations, where an unknown function of time is intended to keep the maximum of the solution to the constant value 0. Our main result is that the full problem has a unique viscosity solution, which is in fact classical. The motivation is a selection-mutation model which, in the limit of small diffusion, exhibits concentration on the zero level set of the solution of the Hamilton-Jacobi equation. Uniqueness is obtained by noticing that, as a consequence of the dynamic programming principle, the solution of the Hamilton-Jacobi equation is classical. It is then possible to write an ODE for the maximum of the solution, and treat the full problem as a nonstandard Cauchy problem.
Mathematics , 2015, Abstract: In this note, we discuss a class of time-dependent Hamilton-Jacobi equations depending on a function of time, this function being chosen in order to keep the maximum of the solution to the constant value 0. The main result of the note is that the full problem has a unique classical solution. The motivation is a selection-mutation model which, in the limit of small diffusion, exhibits concentration on the zero level set of the solution of the Hamilton-Jacobi equation. The uniqueness result that we prove implies strong convergence and error estimates for the selection-mutation model.
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Cody
# Problem 26. Determine if input is odd
Solution 610585
Submitted on 2 Apr 2015 by Przyczajony Tygrys, Ukryty Smok
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 1; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
2 Pass
%% n = 2; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
3 Pass
%% n = 28; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
4 Pass
%% n = 453; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
5 Pass
%% n = 17; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
6 Pass
%% n = 16; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
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Cosmological Assumptions
Isotropy and Homogeneity
We will now discuss some of the assumptions that underlie the theory of relativistic cosmology. The main principle that is adopted in mathematically describing the global structure of the universe is called the Cosmological Principle. It is defined as follows.
Assume that there exists a cosmological time scale. Then the Cosmological Principle is formulated in each constant time slice. None of these time slices have privilege points. In this case we say that the time slice is Homogeneous. A spacelike hypersurface (i.e. the time slice) is said to be homogeneous if the geometry admits 3-D space coordinate translations such that the metric is unaffected by those 3-D space translations. Homogeneity requires invariance under transformations like:
(27.1)
The cosmological Principle also requires that there be no privileged directions on the epoch time slice as well as there being no privileged points. A manifold that has no privileged directions about a point is called Isotropic. Implicit in this statement is the restriction that about the point the manifold is spherically symmetric as well. If we insist on global isotropy then we are constraining the universe manifold to be isotropic about every point. This global constraint can be shown to imply that the universe is homogeneous. The opposite case of this however does not follow. If the universe is globally homogeneous, it is not necessarily globally isotropic. In more technical language the Cosmological Principle can be written as follows:
As stated above the assumption of the Cosmological Principle allows us to develop mathematical models describing the global characteristics of the entire universe. From a phenomenological point of view cosmic homogeneity implies the existence of a smeared out universe at the largest length scales (well over 100 million light-years). Remember 1 light-year is approximately
Isotropy of the universe has good data supporting it. Here are some isotropy determinations from astronomy observations:
1. The visible galaxy distribution is isotropic to within 30%.
2. The Hubble Expansion Constant is isotropicly equal to with 25%.
3. The distribution of radio galaxies is isotropic to within 5%.
4. Cosmic X-Rays are isotropic to within 5%.
5. and, finally, the thing that really nails down the isotropic assumption: the Cosmic Background Radiation which is isotropic to within 2 parts in 100,000.
For more information about the Cosmological Principle and relevant diagrams go to the cosmology.uwinnipeg.ca website and look at the webpage called Isotropy-Homogeneity.
People have studied inhomogeneous and anisotropic solutions of the Einstein Equations in the context of the very early universe since it seems reasonable that the universe would be less symmetrical at the earliest times. Inhomogeneous solutions can be quite globally complicated but anisotropic solutions can be simply described. As an example of an anisotropic metric consider the vacuum solution called the Kasner solution. In c=1 units the line element for this solution looks like the following:
(27.2)
(27.3)
A specific Kasner metric is formed when we pick
The line element is then given by
(27.4)
It is easy to see that this gravitational field is anisotropic in the z direction. As the universe expands with time in the x and y directions, the universe contracts in the z direction. The entire universe model resembles a planar pancake gravitational distribution as time proceeds.
Weyl's Postulate
In 1923 Hermann Weyl thought about the apparent contradiction in applying a theory like General Relativity, that was set up to be generally covariant, to one particular set of circumstances, that of describing just one universe, our universe. In a universe that is expanding there seems to be a preferred coordinate system, that coordinate system that is comoving with the background expansion flow. Weyl decided that, in the application of GR to a unique symmetrical system like the universe, there must be underlying phenomenologically based postulates that are formulated from local observations. He reasoned that there should be a privileged class of observers that is comoving with the smeared out motion f the galaxies. Weyl postulated a substratum (i.e. a fluid) pervading all of space. In this fluid galaxies move like fundamental particles. A special motion for these galaxies is then assumed.
This postulate implies that there is one, and only one, geodesic passing through each point of spacetime. Hence the matter at any point possesses a unique velocity. The essence of Weyl's Postulate is that we should be able to take the substratum to be represented as a perfect fluid. In reality galaxies will not follow the postulated cosmological motion exactly. There will be random directional motion of the galaxies usually much less than the speed of light ( ). From a global cosmological perspective this velocity is smeared out to be overall negligible and furthermore it will be almost always nonrelativistic. This is in distinction to the velocity of the flow along with the universe's expansion, which can be at quite high speeds. It is, also, always unidirectional (i.e. _ away from you).
Application of the 3 Assumptions to Relativistic Cosmology
In order to describe the universe as a whole even though we have access to only local observations we have to base our cosmology on three main assumptions:
Weyl's Postulate demands that the fluid geodesics be orthogonal to a family of spacelike hypersurfaces. We introduce a set of coordinates . In each spacelike hypersurface corresponding to some time , the 3-D space coordinates of the fluid particle are constant. This is what we mean by saying that we are adopting a comoving coordinate system. We are moving with the expansion flow so that at any given cosmological instant we do not see the expansion motion. We can ensure orthogonality of the spacelike hypersurfaces by adopting the following form for our metric.
(27.5)
With this explicit orthogonality structure we are guaranteed a cosmic interpretation of time.
This global concept of time implies that simultaneity exists along the orthogonal spacelike hypersurface. We define the terms World Map and the World Picture in terms of this cosmic simultaneity.
In a comoving frame the large scale patterns stay the same. As the universe expands we don't see the universe expansion effects. We see our relationship to other parts of the universe stay the same. Consider the situation shown below where points at the endpoints of the triangle pattern always see themselves in the same triangular pattern no matter how much the space between the points has increased.
The Cosmological Principle implies that the 1st triangle must be geometrically similar to the second with respect to time. The cosmological aspect ratio is fixed. This forces the magnification factor due to expansion to be independent of the position of the triangle within the spacelike hypersurface. Mathematically, time can enter into the 3-D metric function only as a common magnification factor for all 3-D points.
(27.6)
Note that the magnification factor for the expansion of the universe between any two distinct times is given by
(27.7)
For this reason the function is called the Scale Factor. This function must be a real number since we want the spacelike/timelike character of the intervals in the hypersurface to be preserved.
We have placed very strict mathematical constraints on the manifold of points that we wish to model the universe with. We have demanded that the manifold be homogeneous and isotropic in any given time slice and from this arises the constraint that the geometrical arrangement the 3-D space points must be independent of the time variation. What this effectively means is that the curvature at every point in our manifold must be constant. Only with constant curvature, is every point guaranteed to be geometrically equivalent. It is worth remembering here that we are making statements about the universe at the present epoch concerning length scales of the order of 500 million light years and higher.
We have therefore arrived at the conclusion that in our phenomenologically- based cosmological model we must adopt a space of constant 3-D curvature.
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# Solved – Serially Uncorrelated but dependence in ARCH model
The title is a bit shaky, but it sums up the question I have in volatility models.
In a text I'm reading, the author gives the ACF plots on the return series of a stock.
He says that if we look at the sample ACF of the returns (The first plot), there are no signs of significant serial correlation.
However, if we look at the ACF of the returns squared (The second plot), it tells us that the return series is indeed serially uncorrelated, but dependent. Could someone explain how he came to this conclusion? And in general, what is the intuition behind looking at the square of the series when we are searching for the "ARCH" effect? [Because we also sometimes do so for the residual of a series after applying an ARMA model]
Contents
This is a common observations for daily returns series. The level is often found to be unpredictable (if not, then we would be able to make a lot of money with a simple ARMA model), while we are able to predict volatility.
To be a bit more explicit, assume a GARCH model:
begin{align} r_t &= varepsilon_t = sigma_t z_t \ sigma_t^2 &= omega + alpha varepsilon_{t-1}^2+ betasigma_{t-1}^2 end{align} where \$z_t\$ is iid with zero mean and unit variance. We have \$E[r_t]=E[sigma_t]E[z_t]=E[sigma_t]cdot 0 = 0\$. Thus, we have that the autocorrelation of returns \$E[r_t r_{t-h}] = E[z_t]E[sigma_t r_{t-h}] = 0\$. However, it is possible to show that begin{equation} corr(varepsilon_{t-1}^2,varepsilon_{t-h}^2) = K(alpha + beta)^h end{equation} Hence, the correlation is proportional to \$(alpha + beta)^h\$ – this also explains why \$alpha + beta\$ is refered to as the persistence in a GARCH process.
The ACF of squared returns shows us that we have higher order dependence that we may model with a GARCH model.
Note that if the ACF of returns are not zeros, then we should employ some dynamics to filter this out, but if not the case one simply proceeds with zero or constant mean.
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https://www.mytutor.co.uk/answers/5530/GCSE/Chemistry/What+is+the+volume+of+carbon+dioxide+released+at+room+temperature+and+pressure+when+6.2+g+of+copper+carbonate+reacts+with+excess+dilute+sulfuric+acid%253F
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520 views
### What is the volume of carbon dioxide released at room temperature and pressure when 6.2 g of copper carbonate reacts with excess dilute sulfuric acid?
Background Information
This is a neutralisation reaction between an acid (sulfuric acid) and a base (copper carbonate). When a carbonate reacts with an acid, the products are a salt, carbon dioxide and water.
carbonate + acid --> salt + carbon dioxide + water
ATTENTION: If you are NOT familiarised with these reactions, I advise first to revise and then try to solve the question.
Example
First thing to do when we tackle problem like this one is to write down the balanced equation (if it is not given)
CuCO3 + H2SO4 --> CuSO4 + CO2 + H2O
RFM (CuCO3) = 124 g mol-1
Number of moles of CuCO3 = 6.2 g / 124 g mol-1 = 0.05 mol
For every 1 mole of CuCO3, 1 mole of CO2 is produced (Using the stoichiometry of the equation)
Number of moles of CO2 = 0.05 mol
Volume of CO2 gas released = 0.05 mol * 24 dm3 = 1.2 dm3 = 12000 cm3
12 months ago
Answered by Andreas, a GCSE Chemistry tutor with MyTutor
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#### Other GCSE Chemistry questions
25cm3 of NaOH (2M) were titrated with 1.25M H2SO4. Write down the balanced reaction equation. Calculate the number of moles of NaOH used in the titration and hence deduce the volume of sulfuric acid used in the titration. Give your answer in dm3.
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Cheatography
# RM Reference Notes Cheat Sheet by Tash23
3 page sheet + matrix for exam (double sided)
### Types of Data
Categorical/Nominal Do not hold numerical meaning (arbitrary) Ordinal Rank ordering, differences not equal Interval Intervals between points on a scale are equal and the same, zero is arbitrary Ratio Zero is NOT arbitrary (an absence)
### Experimental Designs
Balanced each cell (each combination of factors) contain the same number of replications (how many measurements Complete every level of one factor combined with every level of the other factor(s) Incomplete Lots of factors or many measurements (nested/block design best) Single subject/repeated measures Subject acts as their own control
Ceiling effects: Test is too easy (100%)
Floor effects: Test is too hard (0%)
Learning effects: subjects improve with more trials
Order effects: test order may have effect on outcome
### Characteristics of Data Sets
Data Shape Frequency distributions are a common way to describe data shape (range of scores) Location finding central tendency or middle of data Spread Variance -> range, SD and IQR Outliers Clustering e.g. bimodal distribution Granularity Data only takes on certain values (e.g. discrete data + rounded continuous) (e.g. discrete data + rounded continuous)
### Types of Sampling
Random Increased ability to generalise to population Systematic Choosing subjects from a population at a regular interval (choosing every second item) Cluster Randomly select a few schools in your sample and have all students as participants Convenience Sample used because it is accessible rather than representative of a population
### Central Limit Theorem
• draw a large enough sample from the population and plot all of those sample means, our sampling distribution will approach normal • Sampling distribution uses sample means • Population mean: mean of all sample means Standard Error - SD of sampling distribution 95% CI = sample mean +- 1.96 x SE
### Pearson's Correlation (r)
Strength Positive Negative Strong .8 to 1 -.8 to -1 Moderate .5 to .7 -.5 to -.7 Weak 0 to .4 0 to -.4
### ANOVA Variance
DF Sum of Squares Mean Sqaure Between Groups no. groups -1 How much data varies between different groups (variance) Average variance between groups Within Groups no. data points - no. of groups How much data varies within each group (variance) Average variance within groups Total no. data points - 1
### Types of ANOVAs
One-way 1 factor/independent variable (categorical) Two-way 2+ factors/IVs (categorical), interactions Repeated Measures Measure the same outcome variable on the same population twice Each subject is now a random factor (rather than fixed factors)
### T-test Types
Test Description DF 1-sample (single) Compares your experimental group with a hypothesised or known value n-1 2-sample (independent) Compares the means for two independent samples (n1-1) + (n2-1) Paired measuring something for the same group of people n-1
One tailed: Directionless -> one group is different from the other group (in pos or neg direction)
Two tailed: Directional -> one group if larger or smaller than the other
### Linear Regression
Beta degree of change in the outcome variable for every 1 unit of change in the predictor variable R-Sqaured Fit of the model and represents how much variance in the DV can be accounted for by the IV Analysis of Variance Adj SS (adjusted sum of squares) -> total variance of data - The error SS is what is left over -> variance that cannot be explained by other factors or variance in the model
Predicting
CI: If we repeated our experiment many times an degenerated a confidence interval each time, 95% of those confidence interval will contain the true population value
Prediction Interval: Predicting future observations from the regression equation
### Assumptions of Parametric Tests
1. Normally distributed data 2. Homogeneity of variance 3. Interval/ratio data 4. Independence This means that you may have to use non-parametric tests when... • your data is better represented by the median (e.g. skewed data like salary or house prices), or • you have a very small sample size, or • you have ordinal (e.g. rating scales, some questionnaire results) or categorical data
### Parametric and Non-Parametric Equivalent Tests
Parametric Non-Parametric 1-sample AND paired t-test Sign test or Wilcoxon signed-rank test 2-sample t-test Mann-Whitney test One-way ANOVA Kruskal-Wallis test Multifactor ANOVA (two-way + repeated measures) N/A N/A Chi-square test
### Types of Qualitative Data
Transcripts (e.g. interview) Allows the researcher to ask about specific things and probe deeply Observation ethnographic studies Pictures Pictures could be photos that the researcher has taken (drawings, rooms etc) Documents Many types (e.g. progress notes) Web content Publicly available (e.g. social media)
### Sampling for a Qualitative Study
Typical Case Average case Extreme case Unusual, unique or distinct case Maximum Variation Looking for the biggest range of perspectives Homogenous Group Minimum variation sampling + Focus on in-depth area of interest Stratified Purpose Selected cases from identified subgroups (e.g. 5 people from 4 age groups) Theoretical Start data collection -> analyse results -> form therapy -> continue sampling Snow Ball One respondent is asked to suggest others. Convinience Recruiting anyone who is at hand
### Qualitative Evidence
Tangibly (concrete) Intangibly Guidelines, protocols understanding what clients want from their clinicians practice recommendations based on qual research broaden knowledge and change behaviours
### Setting up a Qualitative Analysis
Deductive (top-down) Inductive (bottom-up) coding will be influenced by the framework you're using coding will be purely based on what the participant has said, without trying to fit it into a framework.
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## Chemistry: The Central Science (13th Edition)
204 $^{\circ}$C 478 K
This question asks us to convert a measurement in Fahrenheit to both Celsius and Kelvin. To convert to Celsius, we use the formula $^{\circ}$C=($^{\circ}$F-32)$\times$$\frac{5}{9}. To convert to Kelvin, we use the formula K=(^{\circ}F+460)\times$$\frac{5}{9}$. When you use both these formulas with 400 $^{\circ}$F, you get 204 $^{\circ}$C and 478 K
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# 309.022 Mechanics of solid bodies, exercises 2 This course is in all assigned curricula part of the STEOP.\$(function(){PrimeFaces.cw("Tooltip","widget_j_id_20",{id:"j_id_20",showEffect:"fade",hideEffect:"fade",target:"isAllSteop"});});This course is in at least 1 assigned curriculum part of the STEOP.\$(function(){PrimeFaces.cw("Tooltip","widget_j_id_22",{id:"j_id_22",showEffect:"fade",hideEffect:"fade",target:"isAnySteop"});}); 2022W 2021W 2020W 2019W 2018W 2017W 2017S 2016S 2015S 2014S 2013S 2012S 2011S 2010S 2009S 2008S 2007S 2006S 2005S 2004S 2003S 2002S
2022W, UE, 2.0h, 2.0EC
## Properties
• Semester hours: 2.0
• Credits: 2.0
• Type: UE Exercise
• Format: Hybrid
## Learning outcomes
After successful completion of the course, students are able to apply the basic concepts and laws of physics taught in the corresponding lecture to solve fundamental problems in rigid body dynamics. With the help of such mathematical solutions, students can obtain concrete statements about the behavior of the basic mechanical systems, control them with respect to plausibility and, if necessary, explain the validity of the solutions.
In particular, students are able to
• determine velocity and acceleration with respect to different systems of reference for any arbitrary point in a system of a kinematic chain with joints as a function of given coordinates and their derivatives and represent those quantities in vector form in different coordinate systems,
• determine the relationship between forces and motion using Newton's and Euler's principles for solid bodies; determine and solve the equations of motion for rigid body systems and investigate the constraining forces, determine the mechanical energy of a rigid-body system and use the relationship between energy, work and power to construct the equation of motion for systems with one degree of freedom,
• analyze the behavior of rotors and rotating machine parts, especially in connection with static and dynamic unbalance;
• apply the elementary impact theory to plane systems of rigid bodies;
• derive and linearize the equation of motion for oscillating mechanical systems with one degree of freedom and analyze their behavior in case of a free oscillation or as harmonically excited system.
## Subject of course
In the exercise course (UE), exercises matching the content of the corresponding lecture are solved. Constitutive components of each solution process are the identification of suitable physical approaches, their mathematical implementation and application, the physical interpretation of the mathematical solutions, and the critical examination of the results.
The students should be able to understand the necessity of a fundamental theoretical knowledge in the field of dynamics. Based on the basic laws, they should be able to develop solution strategies for more complex problems by using suitable combinations of these laws.
## Teaching methods
Exercise
Students work out solutions independently for a given task and get support from the lecturer. In addition to that, tutors accompany the exercises in order to support the students in finding correct solutions and to answer any questions that may arise. At the end of the lecture, for each of the examples, a possible solution is presented, whereby correlations to the theoretical principles are established and alternative solution strategies are also given.
Homeworks have to be done during the semester to monitor the learning progress. The feedback on the solutions found comes from fellow students in the form of peer reviews. Studying on the homeworks independently enables students to check their current learning progress. The subsequent peer review also provides an insight into different approaches and improves the understanding of the subject area.
## Mode of examination
Immanent
A preliminary discussion via ZOOM will be held together with the lecture Mechanik 2 VO.
Registration in TISS is required to participate in the exercise lecture . This will automatically direct you in the TUWEL course of this course.
Please consult the FAQs of this course for general questions. Please post your questions in the TUWEL discussion forum for this course if your question is not answered by the FAQs.
## Course dates
DayTimeDateLocationDescription
Tue14:00 - 15:0004.10.2022 Freihaus Hörsaal 1, oder per Live-StreamVorbesprechung für das Modul Mechanik 2 (gemeinsam mit der zugehörigen Vorlesung) - Link: https://tuwel.tuwien.ac.at/mod/livestream/view.php?id=1668379
## Examination modalities
The assessment consists of two stages:
# 1. Peer review
Four peer review sessions are held throughout the semester to monitor your learning progress. For a positive assessment of the exercise it is necessary to have completed three of the four peer-review sessions.
A peer review is considered completed when both the own solution and the assigned feedback have been fully submitted. Incomplete submissions or feedbacks or identical submissions by several participants will not be accepted. In these cases, the assessments done during the peer reviews will not be taken into account in the grading.
Submissions in the context of the peer review system are assessment relevant, this means that you will be issued a certificate. If you unsubscribe from the course before the first peer review, you will not be issued a certificate – not even a negative one.
# 2. Test
The positive assessment of the peer review is a prerequisite for taking the final test. This test covers the entire material of the exercise and serves to determine the final grade. A replacement test is offered for students who have failed in the test.
For a positive evaluation, the following points must be documented on the solution sheet for the individual tasks:
1. All the physical-mathematical approaches required to solve the task, including the sketches to understand the respective approach.
2. The main steps of the solution.
3. The final result, expressed in the dimensions given by the question (unless stated otherwise).
The following points should also be noted:
• A final result is evaluated only if the physical approach is completely correct.
• An essential feature is the implementation of the task in a mathematical formulation. In this context, it is pointed out again that the consideration of positive counting directions and the correct signing of the equations are essential.
• Proper mathematical treatment of the equations is required. This means that there are no points for the mathematically correct treatment itself. This only needs to be traceable in the essential steps, but not documented in detail on the solution sheet. On the other hand, the problem cannot be solved without the correct application of mathematics.
• The determined solution (end result) must be dimensionally correct and plausible.
• Tasks that require the successful completion of previous subtasks will only be evaluated if the previous tasks have also been solved correctly. (Basic skills that are queried in preparatory sub-tasks must be mastered).
• The points may be divided according to the degree of difficulty and weighting of the individual questions.
The duration of the test is 60 minutes, tools such as calculators, formulas, scripts etc. are not allowed. Totally eight points can be obtained, whereby four points are required for a positive evaluation of the exercise.
## Exams
DayTimeDateRoomMode of examinationApplication timeApplication modeExam
Fri12:00 - 14:0027.01.2023FH Hörsaal 1 - MWB assessedno application-Übungstest (Paralleltermin)
Fri12:00 - 14:0027.01.2023GM 5 Praktikum HS- TCH assessedno application-Übungstest (Paralleltermin)
Fri12:00 - 14:0027.01.2023GM 2 Radinger Hörsaal - TCH assessedno application-Übungstest (Paralleltermin)
Fri12:00 - 14:0027.01.2023Informatikhörsaal - ARCH-INF assessedno application-Übungstest (Paralleltermin)
Fri12:00 - 14:0027.01.2023EI 7 Hörsaal - ETIT assessedno application-Übungstest (Paralleltermin)
Fri12:00 - 14:0027.01.2023GM 1 Audi. Max.- ARCH-INF assessedno application-Übungstest (Paralleltermin)
Fri10:00 - 12:0017.02.2023GM 5 Praktikum HS- TCH assessedno application-Ersatztest
Fri10:00 - 12:0017.02.2023FH Hörsaal 1 - MWB assessedno application-Ersatztest
Fri10:00 - 12:0017.02.2023Informatikhörsaal - ARCH-INF assessedno application-Ersatztest
Fri10:00 - 12:0017.02.2023GM 1 Audi. Max.- ARCH-INF assessedno application-Ersatztest
## Group dates
GroupDayTimeDateLocationDescription
Gruppe 01 - MO 09:00-11:00Mon09:00 - 11:0017.10.2022 - 23.01.2023Seminarraum BA 05 - MB 309.022 Mechanics of solid bodies, exercises 2 Gruppe 01 - MO 09:00-11:00
Gruppe 02 - MO 09:00-11:00Mon09:00 - 11:0017.10.2022 - 23.01.2023Seminarraum BA 02A 309.022 Mechanics of solid bodies, exercises 2 Gruppe 02 - MO 09:00-11:00
Gruppe 03 - DI 09:00-11:00Tue09:00 - 11:0018.10.2022 - 24.01.2023Seminarraum BA 05 - MB 309.022 Mechanics of solid bodies, exercises 2 Gruppe 03 - DI 09:00-11:00
Gruppe 04 - DI 09:00-11:00Tue09:00 - 11:0018.10.2022 - 24.01.2023Seminarraum BA 02A 309.022 Mechanics of solid bodies, exercises 2 Gruppe 04 - DI 09:00-11:00
Gruppe 05 - DI 16:00-18:00Tue16:00 - 18:0018.10.2022 - 24.01.2023Seminarraum BA 02B 309.022 Mechanics of solid bodies, exercises 2 Gruppe 05 - DI 16:00-18:00
Gruppe 06 - MI 09:00-11:00Wed09:00 - 11:0019.10.2022 - 25.01.2023Seminarraum BA 05 - MB 309.022 Mechanics of solid bodies, exercises 2 Gruppe 06 - MI 09:00-11:00
Gruppe 07 - MI 16:00-18:00Wed16:00 - 18:0019.10.2022 - 25.01.2023Seminarraum BA 05 - MB 309.022 Mechanics of solid bodies, exercises 2 Gruppe 07 - MI 16:00-18:00
Gruppe 08 - DO 09:00-11:00Thu09:00 - 11:0020.10.2022 - 26.01.2023Seminarraum BA 05 - MB 309.022 Mechanics of solid bodies, exercises 2 Gruppe 08 - DO 09:00-11:00
Gruppe 09 - DO 09:00-11:00Thu09:00 - 11:0020.10.2022 - 26.01.2023Seminarraum BA 02A 309.022 Mechanics of solid bodies, exercises 2 Gruppe 09 - DO 09:00-11:00
Gruppe 10 - DO 16:00-18:00Thu16:00 - 18:0020.10.2022 - 26.01.2023Seminarraum BA 02B 309.022 Mechanics of solid bodies, exercises 2 Gruppe 10 - DO 16:00-18:00
Gruppe 11 - DO 16:00-18:00Thu16:00 - 18:0020.10.2022 - 26.01.2023GM 8/9 - Hörsaal des Internationalen Wiener Motorensymposiums 309.022 Mechanics of solid bodies, exercises 2 Gruppe 11 - DO 16:00-18:00
Gruppe 12 - Sammelgruppe MO 16:00-18:00Mon16:00 - 18:0017.10.2022 - 23.01.2023GM 2 Radinger Hörsaal - TCH 309.022 Mechanics of solid bodies, exercises 2 Gruppe 12 - Sammelgruppe MO 16:00-18:00
Gruppe 13 - Online-GruppeMon16:00 - 18:0017.10.2022 - 23.01.2023 Online - Abwicklung über den TUWEL-Kurs. Anmeldung nur mit Begründung unter mechanik2ue@tuwien.ac.at309.022 Mechanics of solid bodies, exercises 2 Gruppe 13 - Online-Gruppe
Reserve-Gruppe 1 Mi 9 - keine Anmeldung möglich!Wed09:00 - 11:0019.10.2022 - 25.01.2023Seminarraum BA 02A 309.022 Mechanics of solid bodies, exercises 2 Reserve-Gruppe 1 Mi 9 - keine Anmeldung möglich!
Reserve-Gruppe 2 Mo 16 - keine Anmeldung möglich!Mon16:00 - 18:0017.10.2022 - 23.01.2023Seminarraum BA 08A - MB 309.022 Mechanics of solid bodies, exercises 2 Reserve-Gruppe 2 Mo 16 - keine Anmeldung möglich!
Reserve-Gruppe 3 Mi 16 - keine Anmeldung möglich!Wed16:00 - 18:0019.10.2022 - 25.01.2023Seminarraum BA 02B 309.022 Mechanics of solid bodies, exercises 2 Reserve-Gruppe 3 Mi 16 - keine Anmeldung möglich!
Reserve-Gruppe 4 MO 16 - keine Anmeldung möglich!Mon16:00 - 18:0017.10.2022 - 23.01.2023Seminarraum BA 02B 309.022 Mechanics of solid bodies, exercises 2 Reserve-Gruppe 4 MO 16 - keine Anmeldung möglich!
## Course registration
Begin End Deregistration end
01.09.2022 00:00 09.11.2022 08:00 19.10.2022 08:00
### Precondition
The student has to be enrolled for at least one of the studies listed below
## Group Registration
GroupRegistration FromTo
Gruppe 01 - MO 09:00-11:0005.10.2022 08:0012.10.2022 12:00
Gruppe 02 - MO 09:00-11:0005.10.2022 08:0012.10.2022 12:00
Gruppe 03 - DI 09:00-11:0005.10.2022 08:0012.10.2022 12:00
Gruppe 04 - DI 09:00-11:0005.10.2022 08:0012.10.2022 12:00
Gruppe 05 - DI 16:00-18:0005.10.2022 08:0012.10.2022 12:00
Gruppe 06 - MI 09:00-11:0005.10.2022 08:0012.10.2022 12:00
Gruppe 07 - MI 16:00-18:0005.10.2022 08:0012.10.2022 12:00
Gruppe 08 - DO 09:00-11:0005.10.2022 08:0012.10.2022 12:00
Gruppe 09 - DO 09:00-11:0005.10.2022 08:0012.10.2022 12:00
Gruppe 10 - DO 16:00-18:0005.10.2022 08:0012.10.2022 12:00
Gruppe 11 - DO 16:00-18:0005.10.2022 08:0012.10.2022 12:00
Gruppe 12 - Sammelgruppe MO 16:00-18:0005.10.2022 08:0012.10.2022 12:00
Gruppe 13 - Online-Gruppe05.10.2022 08:0012.10.2022 12:00
Reserve-Gruppe 1 Mi 9 - keine Anmeldung möglich!05.10.2022 08:0012.10.2022 12:00
Reserve-Gruppe 2 Mo 16 - keine Anmeldung möglich!05.10.2022 08:0012.10.2022 12:00
Reserve-Gruppe 3 Mi 16 - keine Anmeldung möglich!05.10.2022 08:0012.10.2022 12:00
Reserve-Gruppe 4 MO 16 - keine Anmeldung möglich!05.10.2022 08:0012.10.2022 12:00
## Curricula
Study CodeSemesterPrecon.Info
033 245 Mechanical Engineering 3. Semester
Course requires the completion of the introductory and orientation phase
033 282 Mechanical Engineering - Management 3. Semester
Course requires the completion of the introductory and orientation phase
700 Mechanical Engineering 2. Semester
740 Industrial Engineering-Management 2. Semester
## Literature
Lehrunterlagen:
Die Angaben der Übungsbeispiele stehen für Sie im TUWEL-Kurs als Download zur Verfügung und können auch in gedruckter Form während der Sekretariatssprechstunden am Institut erworben werden - Preis: 2€.
Die Lösungen zu den Beispielen werden sukzessive im TUWEL-Kurs freigeschalten.
Weiterführende Literatur:
Gamer, U.; Mack, W.:
Mechanik – Ein einführendes Lehrbuch für Studierende der technischen Wissenschaften.
Springer Verlag Wien, 1999. ISBN: 3-211-82854-0.
Parkus, H.:
Mechanik der festen Körper.
Springer, 2005. ISBN: 978-3-211-80777-4.
Magnus, K.; Müller-Slany H.H.:
Grundlagen der Technischen Mechanik
. Teubner Stuttgart, 2006. ISBN: 978-3-8351-0007-7.
Müller-Slany H.H.:
Aufgaben und Lösungsmethodik Technische Mechanik. Springer Wiesbaden, 2018. ISBN: 978-3-658-22419-6.
Lehmann, T.:
Elemente der Mechanik: 3. Kinetik.
Vieweg Braunschweig, 1977. ISBN: 3-528-19197-X.
Lugner, P.; Desoyer, K,: Novak, A.:
Technische Mechanik – Aufgaben und Lösungen
, Springer Verlag Wien, 1992. ISBN: 3-211-81717-4.
Gross, D.; Hauger, W.; Schröder, J. ; Wall, W. A.:
Technische Mechanik 3: Kinetik.
Springer Berlin Heidelberg, 2019. ISBN: 978-3-662-59550-3.
Gross, D.; Ehlers, W.; Schröder, J.; Müller, R.:
Formeln und Aufgaben zur Technischen Mechanik 3: Kinetik, Hydrodynamik. Springer Berlin Heidelberg, 2019. ISBN: 978-3-662-59681-4.
## Previous knowledge
Attending lecture 309.020 VO “Mechanics of solid bodies 2” preferably in the same semester.
German
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https://astronomy.stackexchange.com/questions/20200/how-many-things-are-wrong-in-this-artist-view-of-the-trappist-1-system
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# How many things are wrong in this "artist view" of the TRAPPIST-1 system?
There is a this poster on the NASA site
(
that irks me. Of course its an artist's fantasy, but since it is on a NASA site, I see people considering it as scientifically accurate. So, beyond the two obvious "In which direction(s) is the sun", and "OMG these two planets are crashing into each other! Before we crash in the debris!", what else is wrong? Tides? Roche's limit? Atmospheres ripped out?
Also, since 1e is an intermediate planet, it is possible to have the other six planets all conveniently placed in the same area of the sky?
• It's an artists view, they don't have to be scientifically accurate in every detail. Feb 26, 2017 at 3:23
• If the planets are sufficiently close to their sun, their brightsides don't have to (and, in fact, can't) all face exactly the same way. Aug 28, 2019 at 2:09
• @sean yes, but they won't be on the same side of the sun. In the illustration the small planet at the top should be close to a half circle IMHO. Aug 28, 2019 at 11:04
• @xenoid: Not if the sun is just out of frame to the right, obscured by the windowframe on that side. Aug 28, 2019 at 20:53
• thank you for the accept! I'd only meant to add supplementary information to the other answers. Now I'm getting more interested in this; it's a fun problem to think about!
– uhoh
Aug 29, 2019 at 0:05
The 2 big planets are probably f & g and they don't look up to scale to me. While f is the size of Earth, or almost 4 times bigger than the Moon, but it's also further away, ~ 1.3 million kilometers, as opposed to 400,000km for the Moon. So I would say f will be somewhat larger than the moon but not anywhere near what the poster shows. G will appear even smaller than f because it's further away. I haven't looked at the orbits in details but I think that sort of event should be possible. Trappist-1e being intermediate planet makes no difference. As for "where is the sun" - it appears to be somewhere to the right, probably below the horizon. That makes sense because any base would probably be along the terminator.
Hominids in the neighborhood is probably the biggest mistake made by the artist.
Sorry to be a little less optimistic than people at NASA, but a journey this far would probably take hundreds of thousands of years.
And we're not in the best condition to prepare for it. The technology needed for such a trip should take time to develop. Too much time, too much ressources... Our time there is limited, and we can difficulty move to another planet to live.
• I think you're missing a particular point of these posters. They're meant to be what travel ads might look like in the future. Presumably, humans, at the time these posters might exist, have mastered space travel and "planet hopping" is as simple as going to the grocery store is now. Feb 27, 2017 at 13:54
• My point is I doubt humanity has enough time to achieve this. Feb 27, 2017 at 13:57
• Are you saying that humanity will go extinct before then? Like I said, presumably in this future time humanity has mastered space travel and a trip to a star only 40 light years away might be a simple 2 week journey for them. Feb 27, 2017 at 13:58
• Yes, it looks so much technology and so many troubles to come before it is available. I hope I'm wrong, though. Feb 27, 2017 at 14:00
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# How to combine SparseArray and If
Because I am dealing with huge matrixes, my computer can not handle it, because of the memory. But, in my matrix, a very small percentage of the elements are non-zero. So I should use SparseArray as far as I am aware. The problem is that, I do not really know how to transform my old way of defining my matrix to the new "SparseArray way". I gonna give a very simple toy model of my old way:
n = 3;
H = ConstantArray[0, {n*n, n*n}];
Do[
i = ix + (iy - 1)*n;
j = jx + (jy - 1)*n;
x = ix;
y = iy;
If[i == n && j == i - 1, H[[i, j]] = t*Exp[I*x],
If[i == n && j == i + n, H[[i, j]] = t,
If[i == n && j == i - (n - 1), H[[i, j]] = t*Exp[-I*x],
If[i == n && j == i + n*(n - 1), H[[i, j]] = t, 0]]]],
{iy, 1, n,1}, {ix, 1, n, 1}, {jy, 1, n, 1}, {jx, 1, n, 1}];
So my matrixes are like this but with many-many If.
Could somebody give me an idea of how to do it?
Thanks a lot
It may not be highly time-efficient (read this) but you can make Part assignments to a SparseArray object just as you would a conventional array, therefore you merely need to replace your ConstantArray with:
SparseArray[{}, {n*n, n*n}]
You can also specify a different background for the array with the third argument:
SparseArray[{}, {n*n, n*n}, 5]
Please take a look at this question, as explicit loops are rarely optimal in Mathematica:
Alternatives to procedural loops and iterating over lists in Mathematica
By the way you should avoid starting user symbols with capital letters (unless you know exactly what you're doing) to avoid conflicts with built-ins, so use h instead of H.
After looking at your code in depth I believe you would be served by something like this:
n = 3;
SparseArray[
{i_, j_} :>
With[{x = Mod[i, n, 1]},
Which[
i != n, 0,
j == i - 1, t*Exp[I*x],
j == i + n, t,
j == i - (n - 1), t*Exp[-I*x],
j == i + n*(n - 1), t,
True, 0
]
],
{n*n, n*n}
]
I observe that all but one row of the resultant array is entirely zero. Of course it will be far more efficient to identify this pattern and program accordingly, but I assume that this code is merely an illustration of the kind of operation you wish to perform and not representative of the operation itself.
• It might be more profitable for OP if he recasts the rules implemented as nested If[]s as actual rules within SparseArray[]... Commented May 16, 2013 at 10:59
• Maybe I could do something similar to this: Commented May 16, 2013 at 11:29
• mat = SparseArray[{i_, j_} /; i == j :> 1, {8, 8}]; Commented May 16, 2013 at 11:29
• But what is the way to put more than one condition as I did, i==j:>1, but many of them ? Commented May 16, 2013 at 11:30
• @Mencia you just need to put them in a list: SparseArray[{{i_, j_} /; i == j -> -2, {i_, j_} /; Abs[i - j] == 1 -> 1, {i_, j_} /; i != j && j == 5 :> i + j}, {10, 10}] // MatrixForm Commented May 16, 2013 at 11:49
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## tandfonline.com – A Web Simulator to Assist in the Teaching of Bayes’ Theorem
tandfonline.com har udgivet en rapport under søgningen “Teacher Education Mathematics”: Abstract Formulae display:?Mathematical formulae have been encoded as MathML and are displayed in this HTML version using MathJax in order to improve their display. Uncheck the box to turn MathJax off. This feature requires Javascript. Click on a formula to zoom. Abstract Teaching some concepts in statistics greatly benefits from individual practice with immediate feedback. In order to provide such practice to a large number of students we have written a simulator based on an historical event: the loss in May 22, 1968, and subsequent search for the nuclear submarine USS Scorpion. Students work on a simplified version of the search and can see probabilities change in response to new evidence. The simulator is designed to assist in the teaching… Continue Reading
## tandfonline.com – Inference and Decision Making for 21st-Century Drug Development and Approval
tandfonline.com har udgivet en rapport under søgningen “Teacher Education Mathematics”: ABSTRACT Formulae display:?Mathematical formulae have been encoded as MathML and are displayed in this HTML version using MathJax in order to improve their display. Uncheck the box to turn MathJax off. This feature requires Javascript. Click on a formula to zoom. ABSTRACT The cost and time of pharmaceutical drug development continue to grow at rates that many say are unsustainable. These trends have enormous impact on what treatments get to patients, when they get them and how they are used. The statistical framework for supporting decisions in regulated clinical development of new medicines has followed a traditional path of frequentist methodology. Trials using hypothesis tests of “no treatment effect” are done routinely, and the p-value < 0.05 is often the… Continue Reading
## tandfonline.com – Why Bayesian Ideas Should Be Introduced in the Statistics Curricula and How to Do So
tandfonline.com har udgivet en rapport under søgningen “Teacher Education Mathematics”: Abstract Formulae display:?Mathematical formulae have been encoded as MathML and are displayed in this HTML version using MathJax in order to improve their display. Uncheck the box to turn MathJax off. This feature requires Javascript. Click on a formula to zoom. Abstract While computing has become an important part of the statistics field, course offerings are still influenced by a legacy of mathematically centric thinking. Due to this legacy, Bayesian ideas are not required for undergraduate degrees and have largely been taught at the graduate level; however, with recent advances in software and emphasis on computational thinking, Bayesian ideas are more accessible. Statistics curricula need to continue to evolve and students at all levels should be taught Bayesian thinking. This… Continue Reading
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# How to Break Coupled ODEs down to first order for Runge-Kutta
My question might seem a bit simple. I am trying to solve a system of ODEs using Runge-Kutta method. I am having difficulty breaking down the equations into a system of first order ones required before applying RK45 to it because there is a third order differential of the same parameter "F" in both equations which I am not used to.
Equation 1 $$\left( {\frac {{\rm d}^{2}}{{\rm d}{\eta}^{2}}}F \left( \eta \right) \right) ^{n-1}{\frac {{\rm d}^{3}}{{\rm d}{\eta}^{3}}}F \left( \eta \right) +{\frac {2\,n+2+ \left( 2\,n-1 \right) kF \left( \eta \right) {\frac {{\rm d}^{2}}{{\rm d}{\eta}^{2}}}F \left( \eta \right) }{n+4}}-{\frac { \left( n+2+ \left( n+1 \right) k \right) \left( {\frac {\rm d}{{\rm d}\eta}}F \left( \eta \right) \right) ^{2 }}{n+4}}+G \left( \eta \right) =0$$
Equation 2
$${\frac { \left( {\frac {{\rm d}^{2}}{{\rm d}{\eta}^{2}}}F \left( \eta \right) \right) ^{n-1} \left( n-1 \right) \left( {\frac {{\rm d}^{3 }}{{\rm d}{\eta}^{3}}}F \left( \eta \right) \right) {\frac {\rm d}{ {\rm d}\eta}}G \left( \eta \right) }{\Pr\,{\frac {{\rm d}^{2}}{{\rm d} {\eta}^{2}}}F \left( \eta \right) }}+ \left( {\frac {{\rm d}^{2}}{ {\rm d}{\eta}^{2}}}F \left( \eta \right) \right) ^{n-1}{\frac { {\rm d}^{2}}{{\rm d}{\eta}^{2}}}G \left( \eta \right) +{\frac {2\,n+2+ \left( 2\,n-1 \right) kF \left( \eta \right) {\frac {\rm d}{{\rm d} \eta}}G \left( \eta \right) }{n+4}}-{\frac {n+2\, \left( n+1 \right) k \left( {\frac {\rm d}{{\rm d}\eta}}F \left( \eta \right) \right) G \left( \eta \right) }{n+4}}=0$$
Boundary Conditions
$$G \left( 0 \right) =1, F \left( 0 \right)=F' \left( 0 \right) =0,G \left( \infty \right) =0,F' \left( \infty \right) =0$$
Any help would be greatly appreciated.
• Out of curiosity, is there any physical meaning behind these equations? They look pretty arbitrary to me. I tried several times to change variables and make it more simpler but it fails with no luck... There are some patterns in these two equations that encouraged me to try make it simpler or write it down as a system of first-order ODEs but at least I can't get it... Are you sure the constants in these two equations are written correctly? For example you have: $\frac{n+2+(n+1)k}{n+4}$ and $\frac{n+2(n+1)k}{n+4}$, which looks pretty similar but not exactly the same... – Alone Programmer Oct 3 '19 at 21:24
• If you are using RK45, you are not going to be able to enforce the boundary conditions at $\infty$. You are going to have to get lucky with your choice of initial data – whpowell96 Oct 3 '19 at 23:43
The usual trick is to add more variables that represent the successive derivatives, as in the equations of motion pf physics written as a set of first order ODE of "2" variables: \begin{align}\dot{x}&=v\\ \dot{v}&=a(x,t)\end{align} instead of a second order ODE of 1 variable $$\ddot{x}=a(x,t) \ .$$ The dot represents the time derivative.
So, applying to your case, you could have Equation 1 as: \begin{align} x & = \frac{d F}{d \eta}\\ y & = \frac{d x}{d \eta} \\ 0 & = y^{n-1}\frac {d y}{d \eta} + \frac{2n+2 + (2n-1) k F y}{n+4} - \frac { (n+2 + (n+1) k) x^2}{n+4} + G(\eta) \end{align} Note that $$y = \frac{d x}{d \eta} = \frac{d^2 F(\eta)}{d \eta^2}$$ and that the last equation can easily be written as $$\frac{d y}{d \eta} = \ldots$$
Equation 2 can have similar substitutions together with the previous ones: \begin{align} u & = \frac{d G}{d \eta}\\ v & = \frac{d u}{d \eta} \\ 0 & = \frac{y^{n-1} (n-1) \frac{d y}{d \eta} u }{\Pr\,y} + y^{n-1} v + \frac {2n+2 + (2n-1) k F u}{n+4} - \frac {n+2 (n+1) k x G}{n+4} \end{align} And the last equation can also be written as $$\frac{d y}{d \eta} = \ldots$$
This means that you will have 4 virtual new variables: $$x$$, $$y$$, $$u$$ and $$v$$.
Edit:
Looking at the number of boundary conditions, since you have 5, you might want to have 5 variables. Depending on your problem, you may write Equation 2 as: \begin{align} u & = \frac{d G}{d \eta}\\ 0 & = \frac{y^{n-1} (n-1) \frac{d y}{d \eta} u }{\Pr\,y} + y^{n-1} \frac{d u}{d \eta} + \frac {2n+2 + (2n-1) k F u}{n+4} - \frac {n+2 (n+1) k x G}{n+4} \end{align} The difference is in the second parcel, where we have $$\frac{d u}{d \eta}$$ and we no longer need to have $$v$$.
And then, plug in the expression you got for $$\frac{d y}{d \eta}$$ in the last equation and solve it for $$\frac{d u}{d \eta} = \ldots$$
Hence, you'll get a system with 5 (first order) ODE and 5 boundary conditions.
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http://www.gurufocus.com/term/dividend_growth_3y/AZO/Dividend%2BGrowth%2BRate/AutoZone%2BInc
| 1,490,829,466,000,000,000
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Switch to:
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3-Year Dividend Growth Rate
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### A new math problem about funcoids
Just a few seconds ago I realized that I have never considered and and even never formulated the following problem: Explicitly describe the set of complemented funcoids. Note that not all principal funcoids are complemented. For example see my book for a…
### My conjecture partially solved
I’ve partially solved my conjecture, proposed Polymath problem described at this page. The problem asks which of certain four expressions about filters on a set are always pairwise equal. I have proved that the first three of them are equal, equality with…
### Partial order funcoids and reloids
Partial order funcoids and reloids formalize such things as “infinitely small” step rotating a circle counter-clockwise. This is “locally” a partial order as every two nearby “small” sets (where we can define “small” for example as having the diameter (measuring along the…
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IBA Final Exam – Flashcards
Unlock all answers in this set
question
To alert Excel that you are entering a formula and not text, type a(n) _____ preceding the formula.
equal sign
question
If the following arithmetic operations all are found in a formula with non parentheses, which one is completed last?
+
question
When you copy a formula with relative cell references down a row, _____.
the row references change in the formula
question
You can select a range using the keyboard, just press the _____ then use the arrow keys to select the desired range.
f8 key
question
You can enter functions______. a. with the keyboard b. the Sum menu c. the Insert Function in the formula d. all of the above
d. all of the above
question
The _____ function is used to determine the lowest number in a given range.
MIN
question
To assign a dollar sign to appear immediately to the left of the first digit with no spaces, use a ______.
floating dollar sign
question
When Excel automatically sets the width of a column based on the widest entry in the column, it is called _____.
best fit
question
Excel has a(n) _____ you can use to check the worksheet for spelling errors.
spell checker
question
______ is the process of finding and correcting errors in a worksheet.
debugging
question
A(n) ______ is a prewritten formula that is built into Excel.
function
question
Values used with a function are called _____.
arguments
question
The order in which calculations are performed in a formula is called _____.
order of operations
question
The ____ function sums the numbers in the specified range and then divides the sum by the number cells with numeric values.
AVERAGE
question
Use the ____ function to display the highest value in a range.
MAX
question
Formatting that appears only when the value in a cell meets specific conditions is called ____.
conditional formatting
question
___ means that the width of a column will be increased or decreased.
best fit
question
T/F- Pressing the F9 key instructs Excel to recalculate all formulas.
true
question
T/F- In the formula =4 8 * 6 - 8 the subtraction operation (-) is completed before the multiplication operation (*).
false
question
T/F- The order of operations is from right to left.
false
question
A unique identifier also is called a(n) ____ key.
primary
question
Each field has a data ___, which indicates what can be stored in the field.
type
question
A field with the ___ data type can store a unique sequential number that Access assigns to a record. Access will increment the number by 1 as each new record is added.
AutoNumber
question
Fields whose data type is Number often require you to change the field ___, which is the storage space assigned to the field by Access.
size
question
The ___ contains a list of all the objects in the database. You use this to open an object.
question
You work on objects such as tables, forms, and reports in the Access ____ area.
work
question
___ orientation means the printout is across the height of the page.
landscape
question
Database ___, also known as metadata, can include such information as the file's author, title, or subject.
properties
question
The simplest approach to recovery involves periodically making a copy of the database (called a __ copy or a save copy).
backup
question
A(n) ____, such as Access, is software that allows you to use a computer to create a database.
database management system
question
The rows in the tables are called ____.
records
question
In Access, all the tables, reports, forms, and queries that you create are stored in a single file called a(n) ____.
database
question
_____ view, is only used to create a table or to modify the structure of the table.
design
question
To see the field size and/or caption for a field, click the field's _____, the small box that precedes the field.
row selector
question
_____ are simply questions, the answers to which are in the database.
queries
question
T/F- Redundancy means storing the same fact in more than one place.
true
question
T/F- Database users typically use standard guidelines for naming tables.
false
question
T/F- To change the default printer that appears inthe Print dialog box, click File on the Ribbon, click the Print tab in the Backstage view, click Print in the Print gallery, then click the Name box arrow and select the desired printer.
true
question
T/F- When importing data, you have two choices. You can create a new table or add the records to an existing table.
true
question
T/F- Report view allows you to make changes, but it does not show you the actual report.
false
question
The ____ slide introduces the presentation to the audience.
title
question
Each presentation template has 12 complementary colors, which collectively are called ____.
scheme
question
The Increase Font Size buttons on the Mini toolbar and in the Font group (__ tab) enlarge the selected characters in predetermined amounts.
home
question
The mouse pointer changed shape by adding a paintbrush to indicate that the Format ___ function is active.
painter
question
Circles, squares, and triangles are among the geometric shapes included in the ____ gallery.
shapes
question
Some ___ are shadows, reflections, glows, bevels, and 3-D rotations.
effects
question
PowerPoint gives you the option to change the line ___, or thickness, starting with 1/4 point.
weight
question
Many PowerPoint transitions have options that you can customize to give your presentation a unique look. To access this feature, click the ___ Options button (Transitions tab | Transition to This Slide group).
effect
question
Checks, currency, and legal documents use ____ to verify their authenticity. These semi-transparent images are visible when you hold the paper up to a light.
watermarks
question
A selected graphic appears surrounded by a selection rectangle, which has small squares and circles, called _____ _____ at each corner and middle locations.
sizing handles
question
_____ is the difference between the darkest and lightest areas of the image.
contrast
question
The ____ slider indicates the amount of opaqueness.
transparency
question
Fill ____ add pattern and texture to a background, which add depth to a slide.
effects
question
A(n) ____ helps letters display prominently by adding a shadow behind the text.
question
The WordArt ____ is the exterior border surrounding each letter or symbol.
outline
question
A(n) ____ edge is angled or sloped and gives the effect of a three- dimensional object.
bevel
question
T/F- People remember at least one-third more information when the document they are seeing or reading contains visual elements.
true
question
T/F- Contrast determines the overall lightness or darkness of the entire image.
false
question
T/F- Gradient fill is one color shade gradually progressing to another shade of the same color or one color progressing to another color.
true
question
T/F- Solid fill is one color used throughout the entire slide.
true
question
A ___ is a program that copies itself repeatedly, for example in memory or on a network, using up resources and possibly shutting down the computer or network.
worm
question
A search ____ is a program that finds Web sites, Web pages, images, videos, news, maps, and other information related to a specific topic.
engine
question
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false
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true
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false
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true
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An addition to a school is 90 feet by 150 feet. How many 18 inch tiles will be needed to do the flooring?
A. 6,000
B. 7,500
C. 12,000
D. 24,000
E. 30,000
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right angled triangle with 2 equal sides
right angled triangle with 2 equal sides
For example, a triangle in which all three sides have equal lengths is called an equilateral triangle while a triangle in which two sides have equal lengths is called isosceles.Triangles classified based on their internal angles fall into two categories: right or oblique. producing a side of a triangle is equal to the sum of the two interior. opposite angles and solving problems using it.Theorem The sum of the three interior angles of a triangle is two right angles. Draw any triangle ABC. Separate the vertices A, B, C as shown in the diagram. Euler diagram of types of triangles, using the definition that isosceles triangles have at least 2 equal sides, i.e. equilateral triangles are isosceles.The inverse trigonometric functions can be used to calculate the internal angles for a right angled triangle with the length of any two sides. It is possible that a right angled triangle can have all sides equal and all angles as right angles!! Shocked? Fake? Any triangle with two sides of equal length is called an isosceles triangle , and any isosceles triangle must contain two equal interior angles. The isosceles triangle of Figure 10 is special because it is also a right-angled triangle. The question says it is a right angled Isosceles traingle , So two sides are equal and one angle is 90So therefore the area of triangle 72m, Hope you understand, Have a great day ! - Bunti 360 !! A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90-degree angle). The relation between the sides and angles of a right triangle is the basis for trigonometry.
Isosceles acute angled triangle: All the angles are acute, two of them being equal, and the other different from these two.Rectangular scalene triangle: it has a right angle and all its sides and angles are different. By the isosceles triangle theorem, the two angles opposite the equal sides are as well equal, but if the third side is different then the third angle will as well not be the same.These two right angle triangles can be solved for using the Pythagoras theorem. Equal sides opposite to equal angles in a triangle (Theorem and Proof) - Duration: 9:35. dostotussigreatho 6,360 views.Identifying the Opposite and Adjacent sides in a right-angled triangle. A special right triangle is a right triangle with some regular feature that makes calculations on the triangle easier, or for which simple formulas exist. For example, a right triangle may have angles that form a simple ratio, such as 45-45-90. This is called an " angle based" right triangle. Let the equal sides be x 1/2x2area20 given x240 xrt40 using the pythagoras theorem c 2a2b2 crt(a2b2) third sidert(4040) rt80 4rt5 cm. The Right Triangles (right-angled triangles) have one right angle (equal to 90).It is possible to have a right isosceles triangle a triangle with a right angle and two equal sides. Two triangles are congruent, if they have accordingly equal: a) two sides and an angle between themTwo right-angled triangles are congruent, if one of the following conditions is valid Right Triangle Congruence.
Isosceles and equilateral triangles arent the only classifications of triangles with special characteristics.All right triangles have two legs, which may or may not be congruent. The legs of a right triangle meet at a right angle. Triangle - 3 sides and 3 angles.When a triangle has all sides equal its called an equilateral triangle. There are different types of triangles. A right-angled triangle has one right angle. Then cut along the diagonal to form two right-angled triangles.This suggests that the area of a triangle is equal to half the area of a rectangle around it. Therefore: In the diagram, we notice that the length of the rectangle is one side of the triangle. This forms two right triangles inside the main triangle, each of whose hypotenuses are "3". The cosine of either of the original acute angles equals 23, or 0.833.Since the two opposite sides on an isosceles triangle are equal, you can use trigonometry to figure out the height. 3 congruent (equal) sides. Classifying Triangles (by angles). Acute Triangle.acute angle. < all angles. 90. Right Triangle. Step 1: This is a right triangle with two equal sides so it must be a 45-45-90 triangle.As long as you know that one of the angles in the right-angle triangle is either 30 or 60 then it must be a 30-60-90 special right triangle. An right angled triangle has one right angle and two equal angles with two equal sides, it is called as an isosceles right angled triangle.The sides of the right isosceles triangle are in a ratio 1 : 1 : sqrt 2. Right Angled Triangle Right Angled Triangle is a triangle with one internal angle equal to 90 degrees (right angle).The other two sides adjacent to the right angle are called legs or catheti. Two equal sides. Scalene Triangle.A right triangle has a right angle. The longest side of the triangle is called the hypotenuse. A triangle in which two of its sides are equal is called isosceles triangle. PQ PR 6 cm.A triangle whose one angles is a right angle (that is 90) is called a right angled triangle or right angle. A special case is an isosceles right triangle that has two equal side lengths, where one angle is equal to 90 degrees and the two acute angles are each equal to 45 degrees. Figure 2. A right-angled triangle with angle B marked. The hypotenuse is as it was in Figure 2, but the other two labels have changed.What is the length of the two equal sides? 3. In a right angled triangle, one angle is 45oand the side next to this angle (not the hy-potenuse) has length 5 cm. A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90-degree angle). The relation between the sides and angles of a right triangle is the basis for trigonometry. Right-angled triangles Concept summary Practice questions. Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides, or. acute isosceles (diagram D) right isosceles (E) also known as a 45 45 90 triangle. obtuse isosceles (F). equilateral (G) all sides are equal and each angle 60, making this the only equiangular triangle. A right-angled triangle (also called a right triangle) is a triangle with a right angle (90) in it.Scalene right-angled triangle. One right angle Two other unequal angles No equal sides. Example: The 3,4,5 Triangle. Isosceles triangles always have two equal sides and two equal angles.A triangle with one 90 angle and two 45 angles is called a right-angled isosceles triangle! The biggest group are the irregular (or scalene) triangles. They do not have any special properties like a right angle or equal sides. These irregular triangles are divided into acute- angled triangles and into obtuse-angled triangles. A triangle is a polygon with 3 sides and 3 angles, the sum of whose measures equals 180.2. Isosceles triangles have 2 congruent sides and 2 congruent angles, 3. Right triangles have a right angle. PYTHAGOREAN TRIPLES. The simplest right angled triangle with sides of integer length is the 3-4-5 triangle.Find the length of one of its altitudes. 13 An isosceles triangle has equal sides of length 8 cm and a base of length 6 cm. Equilateral, Isosceles, Right Triangles. In any equilateral triangle, all sides are congruent and all angles are triangle is 1: 3 :2 and w e are given the sideFigure 2. A right-angled triangle with angle is an angle which has a cosine equal to 5. A triangle with no equal sides. A triangle with one 90 angle.Right Triangle. Equilateral triangle properties: 1) All sides are equal.The longest side of the right triangle(the side opposite the right angle) is called the hypotenuse(or hypothenuse) and the two short sides are legs. If a triangle has two angles equal to each other, the sides which subtend the equal angles will also be equal to one another. Hence, by definition, such a triangle will be isosceles. In the words of Euclid: If in a triangle two angles be equal to one another A triangle with two sides equal in size.A triangle with a right angle. c is the hypotenuse, a and b are the legs of the triangle. Application with two right-angled triangles. A cable 100 m long makes an angle of elevation of 41 with the top of a tower.
a Find the height h of the tower, to the nearest.(Remember, an equilateral triangle has all sides of equal length.) Side c. Angle A or B.This calculator is designed to give the two unknown factors in a right triangle, assuming two factors are known. Isosceles triangle 2 equal sides 2 equal angles.Right-angled triangle One of its angles is a right angle (90). Quadrilaterals Quadrilaterals have four sides. Here are some special quadrilaterals Right Angle Triangle. Triangles are those closed convex polygons having three sides.The longest side of the right triangle is called its Hypotenuse and the other two sides including the right angle are called as its legs or the base and the perpendicular. Area of a triangle - "side angle side" (SAS) method.Although it uses the trigonometry Sine function, it works on any triangle, not just right triangles. where a and b are the lengths of two sides of the triangle C is the included angle (the angle between the two known sides). Solving right triangles. This is a topic in traditional trigonometry. It does not come up in calculus. To SOLVE A TRIANGLE means to know all three sides and all three angles. When we know the ratios of the sides, we use the method of similar figures. Triangle with 2 equal sides 2 equal angles? isosceles.Does a right angle triangle have 2 equal sides? Not necessarily. A triangle with two equal sides is called an isoceles triangle. right-angled triangle ( plural right-angled triangles) File:Triangle. Right.svg A right-angled triangle.en The lowest (neutral scalar) mass has a value of about 44 GeV in all three cases it may be interpreted as the angle opposite w in a right triangle with the other two sides equal to the It has two equal sides, two equal angles, and one right angle.Method 2. Using the Pythagorean Theorem and the fact that the legs of this right triangle are equal Note: The above method is used for constructing a right angled triangle with 2 sides given. 6.4.3 Congruency of Triangles.Since two triangles have corresponding angles equal and common sides equal, by ASA postulate, these triangles are congruent. Triangle. Что значит a right-anglIts similar to triangle. это звучит нормально? Other types of questions. How do you express that triangle ABC and triangle DEF are congruent in mathematics?
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Post subject: A closed form of a hypergeometric seriesPosted: Wed Aug 16, 2017 12:09 pm
Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
The following result is new and is going to be published on Arxiv.org in the upcoming days with many more interesting results by Jacopo D' Aurizio who has actually proved it. Nevertheless , I am posting it here since it is interesting , challenging as well as approachable using only elementary tools.
Prove that
\begin{align*}
{}_4 F_3 \left ( 1, 1, 1 , \frac{3}{2} ; \frac{5}{2} , \frac{5}{2} , \frac{5}{2} ; 1 \right ) &= 27 \sum_{n=0}^{\infty} \frac{16^n}{\left ( 2n+3 \right )^3 \left ( 2n+1 \right )^2 \binom{2n}{n}^2} \\
&= \frac{27}{2} \bigg( 7 \zeta(3) + \left ( 3 - 2 \mathcal{G} \right ) \pi - 12 \bigg)
\end{align*}
where $\mathcal{G}$ denotes the Catalan's constant.
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Home / Area Conversion / Convert Township to Square Micrometer
# Convert Township to Square Micrometer
Please provide values below to convert township to square micrometer [µm^2], or vice versa.
From: township To: square micrometer
### Township to Square Micrometer Conversion Table
TownshipSquare Micrometer [µm^2]
0.01 township9.3239571972096E+17 µm^2
0.1 township9.3239571972096E+18 µm^2
1 township9.3239571972096E+19 µm^2
2 township1.8647914394419E+20 µm^2
3 township2.7971871591629E+20 µm^2
5 township4.6619785986048E+20 µm^2
10 township9.3239571972096E+20 µm^2
20 township1.8647914394419E+21 µm^2
50 township4.6619785986048E+21 µm^2
100 township9.3239571972096E+21 µm^2
1000 township9.3239571972096E+22 µm^2
### How to Convert Township to Square Micrometer
1 township = 9.3239571972096E+19 µm^2
1 µm^2 = 1.0725059959512E-20 township
Example: convert 15 township to µm^2:
15 township = 15 × 9.3239571972096E+19 µm^2 = 1.3985935795814E+21 µm^2
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### Flamire's blog
By Flamire, history, 4 weeks ago,
2002A — Distanced Coloring
idea & solution: xcyle
Hint 1
Hint 2
Tutorial
Solution
2002B — Removals Game
idea & solution: xcyle
Hint
Tutorial
Solution
2002C — Black Circles
idea: Flamire, solution: le0n
Hint
Tutorial
Solution
idea & solution: xcyle
Hint
Tutorial
Solution (Check 1)
Solution (Check 2, LipArcanjo)
2002E — Cosmic Rays
idea: le0n, solution: Flamire
Hint 1
Hint 2
Tutorial
Solution
Solution (priority_queue)
2002F1 — Court Blue (Easy Version)
idea: Flamire, solution: le0n
Hint 1
Hint 2
Tutorial
Solution
2002F2 — Court Blue (Hard Version)
idea: le0n, solution: xcyle
Hint
Hint (alternate version)
Tutorial
Solution
Solution (dfs)
2002G — Lattice Optimizing
idea & solution: xcyle
We apologize for unintended solutions passing, and intended solutions failing with large constants. Brute force runs very fast on $n=18$, which forced us to increase constraints.
Hint 1
Hint 2
Tutorial
Solution
Solution (trie, LipArcanjo)
2002H — Counting 101
idea: le0n, xcyle, solution: le0n, xcyle
Hint 1
Hint 2
Hint 3
Tutorial
Solution (orzdevinwang)
• +125
» 4 weeks ago, # | +15 Woah, I solved A,B,C all through guessing and became purple!
• » » 4 weeks ago, # ^ | 0 You really solved E first... :skull:
• » » » 4 weeks ago, # ^ | 0 I got stuck for an hour trying to prove the earlier problems XD
• » » 4 weeks ago, # ^ | +2 I created video editorial for E: Cosmic Rays.
» 4 weeks ago, # | -15 Why this solution of mine for C is giving WA on test5? 275844339
• » » 4 weeks ago, # ^ | +6 Check for integer overflow sqrt can cause precision loss, i recommend you eliminate it completely
• » » » 4 weeks ago, # ^ | 0 how will we check then?My approach : If after distance(d) time, if our destination points becomes part of any of the circle then the answer is NO otherwise YES.
• » » » » 4 weeks ago, # ^ | 0 you can just compare the squared distance, you end up squaring it again after taking the square root anyways
• » » » » » 4 weeks ago, # ^ | 0 thank you! it worked :)by the way, just want to understand how to identify such things?
• » » » » » 4 weeks ago, # ^ | 0 For me, I used long double in C++ and it worked.
» 4 weeks ago, # | ← Rev. 2 → +5 lol, you can solve D1 in $O(N*Q)$ with pragmas: 275797240Also you can use segment tree too for E: 275831791
• » » 4 weeks ago, # ^ | 0 I also tried spamming pragmas but I got TLE on pretest 13 :) https://mirror.codeforces.com/contest/2002/submission/275836926
• » » 4 weeks ago, # ^ | 0 diko hacked it. Thanks
• » » 4 weeks ago, # ^ | +8 I solved E using stack, I think E is much easier than D1, D2 took me only 15 mins to solve E. https://mirror.codeforces.com/contest/2002/submission/276392625
» 4 weeks ago, # | 0 Can someone try hacking my solution to E? I solve for each value separately and use RMQ to keep track of merging blocks, but I believe my code is O(N^2 log N) if one value appears lots of times, the first and last occurrence of that value have high a_i, and all intermediate occurrences have very low a_i. 275860315
» 4 weeks ago, # | -29 Only solve A, B.I am too weak. T^T
» 4 weeks ago, # | +1 In problem D2 check 1, can someone explain how the merge step works? ("merge the subtree of u into a large node with size siz_u") I don't understand why it is sufficient to just maintain "bad" children, instead of maintaining information for the entire subtree.
• » » 4 weeks ago, # ^ | +2 Let's note the following $pos_x$ : position of $x$ in $p$$sub_x$ subtree size of xFor D1 consider some node $x$ , we call node $x$ valid if the positions of its children in the permutation are ${pos_x +1 , pos_x + t + 1}$ where t is the subtree size of one of the children.A dfs order is valid iff all nodes are valid and $p_1 = 1$So we can maintain a set for bad nodes $bad$ When we swap two values $p_x , p_y$ it only affects $p_x , p_y$ and their parents (because $pos_{p_x}$ becomes $y$ and vice versa)So we can check easily in D1For D2 you should maintain a set {$pos_{child} , child$} for each node $x$ . Node $x$ is valid iff (*) for each two adjacent positions of children $c_i$ , $c_{i+1}$ in the set $pos_{c_{i+1}} - pos_{c_i} = sub_{c_i}$ and $pos_{c_1} - pos_x = 1$ Thus we can maintain also a set of bad children (who don't satisfy (*)) in each node And when we update we only remove and insert new values in the setFinally if the set of bad children of a node is empty than erase this node from $bad$ otherwise insert it.
• » » » 4 weeks ago, # ^ | 0 Thanks! While going through others' implementations I discovered that Um_nik's submission implements this idea nicely.
• » » 4 weeks ago, # ^ | ← Rev. 2 → 0 I suppose, this part is just about proving that subtree dfs log places next to the parent node dfs log. So there's no merging part when talking about solution.Actually, I don't understand how to keep tracking "bad" nodes with sets on D2. I would be glad if someone would explain this part. Can't understand author's code ideas.P.S.: Sorry, don't refresh the page too long , thaks for explaining!
» 4 weeks ago, # | 0 the image in c's tutorial is not visible
» 4 weeks ago, # | ← Rev. 3 → +1 Why this solution of mine for D1 is giving TLE on test9 and it's O(n*q)? My Solution
• » » 4 weeks ago, # ^ | +5 Because n*q is too big
• » » » 4 weeks ago, # ^ | 0 i thought that too, but i asked bcos i see people saying they solved it in O(n * q)
» 4 weeks ago, # | 0 I am really dumb .But Why CD = AD in tutorial of C . I don't understand thatcould anyone explain that pls.
• » » 4 weeks ago, # ^ | 0 I think that CD = AD represent the point where we intersect with a circle since our starting point A has the same distance to D as does the circle's center C.
• » » 4 weeks ago, # ^ | 0 understand it in a way that the speed of expansion of circles is same as walking speed of the guy and thus the circles radius at any instant would be same as the distance covered by the guy till that instant which implies AD=CD
» 4 weeks ago, # | ← Rev. 5 → +3 Alternative (possibly wrong?) solution for D2: Firstly, check that for each $i$ from $2$ to $n$ $p_{i - 1}$ is the parent of $p_i$ except the case when $p_{i - 1}$ is a leaf. Also check that for each $i$ the position of $i$ in $p$ is greater than the position of its parent. Check that the multiset $S = \{ LCA(p_1, p_2), LCA(p_2, p_3), \cdots, LCA(p_{n - 1}, p_n) \}$ matches with such multiset of any valid DFS order (intuition: virtual trees). It can be easily checked by maintaining the multiset hash. Here is my implementation: 275842859. Feel free to uphack it.UPD. It seems that the second condition is unnecessary: 275927827. Now my solution looks similar to the Check 2 in the editorial as songhongyi pointed below (thanks for it). I think the first condition is a "weaker" version of Check 2 and the third condition makes my solution work somehow by making the first one "stronger". I still don't know how to prove it though.UPD 2. Actually the second condition is necessary but the first one isn't. Now it seems less similar to Check 2.
• » » 4 weeks ago, # ^ | ← Rev. 2 → +8 This solution is very similar to check2. I suspect it's correct and should be provable in a similar way. Factle0n failed to hack it so it must be true.
• » » 4 weeks ago, # ^ | +8 The check2 is actually equivalent to $p_1=1$ and $\operatorname{LCA}(p_i,p_{i+1})=fa(p_{i+1})$ forall $1\le i\le n$. Your third condition is actually $\{ {\operatorname{LCA}(p_i,p_{i+1})} \} =\{fa(p_{i+1})\}$, which is obviously weaker than check2. But I'm sure it is equivalent to check2 with your second condition.
» 4 weeks ago, # | 0 Could anyone write out the proof by induction in B? I'm not sure how to prove it...
• » » 4 weeks ago, # ^ | +5 I didn't really understand the subarray thing in the editorial. The way I see it is:Suppose Bob can mimic Alice's move when there are $n$ elements. There are only two ways for this to happen (let $A = a_1 \dots a_n$, $B = b_1 \dots b_n$): We have $a_1 = b_1$ and $a_n = b_n$. If Alice takes $a_1$ Bob can take the same element, i.e., $b_1$. If Alice takes $a_n$ Bob can take $b_n$. We have $a_1 = b_n$ and $a_n = b_1$. If Alice takes $a_1$ Bob can take the same element, i.e., $b_n$. If Alice takes $a_n$ Bob can take $b_1$. If any of the above cases happen, nothing has changed in the game and we keep going, now with $n-1$ elements.If at any moment none of the cases above match, it means Alice has at least one endpoint element which Bob does not (let it be $x$). Bob cannot pick $x$ right after Alice, so he takes any one of his endpoint elements (let it be $y \neq x$). Now Alice can easily win by picking all her left elements except $y$, because at the end of the game, Alice will have $y$ left but Bob already deleted it, so the last elements cannot be equal.If the game keeps going with one of the two cases described initially, Bob can always mimic Alice's move, so Alice cannot win. These two cases can only happen when the whole arrays $A$ and $B$ are equal or one is the reverse of the other.
• » » » 4 weeks ago, # ^ | 0 Thank you so much!
• » » 4 weeks ago, # ^ | ← Rev. 2 → 0 I wish they had elaborated why the interval condition was "intuitive to see".Here is my proof. If there is a pair of neighbors in array A that are not neighbors in B then Alice wins. She just needs to remove the elements until only those two elements remain. Then after Bob's move his remaining two elements would not be the same as the two Alice's elements, and on the following move Alice can leave an element that Bob does not have. So, for Bob to win every pair of neighbors in A need to be neighbors in B. It is easy to show that it can only happen if B is equal to A or its reverse, depending on the order in B of the first two elements of A.
» 4 weeks ago, # | +14 What is $fa$ mentioned in Problem-D Check 2? Ref$fa(p_{i+1})$ must be ancestor of $p_i$.
• » » 4 weeks ago, # ^ | +2 the father(parent) of $p_{i+1}$ on the tree.
» 4 weeks ago, # | ← Rev. 2 → +6 My Insights for A,B,C AObserved that answer is $\min(n,k) \cdot \min(m,k)$ Cases ObservationInput Output ----- ------ 3 3 2 4 =(2*2)=k*k 5 1 10000 5 =(1*5)=n*m 7 3 4 12=(4*3)=m*k 3 2 7 6=(2*3)=n*m 8 9 6 36=(6*6)=k*k 2 5 4 8=(2*4)=n*k Total complexity is $\mathcal{O}(1)$ Codefrom math import * def solve(): n,m,k=map(int,input().split()) print(min(n,k)*min(m,k)) for _ in range(int(input())): solve() BSince We're performing optimally we'll keep removing from left or right thus $\displaystyle [ \color{red}{a_1,a_2,.....,} \color{blue}{a_{n/2}} \color{red}{,....,a_n} \color{black}]$ $\displaystyle [ \color{red}{b_1,b_2,.....,} \color{blue}{b_{n/2}} \color{red}{,....,b_n} \color{black}]$Thus Bob can win only and only if $a$ is the same as $b$ or $a=b^{-1}$ i.e. $b$ reversed.Total complexity is $\mathcal{O}(n)$ Codefrom math import * def solve(): n=int(input()) a=list(map(int,input().split())) b=list(map(int,input().split())) if(a==b or a==b[::-1]): print("Bob") else: print("Alice") for _ in range(int(input())): solve() CThe $mindist$ that is possible is $\min_{i=1}^n \sqrt{x_i^2+y_i^2}$ overall points , and the distance needed to connect $(x_s,y_s)$ and $(x_t,y_t)$ is $\sqrt{(x_s-x_t)^2+(y_s-y_t)^2}$ , let's call it $d$ , we must have $\text{mindist} \le d$ so that it's possible to connect $(x_s,y_s)$ and $(x_t,y_t)$ without intersection of circles.Don't use square root for avoiding square root errors , it's enough to store $(\Delta x)^2+(\Delta y)^2$Total complexity is $\mathcal{O}(n)$ Codefrom math import * def solve(): md=float('inf') n=int(input()) p=[] for i in range(n): x,y=map(int,input().split()) p.append([x,y]) x1,y1,x2,y2=map(int,input().split()) for a,b in p: d=(a-x2)**2+(b-y2)**2 md=min(md,d) sd=(x1-x2)**2+(y1-y2)**2 if(md<=sd): print("NO") else: print("YES") for _ in range(int(input())): solve() Fun FactI didn't solve any of BC without guessing (I proved them) ; If I guessed them then specialist was in hand for real.
• » » 4 weeks ago, # ^ | 0 How did you come up with the conclusion of A? I turn to solve B/C instead.
• » » » 4 weeks ago, # ^ | 0 I tried to come up with something related about multiplication of numbers You can see the cases like this Cases ObservationInput Output ----- ------ 3 3 2 4 =(2*2)=k*k 5 1 10000 5 =(1*5)=n*m 7 3 4 12=(4*3)=m*k 3 2 7 6=(2*3)=n*m 8 9 6 36=(6*6)=k*k 2 5 4 8=(2*4)=n*k I think the observing the thing described in the spoiler is enough to come up with $\displaystyle \min(n,k) \times \min(m,k)$
• » » » » 4 weeks ago, # ^ | 0 smart!
» 4 weeks ago, # | ← Rev. 4 → 0 My idea for D1 was that it looks like a segment tree: Store in each node if that subtree makes sense or not, and the maximum index in that subtree, and of course the index of the node in the permutation. Then on update query, it takes at most O(log n) updates like a segment tree, by moving up and calling combine function on the 2 children of the current node until reaching root. Then the root has the answer, by just checking if it makes sense. Combine function: to get maximum index just max the maximum index of both children and your index, to see if it makes sense one of the children's index must be 1 more than the node's index, and then the other's index must be 1 more than the maximum index of the first child so it forms a dfs order that makes sense, and of course both subtrees must make sense (AND them together). Time complexity: O(n + q log n)
» 4 weeks ago, # | +4 In D check 1 the editorial specifies this condition: For every $u$, all of its children $v$ satisfy $[pos_v, pos_v+siz_v-1] \in [pos_u, pos_u+siz_u-1]$ In the code solution, a different, easier check is made: int chk(int x) { return son[x].empty() ? 1 : (q[x] < *son[x].begin() && *--son[x].end() + siz[p[*--son[x].end()]] <= q[x] + siz[x]); } Here son[x] includes ordered indices of $x$'s children in the permutation. Therefore, it is only checking the editorial condition for the furthest child in the permutation, and that the closest child in the permutation has an index not less than the one $x$ itself has. Can someone explain how these conditions are equivalent to the editorial? In fact I tried to check the editorial condition directly in my code but got TLE probably due to constant factor.
• » » 4 weeks ago, # ^ | 0 editorial should have cared to explain that. Nevertheless, here we go. Lets prove it for any node $u$. Assume that condition is satisfied for all immediate children $v$ of $u$. it means, for any immediate child $v$ of $u$, the range $[posv,posv+sizv−1]$ will contain whole subtree of $v$, and nothing else. So, the point is, for any two immediate child $v1$ and $v2$, their range is non intersecting. ($[posv1,posv1+sizv1−1]$ & $[posv2,posv2+sizv2−1]$). If the ranges are non intersecting, then we can just check the first and last range, and if they are contained within $[posu,posu+sizu−1]$, then all other are forced to contain within it.
• » » » 4 weeks ago, # ^ | 0 Please correct me if I'm wrong, but I think you're only proving one side of the equivalence, namely editorial check implies code check. In fact we would be more interested in the other implication ( code check implies editorial check ), since the code check at first glance looks like a weaker condition. That is, for a single node it doesn't make much sense for the checks to be equivalent, but rather the fact that this weaker check works for all nodes at the same time might be what actually makes it equivalent.
• » » » » 4 weeks ago, # ^ | +3 May be, I should have also cared to explain more., I didnt prove editorial check implies code check, that is noobest thing i can comment. what I have proved above?Given that code check is satisfied for any $u$, and all nodes in its subtree, then it implies that Exactly all nodes in subtree of u are within [posu,posu+sizu−1], and nothing else. And this is not just for u, but also for all nodes in subtree of u. Which also equivalent to editorial check. How I have proved that?By induction. Lets say code check is satisfied for all nodes in subtree of $u$, except $u$, we didnt check for $u$ yet. So we assume that for each immediate child $v$ of $u$, all nodes in their subtree are within $[posv,posv+sizv−1]$, and nothing else. which means each of those ranges by immediate child of $u$ are non intersecting and continous. This is the key part. Now we are coming for $u$, Say $vl$ as leftmost position having $v$ out of all immediate child of $u$ and $vr$ as rightmost. And if $posvl$ and $posvr+sizvr-1$ are within $[posu,posu+sizu−1]$, then it indirectly implies that all other $[posvm,posvm+sizvm−1]$ are also within the $u$ range, as $posvl$ <= Unable to parse markup [type=CF_MATHJAX] <= $posvr$ and $posvm+sizvm-1$ cant intersect and cross the $vr$. Now, due to size constraint, if all $v$ ranges are within $[posu,posu+sizu−1]$, it means they are completely filling the range $[posu,posu+sizu−1]$ and nothing else can be inside it.By, induction this will propagate until root node. editorial checkchecking for all immediate child $v$ of $u$, whether $[posv,posv+sizv−1]$ contained within $[posu,posu+sizu−1]$ code checkchecking only for first and last $v$.
• » » » » » 4 weeks ago, # ^ | +3 Ok I guess I didn't understand the induction right. I got it now. Thank you!
» 4 weeks ago, # | -8 How is F2 harder than F1? I don't see any difference, only tighter time limits maybe. Can someone explain why F2 is considered harder than F1?
» 4 weeks ago, # | ← Rev. 3 → 0 Quoting editorial for D2: Then, for each pair of adjacent elements $p_i,p_{i+1}$, $fa(p_{i+1})$ must be an ancestor of $p_i$ Can you tell me what does the notion $fa(p_{i+1})$ mean? I don't think I have seen it being declared anywhere in the tutorial. Thank you in advance.
• » » 4 weeks ago, # ^ | +4 It means the father of $p_{i+1}$. Apparently it's not as widespread as I thought. I've added an explanation.
• » » » 4 weeks ago, # ^ | 0 I see, thanks.
» 4 weeks ago, # | 0 Can anybody explain the proof for C? I get the part where CD = AD and the CD > CB-DB part, but everything else kinda just falls apart... I've forgotten everything about proofs from geometry class
• » » 3 weeks ago, # ^ | 0 I can't understand this proof either, but if you look at the conclusion directly, it will be obvious that if the distance from the center of the circle to the target is less than or equal to the distance from the starting point to the target, it is obvious that the boundary of the circle will be touched before (and when you reach this point).
» 4 weeks ago, # | 0 Solved ABCD1 and became specialist.
» 4 weeks ago, # | ← Rev. 3 → 0 what does this line mean in editorial of $H$? Let b_i be the number of operations that has the element equal to v after block y_i as its center I couldn't understand the rest of the editorial because of that
» 4 weeks ago, # | 0 WOW!You are very good! But could you give me some exegesis in the H code?
» 4 weeks ago, # | +3 why was E placed after D?
• » » 4 weeks ago, # ^ | 0 https://mirror.codeforces.com/blog/entry/132346?#comment-1182297D1 is much easier than E, D2 is also not much harder imo but D1 is sufficient reason anyways
• » » » 4 weeks ago, # ^ | 0 but shouldn't the point sums be nondecreasing? just by convention maybe there's definitely been instances where F1 is easier than E and such
» 4 weeks ago, # | ← Rev. 3 → +1 for the following test case for D2, 1 7 2 1 1 7 2 3 3 1 3 7 4 6 2 5 3 5 5 3 check 1 outputs NO NO but check 2 (and other ACs) output NO YES and i think it should be NO YES. the way siz is calculated in check 1 makes siz[1] = 6 and siz[3] = 3 but shouldn't they be siz[1] = 7 and siz[3] = 4?Edit: nvm the constraint ai < i is not satisfied for this tree so it's a wrong tc
• » » 4 weeks ago, # ^ | 0 $1 \le a_i \lt i$, but in your case $a_3 = 7$ so it's invalid.
» 4 weeks ago, # | 0 Solution 3 in problem F2 is interesting. Despite $L = 50$ is already hacked, higher $L$ should still yield an AC (with the cost of praying that one's code wouldn't TLE).A loose upperbound by me, with $L = 125$: 276065570I wonder how far could the uphack raise up "cheeseable" lowerbound value for $L$.
» 4 weeks ago, # | 0 Could anyone explain how we are supposed to preprocess GCD as mentioned in F1 solution?
• » » 3 weeks ago, # ^ | 0 I didn't preprocess GCD, my solution got passed
• » » » 3 weeks ago, # ^ | +8 My F1 also passed using the normal one. But for F2, I had to use custom binary gcd function (which I copied from one of the CF blogs). That's why I asked if it was possible to preprocess the gcd and then find them in constant time. It will be a huge optimization, nearly 20 times for the given constraints.
» 4 weeks ago, # | +14 For problem D, the following check is also sufficient: for every $i$ : $lca(p[i-1], p[i]) = parent(p[i])$
• » » 4 weeks ago, # ^ | +12 It's just check2 written in another form.
• » » 4 weeks ago, # ^ | +3 Hey, can you explain me how can I use segment tree to solve the problem D2. The authors solution mentions that we can impose some check on each node in the permutation.Specifically, For every u, all of its children v satisfies $[pos_v, pos_v + siz_v-1] \subseteq [pos_u, pos_u + siz_u-1]$And, we can maintain this check by keeping track of the number of u which violates this condition, and check for each u using sets.First of all, how can we do this using sets, and how can we do this using segment tree. I tried thinking a lot, but I could only think of maintaining a segment tree using dfs entry time. So each segment node in the segment tree contains a set/vector containing entry times for each tree node in this segment. And for swapping part, I could do it like replacing the entry time for swapped nodes in the set of their respective segments.Then how can I perform verifying this check for each node using this segment tree? This is where I'm getting stuck. Can you help me out, Please.
• » » » 4 weeks ago, # ^ | 0 Editorial should have mentioned that, they compare only smallest and largest $posv$ to see if they are within $[posu,posu+sizu−1]$, rather than all its childs. see: https://mirror.codeforces.com/blog/entry/132569?#comment-1182635
• » » » » 4 weeks ago, # ^ | 0 Yeah, I get it that checking the first and last child range is equivalent to checking all of children of u. Because the set stores the position of each node in sorted order.And we can easily perform swapping by first deleting the position of a node from its parent's children set and then add the swapped node position to it, and again the set would maintain all the positions in sorted order. The number of good nodes are decreased for now, as we need to check again after swap if they become good nodes.Now after swap, we only need to check for the (parent(a), a) and (a, children(a)) for all children of a. Similarly for (parent(b), b) and (b, children(b)) for all the children of b, for the condition of a good node. Because we want to see if after swap, the node is still contained in its parent range, and the node still contains all of its children within its range. If yes, then the number of good nodes increases by one each time for all these 4 pairs, otherwise not. And after each query, we check if all n nodes are good nodes or not. And we print the answer accordingly.
» 4 weeks ago, # | ← Rev. 2 → 0 Can someone explain why this solution for D is timing out? 276371875I am trying to implement Um_nik's idea from his submission: 275817851
• » » 4 weeks ago, # ^ | 0 Interesting, submitting your solution in GCC 7-32 yielded TLE while GCC 13-64 yielded RTE.Huge red flag of undefined behaviors right here, though I can't yet tell where it was.
» 3 weeks ago, # | 0 Alternative (maybe easier) solution for D : let s[x] = {{j1,u1}, {j2,u2},... } be the set of indices and node id of the children of the node x, pos[x] the position of node x in P, and siz[x] the size of the subtree of node x.It suffices to check that for all i from 1 to n : pos[x] = min(s[x]) + 1 , and for all i from 1, s[x].size() — 1 : Ji+1 — Ji == siz[Ui] meaning the positions of the children of x when sorted should have the difference of siz[u] where u is the node with smaller pos.
» 10 days ago, # | 0 Ask a question to the time complexity of G.this formula $2^{2B}+2^{2-B}$but you use some calculus you know B=1/3 it's minimum. not B=2/3
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Martin
TU Wien
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Symmetry of vector
Determine whether the vector is symmetric or not (vector could be even or odd in length). For example: x = [1 2 3 3 2 1] ...
meer dan 2 jaar ago
Solved
Change the sign of even index entries of the reversed vector
change the signs of the even index entries of the reversed vector example 1 vec = [4 -1 -2 9] ans = [9 2 -1 -4] example2...
meer dan 2 jaar ago
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https://www.merlot.org/merlot/materials.htm?page=3&hasAwards=false&hasComments=false&hasCourses=false&filterTypesOpen=false&dateRange=0&hasEtextReviews=false&hasPeerReviews=false&fromContentBuilderSawDialog=false&isLeadershipLibrary=false&hasCollections=false&filterOtherOpen=false&modifiedDays=7&isContentBuilder=false&filterSubjectsOpen=true&hasAccessibilityForm=false&hasAssignments=false&filterPartnerAffiliationsOpen=true&hasRatings=false&hasSercActivitySheets=false&days=7&filterMobileOpen=false&category=2566&modifiedDateRange=0&hasEditorReviews=false
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# MERLOT Materials
#### Filter by
49-72 of 222 results for: MERLOT Materials
#### Apollonius App for iOS
'Apollonius is the first (and so far the only) Interactive Geometry Software (IGS) for the iPhone and iPod Touch. It... see more
#### Area of 2D Figures
Interactive presentation for middle and high school students forming connections between the areas of two-dimensional... see more
#### Area of Regular Polygons
The goal for this lesson is to provide students with an understanding of how to find the area of any regular polygon.... see more
#### Centroid
Visual demonstration of existence of centroid of a triangle.
#### Combining Vectors Using Right... Combining Vectors Using Right Triangles
Course material for contextualized math for CCCS COETC grant funded course development
#### Common Tangents of 3 Circles
Illustrates the relationship between exterior common tangents of 3 cirlces.
#### Comparison Geometry
This is a free, online textbook that is comprised of articles from a variety of authors. "Comparison Geometry asks: What... see more
#### Course of Differential Geometry
This is a free, online textbook that is designed "for the basic course of differential geometry. It is recommended as an... see more
#### Cross ratios and harmonic conjugate... Cross ratios and harmonic conjugate elements
A tutorial on cross ratios and harmonic conjugate elements.
#### Curved Spaces
Curved Spaces is a flight simulator for multiconnected universes. Such toy universes are the 3D analog of the torus and... see more
#### Curves Visualization (within... Curves Visualization (within GRAPHICA)
Visualization of different models of curves in computer graphics, while requiring learners to construct specific shapes... see more
#### Diamond Theory
Symmetry properties of the 4x4 array. The invariance of symmetry displayed in the author's Diamond 16 Puzzle (online)... see more
#### Euclid, "The Elements," Books I-IV
A clear, graphical walk-through of Euclid's "Elements", books I-IV. The site also includes explanations of the... see more
#### Flatland: A Romance of Many... Flatland: A Romance of Many Dimensions
This is Edwin Abbott's novel about a two-dimensional being who visits one-dimensional and three-dimensional worlds.... see more
#### Fractals Simulators
A series of applets for teaching Fractal Geometry. Includes: L-Systems; Box-Counting Fractal Dimension; Cellular... see more
#### Gallery of Interactive Geometry
The site offers a variety of interactive applications designed to teach various aspects of geometry. Some of the choices... see more
#### Geometry (Mathomatix) App for iOS
"Punflay’s Mathomatix series is back! Kindergarteners will enjoy learning about basic geometrical shapes such as... see more
#### Geometry Gallery
A collection of geometry applets.
#### Hypercube
Visualizing a hypercube, one step at a time.
#### Interactive Airplane Landing -... Interactive Airplane Landing - Application of Trigonometry
Use principles of trigonometry to calculate the correct angle of decent for an aircraft during an emergency landing on an... see more
#### Interactive mathematics on the... Interactive mathematics on the internet
This is a collection of computational and graphical applets as well as on-line exercises.
#### Intro to Vectors
Course material for contextualized math for CCCS COETC grant funded course development
#### Isosceles Triangle Theorem
Illustrates an isosceles triangle theorem.
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https://metric2011.wordpress.com/2014/10/15/notes-of-damien-gaboriaus-lecture-nr-1/
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## Notes of Damien Gaboriau’s lecture nr 1
${\ell^2}$ Betti numbers
Started with M. Atiyah (1976) for manifolds. Very soon (1979), Connes obtained a version for foliations. In 1986, Cheeger and Gromov extended the notion to arbitrary countable groups.
1. Von Neumann dimension
Murray-von Neumann 1944. Let ${V}$ be a closed ${\Gamma}$-invariant subspace in ${\bigoplus_{i=1}^k \ell^2(\Gamma)}$. A real number ${dim_\Gamma(V)}$ can be defined in such of way that
1. ${dim_\Gamma(V)=0\Leftrightarrow V=0}$.
2. ${dim_\Gamma(\ell^2(\Gamma))=1}$.
3. If ${f:V\rightarrow W}$ is bounded and equivariant, ${dim_\Gamma(V)=dim_\Gamma(ker(f))+dim_\Gamma(im(f))}$.
1.1. ${\Gamma}$-trace
Let ${a}$ be an operator which commutes with the left regular representation. Express it in the canonical basis ${\{1_g\,;\,g\in\Gamma\}}$. The diagonal elements ${\langle a1_g,1_g\rangle}$ are all equal. Denote this value by ${\tau(a)}$. Check that ${\tau(ab)=\tau(ba)}$.
1.2. Case ${V\subset\ell^2(\Gamma)}$
Orthogonal projection onto ${V}$ is an element ${p}$ which commutes with the left regular representation. So define ${dim_\Gamma(V)=\tau(p)}$.
Since ${p}$ is an orthogonal projector,
$\displaystyle \langle p1_e,1_e\rangle=\langle p^2 1_e,1_e\rangle=\langle p1_e,p1_e\rangle=|p1_e|^2.$
Therefore ${dim_\Gamma(V)=0}$ implies ${p1_e=0}$, and thus ${p=0}$, ${V=0}$.
1.3. Exercise
If ${\Gamma={\mathbb Z}}$, Fourier transform maps ${\ell^2({\mathbb Z})}$ to ${L^2(S^1)}$, the ${{\mathbb Z}}$ action is by multiplication with function ${z^k}$. ${{\mathbb Z}}$-equivariant operators are multiplication operators with bounded functions ${M_h}$. ${1_e}$ corresponds to constant function 1. ${\langle M_h 1_e,1_e\rangle=\int_{S^1}h(x)\,dx}$. Orthogonal projections are indicator functions of Borel sets ${B}$, invariant subspaces are of the form ${L^2(B)\subset L^2(S^1)}$, ${\tau(1_B)=}$ measure of ${B}$.
1.4. General case
View projector on ${\bigoplus_{i=1}^k \ell^2(\Gamma)}$ as a matrix in block form. Diagonal blocks turn out to be ${\Gamma}$-equivariant. Define ${Trace (p)=\sum_{i=1}^k \tau(p_{ii})}$.
1.5. Homology
Let ${L}$ be a countable simplicial complex with free cocompact ${\Gamma}$-action. ${\ell^2}$ chains make sense, therefore ${\ell^2}$-homology is defined. Reduced ${\ell^2}$-homology is isomorphic to the subspace of ${\ell^2}$ harmonic ${n}$-chains ${\mathcal{H}^{(2)}_n}$. This can be viewed as a ${\Gamma}$-invariant subspace in finitely many copies of ${\ell^2(\Gamma)}$. Define
$\displaystyle \begin{array}{rcl} \beta_n(L,\Gamma)=dim_\Gamma(\mathcal{H}^{(2)}_n). \end{array}$
Observe that
$\displaystyle \begin{array}{rcl} \beta_n(L,\Gamma)=dim_\Gamma(ker(\partial_n))-dim_\Gamma(im((\partial_{n-1})). \end{array}$
Proposition 1 For all ${p}$-connected ${L}$, the ${\ell^2}$-Betti numbers are the same up to ${p}$. This defines ${\beta_n(\Gamma)}$.
1.6. Cheeger-Gromov’s definition
For a general group ${\Gamma}$, an infinite dimensional contractible simplicial complex ${L}$ (with free action of ${\Gamma}$) may be needed. Exhaust it with co-finite ones, and take a limit of ${dim_\Gamma(im (H_n(L_j)\rightarrow H_n(L_{j+1})))}$. Does not depend on exhaustion.
2. Some values of ${\ell^2}$ Betti numbers
Finite groups: all 0 but the first, ${\beta_0=\frac{1}{|\Gamma|}}$.
Infinite groups: ${\beta_0=0}$.
Free group ${\mathbf{F}_n}$: ${0,n-1,0,\ldots}$.
${\Gamma}$ generated by ${g}$ elements: ${\beta_1(\Gamma)\leq g}$.
${\Gamma}$ infinite amenable: ${0,0,\ldots}$ (I will prove this).
Surface group of genus ${g}$: ${0,2g-2,0,0,\ldots}$.
Lattices in ${U(p,q)}$, ${Sp(p,q)}$,…: exactly one nonzero ${\ell^2}$ Betti number.
2.1. Some results
Passing to finite index subgroup multiplies ${\ell^2}$ Betti numbers by index.
Euler characteristic ${\chi(\Gamma)=\sum(-1)^n\beta_n(\Gamma)}$.
Lück’s approximation theorem: see lecture nr 3.
2.2. Atiyah’s conjecture
Atiyah: if torsion free, are ${\ell^2}$ Betti numbers integers ?
The generalized form (under restrictions on torsion) has been disproved, at least for actions. Grigorchuk-Zuk calculation for the lamplighter group.
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https://www.numbersaplenty.com/119904624
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Search a number
119904624 = 2434713217
BaseRepresentation
bin1110010010110…
…01100101110000
322100121210020000
413021121211300
5221143421444
615521550000
72654113110
oct711314560
9270553200
10119904624
1161757134
12341a5300
131bac26a8
1411cd3040
15a7d7469
hex7259970
119904624 has 100 divisors (see below), whose sum is σ = 396645744. Its totient is φ = 34255872.
The previous prime is 119904619. The next prime is 119904677. The reversal of 119904624 is 426409911.
It is a happy number.
119904624 is a `hidden beast` number, since 1 + 199 + 0 + 462 + 4 = 666.
It is a Harshad number since it is a multiple of its sum of digits (36).
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 19 ways as a sum of consecutive naturals, for example, 2464 + ... + 15680.
Almost surely, 2119904624 is an apocalyptic number.
119904624 is a gapful number since it is divisible by the number (14) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 119904624, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (198322872).
119904624 is an abundant number, since it is smaller than the sum of its proper divisors (276741120).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
119904624 is a wasteful number, since it uses less digits than its factorization.
119904624 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 13244 (or 13229 counting only the distinct ones).
The product of its (nonzero) digits is 15552, while the sum is 36.
The square root of 119904624 is about 10950.0969858719. The cubic root of 119904624 is about 493.1117038732. Note that the first 3 decimals are identical.
The spelling of 119904624 in words is "one hundred nineteen million, nine hundred four thousand, six hundred twenty-four".
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https://www.xpmath.com/forums/arcade.php?s=748bac68776d06df0a4177d5fa0dfea7&categoryid=4
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Publish
The topic you requested is included in another documentation set. For convenience, it's displayed below. Choose Switch to see the topic in its original location.
# Decimal.ToSingle Method
.NET Framework 4.5
Converts the value of the specified Decimal to the equivalent single-precision floating-point number.
Namespace: System
Assembly: mscorlib (in mscorlib.dll)
## Syntax
```public static float ToSingle(
decimal d
)
```
#### Parameters
d
Type: System.Decimal
The decimal number to convert.
#### Return Value
Type: System.Single
A single-precision floating-point number equivalent to the value of d.
## Remarks
This operation can produce round-off errors because a single-precision floating-point number has fewer significant digits than a Decimal.
## Examples
The following code example converts Decimal numbers to Single values using the ToSingle method.
```// Example of the decimal.ToSingle and decimal.ToDouble methods.
using System;
class DecimalToSgl_DblDemo
{
static string formatter = "{0,30}{1,17}{2,23}";
// Convert the decimal argument; no exceptions are thrown.
public static void DecimalToSgl_Dbl( decimal argument )
{
object SingleValue;
object DoubleValue;
// Convert the argument to a float value.
SingleValue = decimal.ToSingle( argument );
// Convert the argument to a double value.
DoubleValue = decimal.ToDouble( argument );
Console.WriteLine( formatter, argument,
SingleValue, DoubleValue );
}
public static void Main( )
{
Console.WriteLine( "This example of the \n" +
" decimal.ToSingle( decimal ) and \n" +
" decimal.ToDouble( decimal ) \nmethods " +
"generates the following output. It \ndisplays " +
"several converted decimal values.\n" );
Console.WriteLine( formatter, "decimal argument",
"float", "double" );
Console.WriteLine( formatter, "----------------",
"-----", "------" );
// Convert decimal values and display the results.
DecimalToSgl_Dbl( 0.0000000000000000000000000001M );
DecimalToSgl_Dbl( 0.0000000000123456789123456789M );
DecimalToSgl_Dbl( 123M );
DecimalToSgl_Dbl( new decimal( 123000000, 0, 0, false, 6 ) );
DecimalToSgl_Dbl( 123456789.123456789M );
DecimalToSgl_Dbl( 123456789123456789123456789M );
DecimalToSgl_Dbl( decimal.MinValue );
DecimalToSgl_Dbl( decimal.MaxValue );
}
}
/*
This example of the
decimal.ToSingle( decimal ) and
decimal.ToDouble( decimal )
methods generates the following output. It
displays several converted decimal values.
decimal argument float double
---------------- ----- ------
0.0000000000000000000000000001 1E-28 1E-28
0.0000000000123456789123456789 1.234568E-11 1.23456789123457E-11
123 123 123
123.000000 123 123
123456789.123456789 1.234568E+08 123456789.123457
123456789123456789123456789 1.234568E+26 1.23456789123457E+26
-79228162514264337593543950335 -7.922816E+28 -7.92281625142643E+28
79228162514264337593543950335 7.922816E+28 7.92281625142643E+28
*/
```
## Version Information
#### .NET Framework
Supported in: 4.5.2, 4.5.1, 4.5, 4, 3.5, 3.0, 2.0, 1.1, 1.0
#### .NET Framework Client Profile
Supported in: 4, 3.5 SP1
#### Portable Class Library
Supported in: Portable Class Library
#### .NET for Windows Store apps
Supported in: Windows 8
#### .NET for Windows Phone apps
Supported in: Windows Phone 8.1, Windows Phone 8, Silverlight 8.1
## Platforms
Windows Phone 8.1, Windows Phone 8, Windows 8.1, Windows Server 2012 R2, Windows 8, Windows Server 2012, Windows 7, Windows Vista SP2, Windows Server 2008 (Server Core Role not supported), Windows Server 2008 R2 (Server Core Role supported with SP1 or later; Itanium not supported)
The .NET Framework does not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.
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https://learn.careers360.com/school/question-explain-solution-rd-sharma-class-12-chapter-12-derivative-as-a-rate-measurer-exercise-multiple-choise-question-12-maths/?question_number=12.0
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#### Explain solution RD Sharma class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 12 maths
$B.\; \; \frac{3}{16\pi }\; cm/sec$$B.\; \; \frac{3}{16\pi }\; cm/sec$
Hint:
Here we use formula of a sphere of radius r is defined by
\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad.....(i) \end{aligned}
Given:
\begin{aligned} &r=2 c m\; \; \frac{d V}{d t}=3 \mathrm{~cm}^{2} / \mathrm{sec} \end{aligned}
Solution:
→ Differentiating (i) with respect to t, we get
\begin{aligned} &\frac{d V}{d t}=\frac{4}{3} \pi r^{2} \times \frac{d r}{d t} \\ &3=4 \pi \times 2^{2} \times \frac{d r}{d t} \\ &\frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}
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https://what.thedailywtf.com/topic/6425/this-is-how-we-get-tdwtf-style-code
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# This is how we get TDWTF style code.
• Right, so I finally stopped procrastinating my programming homework, and opened up the book to find this snippet of example code (with minor syntax fixes applied, since the purpose of the corresponding exercise was to fix those):
`void taxCalculations(num income) num count, tax const num NUMBRKTS = 5 num brackets[NUMBRKTS] = 15000, 22000, 40000, 70000, 100000 num rates[NUMBRKTS] = 0, 0.15, 0.18, 0.22, 0.28, 0.30 while count < NUMBRKTS and income > brackets[count] count = count + 1 endwhile tax = income * rates[count]return tax`
I'm still not sure exactly how that while loop works, but I translated it into Python and it appears to have the intended effect. Other WTFs, most present throughout the book: using completely ridiculous names for constants complying with an arbitrary law restricting variable names to one vowel; blithely ignoring the (important) difference between int and float; reusing the counter variable of a loop outside of that loop; and camel case (I know, not really a WTF but I personally can't stand it). (Side note: Java's println() is cited as an example of camel case. Fail.)
And these students (all 6 of us, the rest dropped the class) will someday be allowed to actually touch code?
• It's not really that bad. I imagine the original code were javascript or a similary language which only have floatingpoints.
The purpose of the loop is to find out which bracket the given income is in. When the loop terminates count is the correct bracket. No magic there, and quite easy to understand. But yes, count is a bad name for that variable, should have been userBracket or something similary. (Well, it really should have been its own function, but I disgress).
ps: I think you miss a count=0; before the loop.
• I understand what the loop does, I just had to stare at it way too long to figure out how it does it. I would have broken it out into a set of if statements. Just as efficient as a loop and a lot clearer (well, at the expense of a few bytes of space.)
(And the book is semi-written for C++ - there's a smaller book that goes with it containing C++ code for all this that I've been too {scared,lazy} to open until this point. It doesn't seem as bad as the pseudocode, but still contains minor WTFs such as a constant called NUM_LOOPS (loops != iterations!) and saying the string "Chicago" (as a std::string) will take up exactly 7 bytes of memory (but "may be different on your system"). They're not major but a resource intended for learning shouldn't contain things like that.
• TRWTF is that tax brackets don't work that way.
• @scgtrp said:
saying the string "Chicago" (as a std::string) will take up exactly 7 bytes of memory (but "may be different on your system"). They're not major but a resource intended for learning shouldn't contain things like that.
Not counting the overhead of the std::string object itself (which is likely 3*sizeof(pointer)) or memory allocation, it can very well take exactly 7 bytes. std::string does not depend on a terminating nil character, and may contain embedded nils. It generally works by storing pointers to the beginning and end of allocated area as well as the end of actual data (the standard does not actually mandate this internal structure, but strongly hints towards it).
So, there's no factual WTF there, although I can't quite decide whether the statement is too vague (no mention of the overhead) or gives too much details (this is intended for relative beginners after all).
• @scgtrp said:
`void taxCalculations(num income) num count, tax const num NUMBRKTS = 5 num brackets[NUMBRKTS] = 15000, 22000, 40000, 70000, 100000 num rates[NUMBRKTS] = 0, 0.15, 0.18, 0.22, 0.28, 0.30 while count < NUMBRKTS and income > brackets[count] count = count + 1 endwhile tax = income * rates[count]return tax`
1. Function is declared void but returns a value
2. count is not initialized (some languages zero-initialize everything by default)
3. rates is declared to be the same size as brackets, but contains one more value
4. If income falls outside the highest bracket, the rates array is accessed past its declared size
5. The tax variable is only assigned once just before returning its value - the expression could have been returned directly instead
6. Using apparent floating-point values for monetary calculations
7. Using a variable of an apparent non-integer type to declare array sizes
8. Tax calculation does not work that way
Did I miss anything?
• @scgtrp said:
I would have broken it out into a set of if statements. Just as efficient as a loop and a lot clearer (well, at the expense of a few bytes of space.)
But now you need to understand the code and change the if-else ladder if you add or remove a tax bracket.
• @scgtrp said:
I understand what the loop does, I just had to stare at it way too long to figure out how it does it.
I put it to you that there are two possible places for the problem to reside in this scenario, and only one of them is the code.
Though I do agree that an if/else chain would be preferable here.
The real WTF though is hard-coding tax brackets in a program. Don't they know they're going to change regularly? Those should be in a database, and then you can just use a very simple SQL statement to pull out the relevant figure without obfuscation or fragility.
• @Iago said:
Though I do agree that an if/else chain would be preferable here.
The real WTF though is hard-coding tax brackets in a program. Don't they know they're going to change regularly? Those should be in a database, and then you can just use a very simple SQL statement to pull out the relevant figure without obfuscation or fragility.
I wonder what your if/else chain would look like when the values are loaded from a database.
• @tdb said:
@Iago said:
Though I do agree that an if/else chain would be preferable here.
The real WTF though is hard-coding tax brackets in a program. Don't they know they're going to change regularly? Those should be in a database, and then you can just use a very simple SQL statement to pull out the relevant figure without obfuscation or fragility.
I wonder what your if/else chain would look like when the values are loaded from a database.
http://thedailywtf.com/articles/soft_coding.aspx ?
• What language is this anyway?
@scgtrp said:
num rates[NUMBRKTS] = 0, 0.15, 0.18, 0.22, 0.28, 0.30
Writing a beginner course in programming in a language that allows discrepancies like this is a WTF in itself
• @tdb said:
@scgtrp said:
`void taxCalculations(num income) num count, tax const num NUMBRKTS = 5 num brackets[NUMBRKTS] = 15000, 22000, 40000, 70000, 100000 num rates[NUMBRKTS] = 0, 0.15, 0.18, 0.22, 0.28, 0.30 while count < NUMBRKTS and income > brackets[count] count = count + 1 endwhile tax = income * rates[count]return tax`
1. Function is declared void but returns a value
2. count is not initialized (some languages zero-initialize everything by default)
3. rates is declared to be the same size as brackets, but contains one more value
4. If income falls outside the highest bracket, the rates array is accessed past its declared size
5. The tax variable is only assigned once just before returning its value - the expression could have been returned directly instead
6. Using apparent floating-point values for monetary calculations
7. Using a variable of an apparent non-integer type to declare array sizes
8. Tax calculation does not work that way
Did I miss anything?
Doesn't 'count < NUMBRKTS' take care of the fourth item in your list?
@tdb said:
@scgtrp said:
`void taxCalculations(num income) num count, tax const num NUMBRKTS = 5 num brackets[NUMBRKTS] = 15000, 22000, 40000, 70000, 100000 num rates[NUMBRKTS] = 0, 0.15, 0.18, 0.22, 0.28, 0.30 while count < NUMBRKTS and income > brackets[count] count = count + 1 endwhile tax = income * rates[count]return tax`
1. Function is declared void but returns a value
2. count is not initialized (some languages zero-initialize everything by default)
3. rates is declared to be the same size as brackets, but contains one more value
4. If income falls outside the highest bracket, the rates array is accessed past its declared size
5. The tax variable is only assigned once just before returning its value - the expression could have been returned directly instead
6. Using apparent floating-point values for monetary calculations
7. Using a variable of an apparent non-integer type to declare array sizes
8. Tax calculation does not work that way
Did I miss anything?
Doesn't 'count < NUMBRKTS' take care of the fourth item in your list?
Not exactly. The last iteration through the loop, count will start off as 4, which is less than NUMBRKTS. The code in the loop will set count to 5.
Then, the assignment to tax will reference rates[5]. So, it depends on which part of the broken declaration of rates you believe. It's declared as an array of size 5, but is initialized with 6 elements.
• @levbor said:
What language is this anyway?
@scgtrp said:
num rates[NUMBRKTS] = 0, 0.15, 0.18, 0.22, 0.28, 0.30
Writing a beginner course in programming in a language that allows discrepancies like this is a WTF in itself
I'm betting pseudocode. The author of the OP did mention in some post that there was a companion book containing C++ translations of the examples.
• It is pseudocode. TRWTF seems to be my bug finding skills when I'm up late doing homework due the next day. Plenty of stuff there I missed
I still think it would have been better to write the textbook in a real language so students can actually try the code. I always debug my code by actually compiling it and testing it, instead of just staring at it until bugs pop out at me. (The C++ companion book only contains language-specific stuff for the main text, not the exercises like this one.)
• @scgtrp said:
It is pseudocode. TRWTF seems to be my bug finding skills when I'm up late doing homework due the next day. Plenty of stuff there I missed
I still think it would have been better to write the textbook in a real language so students can actually try the code. I always debug my code by actually compiling it and testing it, instead of just staring at it until bugs pop out at me. (The C++ companion book only contains language-specific stuff for the main text, not the exercises like this one.)
you could always just try writing it in C/C++ and testing it that way. The idea of debugging pseudocode has always struck me as silly.
• @DescentJS said:
@scgtrp said:
It is pseudocode. TRWTF seems to be my bug finding skills when I'm up late doing homework due the next day. Plenty of stuff there I missed
I still think it would have been better to write the textbook in a real language so students can actually try the code. I always debug my code by actually compiling it and testing it, instead of just staring at it until bugs pop out at me. (The C++ companion book only contains language-specific stuff for the main text, not the exercises like this one.)
you could always just try writing it in C/C++ and testing it that way. The idea of debugging pseudocode has always struck me as silly.
Even more, the task was to correct the syntax errors in the pseudocode.
• @morbiuswilters said:
TRWTF is that tax brackets don't work that way.
Yah, whatever prof or TA wrote this example is in for a fun time with the IRS, come April 15th.
• @blakeyrat said:
@morbiuswilters said:
TRWTF is that tax brackets don't work that way.
Yah, whatever prof or TA wrote this example is in for a fun time with the IRS, come April 15th.
Why would the IRS care? The error is in their favor. The way the brackets really work is that the first X dollars are taxed at rate 1, the next Y dollars at rate 2, the next Z dollars at rate 3. Taxing the entire amount at rate 3 would be grossly OVER-taxed.
If tax brackets worked the way they do in this "program," then it would be possible to actually end up with a SMALLER paycheck when your boss gives you a RAISE (because you've bumped into a higher bracket). Obviously, people would overthrow the government if it worked like that.
• [QUOTE] it would be possible to actually end up with a SMALLER paycheck when your boss gives you a RAISE (because you've bumped into a higher bracket).[/QUOTE]
I've had this happen. The jerk bumped me just over the tax line. from just under it. Wise-ass found out the meaning of tax bracket pretty fast at my next pay check.
• @morbiuswilters said:
TRWTF is that tax brackets don't work that way.
Only in the more enlightened parts of tax codes.
UK Stamp Duty Land Tax (a percentage tax on the value of land or buildings bought) works the way the program works. Go £1 into the next band up, and you get charged the percentage in that band on the whole price, not just the marginal bit.
Naturally this constricts prices to below the band limits in cases where normally they'd be a certain amount above them.
Income tax works on marginal rates, so the whole tax code isn't completely barking. At least not in that way.
• @belgariontheking said:
Even more, the task was to correct the syntax errors in the pseudocode.
Ding, ding, ding! Gentlemen, we have a winner. Pseudocode has no defined syntax. Its sole purpose is to allow you to document your logic in a human-readable way without having to worry about any particular syntax.
• @joelkatz said:
@belgariontheking said:
Even more, the task was to correct the syntax errors in the pseudocode.
Ding, ding, ding! Gentlemen, we have a winner. Pseudocode has no defined syntax. Its sole purpose is to allow you to document your logic in a human-readable way without having to worry about any particular syntax.
Not entirely true, I have a textbook that uses a pseudocode with a definined syntax. It's more that the code is not compilable and may use odd symbols. Such as what the textbook uses for variable assignment "<FONT size=4>←</FONT>"
• @DescentJS said:
the textbook uses for variable assignment "<font size="4">←</font>"
What the fuck.
• @DescentJS said:
Such as what the textbook uses for variable assignment "<font size="4">←</font>"
Sounds like an early sign of APL Syndrome. Burn it quickly before it infects something else.
• @Someone You Know said:
@DescentJS said:
Such as what the textbook uses for variable assignment "<font size="4">←</font>"
Sounds like an early sign of APL Syndrome. Burn it quickly before it infects something else.
Too late. That's what wikipedia pseudocode uses for variable assignment, too.
• @dhromed said:
@DescentJS said:
the textbook uses for variable assignment "<font size="4">←</font>"
What the fuck.
Eh, Knuth?
• @Obfuscator said:
@dhromed said:
@DescentJS said:
the textbook uses for variable assignment "<font size="4">←</font>"
What the fuck.
Eh, Knuth?
Gesundheit!
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# Extension of regular funtion
This is an exercise of Hartshorne's book.\ For a quasi projective variety $Y$ with dimension $\geqq 2$ and $p \in Y$ a normal point. If $f$ is regualr on $Y-\{p\}$ then, f can extend to a regular function on $Y$.\ I want to get some hint to prove this problem....
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Since the problem is local around $p$, you can assume that $Y=Spec(A)$ where $A$ is a noetherian domain (quasi-projectiveness is irrelevant).
Clearly, every point $\mathfrak q \in Spec(A)$ of height one is distinct from $p$ (since $\mathfrak q$ it corresponds to a subvariety of codimension $1$). So every function $f$ defined on $Spec(A)\setminus \lbrace p\rbrace$ is defined at $\mathfrak q$.
You can then conclude that $f\in A$, that is $f$ extends regularly through $p$, thanks to the formula valid for a noetherian normal domain (Matsumura, Commutative ring theory, Theorem 11.5, page 81)
$$A=\bigcap_{ht(\mathfrak q)=1} A_ \mathfrak q$$
A general result in this vein is that if $X$ is a locally noetherian normal integral scheme and $Y\subset X$ a closed subset of codimension $\geq 2$, the restriction morphism $\mathcal O_X(X)\to O_X(X\setminus F)$ is bijective.
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Thank Georges, But, This problem is one of exercise of Hartshorne's book "Algebraic geometry" chapterI.1.3.... Maybe there is s proof with out scheme theory...?? – Sang Cheol Lee Mar 8 '12 at 15:36 @Georges Elencwajg : Could you please show me where to find the proof of the above general result ? – Arsenaler Jan 27 at 18:46
This is a simple corollary of so called, algebraic Hartog's lemma. And it requires fair amount of commutative algebra......You can find more informations and geometric intuition in the Vakil's lecture note 12.3.10
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# H Bridge Diagram
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Installing a tachometer on your Vehicles can assist in preventing critical repair problems, however. You might have a weak ground issue. The way the brain learns is a subject that still requires a good deal of study. How it learns can be associated by how it is able to create memories.
In a parallel circuit, each unit is directly linked to the power supply, so each system gets the exact voltage. There are 3 basic sorts of standard light switches. The circuit needs to be checked with a volt tester whatsoever points. H Bridge Diagram. Each circuit displays a distinctive voltage condition. You are able to easily step up the voltage to the necessary level utilizing an inexpensive buck-boost transformer and steer clear of such issues. The voltage is the sum of electrical power produced by the battery. Be sure that the new fuse isn't blown, and carries the very same amperage.
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When you have just a single cable going into the box, you're at the close of the run, and you've got the simplest scenario possible. All trailer plugs and sockets are extremely easy to wire. The adapter has the essential crosslinks between the signals. Wiring a 7-pin plug on your truck can be a bit intimidating when you're looking at it from beyond the box.
The control box may have over three terminals. After you have the correct size box and have fed the cable to it, you're almost prepared to permit the wiring begin. Then there's also a fuse box that's for the body controls that is situated under the dash. H Bridge Diagram. You will find that every circuit has to have a load and every load has to have a power side and a ground side. Make certain that the transformer nameplate power is enough to supply the load that you're connecting.
The bulb has to be in its socket. Your light can be wired to the receiver and don't require supply additional capacity to light as it can get power from receiver. In the event the brake lights aren't working, a police officer may block the vehicle and issue a warning to create the repair within a particular time limit. Even though you would still must power the relay with a power source or battery. Verify the power is off before trying to attach wires. In case it needs full capacity to begin, it won't operate in any way.
Replacing thermostat on your own without a Denver HVAC technician can be quite harrowing if you don't hook up the wiring correctly. After the plumbing was cut out, now you can get rid of the old pool pump. It's highly recommended to use a volt meter to make sure there is no voltage visiting the motor, sometimes breakers do not get the job done properly, also you might have turned off the incorrect breaker. Remote distance is left up to 500m. You may use a superior engine ground. The second, that's the most frequently encountered problem, is a weak ground in the computer system. Diagnosing an electrical short can be extremely tough and costly.
H-bridge chip diagram basic bridge diagram h-bridge circuit diagram h-bridge block diagram h bridge design h-bridge motor driver circuits bi-directional h-bridge motor circuit h-bridge circuit diagram pin.
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# Accounting
posted by Lyra
What is the value of a stock that grows at a supernormal rate of 18% for the first four years, and then slows down to a constant growth rate of 10%? An annual dividend of \$2.00/share was just paid, and the rate of return on common stock is 13%.
1. Lyra
Would anyone be able to help me get started with the formula?
2. Lyra
I think this is the solution. Please let me know what you think.
Supernormal Growth:
D0 = \$2.00
D1 = 2.00(1+0.18) = \$2.36
D2 = 2.36(1+0.18) = \$2.78
D3 = 2.78(1+0.18) = \$3.28
D4 = 3.28(1+0.18) = \$3.87
Constant Growth:
D5 = [3.87(1+0.10)] / [0.13 – 0.10] = 4.257 / 0.03 = \$141.90
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SALAH ALRABEEI
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Find smallest Eigenvalue and the corresponding eigenvector.
You can find it here
Why my graph not same as research paper?
You have two mistakes here s=(exp(p_0)*T); The three s in the code should be this s=(exp(-p_0*T));
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If I got you correctly, I think you want you want this clear A = magic(4) [n,m]=size(A) [As,Cur_ind]=sort(A,'ascend') Or...
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Create a vector using square brackets going from 1 to the given value x in steps on 1. Hint: use increment.
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Times 2 - START HERE
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compare groups of items regarding overlaps
Use [val,ndxA,ndxB] = intersect(A,B) It will give you the overlapping val and its index in both groups A and B
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How to get rid of for loop?
function [x_vals,y_vals,I,A]=monte_carlo_syms(x,y,f,g) x_vals=unifrnd(0,1,[200,1]); y_vals=unifrnd(0,1,[200,1]); I = nan*leng...
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yline(put the value)
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# Exaggerate
Why is “exaggerate” spelled with two g’s?
How should I write the above sentence?:
-Why is “exaggerate” spelled with two g’s?
-Why is “exaggerate” spelled with two “g”'s?
or
-Why is “exaggerate” spelled with two "g"s?
Ahhhhhhhhhhhhhh!
I spelled it wrong today in my adult class today and a student corrected me. I told him that he was wrong and we looked it up in the dictionary. How embarassing.
I just figured it out. With two g’s, the it’s a short “a”. With one it’s a long “a”. The pronunciation of “exa” overrides the “g” factor and to keep the original “exa” pronunciation, you need two g’s.
Or something like that.
Today wasn’t a good day. I argued with my boss about what he wants to pay me to do some 4-hour/day OT class.
Ahhhhhhhhhh!
And that’s not all - ‘embarrassing’ is spelled with two r-s.
Two arses I thought that was.
You see. I’m just going downhill. Or is it down hill.
Calgon! Take me away!
[quote=“j99l88e77”]Why is “exaggerate” spelled with two g’s?
How should I write the above sentence?:
-Why is “exaggerate” spelled with two g’s?
-Why is “exaggerate” spelled with two “g”'s?
or
-Why is “exaggerate” spelled with two "g"s?
Ahhhhhhhhhhhhhh!
I spelled it wrong today in my adult class today and a student corrected me. I told him that he was wrong and we looked it up in the dictionary. How embarassing. [/quote]
Well, it’s probably spelled that way because it’s derived from a Latin word that was spelled that way, which in turn was derived from another Latin word that was spelled that way (aggerare, “to heap up,” which is the same root as in aggregate, per the OED). “Exaggerate” came to English via French, which is why it has the soft g instead of the hard g, and French does now spell it with one g, but in medieval Fr. it had both. Personally I’d think that both gs are necessary for the pronunciation, as the consonant sound is pretty long – sounds like “exaj jer ate” instead of “exa jer ate” but that could be just because I’m thinking about it right now.
Web sources (Brian Forte, dictionary.com) seem to indicate that the plural of single letters is formed by taking the apostrophe – g’s – but this is one of those cases where I’ll insist the entire rest of the world is wrong, because it just plain looks like a possessive to me. Write Gs on my tombstone, there’s no justification for the apostrophe.
I just have to face it. I made a boo boo. Like when you spell “afraid” “affraid”. Ahhhhhhhhhhhhh!
[quote=“yisha’ou”][quote=“j99l88e77”]Why is “exaggerate” spelled with two g’s?
How should I write the above sentence?:
-Why is “exaggerate” spelled with two g’s?
-Why is “exaggerate” spelled with two “g”'s?
or
-Why is “exaggerate” spelled with two "g"s?
Ahhhhhhhhhhhhhh!
I spelled it wrong today in my adult class today and a student corrected me. I told him that he was wrong and we looked it up in the dictionary. How embarassing. [/quote]
Well, it’s probably spelled that way because it’s derived from a Latin word that was spelled that way, which in turn was derived from another Latin word that was spelled that way (aggerare, “to heap up,” which is the same root as in aggregate, per the OED). “Exaggerate” came to English via French, which is why it has the soft g instead of the hard g, and French does now spell it with one g, but in medieval Fr. it had both. Personally I’d think that both gs are necessary for the pronunciation, as the consonant sound is pretty long – sounds like “exaj jer ate” instead of “exa jer ate” but that could be just because I’m thinking about it right now.
Web sources (Brian Forte, dictionary.com) seem to indicate that the plural of single letters is formed by taking the apostrophe – g’s – but this is one of those cases where I’ll insist the entire rest of the world is wrong, because it just plain looks like a possessive to me. Write Gs on my tombstone, there’s no justification for the apostrophe.[/quote]
:bravo: :bravo:
You talk purty.
:bravo: :bravo:
You talk purty.[/quote]
Thanks!
Speak for yourself.
Speak for yourself.[/quote]
You see. I know these things as I’m typing them, but sometimes I don’t go back and correct them. Then it comes back to haunt me.
Oh well. I got a raise today anyway. 6500 NT/month (200,000 Won). Wahoo! Think I’ll go spend it. Maybe.
How about if the apostrophe is being used to form a contraction?
Perhaps the full form should be "G"s, and the apostrophe is taking the place of the opening and closing quotes, thus g’s?
And what if you are trying to specify lowercase g’s? Lowercase gs? Looks like an abbreviation. What about As? Looks like a word. Some of the other capital letter + ‘s’ combinations look like they could belong on the periodic table of elements. Not to mention that Gs is a reserved term meaning the normal gravity of the earth.
Not that I’m passionate on the subject. Afterall,
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Commit 1bed793b by Dan Frumin
### Example: Landin's knot is equivalent to the Y combinator
parent fab4fb46
... ... @@ -27,6 +27,7 @@ theories/logrel/soundness_binary.v theories/logrel.v theories/examples/lock.v theories/examples/ticket_lock.v theories/examples/bot.v theories/examples/counter.v theories/examples/lateearlychoice.v theories/examples/par.v ... ...
From iris.proofmode Require Import tactics. From iris_logrel Require Export logrel. From iris_logrel Require Export logrel examples.bot. (* Semantic typeability of the Y combinator *) (* Semantic typeability of the Y combinator and the Landin's knot *) Definition Y : val := λ: "f", (λ: "x", "f" ("x" "x")) (λ: "x", "f" ("x" "x")). Definition Knot : val := λ: "f", let: "z" := ref bot in "z" <- (λ: "x", "f" (!"z" #()));; !"z" #(). Section contents. Context `{logrelG Σ}. Lemma Y_semtype Δ Γ A : ... ... @@ -20,6 +24,40 @@ Section contents. iApply (bin_log_related_app with "Hff"). by iApply "IH". Qed. Lemma KNOT_Y Δ Γ A : {⊤,⊤;Δ;Γ} ⊨ Knot ≤log≤ Y : TArrow (TArrow A A) A. Proof. unlock Y Knot. simpl. iApply bin_log_related_arrow; eauto. iAlways. iIntros (f1 f2) "#Hff". rel_let_l. rel_let_r. rel_alloc_l as z "Hz". rel_let_l. rel_store_l. rel_let_l. iLöb as "IH". rel_let_r. rel_load_l. rel_let_l. iApply (bin_log_related_app with "Hff"). by iApply "IH". Qed. Lemma Y_KNOT Δ Γ A : {⊤,⊤;Δ;Γ} ⊨ Y ≤log≤ Knot : TArrow (TArrow A A) A. Proof. unlock Y Knot. simpl. iApply bin_log_related_arrow; eauto. iAlways. iIntros (f1 f2) "#Hff". rel_let_l. rel_let_r. rel_alloc_r as z "Hz". rel_let_r. rel_store_r. rel_let_r. iLöb as "IH". rel_let_l. rel_load_r. rel_let_r. iApply (bin_log_related_app with "Hff"). by iApply "IH". Qed. End contents. Theorem Y_typesafety f e' τ thp σ σ' : ... ...
From iris.proofmode Require Import tactics. From iris_logrel Require Export logrel. Definition bot : val := rec: "bot" <> := "bot" #(). Lemma bot_typed Γ τ : Γ ⊢ₜ bot : TArrow TUnit τ. Proof. solve_typed. Qed. Hint Resolve bot_typed : typeable. Section contents. Context `{logrelG Σ}. Lemma bot_l ϕ Δ Γ E K t τ : (ϕ -∗ {E;Δ;Γ} ⊨ fill K (bot #()) ≤log≤ t : τ) -∗ Maintainer Isn't this premise redundant? Isn't this premise redundant? Maintainer It most certainly is. I guess I had the lemma to be in this shape for the ease of some proof, but it is no longer needed. It most certainly is. I guess I had the lemma to be in this shape for the ease of some proof, but it is no longer needed. Maintainer I've fixed it in 7546cf67 I've fixed it in 7546cf67316c4749474372b29c4b5bfb18cae23a Please register or sign in to reply {E;Δ;Γ} ⊨ fill K (bot #()) ≤log≤ t : τ. Proof. iIntros "Hlog". iLöb as "IH". rel_rec_l. unlock bot; simpl_subst/=. iApply ("IH" with "Hlog"). Qed. End contents.
... ... @@ -15,7 +15,7 @@ where every v_i is well-typed Unit -> Unit *) From iris.proofmode Require Import tactics. From iris_logrel Require Export logrel examples.various (* for bot *). From iris_logrel Require Export logrel examples.bot. Definition or : val := λ: "e1" "e2", let: "x" := ref #0 in ... ...
... ... @@ -75,10 +75,6 @@ Definition nth : val := rec: "nth" "l" "n" := else "nth" (Snd "xs") ("n" - #1) end. Lemma bot_typed Γ τ : Γ ⊢ₜ bot : TArrow TUnit τ. Proof. solve_typed. Qed. Hint Resolve bot_typed : typeable. Lemma nth_typed Γ τ : Γ ⊢ₜ nth : TArrow (LIST τ) (TArrow TNat τ). Proof. ... ...
... ... @@ -4,7 +4,7 @@ *) From iris.proofmode Require Import tactics. From iris.algebra Require Import csum agree excl. From iris_logrel Require Import logrel examples.lock examples.counter. From iris_logrel Require Export logrel examples.lock examples.counter examples.bot. Section refinement. Context `{logrelG Σ}. ... ... @@ -219,18 +219,6 @@ Section refinement. rel_vals; eauto. } Qed. Definition bot : val := rec: "bot" <> := "bot" #(). Lemma bot_l ϕ Δ Γ E K t τ : (ϕ -∗ {E;Δ;Γ} ⊨ fill K (bot #()) ≤log≤ t : τ) -∗ {E;Δ;Γ} ⊨ fill K (bot #()) ≤log≤ t : τ. Proof. iIntros "Hlog". iLöb as "IH". rel_rec_l. unlock bot; simpl_subst/=. iApply ("IH" with "Hlog"). Qed. (* /Sort of/ a well-bracketedness example. Without locking in the first expression, the callback can reenter the body in a forked thread to change the value of x ... ...
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# Co-efficient of Friction related to Fasteners
B
#### BUVANESWARI
Hello Friends!!
1)Define Co-efficient of Friction in related to fasteners.
2)Factors affecting Coefficient of Friction
3)How is it determined?
4)How co-efficient of Friction affects parts failure in regards to ELV compliance?
#### Miner
##### Forum Moderator
Staff member
Super Moderator
Re: Co-efficient of friction problem
Hello Friends!!
1)Define Co-efficient of Friction in related to fasteners.
2)Factors affecting Coefficient of Friction
3)How is it determined?
4)How co-efficient of Friction affects parts failure in regards to ELV compliance?
1) The definition of coefficient of friction does not change for a fastener. Picture a threaded fastener as an inclined plane that has been wrapped around a cylinder. The flanks of the thread form the surface of this inclined plane. While the geometry is a little more complex than this (e.g., the surface is tilted), it essentially holds true.
2) lubrication, plating, type of plating, contamination, etc.
4) In threaded fasteners, there are two basic issues that could occur. If a threaded fastener is contaminated with oil the friction is reduced. This means that at a given torque, the fastener will turn more times than it would have under typical friction. This places a higher tensile load on the fastener that could lead to immediate tensile failure, or a longer term failure. Another situation is a threaded fastener with a heavy plated coating. This could increase the friction leading to fewer turns under the same torque, which in turn places a lower tensile force on the fastener. This may lead to failures from inadequate clamp forces or long term loosening of the fastener.
U
#### Umang Vidyarthi
Hello Friends!!
1)Define Co-efficient of Friction in related to fasteners.
2)Factors affecting Coefficient of Friction
3)How is it determined?
4)How co-efficient of Friction affects parts failure in regards to ELV compliance?
Miner has guided you well on your querry. Have a look here for 'factors affecting friction and coefficient of friction for nut & bolts.
Here are some good vidieos on Friction.
Attached is a study on 'Friction coefficient of fastners' , conducted by SAKAI Tomotsugu of Toyota motor co. This will put you into the picture.
To the best of my knowledge, ELV compliance for fasteners is a change of passivation in zinc plating. How much coeff. of friction affects is beyond me.
Hope this helps
Umang
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#### harry
Super Moderator
In the context of RoHS, ELV, fasteners and coefficient of friction, it is worth noting the following:
Cr6+ (Hexavalent chrome) is banned. Hexavalent chrome has good properties such as low coefficient of friction, anti galling and high hardness not found in the many other substitutes such as Cr3+ (Trivalent chrome).
To overcome this, the use of an ‘organo-mineral type of lubricant topcoat’ is often employed as a friction modifier to help improve the coefficient of friction (most automotive specification called for coefficient of friction for fasteners of between 0.10-0.15).
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# math
posted by on .
Factor each polynomial completely, if a polynomial is prime say so
9x^2+4y^2
• math - ,
This cannot be factored without using imaginary numbers.
The answer is (3x + 2i y)(3x - 2i y)
where i is the square root of -1
• math - ,
if i is square root of -1
would the anwer be than
(3x+1/2y)(3x+1/2y)
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Re: boolean function, interpolation
• To: mathgroup at smc.vnet.net
• Subject: [mg61567] Re: boolean function, interpolation
• From: Maxim <ab_def at prontomail.com>
• Date: Sat, 22 Oct 2005 03:24:06 -0400 (EDT)
• References: <dj2719\$bgb\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```On Tue, 18 Oct 2005 07:08:25 +0000 (UTC), edi <esolakster at gmail.com> wrote:
> Hi,
>
> I would like to interpolate vectors (dimension:20-30 bits) over GF(2)
> and values (one bit)to a function (boolean). Is there easy way to this
> in Mathematica?
>
> Thanks.
>
If you just want to create a function which takes given values on given
boolean vectors, it is not hard:
In[1]:=
boolint[Lvecval_] := Module[
{conj},
If[# == 0, 1 - #2, #2]&,
{Lbool, Array[Slot, Length@ Lbool]}];
Total@ Cases[Lvecval, {v_, 1} :> conj[v]] //
Evaluate // Function
]
In[2]:=
boolint[{{{0, 0, 0}, 1}, {{0, 1, 0}, 0}, {{1, 1, 1}, 1}}]
Out[2]=
(1 - #1)*(1 - #2)*(1 - #3) + #1*#2*#3&
We have constructed a function f such that
f[0, 0, 0] == 1
f[0, 1, 0] == 0
f[1, 1, 1] == 1
Maxim Rytin
m.r at inbox.ru
```
• Prev by Date: Re: Eliminating parameters
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Helper III
## Distinct Count based on multiple column
Hi,
I want to display value which is as described below.
My calculation is like this
var a =Distinct count of value 1
var b= Count of Value 2
var ans=b/a
I want to display ans as my result.
But Problem is there I have multiple records for all records. That means for value 1 (distinct) I got multiple values. So my calculation is not properly set as my requirement. So please help me in this case. If Found Unique count that is based on multiple column(value 1 and value 2) for var b.
Thanks.
1 ACCEPTED SOLUTION
Microsoft
Hi @Krunalbpatel,
Please change the formula of var b to the following formula:
```var b = COUNTROWS(
SUMMARIZE(Table,Table[Name],Table[Value1],Table[Value2])
)```
I create a table visual to display your expected result. For more details, you can review the example in this attached PBIX file:https://1drv.ms/u/s!AhsotbnGu1NogXByWhS2F551DemE.
Thanks,
Lydia Zhang
Community Support Team _ Lydia Zhang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
6 REPLIES 6
Community Champion
Hi Krunal,
Happy to help you with this. Please provide a snapshot of data and what are you trying to achieve.
Regards
Bhavesh
Thanks & Regards,
Bhavesh
Love the Self Service BI.
Helper III
Hi,
I have data like this
Id Name Value1 Value2
1 abc 1 1
2 abc 1 2
3 abc 1 3
4 pqr 2 1
5 pqr 2 2
6 abc 1 1
7 abc 1 2
8 mnq 3 1
9 pqr 2 1
10 xyz 4 2
Now my calculation is like this
var a = DISTINCTCOUNT(value1)
var b = COUNT(Value2)
My result wold give me like
abc 1 - 5
pqr 1 - 3
mnq 1 - 1
xyz 1 - 1
But I need
abc 1 - 3
pqr 1 - 2
mnq 1 - 1
xyz 1 - 1
For this I need to count distinct on multiple column so
1 abc 1 1
6 abc 1 1
this will result only 1
Thanks And Regards
Krunal Patel
Microsoft
Hi @Krunalbpatel,
Please change the formula of var b to the following formula:
```var b = COUNTROWS(
SUMMARIZE(Table,Table[Name],Table[Value1],Table[Value2])
)```
I create a table visual to display your expected result. For more details, you can review the example in this attached PBIX file:https://1drv.ms/u/s!AhsotbnGu1NogXByWhS2F551DemE.
Thanks,
Lydia Zhang
Community Support Team _ Lydia Zhang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Helper III
Hi Lydia Zhang,
Thank you for your reply. I got my solution and its works fine for me.
Thanks and Regards,
Krunal Patel
Resolver I
Hi Krunalbpatel,
Test= SUMX(DISTINCT('Table'[Value2]),CALCULATE( VALUES('Table' [Value2])))
I hope this helps,
Christian
Helper III
Hi cosborn1231,
No This will not help me. In this case I got the sum of distinct values but I want the distinct value count against value 1.
Thanks and Regards,
Krunal Patel
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IBPS Exam Guru
free Online Practice
Online Practice: IBPS Clerk PO Specialist Officer &RRB Prepare
Exercise
Bank Clerk :: CN Test 131
Home > Bank Clerk > CN Test 131 > General Questions
1 .
If the numerator of a fraction is increased by 300% and the denominator is increased by 260% the resultant fraction is 9/11. What is the original fraction ?
8/55 19/55 39/55 None of these
2 .
What is the average of the following set of members ?
205, 336, 425, 527, 445, 562, 720
560 460 660 540
3 .
24 men complete a work in 12 days. In how many days can 18 men complete the same piece of work ?
16 days 12 days 9 days 15 days
4 .
The product of two consecutive even numbers is 15624. Which is the larger number ?
116 124 126 128
5 .
Mohan invests an amount of Rs 7690 at the rate of 7% p.a. for 2 years. What approximate amount of compound interest will be obtained after 2 years ?
Rs 1114 Rs 1118 Rs 2114 Rs 1211
6 .
Tripti, Rani and Madhulika begin to jog around a circular stadium. They complete their revolution in 64 sec, 36 sec and 43 sec respectively. After how many seconds will they be together at the starting point ?
24768 25768 26768 23768
7 .
What approximate value should come in the place of (?) in the following equation ?
V936 * 20.6 + 216.24 = ?
735 935 835 1015
8 .
Ravi has some hens and some cows. If the total number of animal heads is 78 and the total numberof feet is 210 is then how many cows does Ravi have ?
37 29 32 27
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# What is a plate appearance in baseball?
## What is a plate appearance in baseball?
A plate appearance refers to a batter’s turn at the plate. Each completed turn batting is one plate appearance. Total plate appearances are used to determine which players have qualified for the batting title; at-bats are not used for this purpose, even though at-bats are used to decipher batting average.
How is baseball plate appearance calculated?
For our purposes, we will define a plate appearance as PA = H + BB + K + HBP + SH + SF + DI + E+ DFO where: PA = Plate Appearance. H = Hit (single, double, triple, or home run) BB = Walk (Four balls before three strikes)…The Infinitely Long MLB Plate Appearance.
Ball – B 0.3611 P(B)
TotPIP – P 0.2222 P(PIP)=1-P(B)-P(S)
### What is the difference between plate appearance and at bat?
Definition. An official at-bat comes when a batter reaches base via a fielder’s choice, hit or an error (not including catcher’s interference) or when a batter is put out on a non-sacrifice. (Whereas a plate appearance refers to each completed turn batting, regardless of the result.)
What is the minimum plate appearance in baseball?
502 plate appearances
Rule 9.22(a) of the Official Baseball Rules make a single allowance to the minimum requirement of 502 plate appearances for the purposes of determining the batting, slugging or on-base percentage title.
## What are the dimensions of a baseball home plate?
The baseball home plate dimensions are 17 inches by 8 1/2 inches by 8 1/2 inches by 12 inches. The shape is somewhat like a pentagon.
What is home plate in baseball field?
home plate. n. A base, usually consisting of a hard rubber slab, at one of the corners of a diamond at which a batter stands when hitting and which a base runner must finally touch in order to score.
### What is a baseball home plate?
Definition of home plate : a 5-sided rubber slab at one corner of a baseball diamond at which a batter stands when batting and which must be touched by a base runner in order to score baseball : the base that a runner must touch in order to score : the base that a baseball runner must touch to score
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https://electronics.stackexchange.com/questions/515961/what-is-the-use-of-second-transistor-in-this-circuit
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# What is the use of second transistor in this circuit?
I have the circuit above for which I am supposed to find the voltage gain $$\V_{out}\over V_{in}\$$, but I don't understand what purpose the second transistor (right one, Q2) has here. If it wasn't present, I would have calculated $$\dV_{out}\over dV_{in}\$$ by differentiating $$\V_{out}=V_{dd}-I_cR_c\$$ where $$\I_c=I_se^{V_{be}\over V_T}\$$. Due to the presence of second transistor, a part of $$\I_e\$$ will flow in it, which will in turn reduce $$\V_e\$$ of the first transistor. Can I get some hints on how to approach this problem? Are such circuits of any practical use?
• The function of Q2 appears to be "go up in smoke." Any appreciable current through its base-emitter junction would cause it to become a short circuit between Vdd and ground.
– JRE
Aug 11, 2020 at 12:19
• It makes a good Active "dummy load" Aug 11, 2020 at 12:27
• Ignoring that this is a useless circuit: What does the presence of Q2 mean for the voltage across $R_E$? When V($R_E$) = 0.5 V, what happens? When V($R_E$) = 0.7 V, what happens? Whoever made this exercise should have connected the collector of Q2 to its base so that Q2 becomes a diode or (more advanced) ask why connecting the collector of Q2 like that isn't a good idea. Aug 11, 2020 at 13:35
• "Crowbar protection" would be one legitimate use. At some value of Vin, the fuse in Vdd blows. I have heard of a similar circuit built into an audio mixing desk, with the express purpose of allowing recording engineers to terminate an out of control session (think "Spinal Tap") because of smoke coming out of the desk...
– user16324
Aug 11, 2020 at 14:46
• @BrianDrummond going up to 12 in this case
– user16222
Aug 11, 2020 at 15:00
There are multiple ways to solve this circuit, one is putting it into a simulator. Considering that this is an assignment, I'd expect that you need to determine the base-emitter resistance of the second transistor which will be in parallel with $$\R_E\$$. The value of this depends on the operating point, so that might require whatever model you were taught in class.
After finding $$\r_b\$$ you can continue with your AC analysis.
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https://www.rdocumentation.org/packages/HH/versions/3.1-19/topics/ci.plot
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# ci.plot
From HH v3.1-19
0th
Percentile
##### Plot confidence and prediction intervals for simple linear regression
The data, the least squares line, the confidence interval lines, and the prediction interval lines for a simple linear regression (lm(y ~ x)) are displayed. Tick marks are placed at the location of xbar, the x-value of the narrowest interval.
Keywords
hplot, regression
##### Usage
ci.plot(lm.object, ...)
## S3 method for class 'lm':
ci.plot(lm.object,
xlim=range(data[, x.name]),
newdata,
conf.level=.95,
data=model.frame(lm.object),
newfit,
ylim,
pch=19,
lty=c(1,3,4,2),
lwd=2,
main.cex=1,
main=list(paste(100*conf.level,
"% confidence and prediction intervals for ",
substitute(lm.object), sep=""), cex=main.cex), ...
)
##### Arguments
lm.object
Linear model for one y and one x variable.
xlim
xlim for plot. Default is based on data from which lm.object was constructed.
newdata
data.frame containing data for which predictions are wanted. The variable name of the column must be identical to the name of the predictor variable in the model object. Defaults to a data.frame containing a vector spanni
conf.level
Confidence level for intervals, defaults to .95
data
data extracted from the lm.object
newfit
Constructed data.frame containing the predictions,confidence interval, and prediction interval for the newdata.
ylim
ylim for plot. Default is based on the constructed prediction interval.
pch
Plotting character for observed points.
lty, lwd
Line types and line width for fit and intervals.
main.cex
Font size for main title.
main
Main title for plot
...
Additional arguments to be passed to panel function.
##### Value
• "trellis" object containing the plot.
##### Note
The predict.lm functions in S-Plus and R differ. The S-Plus function can produce both confidence and prediction intervals with a single call. The R function produces only one of them in a single call. Therefore the default calculation of newfit within the function depends on the system.
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If you get a normal polygon and you would like to earn a similar shape with sides which are twice as long, how much larger will the region of the new shape be. This huge triangle is made up of the original trapezoid and a more compact triangle, very similar to the larger one. Let the duration of side be a.
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There continue to be prerequisites. Listed following are a few of the major highlights of the class. I am presently studying mathematics.
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A Startling Fact about Mathematics for Machine Learning Uncovered
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What the In-Crowd Won’t Tell You About Mathematics for Machine Learning
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This usually means that it’s possible for you to construct a superior predictive model without nearly every understanding of calculus or linear algebra. Each math topic has many unique types of math worksheets to cover various types of problems you may choose to work on. This is a fast whip around the topics in linear algebra you need to be familiar with.
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All the faces are composed of polygons. The 2 rectangles aren’t similar. Similar triangles are triangles that have exactly the same form but possibly various size.
Learning is a rather important aspect. For beginners, you don’t require a lot of Mathematics to begin doing Machine Learning. Clearly, Machine Learning is an amazingly strong tool.
There are similarities between both. There are two sorts of time complexity success. There are a lotof math concepts you need to comprehend in order to understand how to tell time.
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12.0 200uA and 10uA (T7 Only) [T-Series Datasheet] | LabJack
# 12.0 200uA and 10uA (T7 Only) [T-Series Datasheet]
## Overview - T7 Only
The T7 has 2 fixed current source terminals useful for measuring resistance (thermistors, RTDs, resistors). The 10UA terminal provides approximately 10 µA and the 200UA terminal provides approximately 200 µA, but the actual values should be read from the calibration constants, or better yet measured in real-time using a fixed shunt resistor.
Using the equation V=IR, with a known current and voltage, it is possible to calculate the resistance of the item in question. Figure 12-1 shows a simple setup measuring 1 resistor.
The factory value of each current source is noted during calibration and stored with the calibration constants on the device. These can be viewed using the Device Info tab in Kipling, or read programmatically. Note that these are fixed constants stored during calibration, not some sort of real-time readings.
##### Constant Current Sources
Name Start Address Type Access CURRENT_SOURCE_200UA_CAL_VALUE Fixed current source value in Amps for the 200UA terminal. This value is stored during factory calibration, it is not a current reading. Using the equation V=IR, with a known current and voltage, it is possible to calculate resistance of RTDs. 1902 FLOAT32 R CURRENT_SOURCE_10UA_CAL_VALUE Fixed current source value in Amps for the 10UA terminal. This value is stored during factory calibration, it is not a current reading. Using the equation V=IR, with a known current and voltage, it is possible to calculate resistance of RTDs. 1900 FLOAT32 R
Example:
To read the factory value of the 200uA current source, perform a read of Modbus address 1902, and the result would be in the form of a floating point number, e.g. 0.000197456 amps.
## Examples Of Measuring Resistance
Multiple resistances can be measured by putting them in series and measuring the voltage across each. Some applications might need to use differential inputs to measure the voltage across each resistor, but for many applications it works just as well to measure the single-ended voltage at the top of each resistor and subtract in software.
Figure 12-1 Figure 12-2
Figure 12-1 shows a simple setup measuring 1 resistor. If R1=3k, the voltage at AIN0 will be 0.6 volts.
Figure 12-2 shows a setup to measure 3 resistors using single-ended analog inputs. If R1=R2=R3=3k, the voltages at AIN0/AIN1/AIN2 will be 1.8/1.2/0.6 volts. That means AIN0 and AIN1 would be measured with the ±10 volt range, while AIN2 could be measured with the ±1 volt range. This points out a potential advantage to differential measurements, as the differential voltage across R1 and R2 could be measured with the ±1 volt range, providing better resolution.
Figure 12-3 Figure 12-4
Figure 12-3 shows a setup to measure 2 resistors using differential analog inputs. AIN3 is wasted in this case, as it is connected to ground, so a differential measurement of AIN2-AIN3 is the same as a single-ended measurement of AIN2. That leads to Figure 12-4, which shows R1 and R2 measured differentially and R3 measured single-ended.
## Remarks
Maximum load resistance: The current sources can drive about 3 volts max, thus limiting the maximum load resistance to about 300 kΩ (10UA) and 15 kΩ (200UA). Keep in mind that high source resistance could cause settling issues for analog inputs.
Using a fixed resistor to calculate actual current: For some applications the accuracy and temperature coefficient of the current sources is sufficient, but for improvement a fixed resistor can be used as one of the resistors in the figures above. The Y1453-100 and Y1453-1.0K from Digi-Key have excellent accuracy and very low tempco. By measuring the voltage across one of these you can calculate the actual current at any time.
Handling load changes resulting in noise: The current sources are not particularly fast in reacting to load changes. This can show up as noise when rapidly sampling multiple channels using the same current source. Improve behavior by adding a 1 µF ceramic capacitor from the current source to GND and/or increasing settling time.
Temperature coefficients: Figures 12-5 and 12-6 show the typical current source output variation over temperature. Both sources typically have low temperature coefficients at or near 25C. Beyond 25C, the temperature coefficient variation may need to be accounted for, depending on application requirements.
Figure 12-5. Typical temperature coefficient values over operating temperature range . Figure 12-6. Typical current source deviation from 25C output over operating temperature range [1].
## Example - PT100 or PT1000 RTD
Assume that R1 in Figure 12-1 is a PT100 RTD. A PT100 RTD is 100 ohms at 0 degC. The response of an RTD is nonlinear, but the linear slope 0.384 ohms/degC works well from about -40 to +150 degC. That leads to the following expression:
R = (0.384 * DegC) + 100
...which can be rearranged to:
DegC = (2.604 * R) - 260.4
We are determining R by measuring the voltage that results from a known current passed through R, that is R = V/I, so we can say:
DegC = (2.604 * V/I) - 260.4
This tells us that the slope is 2.604/I and the offset is -260.4. To determine I, you can just use 0.0002 amps, or use the factory calibration value read from CURRENT_SOURCE_200UA_CAL_VALUE, or use a precision fixed resistor as mentioned above to measure I in real time. Assume we read the factory calibration value as 0.000200 amps, and thus use a constant slope of 2.604/0.0002 = 13020. We can now use the AIN-EF Offset and Slope feature to apply this slope and offset:
AIN0_EF_INDEX = 1 // feature index for Offset and Slope
AIN0_EF_CONFIG_D = 13020.0 // slope
AIN0_EF_CONFIG_E = -260.4 // offset
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(%42%) Cash Flow Analysis: NPV, IRR, MIRR
06-18-2020, 07:00 PM
Post: #1
Werner Senior Member Posts: 514 Joined: Dec 2013
(%42%) Cash Flow Analysis: NPV, IRR, MIRR
NPV, IRR and MIRR calculations involving multiple uneven cash flows at regular intervals.
This program replaces (and adds on to) my previous entry here.
It will run in all flavours of the 42: HP42S, Emu42, Free42, DM42.
No local variables were used.
1. Interactive mode
----------------------
usage: XEQ "CFA" -> [CFj] [IRR%] [NPV] [+%] [-%] [MIRR%]
CFj : matrix of cashflows CF0..CFn in column 1 and (optionally in column 2) their corresponding frequency Nj
IRR% : Internal Rate of Return, or discount rate for NPV calculation
NPV : Net Present Value
+% : re-investment rate applied to positive cashflows
-% : financing rate applied to negative cashflows
MIRR% : Modified Internal Rate of Return
example 1
---------
j CFj
0 -100 000
1 18 000
2 -50 000
3 25 000
4 25 000
5 225 000
1.XEQ "CFA"
2.create the matrix
6 ENTER 1 NEWMAT EDIT
[[-100000 ]
[ 18000]
[-50000]
[ 25000]
[ 25000]
[225000 ]]
3.EXITALL R/S (EXITALL also exits from the menu, use R/S to get back)
4. [CFj] -> CFj=[ 6x1 Matrix ]
5. 10 [IRR%] -> IRR%=10
6. [NPV] -> 50606.83
7. [IRR%] -> 19.33
8. 10 [+%] -> +%=10
9. 5 [-%] -> -%=5
10. [MIRR%] -> 16.29
example 2
---------
j CFj
0 -100000
1 -100000
2 40000
3 40000
4 40000
5 40000
6 -10000
7 35000
8 35000
9 35000
10 35000
11 35000
12 -10000
13 30000
14 30000
15 30000
16 30000
17 30000
18 -10000
now create the 7x2 matrix
[[-100000 2]
[ 40000 4]
[ -10000 1]
[ 35000 5]
[ -10000 1]
[ 30000 5 ]
[ -10000 1]]
and determine
- NPV with discount rate 8% (52933.21)
- IRR% (12.19)
- MIRR% with reinvestment rate 8% and financing rate 6% (9.25)
2 .Program mode
-------------------
The routines NPV, IRR% and MIRR% can also be called from within a program.
They all need the variable CFj to hold the cash flows (and frequencies), and additionally:
NPV: IRR% discount rate
MIRR%: the re-investment rate +% and financing rate -%
Program listing:
Code:
00 { 445-Byte Prgm } 01▸LBL "CFA" 02 MVAR "CFj" 03 MVAR "IRR%" 04 MVAR "NPV" 05 MVAR "+%" 06 MVAR "-%" 07 MVAR "MIRR%" 08▸LBL 10 09 VARMENU "CFA" 10 RTN 11 FC? 47 @ VARMENU no longer active 12 GTO 10 13 ALENG 14 ASTO ST Y 15 XEQ IND ST X 16 GTO 10 17▸LBL 03 18 "CFj" 19 ASTO ST X 20▸LBL 05 21 X≠Y? 22 GTO IND ST Y 23▸LBL 02 24 RCL IND ST Y 25 RTN 26▸LBL 04 27▸LBL "IRR%" 28 PGMSLV "NPV" 29 CLX 30 STO "IRR%" 31 100 32 SOLVE "IRR%" 33 RTN 34▸LBL "NPV" 35 MVAR "CFj" 36 MVAR "IRR%" 37 1 38 RCL "IRR%" 39 % 40 + @ R 41 0 @ sum 42 INDEX "CFj" 43 J- 44 J- 45 J+ 46 FC? 76 47 GTO 07 48▸LBL 06 @ all Nj=1, use Horner scheme 49 RCLEL 50 + 51 RCL÷ ST Y @ R 52 J- 53 FC? 77 54 GTO 06 55 GTO 04 56▸LBL 07 @ CFj Nj, index at Nj 57 RCLEL 58 J- 59 ABS 60 DSE ST L 61 GTO 00 62 CLX @ Nj=1 63 RCLEL 64 + 65 RCL÷ ST Y @ R 66 GTO 01 67▸LBL 00 @ Nj>1, group CFj 68 RCL ST Z @ R Nj sum R 69 X<>Y 70 Y↑X 71 STO÷ ST Y 72 X<> ST L 73 +/- 74 0.01 75 RCL× "IRR%" 76 XEQ "Ni" 77 RCLEL 78 × 79 - 80▸LBL 01 81 J- 82 FC? 77 83 GTO 07 84▸LBL 04 85 × 86 RTN 87▸LBL "MIRR%" 88 MVAR "CFj" 89 MVAR "+%" @ reinvestment rate, apply to positive CFj 90 MVAR "-%" @ financing rate, apply to negative CFj 91 CLX @ FV 92 ENTER @ PV 93 INDEX "CFj" 94 J+ 95 J- 96 FC? 76 97 GTO 09 98▸LBL 08 @ X:PV Y:FV, all Nj=1 99 RCL "-%" 100 % 101 + 102 X<>Y 103 RCL "+%" 104 % 105 + 106 X<>Y 107 RCLEL 108 X<0? 109 STO+ ST Y 110 X>0? 111 STO+ ST Z 112 R↓ 113 J+ 114 FC? 77 115 GTO 08 116 +/- 117 ÷ 118 RCL "CFj" 119 DIM? 120 - 121 GTO 04 122▸LBL 09 @ PV FV, CFj Nj, positioned on CFj @ FV := FV*(1+'+%'/100)^Nj @ PV := PV*(1+'-%'/100)^Nj 123 J+ 124 1 125 RCL "+%" 126 % 127 + 128 RCLEL 129 Y↑X 130 STO× ST Z 131 SIGN 132 RCL "-%" 133 % 134 + 135 RCLEL 136 Y↑X 137 STO× ST Y 138 X<> ST L @ Nj 139 J- 140 RCLEL 141 X<0? 142 GTO 00 143 CLX 144 0.01 145 RCL× "+%" 146 XEQ "Ni" 147 RCLEL 148 × 149 STO+ ST Z 150 R↓ 151 GTO 01 152▸LBL 00 153 CLX 154 0.01 155 RCL× "-%" 156 XEQ "Ni" 157 RCLEL 158 × 159 + 160▸LBL 01 161 I+ 162 FC? 76 163 GTO 09 164 +/- 165 ÷ 166 RCL "CFj" @ determine sum Nj 167 TRANS 168 RSUM 169 EDIT 170 → 171 ENTER 172 EXITALL 173 R↓ 174 DSE ST X 175▸LBL 04 176 1/X 177 Y↑X 178 RCL "-%" 179 % 180 + 181 100 182 STO× ST Y 183 - 184 RTN 185▸LBL "Ni" @ ((1+i)^N - 1)/i 186 X=0? 187 GTO 00 188 LN1+X 189 STO× ST Y 190 X<> ST L 191 X<>Y 192 E↑X-1 193 X<>Y 194 ÷ 195 RTN 196▸LBL 00 197 + 198 END
Cheers, Werner
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1-617-275-8164
# Variation of Pressure in a Fluid at RestPhysics Assignment Help & Live Online Tutoring From Our Expert Physics Tutors
## PhysicsFluid Mechanics and Thermal Physics Fluid Statics
Variation of Pressure in a Fluid at Rest When a fluid is at rest, then all points at the same depth must be at same pressure. That is the pressure at the same dept of a fluid at rest must be same. However, it varies with depth since it will have weight of the fluid on the top of it. Po is the atmospheric pressure = 1.00 atm = 1.013 × 10^5 Pa g is the acceleration due to gravity ρ is the density of the fluid h is the depth of the point at which the pressure P is to be calculated A is the cross sectional area Consider the above figure. Since the fluid is at rest the net force is zero: PA – mg – PoA = 0 However Mass (M) = Density (ρ) × Volume (V) = Density × Area × Height = ρAh. Thus, we can write: PA – PoA – ρAhg = 0 On simplifying we get: P = Po + ρgh Hence, the pressure P at a depth h below the surface of a fluid open to the atmosphere is greater than the atmospheric pressure by an amount ρgh. The difference in pressure between two points of unequal depth causes buoyant force.
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Cody
# Problem 1087. Magic is simple (for beginners)
Solution 1835898
Submitted on 3 Jun 2019 by Monika Phadnis
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 3; y_correct = 15; assert(isequal(magic_sum(n),y_correct))
2 Pass
n = 5; y_correct = 65; assert(isequal(magic_sum(n),y_correct))
3 Pass
n = 7; y_correct = 175; assert(isequal(magic_sum(n),y_correct))
4 Pass
n = 8; y_correct = 260; assert(isequal(magic_sum(n),y_correct))
5 Pass
n = 20; y_correct = 4010; assert(isequal(magic_sum(n),y_correct))
6 Pass
n = 100; y_correct = 500050; assert(isequal(magic_sum(n),y_correct))
7 Pass
n = 200; y_correct = 4000100; assert(isequal(magic_sum(n),y_correct))
8 Pass
n = 1000; y_correct = 500000500; assert(isequal(magic_sum(n),y_correct))
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• anonymous
Need more help will Fan and Medal!! The vertices of a triangle are P(2, -4), Q(-5, 3), and R(-1, -2) What are the vertices of the image reflected across the y-axis? A. P'(2, -4), Q'(-5, 3), and R'(-1, -2) B. P'(2, 4), Q'(-5, -3), and R'(-1, 2)***my guess C. P'(-2, -4), Q(5, -3), and R'(1, -2) D. P'(-2, 4), Q'(5, -3), and R'(1, 2)
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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Looking for something else?
Not the answer you are looking for? Search for more explanations.
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# Calendar
In Calendar, questions are mainly based on finding the day of the week if we are given a date. For example, we may be asked to find the day on 2 February, 1981.
• Finding day from date is based on calculating number of odd days. By odd days, we mean number of days more than complete number of weeks. For example,
Number of days in a non – leap year = 365
365 mod 7 = 1
So, number of odd days in a non – leap year = 1
• Number of days in a leap year = 366
=> Number of odd days in a leap year = 366 mod 7 = 2
• Number of odd days in 100 years (76 non – leap years + 24 leap years) = [(76 x 1) + (24 x 2)] mod 7 = (76 + 48) mod 7 = 124 mod 7 = 5 days
• Number of odd days in 200 years = (2 x Number of odd days in 100 years) mod 7 = 10 mod 7 = 3
• Number of odd days in 300 years = (3 x 5) mod 7 = 1
• Number of odd days in 400 years = (4 x 5 + 1) mod 7 = 21 mod 7 = 0
Note that here, we have added 1 day extra because 400th year would itself be a leap year.
• To check if a non – centennial year is a leap year, we divide it by 4. If the remainder is 0, the year is a leap year. For example, 2016 mod 4 = 0. Thus, we can safely deduce that 2016 is a leap year.
• To check if a centennial year is a leap year, we divide it by 400. If the remainder is 0, the year is a leap year. For example, 1700 mod 400 = 100. So, it was not a leap year. But 1600 mod 400 = 0. Thus, we can safely deduce that 1600 was a leap year.
• Number of odd days = 0, Day = Sunday
Number of odd days = 1, Day = Monday
Number of odd days = 2, Day = Tuesday
Number of odd days = 3, Day = Wednesday
Number of odd days = 4, Day = Thursday
Number of odd days = 5, Day = Friday
Number of odd days = 6, Day = Saturday
### Sample Problems
Question 1 : What was the day on 14 April, 2000 ?
Solution : 1600 will have 0 odd days.
300 years will have 1 odd day.
Now, in the next 99 years, we would be having 75 non – leap years and 24 leap years.
=> Number of odd days = (75 x 1) + (24 x 2) = 75 + 48 = 123 mod 7 = 4 odd days
Total odd days till now = 1 + 4 = 5
Number of odd days in January = 31 mod 7 = 3
Number of odd days in February (2000 is a leap year) = 29 mod 7 = 1
Number of odd days in March = 31 mod 7 = 3
Number of odd days till 14 April, 2000 in the month of April= 14 mod 7 = 0
So, total number of odd days = 5 + 3 + 1 + 3 = 12 mod 7 = 5
Thus, 14 April, 2000 was Friday (odd days = 5 => Friday)
Question 2 : What was the day on 16 August, 1947 ?
Solution : 1600 will have 0 odd days.
300 years will have 1 odd day.
Now, in the next 46 years, we would be having 35 non – leap years and 11 leap years.
=> Number of odd days = (35 x 1) + (11 x 2) = 35 + 22 = 57 mod 7 = 1 odd days
Total odd days till now = 1 + 1 = 2
Number of odd days in January = 31 mod 7 = 3
Number of odd days in February (1947 is a non – leap leap year) = 28 mod 7 = 0
Number of odd days in March = 31 mod 7 = 3
Number of odd days in April = 30 mod 7 = 2
Number of odd days in May = 31 mod 7 = 3
Number of odd days in June = 30 mod 7 = 2
Number of odd days in July = 31 mod 7 = 3
Number of odd days till 16 August, 1947 = 16 mod 7 = 2
So, total number of odd days = 2 + 3 + 0 + 3 + 2 + 3 + 2 + 3 + 2 = 20 mod 7 = 6
Thus, 16 August, 1947 was Saturday (odd days = 6 => Saturday)
Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.
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Solved
# Converting C to Assembly Language
Posted on 2007-11-14
780 Views
Is it possible to convert a program written in c to assembly language that will solve quadratic equation? Below is a sample program.
// Program to calculate roots of a quadratic equation ie ax2+bx+c=0 //
# include <stdio.h>
# include <conio.h>
#include <math.h>
void main()
{
float a, b, c, d, x1,x2;
clrscr();
printf("Enter coefficients of quadratic equation ie a, b and c: ");
scanf("%f%f%f", &a, &b, &c);
d=b*b-4*a*c;
if (d>0)
{
x1= (-b+(sqrt(d)))/(2*a);
x2= (-b-(sqrt(d)))/(2*a);
printf("Roots x1= %f and x2= %f are real and unequal", x1, x2);
}
else if (d==0)
{
x1= (-b/2*a);
x2= x1;
printf("Roots x1= %f and x2= %f are real and equal", x1, x2);
}
else
{
x1= (-b/2*a);
x2= sqrt(abs(d))/(2*a);
printf("Real part= %f and imaginary part= %f are complex conjugates", x1,x2);
}
printf( \n\n\t\t Press any key to continue& );
getch();
}
0
Question by:dnice143
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LVL 53
Accepted Solution
Infinity08 earned 250 total points
ID: 20283774
Almost all compilers have a flag to generate assembler output. gcc for example uses the -S flag :
gcc -S -c source.c
and the assembler will be in source.s
0
LVL 84
Expert Comment
ID: 20283822
cc -S
0
Author Comment
ID: 20284341
so using a gcc compiler will convert the text for me?
0
LVL 53
Expert Comment
ID: 20284404
No, gcc is just a common compiler. You might be using a different compiler ... Are you ?
0
LVL 53
Expert Comment
ID: 20284419
>> No, gcc is just a common compiler.
What I meant by that is : you CAN use gcc to do it, but you don't HAVE to. Most compilers have an option to generate assembler output.
0
Author Comment
ID: 20284457
So you are saying that it is possible to utilize assembly language to find roots to the quadratic equation. I downloaded Miricale Workbench on yesterday and It gave me an error when trying to link and compile.
0
LVL 53
Expert Comment
ID: 20284485
>> So you are saying that it is possible to utilize assembly language to find roots to the quadratic equation.
That was not what I was saying, but yes it's possible, although not very easy. Why would you want to do that anyway ? Just write the code in C, and let the compiler do its work and generate an executable. Do you have a specific reason ?
>> I downloaded Miricale Workbench on yesterday and It gave me an error when trying to link and compile.
That's most likely completely unrelated to the subject of this question ... Can you give a bit more background about what exactly you are trying to accomplish ?
0
LVL 40
Assisted Solution
evilrix earned 250 total points
ID: 20287566
>> Is it possible to convert a program written in c to assembly language that will solve quadratic equation?
Do you you want to see disassembly? If do you can see this in Visual Studio if you put a break-point in the code you are interested in and then open the disassembly window from the debug menu.
The attached is just a snippet for your code, the full output is HUGE because it also includes all the standard libraries.
I hope this helps.
-Rx.
``````
--- c:\temp\testr\testr\testr.cpp ----------------------------------------------
// testr.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
// Program to calculate roots of a quadratic equation ie ax2+bx+c=0 //
# include <stdio.h>
# include <conio.h>
#include <math.h>
void main()
{
00401000 push ebp
00401001 mov ebp,esp
00401003 and esp,0FFFFFFC0h
00401006 sub esp,3Ch
00401009 push esi
float a, b, c, d, x1,x2;
// clrscr();
printf("Enter coefficients of quadratic equation ie a, b and c: ");
0040100A mov esi,dword ptr [__imp__printf (4020A4h)]
00401010 push offset string "Enter coefficients of quadratic "... (4020F4h)
00401015 call esi
scanf("%f%f%f", &a, &b, &c);
0040101A lea eax,[esp+38h]
0040101E push eax
0040101F lea ecx,[esp+38h]
00401023 push ecx
00401024 lea edx,[esp+38h]
00401028 push edx
00401029 push offset string "%f%f%f" (402130h)
0040102E call dword ptr [__imp__scanf (40209Ch)]
d=b*b-4*a*c;
00401034 fld dword ptr [esp+44h]
00401038 fld st(0)
0040103D fmul st(0),st
0040103F fld dword ptr [esp+30h]
00401043 fld qword ptr [__real@4010000000000000 (402200h)]
00401049 fmul st,st(1)
0040104B fmul dword ptr [esp+38h]
0040104F fsubp st(2),st
00401051 fxch st(1)
00401053 fstp dword ptr [esp+3Ch]
if (d>0)
00401057 fldz
00401059 fld dword ptr [esp+3Ch]
0040105D fcom st(1)
0040105F fnstsw ax
00401061 test ah,41h
00401064 jne main+0B7h (4010B7h)
00401066 fstp st(3)
00401068 fstp st(1)
0040106A fstp st(0)
{
x1= (-b+(sqrt(d)))/(2*a);
x2= (-b-(sqrt(d)))/(2*a);
0040106C call _CIsqrt (401930h)
00401071 fstp dword ptr [esp+3Ch]
00401075 fld dword ptr [esp+30h]
printf("Roots x1= %f and x2= %f are real and unequal", x1, x2);
00401079 sub esp,10h
0040107E fld dword ptr [esp+44h]
00401082 fld st(0)
00401084 fchs
00401086 fld dword ptr [esp+4Ch]
0040108A fld st(0)
0040108C fsubp st(2),st
0040108E fxch st(1)
00401090 fdiv st,st(3)
00401092 fstp dword ptr [esp+4Ch]
00401096 fld dword ptr [esp+4Ch]
0040109A fstp qword ptr [esp+8]
0040109E fsubrp st(1),st
004010A0 fdivrp st(1),st
004010A2 fstp dword ptr [esp+4Ch]
004010A6 fld dword ptr [esp+4Ch]
004010AA fstp qword ptr [esp]
004010AD push offset string "Roots x1= %f and x2= %f are real"... (402138h)
004010B2 jmp main+13Dh (40113Dh)
}
else if (d==0)
004010B7 fucom st(1)
004010B9 fnstsw ax
004010BB fstp st(1)
004010BD test ah,44h
004010C0 jp main+0E9h (4010E9h)
004010C2 fstp st(0)
x2= (-b-(sqrt(d)))/(2*a);
printf("Roots x1= %f and x2= %f are real and unequal", x1, x2);
}
else if (d==0)
{
x1= (-b/2*a);
x2= x1;
printf("Roots x1= %f and x2= %f are real and equal", x1, x2);
004010C4 sub esp,10h
004010C7 fxch st(1)
004010C9 fchs
004010CB fmul qword ptr [__real@3fe0000000000000 (4021F8h)]
004010D1 fmulp st(1),st
004010D3 fstp dword ptr [esp+4Ch]
004010D7 fld dword ptr [esp+4Ch]
004010DB fst qword ptr [esp+8]
004010DF fstp qword ptr [esp]
004010E2 push offset string "Roots x1= %f and x2= %f are rea"... (402168h)
}
else
004010E7 jmp main+13Dh (40113Dh)
}
else if (d==0)
004010E9 fstp st(2)
004010EB fstp st(0)
{
x1= (-b/2*a);
x2= sqrt(abs(d))/(2*a);
004010ED fabs
printf("Real part= %f and imaginary part= %f are complex conjugates", x1,x2);
004010EF fstp dword ptr [esp+3Ch]
004010F3 fld dword ptr [esp+3Ch]
004010F7 call _CIsqrt (401930h)
004010FC fstp dword ptr [esp+3Ch]
00401100 fld dword ptr [esp+3Ch]
00401104 sub esp,10h
00401107 fld dword ptr [esp+40h]
0040110B fld st(0)
0040110F fdivp st(2),st
00401111 fxch st(1)
00401113 fstp dword ptr [esp+4Ch]
00401117 fld dword ptr [esp+4Ch]
0040111B fstp qword ptr [esp+8]
0040111F fld dword ptr [esp+44h]
00401123 fchs
00401125 fmul qword ptr [__real@3fe0000000000000 (4021F8h)]
0040112B fmulp st(1),st
0040112D fstp dword ptr [esp+4Ch]
00401131 fld dword ptr [esp+4Ch]
00401135 fstp qword ptr [esp]
00401138 push offset string "Real part= %f and imaginary part"... (402194h)
0040113D call esi
}
printf(" \n\n\t\t Press any key to continue&");
00401142 push offset string "\x1c\n\n\t\t Press any key to continue&"... (4021D0h)
00401147 call esi
getch();
0040114C call dword ptr [__imp___getch (4020A8h)]
}
00401152 xor eax,eax
00401154 pop esi
00401155 mov esp,ebp
00401157 pop ebp
00401158 ret
``````
0
LVL 40
Expert Comment
ID: 20287568
BTW: Is this code snippet thing new? I've not seen it until just now!!! :)
0
LVL 53
Expert Comment
ID: 20287603
>> BTW: Is this code snippet thing new? I've not seen it until just now!!! :)
Seems so ... First time I see it too - too bad you can only attach it at the end of the post.
evilrix, check this other question by the same asker : it clarifies a bit more what he's trying to do :
http://www.experts-exchange.com/Programming/Languages/C/Q_22961507.html
0
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ID: 20287693
Thanks "I8" -- you're name's too long to type so I have adopted this pseudo name -- I hope that's ok :)
0
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ID: 20287703
Erg -- assembly - yuk! :)
Actually, I8, I'm keen to get more into this assembly lark (I think a good C++ programmer should have some grounding) are you aware of any good books for complete idiots? I've done some scouring but found nothing useful. Of course, I'm more interested in theory than platform specific stuff.
BTW: No points for answering this -- but if you want I could open it as a proper Q... if you like :-p
0
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Expert Comment
ID: 20287728
>> Thanks "I8" -- you're name's too long to type so I have adopted this pseudo name -- I hope that's ok :)
Call me 8 lol
For assembly : to be honest, I'm mostly self-taught. Since there are so many different platforms that each have their own version of assembler, it's pretty much impossible to have one book that covers all. Most of the time, the official architecture manuals are the most interesting. Here's a nice collection for Intel for example :
http://www.x86.org/intel.doc/inteldocs.htm
0
LVL 40
Expert Comment
ID: 20287760
>> Call me 8 lol
Tsk.. that would just be rude -- ok, if you insist, 8 :)
>> For assembly : to be honest, I'm mostly self-taught.
I like to think I am a damn good C++ programmer (there's nothing like a self inflated ego eh?) but frankly I wouldn't know a jmp from a cmp. I guess I shall just plug on with MASM and the few tutorials I have found.
I suppose I could also follow some of the threads in the Assembly zone -- like the bomb one you diffused :)
Many thanks 8!
-Rx.
0
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Expert Comment
ID: 20287851
>> like the bomb one you diffused
Those type of exercises are indeed fun challenges that teach you a lot about assembler. Just dive into the deep ... It's the best way to learn how to swim ... maybe lol
0
LVL 18
Expert Comment
ID: 20454501
Hi,
Not exactly what you are looking for, but I'd like to share an interesting research I did when working with IBM.
AIX, the IBM's Unix implementation for the RISC/6000 servers and workstations, had 3 compilers, xlc, xlp and xlf, respectively C, Pascal and Fortran. I have compiled a very single 'for' loop with a single instruction to increment an integer. I have used a command line option, like the proposed here, to generate Assembly code from the 3 languages . Surprisely, the result was absolutely the same for the 3 codes. This was around 1991. I think nowadays the compilers, mainly C or C++, are still better, and capable of very optimized assembly. So, it could be very instructive to watch ASM code from these compilers.
Anyway you'll notice that the math part could be human readable code. But the code for printf, for instance, would be really hard to understand...
Now, related to your question (I hope...)
To create an ASM code for printf you must prepare the pointers to your text before call an Operating System API function to send it to the display. So, besides your pure ASM code to the math, you must call Windows (or Linux if you work in that environment) to do interfaces and I/O.
Also, you'll spend a huge time to develop your own formatting routine to transform the float number at memory to characters in the screen (the '%f' printf formatting directive, for instance).
Then, if you are looking for SPEED, it could make sense to code the math in Assembler, but not the printf. You'll have no gain in the printf and will have a huge work for nothing.
Jose
0
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ID: 20454550
0
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Expert Comment
ID: 20946091
Same as his other question ... Some good advice was given, and then dnice143 disappeared ...
0
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Expert Comment
ID: 20946426
The following both answer the OPs Q: -
{ http:#20283774 }
{ http:#20287566 }
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# Inequality derived from a famous problem
Let's have the following inequality: $$\frac{2}{3}(\sqrt 5-1)^3\lessgtr\sqrt[3]{2}$$.
Which part is greater, the left or the right? No calculator solutions are accepted.
• The doubling of the cube problem remained unsolved for two thousand three hundred years. One hundred fifty years ago someone proved that the doubling of the cube is impossible with numbers constructible with compass and straight edge. But when we allow other regular solids, such as cones and spheres, then the doubling of the volume by compass and straight edge is achievable. Let's have a cone with $r =(\sqrt{5}-1)$ and $h=2(\sqrt{5}-1)$, the $V_c=\frac{2}{3}(\sqrt{5}-1)^3\pi$. A sphere with the same radius has $V_s=\frac{4}{3}(\sqrt{5}-1)^3\pi$ which is exactly twice the volume of the cone. Commented May 24, 2021 at 0:43
The LHS is
smaller
than the RHS. Here's how we're going to prove it.
First, we'll turn the requested comparison into one involving the famous number $$\frac{1+\sqrt5}2$$. Call this $$\phi$$ and write $$\bar\phi=\frac{1-\sqrt5}2$$; we have the following key facts. (1) $$\phi\cdot\bar\phi=-1$$ (which is how we turn the original comparison into one involving $$\phi$$). (2) $$\phi,\bar\phi$$ are what are called conjugates, which has the consequence that various things built from them are integers. (3) Specifically, the numbers $$\phi^n+\overline\phi^n$$ are integers when $$n$$ is an integer, and there's a nice way to calculate them. (4) $$\bar\phi$$ is small, which means that the values of $$\phi^n$$ are well approximated by these integers. Putting these facts together will lead us where we need to go.
OK, let's get started. We want to know whether $$\frac23\left(\sqrt5-1\right)^3<\sqrt[3]2$$. So,
get it in terms of $$-\bar\phi=\frac{\sqrt{5}-1}2$$ by pushing some factors of 2 around: $$\frac{16}3\left(\frac{\sqrt5-1}2\right)^3<\sqrt[3]2$$;
exploit the relationship between $$\phi,\bar\phi$$: $$\frac{16}3\left(\frac{\sqrt5+1}2\right)^{-3}<\sqrt[3]2$$;
multiply both sides by that thing involving $$\phi$$: $$\frac{16}3<\sqrt[3]2\left(\frac{\sqrt5+1}2\right)^3$$;
get rid of the cube root by cubing both sides: $$\frac{4096}{27}<2\left(\frac{\sqrt5+1}2\right)^9$$;
cancel a factor of 2 from both sides: $$\frac{2048}{27}<\left(\frac{\sqrt5+1}2\right)^9$$.
That's the first step completed. Now
let's look at the thing we now have on the RHS. It's $$\phi^9$$. So, continuing with our plan, write $$u_n=\phi^n+\bar\phi^n$$. If you replace $$\phi,\bar\phi$$ with their explicit expressions in terms of $$\sqrt5$$ and expand out the n'th powers using the binomial theorem, you'll see that all the terms with an odd number of factors of $$\sqrt5$$ cancel out, so $$u_n$$ is rational. In fact, we can do better than that; a quick calculation shows that $$u_n$$ obeys the same recurrence relation as the Fibonacci numbers do ($$u_{n+2}=u_{n+1}+u_n$$), and that $$u_0=2,u_1=1$$ -- so in fact all the $$u_n$$ are integers, and we can easily calculate them with that recurrence relation. For obvious reasons we are interested in $$u_9$$ which comes out to be 76.
Nearly there. Finally,
let's see how far away from 76 $$u_9$$ actually is. The difference is $$\bar\phi^9$$. Remember that $$\left|\bar\phi\right|<1$$; in fact $$\bar\phi\cong-0.618$$. So its 9th power should be pretty small. But we can be more precise than that; remember that $$\phi\cdot\bar\phi=-1$$, so $$\left|\bar\phi^9\right|$$ is the reciprocal of $$\phi^9$$ -- which, since certainly $$\left|\bar\phi^9\right|<1$$, is bigger than $$u_9-1=75$$. Hence in fact $$\left|\bar\phi^9\right|<1/75$$, which means that $$|\phi^9-76|<1/75$$. And now we really are done, because 2048/27=76-4/27 which is certainly less than 76 by more than 1/75. So the LHS is smaller than the RHS.
(The above is fairly long, but only because I've gone into quite a lot of detail. The actual calculation is rather quick, and for those familiar with these ideas each step is more or less the "obvious" one.)
• That conjugate trick is actually quite pretty. Commented May 24, 2021 at 0:34
• Glad you like it! :-) Commented May 24, 2021 at 0:42
• Could you elaborate on the steps a bit more? I'm sure this solution is correct but the explanation is hard to follow. Commented May 24, 2021 at 1:23
• How about now? I've been more explicit about several of the steps, and put a sort of outline of the strategy and key ideas at the start. Commented May 24, 2021 at 2:03
• Thank you, the explanation is very educational and is much easier to follow! Commented May 24, 2021 at 2:17
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Go to content
A car moves with constant acceleration. The green control panel contains text fields where you can vary the values of initial position, initital velocity and acceleration (don't forget to press the "Enter" key!). By using the buttons at the top right you can bring back the car to its initial position or stop and resume the simulation. If you choose the option "Slow motion", the movement will be ten times slower.
Three digital clocks indicate the time elapsed since the start. As soon as the car has reached the green respectively red light barrier with its front bumper, the corresponding clock will stop. Both light barriers are adjustable by dragging the mouse with pressed mouse button.
Three diagrams illustrate the motion of the vehicle:
* Position x versus time t
* Velocity v versus time t
* Acceleration a versus time t
Based on free Java applets from Java Applets on Physics
## Glossary
### acceleration
the rate of change of velocity with time. It is a vector quantity with magnitude and direction.
### constant
a value in a formula or equation that cannot change.
### light
having negligible mass.
### velocity
the rate of change of displacement. It is a vector quantity with magnitude and direction.
Full Glossary List
## This question appears in the following syllabi:
SyllabusModuleSectionTopicExam Year
AQA A-Level (UK - Pre-2017)M1KinematicsConstant acceleration-
AQA AS Maths 2017MechanicsKinematics and UnitsConstant Acceleration Formulae-
AQA AS/A2 Maths 2017MechanicsKinematics and UnitsConstant Acceleration Formulae-
CCEA A-Level (NI)M1KinematicsConstant acceleration-
CIE A-Level (UK)M1KinematicsConstant acceleration-
Edexcel A-Level (UK - Pre-2017)M1KinematicsConstant acceleration-
Edexcel AS Maths 2017MechanicsConstant Acceleration MotionConstant Acceleration Formulae-
Edexcel AS/A2 Maths 2017MechanicsConstant Acceleration MotionConstant Acceleration Formulae-
OCR A-Level (UK - Pre-2017)M1KinematicsConstant acceleration-
OCR AS Maths 2017MechanicsWorking with MotionConstant Acceleration Formulae-
OCR MEI AS Maths 2017MechanicsWorking with MotionConstant Acceleration Formulae-
OCR-MEI A-Level (UK - Pre-2017)M1KinematicsConstant acceleration-
Pre-U A-Level (UK)MechKinematicsConstant acceleration-
Universal (all site questions)KKinematicsConstant acceleration-
WJEC A-Level (Wales)M1KinematicsConstant acceleration-
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# How Many Solar Panels For 5Kw Inverter?
How many solar panels will you need for 5kW? – To make up a 5kW solar system, you need 14 solar panels, assuming you use 370W panels – that will actually give you 5.18kW. Each panel will be about 1.8 metres x 1 metre, so you’ll need at least 25.2m² of roof space. To give you a feel for how big 25.2m² is, this picture may help:
## How many solar panels are in a 5kW system?
How Much Energy Does a 5kW Solar Panel System Produce? – A 5kW solar panel system generates approximately 20 kWh on a good day with sunshine and around 4,500 kWh of electricity is generated throughout the year. The real power generated will depend on various factors such as the location, performance of the equipment and the installation.
### How many solar panels can be connected to an inverter?
How many solar panels can I connect to my inverter? – The number of solar panels you can connect to your inverter is identified by its wattage rating. For example, if you have a 5,000 W inverter, you can connect approximately 5,000 watts (or 5 kW) of solar panels. Using 300 W solar panels, you could then connect roughly 17 solar panels (5000 W / 300 W per panel).
You might be interested: How Long Do Solar Flares Last?
#### How many 300W solar panels do I need for 5KVA inverter?
To make up a 5kW solar system, you need 17 solar panels, assuming you use 300W panels – that will actually give you 5.1kW.
### How long can 5kVA inverter last?
Product Description – 5kVA inverter with 4 pieces of 200 Ah battery is most inverter combo set which is designed for home where power cuts are frequent and for long hours, 15- 12 hours.5 kVA inverter is good enough for 10- 14 bhk home for running 20-25 led bulbs,3-5 fans, 4 television, 1 mixer, and small 160-liter refrigerator during power cuts.
## How many solar panels do I need for a 5kVA 48v inverter?
How many solar panels will you need for 5kW? – To make up a 5kW solar system, you need 14 solar panels, assuming you use 370W panels – that will actually give you 5.18kW. Each panel will be about 1.8 metres x 1 metre, so you’ll need at least 25.2m² of roof space. To give you a feel for how big 25.2m² is, this picture may help:
### How much load can a 5kVA inverter take?
Prices Of 5kVA Inverters In India – The cost of a solar inverter is roughly 25-30% of the total cost of a solar panel system. Solar inverter prices are influenced by several factors, including inverter technology, efficiency, warranty period, and brand quality.
• The overall 5kVA inverter price also includes transportation, installation,, and all additional expenditures.
• Investing in a solar inverter might seem an expensive move initially, but it is worth the cost in the longer run.
• The 5kVA solar inverter price varies according to the specification of the inverter.
You might be interested: Yang Merupakan Galaksi Tempat Tata Surya Kita Berada Adalah?
A normal inverter is a non-solar inverter with a starting voltage range of 650VA to 1800VA and a price range of ₹ 3,500 to ₹ 9,500, depending on brand, performance, and warranty. The overall price of a solar inverter ranges from ₹7,000 to ₹ 1,50,000.
### How many solar panels do I need for a 5.5 kW inverter?
Did you know that 5.5kW solar power systems can consist of a different number of panels depending on the size of the solar panels? Here are some common panel sizes which could make up a 5.5kW system:
330W (17 x solar panels to make 5.61kW) 350W (16 x solar panels to make 5.60kW) 370W (15 x solar panels to make 5.55kW) 390W (14 x solar panels to make 5.46kW) 400W (14 x solar panels to make 5.60kW) 420W (13 x solar panels to make 5.46kW) 450W (12 x solar panels to make 5.40kW) 480W (12 x solar panels to make 5.76kW) 500W (11 x solar panels to make 5.50kW)
## What load can 5kVA inverter carry?
Prices Of 5kVA Inverters In India – The cost of a solar inverter is roughly 25-30% of the total cost of a solar panel system. Solar inverter prices are influenced by several factors, including inverter technology, efficiency, warranty period, and brand quality.
The overall 5kVA inverter price also includes transportation, installation,, and all additional expenditures. Investing in a solar inverter might seem an expensive move initially, but it is worth the cost in the longer run. The 5kVA solar inverter price varies according to the specification of the inverter.
A normal inverter is a non-solar inverter with a starting voltage range of 650VA to 1800VA and a price range of ₹ 3,500 to ₹ 9,500, depending on brand, performance, and warranty. The overall price of a solar inverter ranges from ₹7,000 to ₹ 1,50,000.
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GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# From 400 to 700?
Author Message
Intern
Joined: 20 Jun 2018
Posts: 1
### Show Tags
27 Jun 2018, 23:29
1
Hi Gmat Brotherhood,
Currently, I am really unmotivated and discouraged that I will do well on the Gmat, as I did my first practice test and only scored 400 Points ...
I will write the GMAT in approx. 3-4 month.
Do you think it is possible to achieve a score over, lets say, at leats 650? Or even better 700?
FYI: Actually I have learnt all the concepts already at some Point in time during my School years, so they are not new for me, I just forgot them somehow.
Wish you the best and thank you for your Input,
CP
examPAL Representative
Joined: 07 Dec 2017
Posts: 841
Re: From 400 to 700? [#permalink]
### Show Tags
28 Jun 2018, 02:48
CaptainPatch wrote:
Hi Gmat Brotherhood,
Currently, I am really unmotivated and discouraged that I will do well on the Gmat, as I did my first practice test and only scored 400 Points ...
I will write the GMAT in approx. 3-4 month.
Do you think it is possible to achieve a score over, lets say, at leats 650? Or even better 700?
FYI: Actually I have learnt all the concepts already at some Point in time during my School years, so they are not new for me, I just forgot them somehow.
Wish you the best and thank you for your Input,
CP
Hey CaptainPatch
I have no idea what your abilities are, so I can't say if 650-700 is realistic, but what I can tell you is that I wouldn't worry too much about the score on the first practice test, if you haven't studied before it.
Everyone has studied the material before, and pretty much everyone forgets it by the time they approach the GMAT. That's totally natural - it is necessary to re-learn it. Even after that, the material is not the main issue: what the the GMAT tests is your ability to answer quickly.
Maybe you'll find this blog post useful:
https://exampal.com/gmat/blog/prepare-gmat-60-days/
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Re: From 400 to 700? [#permalink]
### Show Tags
28 Jun 2018, 05:51
CaptainPatch wrote:
Hi Gmat Brotherhood,
Currently, I am really unmotivated and discouraged that I will do well on the Gmat, as I did my first practice test and only scored 400 Points ...
I will write the GMAT in approx. 3-4 month.
Do you think it is possible to achieve a score over, lets say, at leats 650? Or even better 700?
FYI: Actually I have learnt all the concepts already at some Point in time during my School years, so they are not new for me, I just forgot them somehow.
Wish you the best and thank you for your Input,
CP
The time you have on hand is enough to bump your score significantly. By how much you can improve will depend on how wisely you utilize the time and the way you study. The more self-aware you are, the more you will be able to identify and work on your weak areas. Good luck!
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Re: From 400 to 700? [#permalink]
### Show Tags
28 Jun 2018, 12:39
Hi CaptainPatch,
Before I can offer you the specific advice that you’re looking for, it would help if you could provide a bit more information on how you've been studying and your goals:
Studies:
1) How long have you studied?
2) What study materials have you used so far?
3) What were the Quant and Verbal Scaled Scores for this 400?
Goals:
4) When are you planning to apply to Business School?
5) What Schools are you planning to apply to?
GMAT assassins aren't born, they're made,
Rich
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Joined: 16 Aug 2015
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Re: From 400 to 700? [#permalink]
### Show Tags
09 Jul 2018, 02:08
Hi CaptainPatch,
You can surely improve your score within 3-4 months! If you want 700, a good mix of 700 is Q49 and V35. If you do better on verbal, you quant score can be lower than 49. Since we are specialized in math, we can give you advice on math only.
Based on the score, it seems like you need to work on your basic math skills first before studying GMAT Math.
It seems like you already know the concepts but you didn't study them for many years so all you need is to go over the basics to refresh your memory
If you would like to study basics first, you can also learn them on our website as we offer them as free lessons.
Also, the other reason for your low score is that you don't seem to understand how GMAT Math works.
In fact, there are two important things you need to know about GMAT Math;
Firstly, GMAT is a logic test, not a general math test. That is, you need to learn the logic to tackle GMAT questions.
You can still solve GMAT questions with the conventional method.
If you aim to hit a high score in quant, the conventional method is time-consuming and tends to be inefficient. You may hit Q45 or above with the conventional way, but it is difficult or very long to study until you hit Q49-51. Hence, we strongly recommend learning the logic, which you can learn from our online course.
The other important factor is to practice the most current GMAT questions. A number of questions you get to practice online tend to be out-of-date and they might not be that relevant to the current types of GMAT questions. If you are not used to the current types of questions, this could also be a reason to why your score is not improving. However, all the questions you get to practice from our online course are developed based on the current type of GMAT questions.
Most importantly, we can help you with the two key factors if youw ant to succeed in GMAT,
You may try our Free Resources including 4-hour video lesson, Diagnostic Test and Mock Test.
www.mathrevolution.com
If you have any further question, you can always reach out to us at info@mathrevolution.com
Best regards,
Jin
Math Revolution
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Re: From 400 to 700? [#permalink]
### Show Tags
10 Jul 2018, 09:29
CaptainPatch wrote:
Hi Gmat Brotherhood,
Currently, I am really unmotivated and discouraged that I will do well on the Gmat, as I did my first practice test and only scored 400 Points ...
I will write the GMAT in approx. 3-4 month.
Do you think it is possible to achieve a score over, lets say, at leats 650? Or even better 700?
FYI: Actually I have learnt all the concepts already at some Point in time during my School years, so they are not new for me, I just forgot them somehow.
Wish you the best and thank you for your Input,
CP
Hi
It is possible to get 700 score with sound strategies and study plan after 400 diagnostic test score. Here is a study plan for you
Prep Material
For Concept Learning
Manhattan Quant Guides
Manhattan Verbal Guides
For CR: The Powerscore GMAT Critical Reasoning Bible
For RC: Aristotle RC Grail
For Practice
The Official Guide for GMAT 2015-18
The Official Guide for GMAT Quantitative Review 2015-18
The Official Guide for GMAT Verbal Review 2015-18
You can start with Quant or Verbal which suits you. If you have started with Quant then Start with the Arithmetic but if started with verbal then start first with Sentence correction. One month for learning Quant concepts and one month for practicing question and same practice for Verbal. During you Practicing question don't forget to make an error log to track your weak areas after practice. Once you know your weak areas revise your Concepts related to those areas and do some more Practice. 6-8 CATs are enough for practice the real tests. Make your Stamina for sitting 3 hours in the test and don't study more than 2 hours in one sit and 4 hours per day
Top CATs for Practice
1. Official GMAC CATs
2. Manhattan CATs
3. Kaplan CATs
4. GMAT Club Quant CATs
Good Luck
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Re: From 400 to 700? [#permalink]
### Show Tags
11 Jul 2018, 09:01
CaptainPatch wrote:
Hi Gmat Brotherhood,
Currently, I am really unmotivated and discouraged that I will do well on the Gmat, as I did my first practice test and only scored 400 Points ...
I will write the GMAT in approx. 3-4 month.
Do you think it is possible to achieve a score over, lets say, at leats 650? Or even better 700?
FYI: Actually I have learnt all the concepts already at some Point in time during my School years, so they are not new for me, I just forgot them somehow.
Wish you the best and thank you for your Input,
CP
Hi CaptainPatch,
Yes it's definitely possible. You should dedicate around 3 months to improve your score. It’s a good thing you have taken a GMAT Mock once. You now know your weaknesses and work on them. If you are willing to study dedicatedly for that period, you are sure to achieve your goal. I think you need to solidify you base and adopt a proper technique to answer the questions. I believe you may benefit from taking a GMATPREP course. If you are willing, there are some great GMAT prep companies that can help you with your preparation.
In order to make an informed decision I would highly encourage you to go to their websites and try on their free trial and decide for yourself which one do you like better. You try out free access to EmpowerGMAT, Magoosh and Optimus Prep as they have great reviews on GMATCLUB.
Also for verbal, I would highly encourage you to consider e-gmat verbal online or the e-gmat verbal live course. They are both amazing courses especially designed for non-natives. They offer almost 25% of their courses for free so you can try out their free trial to decide which one you want to go for. Plus the e-gmat Scholaranium which is included in both the courses is one of the best verbal practice tools in the market. You can easily track your progress in that you can identify your strengths and analyze and improve on your weak areas.
You can also try out the MGMAT guides they are phenomenal and cover the entire syllabus really well. Just by going through these guides and solving the OG will help you reach 600+.I must add that if you are particularly looking to discover and improve on your weak areas in Quant; a subscription to GMATCLUB tests is the best way to do that. They are indeed phenomenal and will not only pinpoint your weak areas but also help you improve on them.
Further taking multiple mocks might help. Apart from the GMATPREP, Manhattan GMAT tests and Veritas Prep Tests in my experience have good verbal and Quant section and will certainly help you point out and improve your weak areas.
Further another advantage of taking many mocks is to build up your stamina. Apart from the GMATPREP tests, taking practise tests of any major GMATPREP company ought to do that.
Lastly, you can check out a very interesting article by Mike McGarry from Magoosh detailing a 3 month study plan
https://magoosh.com/gmat/2012/3-month-g ... -students/. You will find it very helpful as it gives out a study plan as per your needs.
Hope this helps. All the best.
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Joined: 25 May 2017
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Location: United States (NY)
GMAT 1: 650 Q42 V38
GRE 1: Q161 V164
Re: From 400 to 700? [#permalink]
### Show Tags
13 Jul 2018, 06:37
CaptainPatch wrote:
Hi Gmat Brotherhood,
Currently, I am really unmotivated and discouraged that I will do well on the Gmat, as I did my first practice test and only scored 400 Points ...
I will write the GMAT in approx. 3-4 month.
Do you think it is possible to achieve a score over, lets say, at leats 650? Or even better 700?
FYI: Actually I have learnt all the concepts already at some Point in time during my School years, so they are not new for me, I just forgot them somehow.
Wish you the best and thank you for your Input,
CP
4 months? no
you're better off cancelling the test, and not rescheduling one until you achieve a three test trailing average of at least 650. otherwise, you'll be wasting your money
_________________
feel free to reach out ⁓
Re: From 400 to 700? &nbs [#permalink] 13 Jul 2018, 06:37
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# Chemistry
Find the pH of mixture of acids. 0.185 M in HCHO2 and 0.225 M in HC2H3O2
Im using an ice chart of weak acid and putting in strong acid in H+ initiAL concentration. I've done the problems many different ways but cannot seem to get the right answer help please?Answer is pH of 2.19.
You said...
You must recognize that this is a mixture of two weak acids; i.e., formic acid and acetic acid. I looked up Ka for both and used 1.77E-4 for Ka HCOOH and 1.8E-5 for Ka CH3COOH.
Calculate the H^+ from the strong acid, then add the H^+ from the weak acid. Formic acid first since it is the stronger. .........HCOOH ==> H^+ + HCOO^-initial..0.185.....0......0 change...-x........x......x equil..0.185-x.....x......x
Ka = (H^+)(HCOO^-)/(HCOOH) Solve for H^+. This is what acts as the common ion (remember Le Chatelier's Principle). This causes the acetic acid to ionize less than it would if were just acetic acid solution.
........CH3COOH ==> H^+ + CH3COO^-initial..0.225.......0......0 change....-x........x........x equil...0.225-x......x........x
Ka acet acid = (H^+)(CH3COO^-)/(CH3COOH) Substitute TOTAL H^+ into the Ka expression for CH3COOH. That will be about 0.00572 from HCOOH from the above calculation plus xfrom this ionization. Solve for x,add this (H^+) to the 0.00572 from HCOOH, then convert to pH. I obtained pH = 2.19
I STILL CAN'T GET 2.19???
I got concentration of HCOOH rxn to be 5.77X10^-3. And the CH3COOH rxn to have concentration of 2.01X10^-3. I added those together to get 7.78X10-3. What do you mean add to ka expression? I keep getting 1.35 now.
1. 👍
2. 👎
3. 👁
1. You should have told me what Ka values you are using. Not all texts have the same values although they are close. I'm using 1.77E-4 for formic acid and I will call that HF. I know that isn't formic acid and you know that, too, but it saves some space on the line. HAc is acetic acid and I'm using Ka for HAc of 1.8E-5.
............HF ==> H^+ + F^-
initial....0.185...0......0
change......-x......x.....x
equil.....0.185-x...x.....x
1.77E-4 = (x)(x)/(0.185-x)
I'm not going to do this step by step but this is the way you set it up. You should get an answer for x = 0.00572 if you assume 0.185-x = 0.185.
.............HAc ==> H^+ + Ac^-
initial......0.225....0.....0
change.......-x.......x......x
equil.....0.225-x.....x.......x
Ka = 1.8E-5 = (0.00572+x)(x)/(0.225-x) and solve for x
(Note:I suspect you didn't substitute the 0.00572 here. That's a common error.)
If I assume 0.00572+x = 0.00572 and 0.225-x = 0.225, then x = 7.2E-4
Then 0.00572 + 7.2E-4 = ? and -log of that is 2.19. Voila!.
That's all I did earlier in the day when I first responded to your post. You may ask what happens if we don't make those assumptions so here is what you get.
For formic acid, x = 0.00563 instead of 0.00572 (hardly worth talking about).
For acetic acid, x = 0.000644 (again, not much difference)
So 0.00563 + 0.000644 = 0.00627 and the pH =2.20
Let me know if you don't understand what I did. The only thing I've omitted is the algebra.
1. 👍
2. 👎
2. Oh okay yeah I didn't add the 5.77x10^-3 to the weaker acid. So I determine concentration of stronger acid then put that H concentration into the weaker acid then I determine the concentration of the H of CH3COOH and get 7.02x10^-4 and add the H+ concentration of 5.77x10^-3 and get 6.47X10^-3 and take pH of it to get 2.19!!! Ah I got it now!!!
But one question, why do we add the 5.77x10^-3 again in the end if we added it to the equilibrium expression? Why do we have to do that?
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3. Since the HAc is the weaker acid, the formic acid acts, according to Le Chatelier's Principle, to shift the weaker acid to the left.
HAc ==>H^+ + Ac^-
Adding H^+ from the other acid make HAc ionize to a smaller extent. You can work out how much it would ionize on its own and that is about 0.002 so you can see that it ionizes in the presence of formic acid much less. Back to the point, so that is done to calculate the amount acid contributed by HAc. Then you add the amount contributed by HAc to the amount contributed by the formic acid and calculate pH from the total H^+.
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•Which of the following solutions will be the best buffer at a pH of 9.26? (Ka for HC2H3O2 is 1.8 x 10–5; Kb for NH3 is 1.8 x 10–5.) a)0.20 M HC2H3O2 and 0.20 M NaC2H3O2 b)3.0 M HC2H3O2 and 3.0 M NH4Cl c)0.20 M NH3 and 0.20
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Woof
Topic Author
Posts: 84
Joined: Tue Jul 21, 2009 9:00 am
### Volumetric Weight - Perfect Goods
I've had to ship lots of things overseas through the years, and I've yet to ship anything whos weight and volume matched the carriers volumetric weight calculations. The items I've shipped have either been much heavier than their volumetric size, or much larger than their volumetric weight.
Today, for example, I've shipped some old cockpit panels from a 737. The package weighed just 2kg, but was 95x30x50cm in size, so I paid for a volumetric weight of 24kg.
So, I'm wondering what single material, generally, would create a volumetric weight of 1:1? I thought this might be water, which weights 1g per cubic centimeter, so 1kg of water would take up 10x10x10cm, but this equates to 0.5kg in volumetric weight... so I have no clue.
Anyone help?
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oly720man
Posts: 5808
Joined: Fri May 21, 2004 7:13 am
### RE: Volumetric Weight - Perfect Goods
Quoting Woof (Thread starter):Today, for example, I've shipped some old cockpit panels from a 737. The package weighed just 2kg, but was 95x30x50cm in size, so I paid for a volumetric weight of 24kg.
From this I get a "density" of 168.42kg/m^3 (= 24/(0.95*0.3*0.5))
So if you had a box that was 1m x 1m x 1m and weighed 168.42kg then you'd be at 1:1.
Ground cork has a density of 160kg/m^3
http://www.simetric.co.uk/si_materials.htm
so a large box full of bits of cork would probably do it. Not a single material I know, but something like cork will be there or thereabouts. Bulk density of cork is 240kg/m^3.
wheat and dairy can screw up your brain
petertenthije
Posts: 3426
Joined: Tue Jul 10, 2001 10:00 pm
### RE: Volumetric Weight - Perfect Goods
Volume/metric weight is also dependant on the mode of transit and even the individual carrier involved. In road transport and ocean (LCL) a volume/metric of 333 KGs per Cu-M is most common in Europe. From what I hear from my US collegues, they usually calculate 250 KGs per Cu-M.
I do not know what the standard is for airfreight.
The reason that, in Europe, the default is 333 KGs is very simple: it's to max out a truck. A typical trailer will carry up to 24000 KGs and has a maximum capacity of 89 Cu-M ("tautliner": 13,6M x 2,45M x ~2,7M). However, with standard pallets you would normally end up loading a truck at ~72 Cu-M. Divide 24000 by 72 and you get 333.
[Edited 2010-09-15 09:19:59]
Attamottamotta!
Woof
Topic Author
Posts: 84
Joined: Tue Jul 21, 2009 9:00 am
### RE: Volumetric Weight - Perfect Goods
Thanks for the replies guys. The moral of the story for me is...
If you are shipping goods that are not the perfect weight / size combination, try and combine them with goods that make the total package as close to the ideal mass as possible. I'm sure this is 'sucking eggs' for those in the know, but it will save me some money in the future.
Deleted
petertenthije
Posts: 3426
Joined: Tue Jul 10, 2001 10:00 pm
### RE: Volumetric Weight - Perfect Goods
Quoting Woof (Reply 3): If you are shipping goods that are not the perfect weight / size combination, try and combine them with goods that make the total package as close to the ideal mass as possible.
Yes, preferably without increasing the dimensions.
So if you got a big box of 1,0 Cu-M (Chg-Wght 333 KGs) that weighs 250 KGs, then you can in essence freely add ~80 KGs as long as the size of the box does not increase. So if you got some unused space in your box, you migt as well use it as your shipping price won't be affected. You might have some additional handling or customs costs, assuming they are charged in real KGs as opposed to Chg-Kgs, but that should be relatively small change.
Note: above example uses road / LCL weights. The chargeable weight for air might be different!
Attamottamotta!
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Government Aircraft Aircraft flying government officials
Helicopters Our large helicopter section. Both military and civil versions
Blimps / Airships Everything from the Goodyear blimp to the Zeppelin
Night Photos Beautiful shots taken while the sun is below the horizon
Accidents Accident, incident and crash related photos
Air to Air Photos taken by airborne photographers of airborne aircraft
Special Paint Schemes Aircraft painted in beautiful and original liveries
Airport Overviews Airport overviews from the air or ground
Tails and Winglets Tail and Winglet closeups with beautiful airline logos
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## Challenge #100: Find Ned!
Director, Customer Enablement
@lminors - link is back online! Sorry for the hiccup!
Alteryx
Got within 0.3m in 3.3 seconds.
My solution:
Spoiler
Tried several different approaches, including buffering the network to refine, but decided that iterating through reducing grid-sizes for the grid furthest from the road network (using find nearest) was the optimal solution.
Another little trick I used to optimise was to use the records from file load within the find nearest tool, which shaved several seconds off!
Fireball
This one took me a while!
Spoiler
Result:
Spoiler
As others did, I used the Find Nearest tool with increasingly smaller grid squares. For each iteration, I kept the 10 closest grid squares to limit the data size.
I used 10-km, then 0.5-km, then 0.05-km (50 m) grids.
Alteryx Certified Partner
Asteroid
Hi! Here my solution :)
Spoiler
I tried it first the way of 'brute force' solution...but after processing +50GB and not even 1% of total, i changed to an approximation
First, I looked for 10x10km areas where there were no roads...then, modelling those areas with points 100m distance from one to another...and again looking for the maximum-minimum distance...
Pulsar
Spoiler
All I can say about my process to get this answer is yikes...
Highlighted
Alteryx Certified Partner
Spoiler
Haven't used spatial tools much , and so I looked for some ideas. In the end, I took @David-Carnes' idea of using a couple of grids and the find nearest tool to get nearer and nearer to Ned's location. Excellent challenge though!
Alteryx Certified Partner
Spoiler
I tried different approaches, but ended up with some iterations. 12.2 seconds seems to be okay ...
Comet
Didn't come up with an efficient method and the method I wanted to take takes too long to solve, so ended up with this which takes over 2 minutes to solve.
Spoiler
So the way I originally approached this was to take the road network and add a buffer around each point, to maximise the chance that it would create a complete network. Then summarised this to get a polygon for the entire road network. Then doing a spatial process with the State polygon and splitting to end up with the holes/regions which I could then use a spatial process to get the centroid of and calculate the distance to nodes on the road network and sorting to find the furthest point but this took too long to solve (over 15 mins and still without an answer).
So instead took a similar approach to others finding smaller grids with no roads in until you get to a solution (worked with 100m grids).
Asteroid
Spoiler
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https://discuss.codechef.com/t/can-not-found-problem-in-my-solution/100680
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# Can not found problem in my solution
Anyone help me to find problem in my solution because it cannot pass hissen test case.
#include <bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
int t;
cin>>t;
while(t–){
long long a,b;
cin>>a>>b;
if(a%2==0&&b%2==0){
cout<<string(a/2,‘a’)+string(b,‘b’)+string(a/2,‘a’)<<endl;
cout<<string(b/2,‘b’)+string(a,‘a’)+string(b/2,‘b’)<<endl;
}
else if(b%2==0){
cout<<string(a/2,‘a’)+string(b/2,‘b’)+‘a’+string(b/2,‘b’)+string(a/2,‘a’)<<endl;
cout<<string(b/2,‘b’)+string(a,‘a’)+string(b/2,‘b’)<<endl;
}
else if(a%2==0){
cout<<string(b/2,‘b’)+string(a/2,‘a’)+‘b’+string(a/2,‘a’)+string(b/2,‘b’)<<endl;
cout<<string(a/2,‘a’)+string(b,‘b’)+string(a/2,‘a’)<<endl;
}
else{
cout<<-1<<endl;
}
}
return 0;
}
Hi there.
For someone to be able to help you, you will need to provide a link to the problem you’re trying to solve. Also, it helps a lot if you format your code properly, like this:
``````#include <bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
int t;
cin >> t;
while (t--) {
long long a, b;
cin >> a >> b;
if (a % 2 == 0 && b % 2 == 0) {
cout << string(a / 2,'a') + string(b,'b') + string(a / 2,'a') << endl;
cout << string(b / 2,'b') + string(a,'a') + string(b / 2,'b') << endl;
} else if (b % 2 == 0) {
cout << string(a / 2,'a') + string(b / 2,'b') +'a'+ string(b / 2,'b') +
string(a / 2,'a')
<< endl;
cout << string(b / 2,'b') + string(a,'a') + string(b / 2,'b') << endl;
} else if (a % 2 == 0) {
cout << string(b / 2,'b') + string(a / 2,'a') +'b'+ string(a / 2,'a') +
string(b / 2,'b')
<< endl;
cout << string(a / 2,'a') + string(b,'b') + string(a / 2,'a') << endl;
} else {
cout << -1 << endl;
}
}
return 0;
}
``````
On the presumption that this is PALINPAIN, your code fails on this test case, for which the expected answer is ‘-1’:
1
2 1
[Edit: linked to wrong problem, sorry]
Try this case:
7
1 1
1 2
2 1
1 3
3 1
1 4
4 1
should all be -1, yours is
-1
bab
bab
aba
aba
-1
-1
bbabb
bbabb
aabaa
aabaa
I did this but this gave me wrong answer???
#include
#include
#include
#include
using namespace std;
int main(){
int t; cin>>t;
while(t–){
int x,y; cin>>x>>y;
char a=‘a’,b=‘b’;
`````` if(x==1 || y==1 || (x%2==1 && y%2==1))
cout<<-1<<endl;
else{
cout<<string((x/2),a)+string(y,b)+string((x/2),a)<<endl;
cout<<string((y/2),b)+string(x,a)+string((y/2),b)<<endl;
}
}
return 0;
``````
}
Please format your code properly. using ‘pre-formatted text’, otherwise the editor mangles it.
Try this input, which should show you where you’re going wrong:
2
5 4
4 5
Check the following test case:
1 2
Your code output is:
bab
bab
while the correct answer is:
-1
as both palindrome strings are equal, and it is impossible to generate two different palindrome strings in this case. The issue with your solution is that when either X or Y is even and the other number is equal to 1, it is impossible to generate two different palindrome strings in this corner case.
Check the palindrome_pair class in the following solution.
Accepted
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https://gomathanswerkey.com/texas-go-math-grade-3-lesson-20-3-answer-key/
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# Texas Go Math Grade 3 Lesson 20.3 Answer Key Spending Decisions
Refer to our Texas Go Math Grade 3 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 3 Lesson 20.3 Answer Key Spending Decisions.
## Texas Go Math Grade 3 Lesson 20.3 Answer Key Spending Decisions
Unlock the Problem
To purchase is to obtain by paying money for an item or a service. When you spend money to purchase something, you give it to someone else in exchange for something.
Carlos wants to purchase a new baseball glove before summer vacation. He has saved $36. The glove costs$48. Carlos plans to save $4 each week. It is 3 weeks before summer vacation starts. Will he be able to save enough money to purchase the baseball glove? Example Planned Spending How much money will Carlos have saved before summer vacation? ________ weeks × ________ = _________$36 + ________ = _________
So, Carlos ________ (will, will not) be able to save enough money to buy the baseball glove.
Planned Spending
How much money will Carlos have saved before summer vacation?
4 weeks × 3 = 12
$36 + 12=Â 48 So, Carlos ________ (will, will not) be able to save enough money to buy the baseball glove. Math Idea Planned spending can help you reach a goal. Unplanned spending is frequently not wise. Unplanned spending may prevent you from reaching a goal. Example Unplanned Spending What if Carlos sees a team tee shirt on sale for$10 next week? Should he make the purchase?
If Carlos buys the tee shirt, he will have
$36 – ________ = _________. In 3 weeks, he will have _________ + _________ = __________ If he buys the tee shirt, Carlos ___________ will , will not have enough money saved to buy the baseball glove before summer vacation starts. Answer: Unplanned Spending What if Carlos sees a team tee shirt on sale for$10 next week? Should he make the purchase?
If Carlos buys the tee shirt, he will have
$36 – 10 = 26. In 3 weeks, he will have 26 + 12 = 38 If he buys the tee shirt, Carlos will not(will , will not )have enough money saved to buy the baseball glove before summer vacation starts. Math Talk Mathematical Processes What if Carlos sees the baseball glove on sale for$39 next week? Would it be a wise or unwise purchase? Explain.
Explanation:
If Carlos sees the baseball glove on sale for $39 next week, It Would not be a wise purchase. Because, He has to buy for 48 and now he will get for only 39 so, he would get loss of 9$
Share and Show
Question 1.
Olga will be teaching a ceramics class, which starts in 6 weeks, and she needs to purchase clay. Fifty pounds of clay costs $54. Olga has saved$21. Her plan is to save $5 each week. Will she be able to save enough money to purchase the clay by the time the class starts? 6 weeks × _________ = ________$21 + _________ = __________
Explanation:
she be not able to save enough money to purchase the clay by the time the class starts.
6 weeks × 5 = 30
$21 + 30 = 51 51 < 54. Question 2. Will Olga have enough money to purchase the clay by the time the class starts if she saves$6 each week?
_________ weeks × ________ = __________
$21 + ________ = __________ Answer: Yes Explanation: Yes, Olga have enough money to purchase the clay by the time the class starts if she saves$6 each week
6 weeks × 6 = 36
$21 + 36 = 57$
57 > 54
Question 3.
Suppose Olga does not have any money saved. How much money would she need to save each week to have enough money to purchase the clay by the time the class starts in 6 weeks? Explain.
Answer: 9$per week Explanation: Olga does not have any money saved. she need to save 9$ each week to have enough money to purchase the clay by the time the class starts in 6 weeks
6 weeks × 9$= 54$
Question 4.
Olga would like to purchase some sunglasses that cost $13. If she uses her savings to buy the sunglasses, and then she saves$8 each week, will she have enough money saved in 6 weeks to buy the clay?
$21 – _________ = _________ 6 weeks × _______ = ________ In 6 weeks, she will have ________ + ________ = _________. Answer: Yes. Explanation: Olga would like to purchase some sunglasses that cost$13.
she uses her savings to buy the sunglasses,
and then she saves $8 each week, she will have enough money saved in 6 weeks to buy the clay$21 – 13 = 8
6 weeks × 8 = 48
In 6 weeks, she will have 48 + 8 =56.
56 > 54
Math Talk
Mathematical Processes
Explain when you should not make an unplanned purchase.
Question 5.
Suppose Olga earns $100 a week. She spends$18 for transportation and $46 for food. She saves$30 each week to pay her rent at the end of the month. Can Olga also save $10 a week to buy clay? Explain. Answer: No Explanation: Total income = 100 Expenditure = 18 + 46 + 30 = 94 No, She cannot buy the chain as she cannot save 10$ per week
100 – 94 = 6$She has only 6$ left.
Problem Solving
Question 6.
H.O.T. Multi-Step Tony is saving for the tablet computer pictured. He has saved $169. How much money does he need to save each week if he plans to use the coupon to buy the tablet? Answer: 20$ a week
Explanation:
The total amount for gadget is 279
If he uses the coupon he get 50 discount
279 – 50 = 220
his savings is 179, 229-169 = 60
60 ÷ 3 = 20$Question 7. Write Math Victoria has$36. She starts saving $10 a week to buy a stuffed bear for her cousin. The bear costs$59. Victoria will visit her cousin in 4 weeks. Would it be a wise unplanned spending decision for Victoria to purchase a pair of boots that cost $30? Explain. Answer: Not a wise decision Explanation: Victoria has$36. She starts saving $10 a week to buy a stuffed bear for her cousin. Victoria will visit her cousin in 4 weeks. 4 x 10 = 40 36 + 40 = 76$
The bear costs $59. 76 – 59 = 17$
Victoria will visit her cousin in 4 weeks. It Would not be a wise unplanned spending decision for Victoria to purchase a pair of boots that cost $30. Because she doesn’t have enough money. Question 8. Tanya has$12 to spend. She will earn $10 a night for babysitting 2 nights next week. She wants to buy a backpack that costs$29. How much can Tanya spend today if she plans to buy the backpack in 2 weeks?
Answer: 3$Explanation: Tanya has$12 to spend.
She will earn $10 a night for babysitting 2 nights next week. The total amount is 12 10 x 2 = 20 20 + 12 = 32$
She wants to buy a backpack that costs $29. 32 – 29 = 3$
Tanya can spend 3$today if she plans to buy the backpack in 2 weeks. Question 9. H.O.T. Miguel wants to take his brother to a movie. The movie tickets cost$6. Miguel is saving $5 a week to buy a caboose for his train set, which costs$62. He has saved $54. Would his unplanned spending decision to go to the movies be wise? Explain. Answer: It is not a wise decision Explanation: Miguel wants to take his brother to a movie. The movie tickets cost$6
Miguel is saving $5 a week to buy a caboose for his train set, which costs$62.
He has saved $54. 54 – 6 = 48 There will not be enough money to buy the caboose for train set. Daily Assessment Task Fill in the bubble for the correct answer choice. Question 10. Multi-Step Sophie plans to spend$10 for a new dog toy. She saves $1 each week for 8 weeks. Then Sophie spends$3 on snacks. How many more weeks will Sophie have to save to have enough money to buy the toy?
(A) 5 weeks
(B) 6 weeks
(C) 3 weeks
(D) 4 weeks
Explanation:
Multi-Step Sophie plans to spend $10 for a new dog toy. She saves$1 each week for 8 weeks.
8 x 1 = 8
Then Sophie spends $3 on snacks. 8 – 3 = 5 5 + 5 = 10 Sophie have to save for more 5 weeks. Question 11. Reasoning Olivia wants to buy a doghouse that costs$150. She saves $4 a week for 5 weeks and then receives a gift of$80. When the doghouse goes on sale for $99, she buys it. Which best describes her spending decision? (A) It is a wise unplanned spending decision, because her savings plus the gift allow her to buy the doghouse on sale. (B) It is an unwise unplanned spending decision because unplanned spending decisions never make sense. (C) It is a wise planned spending decision because she has already saved$150 in 5 weeks to buy the doghouse.
(D) It is an unwise unplanned spending decision because the gift plus her savings are not enough to buy the doghouse.
Explanation:
It is a wise unplanned spending decision, because her savings plus the gift allow her to buy the doghouse on sale.
Texas Test Prep
Question 12.
Keesha plans to save $5 a week for a new skateboard. The skateboard costs$47, and she has $22 saved. Which will result in her NOT being able to buy the skateboard? (A) She continues to save and makes no unplanned purchases for 5 weeks. (B) She spends$3 and then saves $6 a week for 5 weeks. (C) She saves$5 a week for 2 more weeks and then uses some money to buy a book.
(D) She saves $5 a week for 2 more weeks and then increases her savings to$8 a week for 2 weeks.
Explanation:
Because, She saves $5 a week for 2 more weeks and then uses some money to buy a book . The unplanned purchase of book result in her NOT being able to buy the skateboard. ### Texas Go Math Grade 3 Lesson 20.3 Homework and Practice Answer Key Question 1. Theresa wants to buy a new skateboard that costs$48 before summer vacation starts in 4 weeks. She has saved $17. Her plan is to save$6 each week. Will she be able to save enough money to buy the skateboard?
4 weeks × ________ = _________
$17 + ________ = ___________ Answer: 4 weeks × 6 = 24$
$17 + 24 = 51 Explanation: 51< 48$
Her plan is to save $6 each week. she will be able to save enough money to buy the skateboard Question 2. Suppose Theresa has only$8 saved. How much money will she need to save each week to have enough money to buy the skateboard in 4 weeks? Explain.
4 weeks × 10 = 408 + 40 = 48$Explanation: Theresa has only$8 saved.
she need to save 10$each week to have enough money to buy the skateboard in 4 weeks. Problem Solving Question 3. Deon wants to buy new car speakers that cost$66. He has saved $20 and he has a coupon to save$10. If he wants to buy the speakers in 4 weeks, how much money will he need to save each week?
Answer: 9$per each week Explanation: Deon wants to buy new car speakers that cost$66.
He has saved $20 and he has a coupon to save$10.
If he wants to buy the speakers in 4 weeks, 66-20 = 46
46 – 10 = 36
4 weeks x 9$= 36$
Question 4.
Jared has $25. He starts saving$9 a week to buy a birthday present for his father that costs $45. Jared’s father’s birthday is in 3 weeks. Would it be a wise unplanned spending decision for Jared to spend$15 on a set of comic books? Explain.
Explanation:
Jared has $25. He starts saving$9 a week to buy a birthday present for his father that costs $45. Jared’s father’s birthday is in 3 weeks. 3 weeks x 9 = 27$
27 + 25 = 52$52 – 45 = 7$
It would not be a wise unplanned spending decision for Jared to spend $15 on a set of comic books Lesson Check Fill in the bubble completely to show your answer. Question 5. Which of the following is true about spending decisions? (A) Unplanned spending always helps you reach a goal. (B) Planned spending can help you reach a goal. (C) Planned spending means you can never purchase what you want. (D) Unplanned spending is always a wise decision. Answer: B Explanation: Planned spending can help you reach a goal. Unplanned spending is frequently not wise. Unplanned spending may prevent you from reaching a goal. Question 6. Which of the following is a planned spending decision? (A) Spending all of your allowance each week (B) Buying a poster of your favorite singer with$10 that you found
(C) Saving $12 a week for a trip, and spending half of it each week (D) Saving$15 a week for 5 weeks to buy a bicycle that costs $75 Answer: D Explanation: Saving$15 a week for 5 weeks to buy a bicycle that costs $75 15 x 5 = 75 so, his planned spending is a wise decision. Question 7. Rosa has$16 saved. She wants to buy a new cell phone in 6 weeks that costs $75. Which plan will help her reach her goal? (A) Save$10 each week for 6 weeks.
(B) Save $14 each week for 4 weeks. Then buy a teeshirt that costs$12.
(C) Save $7 each week for 6 weeks. (D) Save$5 each week for 11 weeks.
Explanation:
Rosa has $16 saved. She wants to buy a new cell phone in 6 weeks that costs$75.
Save $10 each week for 6 weeks will help her to reach the goal 10 x 6 = 60 60 + 16 = 76$
Question 8.
Carson has $13 to spend. He will earn$18 washing cars for his neighbors. He plans to buy a pair of jeans that cost $25. How much money will he have left after he buys the jeans? (A)$23
(B) $9 (C)$33
(D) $6 Answer: D Explanation: Carson has$13 to spend.
He will earn $18 washing cars for his neighbors. 13 + 18 = 31$
He plans to buy a pair of jeans that cost $25. 31 – 25 = 6$ will be left after buying the pair of jeans.
Question 9.
Multi-Step Chelsea plans to buy a gift for her sister that costs $15. She saves$3 each week for 4 weeks. Then she spends $6 on school supplies. How many more weeks will Chelsea need to save to buy the gift? (A) 3 weeks (B) 6 weeks (C) 12 weeks (D) 9 weeks Answer: A Explanation: Chelsea plans to buy a gift for her sister that costs$15.
She saves $3 each week for 4 weeks. 3 x 4 = 12 Then she spends$6 on school supplies.
12 – 6 = 6
3 weeks x 3$= 9 9 + 6 = 15 Question 10. Multi-Step Kalon has$80 saved. He earns $115 a week and spends$75 each week on gas and groceries. He plans to buy new tires for his car in 6 weeks that cost $140. How much should he plan to save each week? (A)$15
(B) $20 (C)$10
(D) $12 Answer: C Explanation: Kalon has$80 saved.
He earns $115 a week and spends$75 each week on gas and groceries.
115 – 75 = 40
He plans to buy new tires for his car in 6 weeks that cost $140. 140 – 80 = 60 so, he has to save 10$ per week.
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physics
posted by .
A ball starts from rest and rolls down a hill with uniform acceleration, traveling 200m during the second 4.8s of its motion. How far did it travel the first 4.8 seconds
• physics -
id
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# Index notation for group multiplication and commutators
Are there any serious errors in the following discussion of notation? I will be grateful for any corrections, suggestions and pointers towards the relevant literature.
The index notation for inner automorphisms: $$x^a \equiv_{\text{def}} a^{-1}xa$$ has proved a useful abbreviation, as also has the use of the index $^{-1}$ to denote the anti-involution taking an element to its inverse.
It seems sensible to allow: $$x^{-a} = (x^{-1})^a$$ Indeed, this abbreviation might assist newcomers to the study of groups to avoid the tempting confusion between $x^{-a}$ and $x^{a^{-1}}$.
We already have analogues of the multiplication laws for indices: $$(x^a)^b = x^{ab}$$ and $$x^ay^a = (xy)^a$$
Can we also introduce a further shorthand using a non-commutative addition?
The proposal (if it is not already in use) would be to set $$x^{a+b} \equiv_{\text{def}} x^ax^b$$ If $n$ is an integer it is interpreted as an ordinary power, so $$x^{1+a} \equiv x\cdot x^a$$ a commutator may be written \begin{align} (x,y) &= x^{-1}y^{-1}xy = y^{-x+1} = x^{-1+y} \end{align} with the corresponding \begin{align} (x,y)^{-1} &= y^{-1}x^{-1}yx = y^{-1+x} = x^{-y+1} \end{align} The triple commutator is a challenge even in linear notation, with $$(x,y,z) \equiv ((x,y),z) = y^{-1}x^{-1}yxz^{-1}x^{-1}y^{-1}xyz$$ Perhaps experimenting with index notation does not have much to offer as a practical tool, but it may assist beginners in learning to parse such expressions. The notation $(x,y,z)$ is admirably concise, but does not immediately conjure up a representation of the subtlety of the operations it encapsulates. $z$, the "last man in" sees the following: \begin{align} (x,y,z) & = (y^{-1}x^{-1}y)\cdot (xz^{-1}x^{-1}) \cdot (y^{-1}xy)z \\ &= x^{-y} z^{-x^{-1}} x^{y}z \\ &= z^{-x^{-1+y}+1} \end{align} From a different vantage point, viewing $y$ as the subject acted upon, we may see the triple commutator as \begin{align} (x,y,z) \equiv ((x,y),z) & = y^{-1}(x^{-1}yx)(z^{-1}x^{-1}y^{-1}xy z) \\ &= y^{-1+x+(-x+1)z} \\ &= x^{-y+1+(-1+y)z} \end{align} To conclude, here is an illustration of how even an ad hoc macro notation can sometimes be of assistance in unscrambling the rather long strings met with in commutator relations. It is a well-known (and rather remarkable) result that: $$(x,y^{-1},z)^y\cdot(y,z^{-1},x)^z\cdot(z,x^{-1},y)^x = 1$$ If we define, temporarily, $$\binom{a}{b,c} = aba^{-1}ca$$ then $$(x,y^{-1},z)^y = \binom{x^{-1}}{y^{-1},z^{-1}} \binom{y}{x,z} \\ (y,z^{-1},x)^z = \binom{y^{-1}}{z^{-1},x^{-1}} \binom{z}{y,x} \\ (z,z^{-1},y)^x = \binom{z^{-1}}{x^{-1},y^{-1}} \binom{x}{z,y}$$ The result now follows, since, as we see from the definition $$\binom{a}{b,c}^{-1} = \binom{a^{-1}}{c^{-1},b^{-1}}$$
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Home Certification Excel for Data Analysis Free Certification
# Excel for Data Analysis Free Certification
0
Excel is a powerful tool for data analysis and is widely used in various industries. Here are some key features and techniques in Excel that can be helpful for data analysis:
Get Excel for Data Analysis Certificate from The Digital Adda which you can share in the Certifications section of your LinkedIn profile, on printed resumes, CVs, or other documents.
Exam Details
• Format: Multiple Choice Question
• Questions: 10
• Passing Score: 8/10 or 80%
• Language: English
Here are the questions and answers :
Where does the result appear when a data analyst performs data analysis on a grouped worksheet?
First worksheet
Last Worksheet
All other worksheets except the first
All other worksheets except the last
What tab leads a data analyst to the analysis tool pak in the Excel 2016?
Home
Review
File
Insert
What type of analysis tool performs linear abalysis by using the “Least Squares” to fit a line through a set of observations?
Regression
Fourier analysis
Linear analysis
Anova
What type of analysis tool fills a range with independent random numbers that are drawn from one several distribuctions?
Resonance
Distribution range
Banoulli
Random number generation
How does a data analyst performs data analysis on the remainder of the worksheets without results?
You can’t work on the rest of the worksheets
Delete the Previous results
Restart the analysis operation
Recalculate the analysis tool for each worksheet
What type of analysis tool creates a sample from a population by treating the input range as a population?
Regression analysis tools
Fourier analysis tools
Anova analysis tools
Sampling analysis tools
What do data analysts call FFT in Fourier analysis?
Fourth Fourier text
Fountain Fourier type
Fast Fourier transform
Fourier Fest Treat
What add-in program serves as the data analyzing tool in Excel?
Excel Analyse
Analyse Tool Excel
Analyz Tools
Analysis Toolpak
What analysis tool type tests for equality of the population means that underlie each sample?
Equality test
T-equaliry
Two-sample t-Test
Equality sample test
Which of these is true about data analysis function?
It can be used on multiple worksheets at a time
It can’t be used offline
It can be used on only one worksheet at a time
It can only be used in excel 2016
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:: Zero Based Finite Sequences :: by Tetsuya Tsunetou , Grzegorz Bancerek and Yatsuka Nakamura :: :: Received September 28, 2001 :: Copyright (c) 2001-2017 Association of Mizar Users :: (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland). :: This code can be distributed under the GNU General Public Licence :: version 3.0 or later, or the Creative Commons Attribution-ShareAlike :: License version 3.0 or later, subject to the binding interpretation :: detailed in file COPYING.interpretation. :: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these :: licenses, or see http://www.gnu.org/licenses/gpl.html and :: http://creativecommons.org/licenses/by-sa/3.0/. environ vocabularies NUMBERS, SUBSET_1, FUNCT_1, ARYTM_3, XXREAL_0, XBOOLE_0, TARSKI, NAT_1, ORDINAL1, FINSEQ_1, CARD_1, FINSET_1, RELAT_1, PARTFUN1, FUNCOP_1, ORDINAL4, ORDINAL2, ARYTM_1, REAL_1, ZFMISC_1, FUNCT_4, VALUED_0, AFINSQ_1, PRGCOR_2, CAT_1, AMISTD_1, AMISTD_3, AMISTD_2, VALUED_1, CONNSP_3; notations TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, RELAT_1, FUNCT_1, ORDINAL1, CARD_1, ORDINAL2, NUMBERS, ORDINAL4, XCMPLX_0, XREAL_0, NAT_1, PARTFUN1, BINOP_1, FINSOP_1, NAT_D, FINSET_1, FINSEQ_1, FUNCOP_1, FUNCT_4, FUNCT_7, XXREAL_0, VALUED_0, VALUED_1; constructors WELLORD2, FUNCT_4, XXREAL_0, ORDINAL4, FUNCT_7, ORDINAL3, VALUED_1, ENUMSET1, NAT_D, XXREAL_2, BINOP_1, FINSOP_1, RELSET_1, CARD_1, NUMBERS; registrations XBOOLE_0, SUBSET_1, RELAT_1, FUNCT_1, ORDINAL1, FUNCOP_1, XXREAL_0, XREAL_0, NAT_1, CARD_1, ORDINAL2, NUMBERS, VALUED_1, XXREAL_2, MEMBERED, FINSET_1, FUNCT_4; requirements REAL, NUMERALS, SUBSET, BOOLE, ARITHM; definitions TARSKI, ORDINAL1, XBOOLE_0, RELAT_1, PARTFUN1, CARD_1, FUNCT_1; equalities ORDINAL1, FUNCOP_1, VALUED_1, NAT_1, CARD_1; expansions TARSKI, ORDINAL1, XBOOLE_0, RELAT_1, CARD_1, FUNCT_1; theorems TARSKI, AXIOMS, FUNCT_1, NAT_1, ZFMISC_1, RELAT_1, RELSET_1, ORDINAL1, CARD_1, FINSEQ_1, FUNCT_7, ORDINAL4, CARD_2, FUNCT_4, ORDINAL3, XBOOLE_0, XBOOLE_1, FINSET_1, FUNCOP_1, XREAL_1, VALUED_0, ENUMSET1, XXREAL_0, XREAL_0, GRFUNC_1, XXREAL_2, NAT_D, VALUED_1, XTUPLE_0; schemes FUNCT_1, SUBSET_1, NAT_1, XBOOLE_0, CLASSES1, FINSEQ_1; begin reserve k,n for Nat, x,y,z,y1,y2 for object,X,Y for set, f,g for Function; :: Extended Segments of Natural Numbers ::\$CT theorem Th1: Segm n \/ { n } = Segm(n+1) proof n in Segm(n+1) by NAT_1:45; then A1: {n} c= Segm(n+1) by ZFMISC_1:31; n <= n + 1 by NAT_1:11; then Segm n c= Segm(n+1) by NAT_1:39; hence Segm n \/ { n } c= Segm(n+1) by A1,XBOOLE_1:8; let x be object; assume A2: x in Segm(n+1); then reconsider x as Nat; now x < n+1 by A2,NAT_1:44; then per cases by NAT_1:22; case x < n; hence x in Segm n by NAT_1:44; end; case x = n; hence x in {n} by TARSKI:def 1; end; end; hence thesis by XBOOLE_0:def 3; end; theorem Th2: Seg n c= Segm(n+1) proof let x be object; assume A1: x in Seg n; then reconsider x as Element of NAT; x<=n by A1,FINSEQ_1:1; then x natural; coherence; end; notation let p; synonym len p for card p; end; registration let p; identify len p with dom p; compatibility proof thus len p = card dom p by CARD_1:62 .= dom p; end; end; definition let p; redefine func len p -> Element of NAT; coherence proof card p = card p; hence thesis; end; end; definition let p; redefine func dom p -> Subset of NAT; coherence proof {i where i is Nat:iNat,P[object,object]}: ex p st dom p = A() & for k st k in A() holds P[k,p.k] provided A1: for k st k in A() ex x being object st P[k,x] proof A2: for x being object st x in A() ex y being object st P[x,y] proof let x be object; assume A3: x in A(); A()={i where i is Nat: iNat,F(object) -> object}: ex p being XFinSequence st len p = A() & for k st k in A() holds p.k=F(k) proof consider f being Function such that A1: dom f = A() & for x being object st x in A() holds f.x=F(x) from FUNCT_1:sch 3; reconsider p=f as XFinSequence by A1,FINSET_1:10,ORDINAL1:def 7; take p; thus thesis by A1; end; theorem z in p implies ex k st k in dom p & z=[k,p.k] proof assume A1: z in p; then consider x,y being object such that A2: z=[x,y] by RELAT_1:def 1; x in dom p by A1,A2,FUNCT_1:1; then reconsider k = x as Element of NAT; take k; thus thesis by A1,A2,FUNCT_1:1; end; theorem dom p = dom q & (for k st k in dom p holds p.k = q.k) implies p = q; Lm1: k < len p iff k in dom p proof thus k < len p implies k in dom p proof assume k < len p; then k in Segm len p by NAT_1:44; hence k in dom p; end; assume k in dom p; then k in Segm len p; hence k < len p by NAT_1:44; end; theorem Th8: ( len p = len q & for k st k < len p holds p.k=q.k ) implies p=q proof assume that A1: len p = len q and A2: for k st k finite; coherence; end; theorem rng p c= dom f implies f*p is XFinSequence proof assume rng p c= dom f; then dom(f*p) = len p by RELAT_1:27; hence thesis by ORDINAL1:def 7; end; theorem Th10: k < len p implies dom(p|k) = k proof assume k < len p; then Segm k c= Segm len p by NAT_1:39; hence dom(p|k) = k by RELAT_1:62; end; :: XFinSequences of D registration let D be set; cluster finite for Sequence of D; existence proof {} is Sequence of D by ORDINAL1:30; hence thesis; end; end; definition let D be set; mode XFinSequence of D is finite Sequence of D; end; theorem Th11: for D being set, f being XFinSequence of D holds f is PartFunc of NAT,D proof let D be set, f be XFinSequence of D; dom f c= NAT & rng f c= D by RELAT_1:def 19; hence thesis by RELSET_1:4; end; registration cluster empty -> Sequence-like for Function; coherence; end; reserve D for set; registration let k be Nat, a be object; cluster Segm k --> a -> finite Sequence-like; coherence; end; ::\$CT theorem Th12: for D being non empty set ex p being XFinSequence of D st len p = k proof let D be non empty set; set y = the Element of D; set p = k --> y; reconsider p = k --> y as XFinSequence; reconsider p as XFinSequence of D; take p; thus thesis by FUNCOP_1:13; end; :: :: :: The Empty XFinSequence :: :: :: theorem len p = 0 iff p = {}; theorem Th14: for D be set holds {} is XFinSequence of D proof let D be set; rng {} c= D; hence thesis by RELAT_1:def 19; end; registration let D be set; cluster empty for XFinSequence of D; existence proof {} is XFinSequence of D by Th14; hence thesis; end; end; registration let D be non empty set; cluster non empty for XFinSequence of D; existence proof set k = 1; consider p being XFinSequence of D such that A1: len p = k by Th12; p <> {} by A1; hence thesis; end; end; definition let x; func <%x%> -> set equals 0 .--> x; coherence; end; registration let x; cluster <%x%> -> non empty; coherence; end; definition let D be set; func <%>D -> XFinSequence of D equals {}; coherence by Th14; end; registration let D be set; cluster <%>D -> empty; coherence; end; definition let p,q; redefine func p^q means :Def3: dom it = len p + len q & (for k st k in dom p holds it.k=p.k) & for k st k in dom q holds it.(len p + k) = q.k; compatibility proof let pq be Sequence; A1: len p +^ len q = len p + len q by CARD_2:36; hereby assume A2: pq = p^q; hence dom pq = len p + len q by A1,ORDINAL4:def 1; thus for k st k in dom p holds pq.k=p.k by A2,ORDINAL4:def 1; let k; assume k in dom q; then pq.(len p +^ k) = q.k & k in NAT by A2,ORDINAL4:def 1; hence pq.(len p + k) = q.k by CARD_2:36; end; assume that A3: dom pq = len p + len q and A4: for k st k in dom p holds pq.k=p.k and A5: for k st k in dom q holds pq.(len p + k) = q.k; A6: now let a be Ordinal; assume A7: a in dom q; then reconsider k = a as Element of NAT; thus pq.(dom p +^ a) = pq.(len p + k) by CARD_2:36 .= q.a by A5,A7; end; for a be Ordinal st a in dom p holds pq.a = p.a by A4; hence thesis by A1,A3,A6,ORDINAL4:def 1; end; end; registration let p,q; cluster p^q -> finite; coherence proof dom (p^q) = (dom p)+^dom q by ORDINAL4:def 1; hence thesis by FINSET_1:10; end; end; theorem len(p^q) = len p + len q by Def3; theorem Th16: len p <= k & k < len p + len q implies (p^q).k=q.(k-len p) proof assume that A1: len p <= k and A2: k < len p + len q; consider m being Nat such that A3: len p + m = k by A1,NAT_1:10; k - len p < len p + len q - len p by A2,XREAL_1:14; then m in dom q by A3,Lm1; hence thesis by A3,Def3; end; theorem Th17: len p <= k & k < len(p^q) implies (p^q).k = q.(k - len p) proof assume that A1: len p <= k and A2: k < len(p^q); k < len p + len q by A2,Def3; hence thesis by A1,Th16; end; theorem Th18: k in dom (p^q) implies (k in dom p or ex n st n in dom q & k=len p + n ) proof assume k in dom(p^q); then k in Segm(len p + len q) by Def3; then A1: k < len p + len q by NAT_1:44; now assume len p <= k; then consider n being Nat such that A2: k=len p + n by NAT_1:10; n + len p - len p < len q + len p - len p by A1,A2,XREAL_1:14; then n in Segm len q by Lm1; hence thesis by A2; end; hence thesis by Lm1; end; theorem Th19: for p,q being Sequence holds dom p c= dom(p^q) proof let p,q be Sequence; dom(p^q) = (dom p)+^(dom q) by ORDINAL4:def 1; hence thesis by ORDINAL3:24; end; theorem Th20: x in dom q implies ex k st k=x & len p + k in dom(p^q) proof assume A1: x in dom q; then reconsider k=x as Element of NAT; take k; k < len q by A1,Lm1; then len p + k < len p + len q by XREAL_1:8; then len p + k in Segm(len p + len q) by NAT_1:44; hence thesis by Def3; end; theorem Th21: k in dom q implies len p + k in dom(p^q) proof assume k in dom q; then ex n st n=k & len p + n in dom(p^q) by Th20; hence thesis; end; theorem Th22: rng p c= rng(p^q) proof A1: dom p c= dom(p^q) by Th19; let x be object; assume x in rng p; then consider y being object such that A2: y in dom p and A3: x=p.y by FUNCT_1:def 3; reconsider k=y as Element of NAT by A2; (p^q).k=p.k by A2,Def3; hence x in rng(p^q) by A2,A3,A1,FUNCT_1:3; end; theorem Th23: rng q c= rng(p^q) proof let x be object; assume x in rng q; then consider y being object such that A1: y in dom q and A2: x=q.y by FUNCT_1:def 3; reconsider k=y as Element of NAT by A1; len p + k in dom(p^q) & (p^q).(len p + k) = q.k by A1,Def3,Th21; hence x in rng(p^q) by A2,FUNCT_1:3; end; theorem Th24: rng(p^q) = rng p \/ rng q proof now let x be object; assume x in rng(p^q); then consider y being object such that A1: y in dom(p^q) and A2: x=(p^q).y by FUNCT_1:def 3; reconsider k=y as Element of NAT by A1; y in Segm(len p + len q) by A1,Def3; then A3: k < len p + len q by NAT_1:44; A4: now assume A5: len p <= k; then consider m being Nat such that A6: len p + m = k by NAT_1:10; m + len p - len p < len p + len q - len p by A3,A6,XREAL_1:14; then A7: m in Segm len q by Lm1; q.(k-len p) = x by A2,A3,A5,Th16; hence x in rng q by A6,A7,FUNCT_1:3; end; now assume not len p <= k; then A8: k in dom p by Lm1; then p.k = x by A2,Def3; hence x in rng p by A8,FUNCT_1:3; end; hence x in rng p \/ rng q by A4,XBOOLE_0:def 3; end; then A9: rng(p^q) c= rng p \/ rng q; rng p c= rng(p^q) & rng q c= rng(p^q) by Th22,Th23; then (rng p \/ rng q) c= rng(p^q) by XBOOLE_1:8; hence thesis by A9; end; theorem Th25: p^q^r = p^(q^r) proof A1: for k st k in dom p holds ((p^q)^r).k=p.k proof let k; assume A2: k in dom p; dom p c= dom(p^q) by Th19; hence (p^q^r).k=(p^q).k by A2,Def3 .=p.k by A2,Def3; end; A3: for k st k in dom(q^r) holds ((p^q)^r).(len p + k)=(q^r).k proof let k; assume A4: k in dom(q^r); A5: now assume not k in dom q; then consider n such that A6: n in dom r and A7: k=len q + n by A4,Th18; thus (p^q^r).(len p + k) =(p^q^r).(len p + len q + n) by A7 .=(p^q^r).(len(p^q) + n) by Def3 .=r.n by A6,Def3 .=(q^r).k by A6,A7,Def3; end; now assume A8: k in dom q; then (len p + k) in dom(p^q) by Th21; hence (p^q^r).(len p + k) = (p^q).(len p + k) by Def3 .=q.k by A8,Def3 .=(q^r).k by A8,Def3; end; hence thesis by A5; end; dom ((p^q)^r) = (len (p^q) + len r) by Def3 .= (len p + len q + len r) by Def3 .= (len p + (len q + len r)) .= (len p + len(q^r)) by Def3; hence thesis by A1,A3,Def3; end; theorem Th26: p^r = q^r or r^p = r^q implies p = q proof A1: now assume A2: p^r = q^r; then len p + len r = len(q^r) by Def3; then A3: len p + len r = len q + len r by Def3; for k st k in dom p holds p.k=q.k proof let k; assume A4: k in dom p; hence p.k=(q^r).k by A2,Def3 .=q.k by A3,A4,Def3; end; hence thesis by A3; end; A5: now assume A6: r^p=r^q; then A7: len r + len p = len(r^q) by Def3 .=len r + len q by Def3; for k st k in dom p holds p.k=q.k proof let k; assume A8: k in dom p; hence p.k = (r^q).(len r + k) by A6,Def3 .= q.k by A7,A8,Def3; end; hence thesis by A7; end; assume p^r = q^r or r^p = r^q; hence thesis by A1,A5; end; registration let p; reduce p^{} to p; reducibility proof A1: for k st k in dom p holds p.k=(p^{}).k by Def3; dom(p^{}) = len p + len {} by Def3 .= dom p; hence p^{} = p by A1; end; reduce {}^p to p; reducibility proof A2: for k st k in dom p holds p.k = ({}^p).k proof let k; assume A3: k in dom p; thus ({}^p).k =({}^p).(len {} + k) .=p.k by A3,Def3; end; dom({}^p) = (len {} + len p) by Def3 .= dom p; hence thesis by A2; end; end; ::\$CT theorem Th27: p^q = {} implies p={} & q={} proof assume p^q={}; then 0 = len (p^q) .= len p + len q by Def3; hence thesis; end; registration let D be set; let p,q be XFinSequence of D; cluster p^q -> D-valued; coherence proof A1: rng q c= D by RELAT_1:def 19; rng(p^q) = rng p \/ rng q & rng p c= D by Th24,RELAT_1:def 19; then rng(p^q) c= D by A1,XBOOLE_1:8; hence thesis; end; end; Lm2: for x1, y1 being set holds [x,y] in {[x1,y1]} implies x = x1 & y = y1 proof let x1, y1 be set; assume [x,y] in {[x1,y1]}; then [x,y] = [x1,y1] by TARSKI:def 1; hence thesis by XTUPLE_0:1; end; definition let x; redefine func <%x%> -> Function means :Def4: dom it = 1 & it.0 = x; coherence; compatibility proof let f be Function; thus f = <%x%> implies dom f = 1 & f.0 = x by CARD_1:49,FUNCOP_1:13,72; assume that A1: dom f = 1 and A2: f.0 = x; reconsider g = { [0,f.0] } as Function; for y,z being object holds [y,z] in f iff [y,z] in g proof let y,z be object; hereby assume A3: [y,z] in f; then y in {0} by A1,CARD_1:49,XTUPLE_0:def 12; then A4: y = 0 by TARSKI:def 1; A5: rng f = {f.0} by A1,CARD_1:49,FUNCT_1:4; z in rng f by A3,XTUPLE_0:def 13; then z = f.0 by A5,TARSKI:def 1; hence [y,z] in g by A4,TARSKI:def 1; end; assume [y,z] in g; then A6: y = 0 & z = f.0 by Lm2; 0 in dom f by A1,CARD_1:49,TARSKI:def 1; hence thesis by A6,FUNCT_1:def 2; end; then f = { [0,f.0] }; hence thesis by A2,FUNCT_4:82; end; end; registration let x; cluster <%x%> -> Function-like Relation-like; coherence; end; registration let x; cluster <%x%> -> finite Sequence-like; coherence by Def4; end; theorem p^q is XFinSequence of D implies p is XFinSequence of D & q is XFinSequence of D proof assume p^q is XFinSequence of D; then rng(p^q) c= D by RELAT_1:def 19; then A1: rng p \/ rng q c= D by Th24; rng p c= rng p \/ rng q by XBOOLE_1:7; then rng p c= D by A1; hence p is XFinSequence of D by RELAT_1:def 19; rng q c= rng p \/ rng q by XBOOLE_1:7; then rng q c= D by A1; hence thesis by RELAT_1:def 19; end; definition let x,y; func <%x,y%> -> set equals <%x%>^<%y%>; correctness; let z; func <%x,y,z%> -> set equals <%x%>^<%y%>^<%z%>; correctness; end; registration let x,y; cluster <%x,y%> -> Function-like Relation-like; coherence; let z; cluster <%x,y,z%> -> Function-like Relation-like; coherence; end; registration let x,y; cluster <%x,y%> -> finite Sequence-like; coherence; let z; cluster <%x,y,z%> -> finite Sequence-like; coherence; end; theorem <%x%> = { [0,x] } by FUNCT_4:82; theorem Th30: p=<%x%> iff dom p = Segm 1 & rng p = {x} proof thus p = <%x%> implies dom p = Segm 1 & rng p = {x} proof assume A1: p = <%x%>; hence dom p = Segm 1 by Def4; dom p = {0} by A1,Def4,CARD_1:49; then rng p = {p.0} by FUNCT_1:4; hence thesis by A1,Def4; end; assume that A2: dom p = Segm 1 and A3: rng p = {x}; 1=0+1; then p.0 in {x} by A2,A3,FUNCT_1:3,NAT_1:45; then p.0 = x by TARSKI:def 1; hence thesis by A2,Def4; end; theorem Th31: p = <%x%> iff len p = 1 & p.0 = x by Def4; registration let x; reduce <%x%>.0 to x; reducibility by Th31; end; theorem Th32: (<%x%>^p).0 = x proof 0 in 1 by CARD_1:49,TARSKI:def 1; then 0 in dom <%x%> by Def4; then (<%x%>^p).0 = <%x%>.0 by Def3; hence thesis; end; theorem Th33: (p^<%x%>).(len p)=x proof A1: dom <%x%> = 1 & 0 in Segm(0+1) by Def4,NAT_1:45; len p + 0 = len p; hence (p^<%x%>).(len p) = <%x%>.0 by A1,Def3 .=x; end; theorem <%x,y,z%>=<%x%>^<%y,z%> & <%x,y,z%>=<%x,y%>^<%z%> by Th25; theorem Th35: p = <%x,y%> iff len p = 2 & p.0=x & p.1=y proof thus p = <%x,y%> implies len p = 2 & p.0=x & p.1=y proof assume A1: p=<%x,y%>; hence len p = len <%x%> + len <%y%> by Def3 .= 1 + len <%y%> by Th30 .= 1 + 1 by Th30 .=2; A2: 0 in {0} by TARSKI:def 1; then A3: 0 in dom <%y%> by Def4,CARD_1:49; 0 in dom <%x%> by A2,Def4,CARD_1:49; hence p.0 = <%x%>.0 by A1,Def3 .= x; thus p.1 = (<%x%>^<%y%>).(len <%x%> + 0) by A1,Th30 .= <%y%>.0 by A3,Def3 .= y; end; assume that A4: len p = 2 and A5: p.0=x and A6: p.1=y; A7: for k st k in dom <%y%> holds p.((len <%x%>)+k)=<%y%>.k proof let k; assume k in dom <%y%>; then A8: k in {0} by Def4,CARD_1:49; thus p.((len <%x%>) + k) = p.(1+k) by Th30 .=p.(1+0) by A8,TARSKI:def 1 .=<%y%>.0 by A6 .= <%y%>.k by A8,TARSKI:def 1; end; A9: for k st k in dom <%x%> holds p.k=<%x%>.k proof let k; assume k in dom <%x%>; then k in {0} by Def4,CARD_1:49; then k=0 by TARSKI:def 1; hence thesis by A5; end; dom p = (1+1) by A4 .= (len <%x%> + 1) by Th30 .= (len <%x%> + len <%y%>) by Th30; hence thesis by A9,A7,Def3; end; registration let x,y; reduce <%x,y%>.0 to x; reducibility by Th35; reduce <%x,y%>.1 to y; reducibility by Th35; end; theorem Th36: p = <%x,y,z%> iff len p = 3 & p.0 = x & p.1 = y & p.2 = z proof thus p = <%x,y,z%> implies len p = 3 & p.0 = x & p.1 = y & p.2 = z proof A1: 0 in {0} by TARSKI:def 1; then A2: 0 in dom <%x%> by Def4,CARD_1:49; A3: 0 in dom <%z%> by A1,Def4,CARD_1:49; assume A4: p =<%x,y,z%>; hence len p =len <%x,y%> + len <%z%> by Def3 .=2 + len <%z%> by Th35 .=2+1 by Th30 .=3; thus p.0 = (<%x%>^<%y,z%>).0 by A4,Th25 .=<%x%>.0 by A2,Def3 .=x; 1 in Segm(1+1) & len <%x,y%> = 2 by Th35,NAT_1:45; hence p.1 =<%x,y%>.1 by A4,Def3 .=y; thus p.2 =(<%x,y%>^<%z%>).(len (<%x,y%>) + 0) by A4,Th35 .= <%z%>.0 by A3,Def3 .= z; end; assume that A5: len p = 3 and A6: p.0 = x and A7: p.1 = y and A8: p.2 = z; A9: for k st k in dom <%x,y%> holds p.k=<%x,y%>.k proof A10: len <%x,y%> = 2 by Th35; let k such that A11: k in dom <%x,y%>; A12: k=1 implies p.k=<%x,y%>.k by A7; k=0 implies p.k=<%x,y%>.k by A6; hence thesis by A11,A10,A12,CARD_1:50,TARSKI:def 2; end; A13: for k st k in dom <%z%> holds p.( (len <%x,y%>) + k) = <%z%>.k proof let k; assume k in dom <%z%>; then k in {0} by Def4,CARD_1:49; then A14: k = 0 by TARSKI:def 1; hence p.( (len <%x,y%>) + k) = p.(2+0) by Th35 .=<%z%>.k by A8,A14; end; dom p = (2+1) by A5 .= ((len <%x,y%>) + 1) by Th35 .= ((len <%x,y%>) + len <%z%>) by Th30; hence thesis by A9,A13,Def3; end; registration let x,y,z; reduce <%x,y,z%>.0 to x; reducibility by Th36; reduce <%x,y,z%>.1 to y; reducibility by Th36; reduce <%x,y,z%>.2 to z; reducibility by Th36; end; registration let x; cluster <%x%> -> 1-element; coherence by Th30; let y; cluster <%x,y%> -> 2-element; coherence by Th35; let z; cluster <%x,y,z%> -> 3-element; coherence by Th36; end; registration let n be Nat; cluster n-element -> n-defined for XFinSequence; coherence; end; registration let n be Nat, x be object; cluster n --> x -> finite Sequence-like; coherence; end; registration let n be Nat; cluster n-element for XFinSequence; existence proof take n --> 0; thus card(n --> 0)= n by FUNCOP_1:13; end; end; registration let n be Nat; cluster -> total for n-element n-defined XFinSequence; coherence proof let s be n-element XFinSequence; card s = n by CARD_1:def 7; hence dom s = n; end; end; theorem Th37: p <> {} implies ex q,x st p=q^<%x%> proof assume p <> {}; then consider n being Nat such that A1: len p = n+1 by NAT_1:6; A2: dom p = Segm(n+1) by A1; reconsider n as Element of NAT by ORDINAL1:def 12; set q=p| n; dom q = len p /\ n & Segm n c= Segm len p by A1,NAT_1:11,39,RELAT_1:61; then A3: dom q = n by XBOOLE_1:28; A4: for x being object st x in dom p holds p.x = (q^<%p.(len p - 1)%>).x proof let x be object; assume A5: x in dom p; then reconsider k = x as Element of NAT; A6: now assume A7: k in n; hence p.k=q.k by A3,FUNCT_1:47 .=(q^<%p.(len p - 1)%>).k by A3,A7,Def3; end; A8: now 0 in Segm(0+1) by NAT_1:45; then A9: 0 in dom <%p.(len p - 1)%> by Def4; assume A10: k in {n}; hence (q^<%p.(len p - 1)%>).k =(q^<%p.(len p - 1)%>).(len q + 0) by A3, TARSKI:def 1 .=<%p.(len p - 1)%>.0 by A9,Def3 .=p.k by A1,A10,TARSKI:def 1; end; k in Segm n \/ {n} by A5,Th1,A2; hence thesis by A6,A8,XBOOLE_0:def 3; end; take q; take p.(len p - 1); dom(q^<%p.(len p - 1)%>) = (len q + len <%p.(len p - 1)%>) by Def3 .= dom p by A1,A3,Th30; hence q^<%p.(len p - 1)%>=p by A4; end; registration let D be non empty set; let d1 be Element of D; cluster <%d1%> -> D -valued; coherence; let d2 be Element of D; cluster <%d1,d2%> -> D -valued; coherence; let d3 be Element of D; cluster <%d1,d2,d3%> -> D -valued; coherence; end; :: Scheme of induction for extended finite sequences scheme IndXSeq{P[XFinSequence]}: for p holds P[p] provided A1: P[{}] and A2: for p,x st P[p] holds P[p^<%x%>] proof defpred P1[Real] means for p st len p = \$1 holds P[p]; let p; consider X being Subset of REAL such that A3: for x being Element of REAL holds x in X iff P1[x] from SUBSET_1:sch 3; for k holds k in X proof A4: 0 in REAL by XREAL_0:def 1; defpred R[Nat] means \$1 in X; for p st len p = 0 holds P[p] proof let p; assume len p = 0; then p = {}; hence thesis by A1; end; then A5: R[0] by A3,A4; A6: for n st R[n] holds R[n+1] proof let n; assume A7: R[n]; A8: n+1 in REAL by XREAL_0:def 1; P1[n+1] proof let p; assume A9: len p = n+1; then p <> {}; then consider w being XFinSequence, x such that A10: p = w^<%x%> by Th37; len p = len w + len <%x%> by A10,Def3 .= len w+1 by Def4; then P[w] by A3,A7,A9; hence P[p] by A10,A2; end; hence thesis by A3,A8; end; thus for k holds R[k] from NAT_1:sch 2(A5,A6); end; then len p in X; hence thesis by A3; end; theorem for p,q,r,s being XFinSequence st p^q = r^s & len p <= len r ex t being XFinSequence st p^t = r proof defpred P[XFinSequence] means for p,q,s st p^q=\$1^s & len p <= len \$1 holds ex t being XFinSequence st p^t=\$1; A1: for r,x st P[r] holds P[r^<%x%>] proof let r,x; assume A2: for p,q,s st p^q=r^s & len p <= len r ex t st p^t=r; let p,q,s; assume that A3: p^q=(r^<%x%>)^s and A4: len p <= len (r^<%x%>); A5: now assume len p <> len(r^<%x%>); then len p <> len r + len <%x%> by Def3; then A6: len p <> len r + 1 by Th30; len p <= len r + len <%x%> by A4,Def3; then A7: len p <= len r + 1 by Th30; p^q=r^(<%x%>^s) by A3,Th25; then consider t being XFinSequence such that A8: p^t = r by A2,A6,A7,NAT_1:8; p^(t^<%x%>) = r^<%x%> by A8,Th25; hence thesis; end; now assume A9: len p = len(r^<%x%>); A10: for k st k in dom p holds p.k=(r^<%x%>).k proof let k; assume A11: k in dom p; hence p.k = (r^<%x%>^s).k by A3,Def3 .=(r^<%x%>).k by A9,A11,Def3; end; p^{} =r^<%x%> by A9,A10; hence thesis; end; hence thesis by A5; end; A12: P[{}] proof let p,q,s; assume that p^q={}^s and A13: len p <= len {}; take {}; thus p^{} = {} by A13; end; for r holds P[r] from IndXSeq(A12,A1); hence thesis; end; definition let D be set; func D^omega -> set means :Def7: x in it iff x is XFinSequence of D; existence proof defpred P[object] means \$1 is XFinSequence of D; consider X such that A1: x in X iff x in bool [:NAT,D:] & P[x] from XBOOLE_0:sch 1; take X; let x; thus x in X implies x is XFinSequence of D by A1; assume x is XFinSequence of D; then reconsider p = x as XFinSequence of D; reconsider p as PartFunc of NAT,D by Th11; p c= [:NAT,D:]; hence thesis by A1; end; uniqueness proof defpred P[object] means \$1 is XFinSequence of D; thus for X1,X2 being set st (for x being object holds x in X1 iff P[x]) & ( for x being object holds x in X2 iff P[x]) holds X1 = X2 from XBOOLE_0:sch 3; end; end; registration let D be set; cluster D^omega -> non empty; coherence proof set f = the XFinSequence of D; f in D^omega by Def7; hence thesis; end; end; theorem x in D^omega iff x is XFinSequence of D by Def7; theorem {} in D^omega proof {} = <%>D; hence thesis by Def7; end; scheme SepXSeq{D()->non empty set, P[XFinSequence]}: ex X st for x holds x in X iff ex p st p in D()^omega & P[p] & x=p proof defpred P1[object] means ex p st P[p] & \$1=p; consider Y such that A1: for x being object holds x in Y iff x in D()^omega & P1[x] from XBOOLE_0:sch 1; take Y; x in Y implies ex p st p in D()^omega & P[p] & x=p proof assume x in Y; then x in D()^omega & ex p st P[p] & x=p by A1; hence thesis; end; hence thesis by A1; end; notation let p be XFinSequence; let i,x be set; synonym Replace(p,i,x) for p+*(i,x); end; registration let p be XFinSequence; let i,x be object; cluster p+*(i,x) -> finite Sequence-like; coherence proof dom (p+*(i,x)) = dom p by FUNCT_7:30; hence thesis by FINSET_1:10; end; end; theorem for p being XFinSequence, i being Element of NAT, x being set holds len Replace(p,i,x) = len p & (i < len p implies Replace(p,i,x).i = x) & for j being Element of NAT st j <> i holds Replace(p,i,x).j = p.j proof let p be XFinSequence; let i be Element of NAT, x be set; set f = Replace(p,i,x); thus len f = len p by FUNCT_7:30; i < len p implies not Segm len p c= Segm i by NAT_1:39; hence i < len p implies f.i = x by FUNCT_7:31,ORDINAL1:16; thus thesis by FUNCT_7:32; end; registration let D be non empty set; let p be XFinSequence of D; let i be Element of NAT, a be Element of D; cluster Replace(p,i,a) -> D -valued; coherence proof per cases; suppose i in dom p; then Replace(p,i,a) = p+*(i.-->a) by FUNCT_7:def 3; then A1: rng Replace(p,i,a) c= rng p \/ rng (i.-->a) by FUNCT_4:17; rng (i.-->a) = {a} by FUNCOP_1:8; then A2: rng (i.-->a) c= D by ZFMISC_1:31; rng p c= D by RELAT_1:def 19; then rng p \/ rng (i.-->a) c= D by A2,XBOOLE_1:8; hence rng Replace(p,i,a) c= D by A1; end; suppose not i in dom p; then Replace(p,i,a) = p by FUNCT_7:def 3; hence rng Replace(p,i,a) c= D by RELAT_1:def 19; end; end; end; :: missing, 2008.02.02, A.K. registration cluster -> real-valued for XFinSequence of REAL; coherence proof let F be XFinSequence of REAL; rng F c= REAL by RELAT_1:def 19; hence thesis by VALUED_0:def 3; end; end; registration cluster -> natural-valued for XFinSequence of NAT; coherence proof let F be XFinSequence of NAT; rng F c= NAT by RELAT_1:def 19; hence thesis by VALUED_0:def 6; end; end; :: 2009.0929, A.T. theorem Th42: for x1, x2, x3, x4 being set st p = <%x1%>^<%x2%>^<%x3%>^<%x4%> holds len p = 4 & p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 proof let x1, x2, x3, x4 be set; assume A1: p = <%x1%>^<%x2%>^<%x3%>^<%x4%>; set p13 = <%x1%>^<%x2%>^<%x3%>; A2: p13 = <%x1, x2, x3%>; then A3: len p13 = 3 by Th36; A4: p13.0 = x1 & p13.1 = x2 by A2; A5: p13.2 = x3 by A2; thus len p = len p13 + len <%x4%> by A1,Def3 .= 3 + 1 by A3,Th30 .= 4; 0 in 3 & 1 in 3 & 2 in 3 by CARD_1:51,ENUMSET1:def 1; hence p.0 = x1 & p.1 = x2 & p.2 = x3 by A1,A4,A5,Def3,A3; thus p.3 = p.len p13 by A2,Th36 .= x4 by A1,Th33; end; theorem Th43: for x1, x2, x3, x4, x5 being set st p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%> holds len p = 5 & p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 proof let x1, x2, x3, x4, x5 be set; assume A1: p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>; set p14 = <%x1%>^<%x2%>^<%x3%>^<%x4%>; A2: len p14 = 4 by Th42; A3: p14.0 = x1 & p14.1 = x2 by Th42; A4: p14.2 = x3 & p14.3 = x4 by Th42; thus len p = len p14 + len <%x5%> by A1,Def3 .= 4 + 1 by A2,Th30 .= 5; 0 in 4 & ... & 3 in 4 by CARD_1:52,ENUMSET1:def 2; hence p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 by A1,A3,A4,Def3,A2; thus p.4 = p.len p14 by Th42 .= x5 by A1,Th33; end; theorem Th44: for x1, x2, x3, x4, x5, x6 being set st p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%> holds len p = 6 & p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 & p.5 = x6 proof let x1, x2, x3, x4, x5, x6 be set; assume A1: p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>; set p15 = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>; A2: len p15 = 5 by Th43; A3: p15.0 = x1 & p15.1 = x2 by Th43; A4: p15.2 = x3 & p15.3 = x4 by Th43; A5: p15.4 = x5 by Th43; thus len p = len p15 + len <%x6%> by A1,Def3 .= 5 + 1 by A2,Th30 .= 6; 0 in 5 & ... & 4 in 5 by CARD_1:53,ENUMSET1:def 3; hence p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 by A1,A3,A4,A5,Def3,A2; thus p.5 = p.len p15 by Th43 .= x6 by A1,Th33; end; theorem Th45: for x1, x2, x3, x4, x5, x6, x7 being set st p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%> holds len p = 7 & p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 & p.5 = x6 & p.6 = x7 proof let x1, x2, x3, x4, x5, x6, x7 be set; assume A1: p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%>; set p16 = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>; A2: len p16 = 6 by Th44; A3: p16.0 = x1 & p16.1 = x2 by Th44; A4: p16.2 = x3 & p16.3 = x4 by Th44; A5: p16.4 = x5 & p16.5 = x6 by Th44; thus len p = len p16 + len <%x7%> by A1,Def3 .= 6 + 1 by A2,Th30 .= 7; 0 in 6 & ... & 5 in 6 by CARD_1:54,ENUMSET1:def 4; hence p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 & p.5 = x6 by A1,A3,A4,A5,Def3,A2; thus p.6 = p.len p16 by Th44 .= x7 by A1,Th33; end; theorem Th46: for x1,x2,x3,x4, x5, x6, x7, x8 being set st p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%>^<%x8%> holds len p = 8 & p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 & p.5 = x6 & p.6 = x7 & p.7 = x8 proof let x1, x2, x3, x4, x5, x6, x7, x8 be set; assume A1: p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%>^<%x8%>; set p17 = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%>; A2: len p17 = 7 by Th45; A3: p17.0 = x1 & p17.1 = x2 by Th45; A4: p17.2 = x3 & p17.3 = x4 by Th45; A5: p17.4 = x5 & p17.5 = x6 by Th45; A6: p17.6 = x7 by Th45; thus len p = len p17 + len <%x8%> by A1,Def3 .= 7 + 1 by A2,Th30 .= 8; 0 in 7 & ... & 6 in 7 by CARD_1:55,ENUMSET1:def 5; hence p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 & p.5 = x6 & p.6 = x7 by A1,A3,A4,A5,A6,Def3,A2; thus p.7 = p.len p17 by Th45 .= x8 by A1,Th33; end; theorem for x1,x2,x3,x4,x5,x6,x7, x8, x9 being set st p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%>^<%x8%>^<%x9%> holds len p = 9 & p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 & p.5 = x6 & p.6 = x7 & p.7 = x8 & p.8 = x9 proof let x1, x2, x3, x4, x5, x6, x7, x8, x9 be set; assume A1: p = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%>^<%x8%>^<%x9%>; set p17 = <%x1%>^<%x2%>^<%x3%>^<%x4%>^<%x5%>^<%x6%>^<%x7%>^<%x8%>; A2: len p17 = 8 by Th46; A3: p17.0 = x1 & p17.1 = x2 by Th46; A4: p17.2 = x3 & p17.3 = x4 by Th46; A5: p17.4 = x5 & p17.5 = x6 by Th46; A6: p17.6 = x7 & p17.7 = x8 by Th46; thus len p = len p17 + len <%x9%> by A1,Def3 .= 8 + 1 by A2,Th30 .= 9; 0 in 8 & ... & 7 in 8 by CARD_1:56,ENUMSET1:def 6; hence p.0 = x1 & p.1 = x2 & p.2 = x3 & p.3 = x4 & p.4 = x5 & p.5 = x6 & p.6 = x7 & p.7 = x8 by A1,A3,A4,A5,A6,Def3,A2; thus p.8 = p.len p17 by Th46 .= x9 by A1,Th33; end; :: K.P. 12.2009 theorem :: FINSEQ_2:7 n proof set pn = p|n; set x=p.n; assume A1: len p = n+1; then A2: n < len p by NAT_1:13; then A3: len pn = n by Th51; A4: now let m be Nat; assume m in dom p; then m).m by A3,Th33; end; case m <> len pn; then m< len pn by A5,XXREAL_0:1; then A6: m in dom pn by Lm1; hence (pn^<%x%>).m = pn.m by Def3 .= p.m by A2,A3,A6,Th50; end; end; hence p.m = (pn^<%x%>).m; end; len (pn^<%x%>) = n + len <%x%> by A3,Def3 .= len p by A1,Def4; hence thesis by A4; end; theorem Th54: :: CATALAN2:1 (p^q)|dom p = p proof set r=(p^q)|(dom p); A1: now let k such that A2: k < len p; A3: k in dom p by A2,Lm1; then A4: (p^q).k=p.k by Def3; k+0=len (p^q); then n>=len p+len q by Def3; then k >= len q by A1,XREAL_1:8; then Segm len q c= Segm k by NAT_1:39; then A3: q|k = q by RELAT_1:68; Segm len(p^q) c= Segm n by A2,NAT_1:39; hence thesis by A3,RELAT_1:68; end; suppose A4: n=len p; m < len (p^q) by A4,A5,A8,XXREAL_0:2; then A13: q.(m-len p)=(p^q).m by A12,Th17; A14: m-len p+len p< len (p^q) by A4,A5,A8,XXREAL_0:2; A15: m-len p is Nat by A12,NAT_1:21; len (p^q)=len p+len q by Def3; then m-len p len p; then Segm len p c= Segm n by NAT_1:39; then A1: p|n=p by RELAT_1:68; p^{}=p; hence thesis by A1; end; suppose n <= len p; then reconsider n1=len p-n as Element of NAT by NAT_1:21; defpred P[Nat] means for k st k= len p-\$1 holds ex q st p=(p|k)^q; A2: for m be Nat st P[m] holds P[m+1] proof let m be Nat such that A3: P[m]; let k such that A4: k = len p-(m+1); consider q such that A5: p=(p|(k+1))^q by A3,A4; k<=k+1 by NAT_1:11; then Segm k c= Segm(k+1) by NAT_1:39; then A6: (p|(k+1))|k =p|k by RELAT_1:74; len p-m<=len p-0 by XREAL_1:10; then len (p | (k+1)) = k+1 by Th51,A4; then p|(k+1)=(p|(k+1))|k^<%(p|(k+1)).k%> by Th53; then p=(p|k)^(<%(p|(k+1)).k%>^q) by A5,A6,Th25; hence thesis; end; p|(len p-0)=p & p^{}=p; then A7: P[0]; A8: for m be Nat holds P[m] from NAT_1:sch 2(A7,A2); n=len p-n1; hence thesis by A8; end; end; hence thesis; end; theorem :: FLANG_1:10 len p = n + k implies ex q1, q2 being XFinSequence st len q1 = n & len q2 = k & p = q1 ^ q2 proof defpred P[Nat] means for p being XFinSequence, i, j be Nat st len p = \$1 & len p = i + j ex q1, q2 being XFinSequence st len q1 = i & len q2 = j & p = q1 ^ q2; A1: now let n; assume A2: P[n]; thus P[n + 1] proof let p be XFinSequence; let i, j be Nat; assume that A3: len p = n + 1 and A4: len p = i + j; per cases; suppose A5: j = 0; take q1 = p; take q2 = {}; thus thesis by A4,A5; end; suppose j > 0; then consider k such that A6: j = k + 1 by NAT_1:6; p <> {} by A3; then consider q being XFinSequence, x such that A7: p = q ^ <%x%> by Th37; A8: n + 1 = len q + len <%x%> by A3,A7,Def3 .= len q + 1 by Th30; n = i + k by A3,A4,A6; then consider q1, q2 being XFinSequence such that A9: len q1 = i and A10: len q2 = k and A11: q = q1 ^ q2 by A2,A8; A12: len (q2 ^ <%x%>) = len q2 + len <%x%> by Def3 .= j by A6,A10,Th30; p = q1 ^ (q2 ^ <%x%>) by A7,A11,Th25; hence thesis by A9,A12; end; end; end; A13: P[0] proof let p be XFinSequence; let i, j be Nat; assume that A14: len p = 0 and A15: len p = i + j; p = {} by A14; then A16: p = {} ^ {}; len {} = i by A14,A15; hence thesis by A15,A16; end; for n holds P[n] from NAT_1:sch 2(A13, A1); hence thesis; end; theorem :: FSM_3:6 <%x%>^p = <%y%>^q implies x = y & p = q proof assume A1: <%x%>^p = <%y%>^q; (<%x%>^p).0 = x by Th32; then x = y by A1,Th32; hence thesis by A1,Th26; end; definition let D be set,q be FinSequence of D; func FS2XFS q -> XFinSequence of D means :Def8: len it=len q & for i being Nat st i < len q holds q.(i+1)=it.i; existence proof deffunc F(Nat) =q.(\$1 +1); ex p being XFinSequence st len p = len q & for k be Nat st k in len q holds p.k=F(k) from XSeqLambda; then consider p being XFinSequence such that A1: len p = len q and A2: for k be Nat st k in Segm len q holds p.k=F(k); rng p c= D proof let y be object; A3: rng q c= D by FINSEQ_1:def 4; assume y in rng p; then consider x being object such that A4: x in dom p and A5: y=p.x by FUNCT_1:def 3; reconsider nx=x as Element of NAT by A4; nx FinSequence of D means len it=len q & for i be Nat st 1<=i & i<= len q holds q.(i-'1)=it.i; existence proof deffunc F(Nat) = q.(\$1-'1); ex p being FinSequence st len p = len q & for k being Nat st k in dom p holds p.k=F(k) from FINSEQ_1:sch 2; then consider p being FinSequence such that A1: len p = len q and A2: for k being Nat st k in dom p holds p.k=F(k); rng p c= D proof let y be object; A3: rng q c= D by RELAT_1:def 19; assume y in rng p; then consider x being object such that A4: x in dom p and A5: y=p.x by FUNCT_1:def 3; reconsider nx=x as Element of NAT by A4; A6: nx in Seg len q by A1,A4,FINSEQ_1:def 3; then 1<=nx by FINSEQ_1:1; then nx-1>=0 by XREAL_1:48; then A7: nx-1=nx-'1 by XREAL_0:def 2; A8: nx-'1r) is XFinSequence of D; definition let D be non empty set; let q be FinSequence of D, n be Nat; assume that A1: n>len q and A2: NAT c= D; func FS2XFS*(q,n) -> non empty XFinSequence of D means len q = it.0 & len it=n & (for i be Nat st 1<=i & i<= len q holds it.i=q.i)& for j being Nat st len q r) as XFinSequence of D; <%x%> ^ (FS2XFS q) <>{} by Th27; then reconsider p0=<%x%> ^ (FS2XFS q)^q5 as non empty XFinSequence of D by Th27; 0 < len <%x%>; then A3: 0 in dom (<%x%>) by Lm1; A4: len <%x%>=1 by Def4; 0 in Segm(len <%x%> + len (FS2XFS q)) by NAT_1:44; then 0 in len (<%x%> ^ (FS2XFS q)) by Def3; then A5: p0.0=(<%x%> ^ (FS2XFS q)).0 by Def3 .=(<%x%>).0 by A3,Def3 .=x; A6: for i st 1<=i & i<= len q holds p0.i=q.i proof let i; assume that A7: 1<=i and A8: i<= len q; i-1>=0 by A7,XREAL_1:48; then A9: i-'1=i-1 by XREAL_0:def 2; i)+len (FS2XFS q)) by A4,Def8; then i in Segm(len (<%x%>)+len (FS2XFS q)) by NAT_1:44; then i in len (<%x%> ^ (FS2XFS q)) by Def3; then p0.i =(<%x%>^(FS2XFS q)).(1+(i-'1)) by A9,Def3 .=(FS2XFS q).(i-'1) by A4,A11,Def3 .=q.(i-'1+1) by A10,Def8 .=q.i by A9; hence thesis; end; A12: n-len q>0 by A1,XREAL_1:50; then A13: n-'len q=n-len q by XREAL_0:def 2; then n-'len q>=0+1 by A12,NAT_1:13; then A14: n-'len q -1>=0 by XREAL_1:48; A15: len q5=(n-'len q-'1) by FUNCOP_1:13; A16: for j being Nat st len q ^ (FS2XFS q)) =len (<%x%>) + len (FS2XFS q) by Def3 .=1+len q by A4,Def8; len q0 by XREAL_1:50; then A21: n-'len q=n-len q by XREAL_0:def 2; then n-len q>=0+1 by A20,NAT_1:13; then n-'len q-1>=0 by A21,XREAL_1:48; then A22: n-'len q-'1 =n-(len q+1) by A21,XREAL_0:def 2; 1+len q<=j by A17,NAT_1:13; then j-(1+len q)>=0 by XREAL_1:48; then A23: j-'(1+len q)=j-(1+len q) by XREAL_0:def 2; j-(len q+1)< n-(len q+1) by A18,XREAL_1:9; then A24: j-'len (<%x%> ^ (FS2XFS q)) in Segm(n-'len q-'1) by A19,A23,A22,NAT_1:44; j =len (<%x%> ^ (FS2XFS q))+(j-'len (<%x%> ^ (FS2XFS q))) by A19,A23; then p0.j=q5.(j-'len (<%x%> ^ (FS2XFS q))) by A15,A24,Def3 .=0 by A24,FUNCOP_1:7; hence thesis; end; len p0=len (<%x%> ^ (FS2XFS q)) + len q5 by Def3 .=len <%x%> + len (FS2XFS q) + len q5 by Def3 .= 1 + len (FS2XFS q) + len q5 by Th30 .=1 + len q + len q5 by Def8 .=1+len q+(n-'len q-'1) by FUNCOP_1:13 .=(n-(len q+1))+(len q+1) by A13,A14,XREAL_0:def 2 .=n; hence thesis by A5,A6,A16; end; uniqueness proof let p1,p2 be non empty XFinSequence of D; assume that A25: len q = (p1.0) and A26: len p1=n and A27: for i st 1<=i & i<= len q holds p1.i=q.i and A28: for j being Nat st len q FinSequence of D means :Def11: for m be Nat st m = p.0 holds len it =m & for i st 1<=i & i<= m holds it.i=p.i; existence proof reconsider m0=p.0 as Element of NAT by A1,ORDINAL1:def 12; deffunc F(set)= p.\$1; ex q being FinSequence st len q = m0 & for k being Nat st k in dom q holds q.k=F(k) from FINSEQ_1:sch 2; then consider q being FinSequence such that A3: len q = m0 and A4: for k being Nat st k in dom q holds q.k=F(k); rng q c= D proof A5: m0 < len p by A2,Lm1; let y be object; assume y in rng q; then consider x being object such that A6: x in dom q and A7: y=q.x by FUNCT_1:def 3; reconsider k0=x as Element of NAT by A6; k0 in Seg m0 by A3,A6,FINSEQ_1:def 3; then k0<=m0 by FINSEQ_1:1; then k0 < len p by A5,XXREAL_0:2; then A8: k0 in dom p by Lm1; y=p.k0 by A4,A6,A7; then rng p c= D & y in rng p by A8,FUNCT_1:def 3,RELAT_1:def 19; hence thesis; end; then reconsider q0=q as FinSequence of D by FINSEQ_1:def 4; A9: dom q = Seg m0 by A3,FINSEQ_1:def 3; for m be Nat st m = (p.0) holds len q0 =m & for i st 1<=i & i<= m holds q0.i =p.i by A4,A9,FINSEQ_1:1,A3; hence thesis; end; uniqueness proof reconsider m2=p.0 as Nat by A1; let g1,g2 be FinSequence of D; assume that A10: for m st m = p.0 holds len g1 =m & for i st 1<=i & i<= m holds g1 .i=p. i and A11: for m st m = p.0 holds len g2 =m & for i st 1<=i & i<= m holds g2 . i=p.i; A12: len g1=m2 by A10; A13: for i be Nat st 1<=i & i<=len g1 holds g1.i=g2.i proof let i be Nat; assume A14: 1<=i & i<=len g1; then g1.i=p.i by A10,A12; hence thesis by A11,A12,A14; end; len g2=m2 by A11; hence thesis by A10,A13,FINSEQ_1:14; end; end; theorem for p being XFinSequence of D st p.0=0 & 0 initial for Function; coherence; end; registration cluster -> initial for XFinSequence; coherence proof let p be XFinSequence; let m,n being Nat such that A1: n in dom p; assume m < n; then m in Segm n by NAT_1:44; hence m in dom p by A1,ORDINAL1:10; end; end; :: following, 2010.01.11, A.T. registration cluster -> NAT-defined for XFinSequence; coherence proof let f be XFinSequence; thus dom f c= NAT; end; end; theorem Th62: for F being non empty initial NAT-defined Function holds 0 in dom F proof let F be non empty initial NAT-defined Function; consider x being object such that A1: x in dom F by XBOOLE_0:def 1; dom F c= NAT by RELAT_1:def 18; then reconsider x as Element of NAT by A1; x = 0 or 0 < x; hence 0 in dom F by A1,Def12; end; registration cluster initial finite NAT-defined -> Sequence-like for Function; coherence proof let F be Function; assume A1: F is initial finite NAT-defined; thus dom F is epsilon-transitive proof let x be set; assume A2: x in dom F; then reconsider i = x as Nat by A1; let y be object; assume y in x; then A3: y in Segm i; then reconsider j = y as Nat; j < i by NAT_1:44,A3; hence y in dom F by A1,A2; end; let x,y be set; assume x in dom F & y in dom F; then reconsider x,y as Ordinal by A1; x in y or x = y or y in x by ORDINAL1:14; hence thesis; end; end; theorem for F being finite initial NAT-defined Function for n being Nat holds n in dom F iff n < card F by Lm1; :: from AMISTD_2, 2010.04.16, A.T. theorem for F being initial NAT-defined Function, G being NAT-defined Function st dom F = dom G holds G is initial by Def12; theorem for F being initial NAT-defined finite Function holds dom F = { k where k is Element of NAT: k < card F } proof let F be initial NAT-defined finite Function; hereby let x be object; assume A1: x in dom F; then reconsider f = x as Element of NAT; f < card F by A1,Lm1; hence x in { k where k is Element of NAT: k < card F }; end; let x be object; assume x in { k where k is Element of NAT: k < card F }; then ex k being Element of NAT st x = k & k < card F; hence thesis by Lm1; end; theorem for F being non empty XFinSequence, G be non empty NAT-defined finite Function st F c= G & LastLoc F = LastLoc G holds F = G proof let F be initial non empty NAT-defined finite Function, G be non empty NAT -defined finite Function such that A1: F c= G and A2: LastLoc F = LastLoc G; dom F = dom G proof thus dom F c= dom G by A1,GRFUNC_1:2; let x be object; assume A3: x in dom G; dom G c= NAT by RELAT_1:def 18; then reconsider x as Element of NAT by A3; A4: LastLoc F in dom F by VALUED_1:30; x <= LastLoc F by A2,A3,VALUED_1:32; then x < LastLoc F or x = LastLoc F by XXREAL_0:1; hence thesis by A4,Def12; end; hence thesis by A1,GRFUNC_1:3; end; theorem Th67: for F being non empty XFinSequence holds LastLoc F = card F -' 1 proof let F be initial non empty NAT-defined finite Function; consider k being Nat such that A1: LastLoc F = k; reconsider k as Element of NAT by ORDINAL1:def 12; LastLoc F in dom F by VALUED_1:30; then k < card F by A1,Lm1; then A2: k <= card F -' 1 by NAT_D:49; per cases by A2,XXREAL_0:1; suppose k < card F -' 1; then k+1 < card F -' 1 + 1 by XREAL_1:6; then k+1 < card F by NAT_1:14,XREAL_1:235; then k+1 in dom F by Lm1; then A3: k+1 <= k by A1,VALUED_1:32; k <= k+1 by NAT_1:11; then k+0 = k+1 by A3,XXREAL_0:1; hence thesis; end; suppose k = card F -' 1; hence thesis by A1; end; end; theorem for F being initial non empty NAT-defined finite Function holds FirstLoc F = 0 by Th62,VALUED_1:35; registration let F be initial non empty NAT-defined finite Function; cluster CutLastLoc F -> initial; coherence proof set G = CutLastLoc F; per cases; suppose G is empty; then reconsider H = G as empty finite Function; H is initial; hence thesis; end; suppose G is non empty; then reconsider G as non empty finite Function; G is initial proof let m,l be Nat such that A1: l in dom G and A2: m < l; set M = dom F; reconsider R = {[LastLoc F, F.LastLoc F]} as Relation; R = LastLoc F .--> (F.LastLoc F) by FUNCT_4:82; then A3: dom R = {LastLoc F} by FUNCOP_1:13; then A4: dom F \ dom R = dom G by VALUED_1:36; then l in dom F by A1,XBOOLE_0:def 5; then A5: m in dom F by A2,Def12; l in M by A4,A1,XBOOLE_0:def 5; then m <> LastLoc F by A2,XXREAL_2:def 8; then not m in {LastLoc F} by TARSKI:def 1; hence thesis by A3,A4,A5,XBOOLE_0:def 5; end; hence thesis; end; end; end; reserve l for Nat; theorem for I being finite initial NAT-defined Function, J being Function holds dom I misses dom Shift(J,card I) proof let I be finite initial NAT-defined Function, J be Function; assume A1: dom I meets dom Shift(J,card I); dom Shift(J,card I) = { l+card I: l in dom J } by VALUED_1:def 12; then consider x being object such that A2: x in dom I and A3: x in { l+card I: l in dom J } by A1,XBOOLE_0:3; consider l such that A4: x = l+card I and l in dom J by A3; l+card I < card I by A2,A4,Lm1; hence contradiction by NAT_1:11; end; :: from SCMPDS_4, 2010.05.14, A.T. theorem not m in dom p implies not m+1 in dom p proof assume not m in dom p; then A1: m >= card p by Lm1; m+1 >= m by NAT_1:11; then m+1 >= card p by A1,XXREAL_0:2; hence thesis by Lm1; end; :: from SCM_COMP, 2010.05.16, A.T. registration let D be set; cluster D^omega -> functional; coherence by Def7; end; registration let D be set; cluster -> finite Sequence-like for Element of D^omega; coherence by Def7; end; definition let D be set; let f be XFinSequence of D; func Down f -> Element of D^omega equals f; coherence by Def7; end; definition let D be set; let f be XFinSequence of D, g be Element of D^omega; redefine func f^g -> Element of D^omega; coherence proof reconsider g as XFinSequence of D by Def7; f^g is XFinSequence of D; hence thesis by Def7; end; end; definition let D be set; let f, g be Element of D^omega; redefine func f^g -> Element of D^omega; coherence proof reconsider f,g as XFinSequence of D by Def7; f^g is XFinSequence of D; hence thesis by Def7; end; end; :: missing, 2010.05.15, A.T. theorem Th71: p c= p^q proof A1: dom p c= dom(p^q) by Th19; for x being object st x in dom p holds (p^q).x = p.x by Def3; hence thesis by A1,GRFUNC_1:2; end; theorem Th72: len(p^<%x%>) = len p + 1 proof thus len(p^<%x%>) = len p + len<%x%> by Def3 .= len p + 1 by Th30; end; theorem <%x,y%> = (0,1) --> (x,y) proof A1: dom<%x,y%> = len<%x,y%> .= {0,1} by Th35,CARD_1:50; A2: <%x,y%>.0 = x; <%x,y%>.1 = y; hence <%x,y%> = (0,1) --> (x,y) by A1,A2,FUNCT_4:66; end; reserve M for Nat; theorem Th74: p^q = p +* Shift(q, card p) proof A1: dom Shift(q, card p) = { M+card p:M in dom q } by VALUED_1:def 12; for x being object holds x in dom(p^q) iff x in dom p or x in dom Shift(q, card p) proof let x be object; thus x in dom(p^q) implies x in dom p or x in dom Shift(q, card p) proof assume A2: x in dom(p^q); then reconsider k = x as Nat; per cases by A2,Th18; suppose k in dom p; hence x in dom p or x in dom Shift(q, card p); end; suppose ex n st n in dom q & k=len p + n; hence x in dom p or x in dom Shift(q, card p) by A1; end; end; assume A3: x in dom p or x in dom Shift(q, card p); per cases by A3; suppose A4: x in dom p; dom p c= dom(p^q) by Th19; hence x in dom(p^q) by A4; end; suppose x in dom Shift(q, card p); then ex M st x = M+card p & M in dom q by A1; hence x in dom(p^q) by Th21; end; end; then A5: dom(p^q) = dom p \/ dom Shift(q, card p) by XBOOLE_0:def 3; for x being object st x in dom p \/ dom Shift(q, card p) holds (x in dom Shift(q, card p) implies (p^q).x = Shift(q, card p).x) & (not x in dom Shift(q, card p) implies (p^q).x = p.x) proof let x be object such that A6: x in dom p \/ dom Shift(q, card p); hereby assume A7: x in dom Shift(q, card p); then reconsider k = x as Nat; consider M such that A8: x = M+card p and A9: M in dom q by A7,A1; set m = k -' len p; A10: len p + m = k by A8,NAT_D:34; hence (p^q).x = q.m by A8,A9,Def3 .= Shift(q, card p).x by A8,A9,A10,VALUED_1:def 12; end; assume not x in dom Shift(q, card p); then x in dom p by A6,XBOOLE_0:def 3; hence (p^q).x = p.x by Def3; end; hence p^q = p +* Shift(q, card p) by A5,FUNCT_4:def 1; end; theorem p +* (p ^ q) = p ^ q & (p ^ q) +* p = p ^ q by Th71,FUNCT_4:97,98; reserve m,n for Nat; theorem Th76: for I being finite initial NAT-defined Function, J being Function holds dom Shift(I,n) misses dom Shift(J,n+card I) proof let I be finite initial NAT-defined Function, J be Function; assume A1: dom Shift(I,n) meets dom Shift(J,n+card I); dom Shift(J,n+card I) = { l+(n+card I): l in dom J } by VALUED_1:def 12; then consider x being object such that A2: x in dom Shift(I,n) and A3: x in { l+(n+card I): l in dom J } by A1,XBOOLE_0:3; dom Shift(I,n) = { m+n:m in dom I } by VALUED_1:def 12; then consider m such that A4: x = m+n and A5: m in dom I by A2; consider l such that A6: x = l+(n+card I) and l in dom J by A3; m < card I by A5,Lm1; then l+(n+card I) < n+card I by A4,A6,XREAL_1:6; hence contradiction by NAT_1:11; end; theorem Th77: Shift(p,n) c= Shift(p^q,n) proof p^q = p +* Shift(q, card p) by Th74; then A1: Shift(p^q,n) = Shift(p,n) +* Shift(Shift(q,card p),n) by VALUED_1:23; Shift(Shift(q,card p),n) = Shift(q,n+card p) by VALUED_1:21; then dom Shift(p,n) misses dom Shift(Shift(q,card p),n) by Th76; hence Shift(p,n) c= Shift(p^q,n) by A1,FUNCT_4:32; end; theorem Th78: Shift(q,n+card p) c= Shift(p^q,n) proof A1: Shift(Shift(q,card p),n) = Shift(q,n+card p) by VALUED_1:21; p^q = p +* Shift(q, card p) by Th74; then Shift(p^q,n) = Shift(p,n) +* Shift(Shift(q,card p),n) by VALUED_1:23; hence thesis by A1,FUNCT_4:25; end; theorem Shift(p^q,n) c= X implies Shift(p,n) c= X proof assume A1: Shift(p^q,n) c= X; Shift(p,n) c= Shift(p^q,n) by Th77; hence Shift(p,n) c= X by A1; end; theorem Shift(p^q,n) c= X implies Shift(q,n+card p) c= X proof assume A1: Shift(p^q,n) c= X; Shift(q,n+card p) c= Shift(p^q,n) by Th78; hence thesis by A1; end; registration let F be initial non empty NAT-defined finite Function; cluster CutLastLoc F -> initial; coherence; end; definition let x1,x2,x3,x4 be object; func <%x1,x2,x3,x4%> -> set equals <%x1%>^<%x2%>^<%x3%>^<%x4%>; coherence; end; registration let x1,x2,x3,x4 be object; cluster <%x1,x2,x3,x4%> -> Function-like Relation-like; coherence; end; registration let x1,x2,x3,x4 be object; cluster <%x1,x2,x3,x4%> -> finite Sequence-like; coherence; end; reserve x1,x2,x3,x4 for object; theorem len<%x1,x2,x3,x4%> = 4 proof thus len<%x1,x2,x3,x4%> = len<%x1,x2,x3%> + 1 by Th72 .= 3 + 1 by Th36 .= 4; end; Lm3: <%x1,x2,x3,x4%>.1 = x2 & <%x1,x2,x3,x4%>.2 = x3 & <%x1,x2,x3,x4%>.3 = x4 proof A1: len<%x1,x2,x3%> = 3 by Th36; then A2: 1 in dom<%x1,x2,x3%> by Lm1; thus <%x1,x2,x3,x4%>.1 =<%x1,x2,x3%>.1 by A2,Def3 .= x2; A3: 2 in dom<%x1,x2,x3%> by A1,Lm1; thus <%x1,x2,x3,x4%>.2 =<%x1,x2,x3%>.2 by A3,Def3 .= x3; thus <%x1,x2,x3,x4%>.3 = x4 by A1,Th33; end; registration let x1,x2,x3,x4 be object; reduce <%x1,x2,x3,x4%>.0 to x1; reducibility proof thus <%x1,x2,x3,x4%>.0 =(<%x1%>^<%x2,x3%>^<%x4%>).0 by Th25 .=(<%x1%>^<%x2,x3,x4%>).0 by Th25 .= x1 by Th32; end; reduce <%x1,x2,x3,x4%>.1 to x2; reducibility by Lm3; reduce <%x1,x2,x3,x4%>.2 to x3; reducibility by Lm3; reduce <%x1,x2,x3,x4%>.3 to x4; reducibility by Lm3; end; ::\$CT theorem k < len p iff k in dom p by Lm1; reserve e,u for object; theorem Segm(n+1) --> e = (Segm n --> e)^<%e%> proof set p = Segm n --> e, q = Segm(n+1) --> e; A1: len p = n by FUNCOP_1:13; A2: dom q = n+1 by FUNCOP_1:13 .= len p + len <%e%> by A1,Th31; A3: for k st k in dom p holds q.k=p.k proof let k; assume A4: k in dom p; Segm n c= Segm(n+1) by NAT_1:63; then p c= q by FUNCT_4:4; hence q.k=p.k by A4,GRFUNC_1:2; end; for k st k in dom<%e%> holds q.(len p + k) = <%e%>.k proof let k such that A5: k in dom<%e%>; dom<%e%> = {0} by Th30, CARD_1:49; then A6: k = 0 by A5,TARSKI:def 1; len p < n+1 by A1,NAT_1:13; then len p + 0 in Segm(n+1) by NAT_1:44; hence q.(len p + k) = <%e%>.k by A6,FUNCOP_1:7; end; hence thesis by A2,A3,Def3; end; theorem Th84: dom Shift(<%e%>,card p) = {card p} proof for u holds u in dom Shift(<%e%>,card p) iff u = card p proof let u; thus u in dom Shift(<%e%>,card p) implies u = card p proof assume u in dom Shift(<%e%>,card p); then u in { m+card p where m is Nat:m in dom <%e%> } by VALUED_1:def 12; then consider m being Nat such that A1: u = m+card p and A2: m in dom <%e%>; m in { 0 } by A2, Def4, CARD_1:49; then m = 0 by TARSKI:def 1; hence u = card p by A1; end; 0 in 1 by CARD_1:49,TARSKI:def 1; then 0 in dom <%e%> by Def4; then 0+card p in dom Shift(<%e%>,card p) by VALUED_1:24; hence thesis; end; hence thesis by TARSKI:def 1; end; theorem dom(p^<%e%>) = dom p \/ {card p} proof thus dom(p^<%e%>) = dom(p +* Shift(<%e%>, card p)) by Th74 .= dom p \/ dom Shift(<%e%>,card p) by FUNCT_4:def 1 .= dom p \/ {card p} by Th84; end; theorem <%x%> +~ (x,y) = <%y%> proof A1: dom(<%x%> +~ (x,y)) = dom<%x%> by FUNCT_4:99 .= Segm 1 by Th30; then <%x%> +~ (x,y) is finite by FINSET_1:10; then reconsider p = <%x%> +~ (x,y) as XFinSequence by A1,ORDINAL1:def 7; A2: rng<%x%> = {x} by Th30; then rng p c= {x} \ {x} \/ {y} by FUNCT_4:104; then rng p c= {} \/ {y} by XBOOLE_1:37; then A3: rng p c= {y}; x in rng <%x%> by A2,TARSKI:def 1; then y in rng p by FUNCT_4:101; then rng p = {y} by A3,ZFMISC_1:33; hence <%x%> +~ (x,y) = <%y%> by A1,Th30; end; theorem for I being non empty XFinSequence holds LastLoc I = card I - 1 proof let I be non empty XFinSequence; A1: card I >= 0+1 by NAT_1:13; thus LastLoc I = card I -' 1 by Th67 .= card I - 1 by A1,XREAL_1:233; end;
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CC-MAIN-2018-30
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en
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http://poj.org/showmessage?message_id=351936
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crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00611.warc.gz
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## Re:楼主求解啊，数据过了却还是WA
Posted by tomb at 2016-09-18 18:15:15 on Problem 1328
In Reply To:给大家个数据，希望有用 Posted by:mickeychen at 2010-04-14 19:53:06
```#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
struct isl
{
int x,y;
};
{
float sta,end;
};
{
if(a.end<b.end)
return true;
else
return false;
};
int main()
{
int n,d;
isl isl[1002];
int ymax=0;
int i=0;
int t=0;
while(cin>>n>>d&&(n!=0||d!=0)&&n<=1000&&n>=1)
{
t++;
for(i=0;i<n;++i)
{
cin>>isl[i].x>>isl[i].y;
if(isl[i].y>ymax)
{ymax=isl[i].y;}
}
cout<<" "<<endl;
cout<<"Case"<<t<<":";
if(ymax>d||d<0)
{
return -1;
continue;
}
float len;
for(i=0;i<n;++i)
{
len=sqrt(1.0*d*d-isl[i].y*isl[i].y);
}
int ans=0;
for(i=0;i<n;)
{
int count=0;
for(int j=i+1;j<n;++j)
{
{++count;}
else
{break;}
}
i+=count+1;
ans++;
}
cout << ans << endl;
}
return 0;
}
```
Followed by:
| 409
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|
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| 3.125
| 3
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CC-MAIN-2021-04
|
latest
|
en
| 0.251431
|
http://stackoverflow.com/questions/13915479/c-get-every-number-separately/13915513
| 1,419,764,194,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-52/segments/1419447556379.119/warc/CC-MAIN-20141224185916-00065-ip-10-231-17-201.ec2.internal.warc.gz
| 92,840,755
| 17,844
|
# C++ get every number separately
I have a range of numbers from 100 to 999. I need to get every number separately of it and check whether it can be divided by 2. For example:
``````232
2 divided by 2 = 1 = true
3 divided by 2 = 1.5 = false
2 divided by 2 = 1 = true
``````
and so on.
To get the first number all I have to do is to divide the entire number by 100.
``````int x = 256;
int k = x/100;
``````
so `x` would hold a value of 2.
Now, is there a way to check those other ones? Because `k = x/10;` would already be 25.
-
Hint: get rid of the 2 before dividing by 10 :) – R. Martinho Fernandes Dec 17 '12 at 13:55
You can use the `%` (modulus) operator to get decide divisibility. – Václav Zeman Dec 17 '12 at 13:56
Smells like homework. – plmaheu Dec 17 '12 at 13:57
@pboy Yet people still provide full code answers. Oh well, what can you do... – Let_Me_Be Dec 17 '12 at 13:58
Or convert the number to string and get it one char at-at-time. – Germann Arlington Dec 17 '12 at 14:01
Try this:
``````int x = 256;
int i = x / 100; // i is 2
int j = (x % 100) / 10; // j is 5
int k = (x % 10); // k is 6
``````
-
maybe look into integer division and the modulo.
``````int k1 = (x / 10) % 10 // "10s"
int k2 = ( x / 100 ) % 10 // "100s"
//etc etc
``````
-
Use modulo to get the last digit of the number, then divide by ten to discard the last digit.
Repeat while the number is non-zero.
-
What you need is the modulus operator `%`. It does a division and returns the reminder.
``````1 % 2 = 1
2 % 2 = 0
3 % 2 = 1
4 % 2 = 0
...
``````
-
eg. take 232:
``````int num = 232;
int at_ones_place = num % 10;
int at_tens_place = ( num /10 ) % 10 ;
int at_hundreds_place = (num /100);
``````
-
Oh my god, what is this abomination? – Let_Me_Be Dec 17 '12 at 14:03
@Let_Me_Be feels like a textbook question "spot the mistakes" – paul23 Dec 17 '12 at 14:04
@Let_Me_Be what do you mean? – ADG Dec 17 '12 at 14:06
@Aditya I'm sorry, but there are so many errors in the code, both syntactic and semantic that there simply isn't enough space in the comment to describe them all. For starters, try to compile and run the code. – Let_Me_Be Dec 17 '12 at 14:08
| 744
| 2,170
|
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| 3.21875
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CC-MAIN-2014-52
|
latest
|
en
| 0.850009
|
https://tyrocity.com/physics-notes/gallileans-telescope-2pm7
| 1,696,230,246,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00791.warc.gz
| 634,444,993
| 17,654
|
TyroCity
Physics XI Notes for Physics Notes
Posted on
Gallilean’s Telescope
Galilieo in 1906 constructed a telescope which provides an erect image of an object only with the help of two lenses. Galilean’s telescope is used to see the object on earth.
Construction of telescope
The Galilean’s telescope consists of two lenses.
• Objective
• Eye piece
- Objective
Objective is a convex lens of large focal length “fo”
- Eye piece
Eye piece is a divergent (concave) lens of short focal length “fe“. Eye piece forms an erect and virtual image at the focus of objective on when adjustment is correct.
Working of telescope
The incident rays from a distinct object falls on the objective as a beam of parallel rays at an angle “ a ” and after refraction from objective lens these rays form an inverted image AB at its focus.
The eye piece is so adjusted that the image formed by the objective lies at a distance “fe“. The rays falling on the eye piece are refracted and emerge as parallel beam. Thus, the final image is formed at infinity.
Magnifying power of telescope
Magnifying power is the ratio of visual angle subtended by the image to the visual angle subtended by the object.
i.e
It is because the object is at infinite distance and hence the angle subtended by the object at eye may be taken as the angle subtended by the object at objective.
M = b / a
Since a and b are small angle, therefore we can assume a= tana & b = tanb
This expression shows that for large magnification focal length of objective must be larger than the focal length of eye piece.
Length of telescope
Distance b/w objective lens and eye piece is called length of telescope.
From figure:
Length of telescope is:
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|
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| 3.078125
| 3
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CC-MAIN-2023-40
|
longest
|
en
| 0.924137
|
https://www.meritnation.com/ask-answer/question/7x-7-12x-3-solve-this/algebra/14128383
| 1,638,168,276,000,000,000
|
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# 7x+7=12x-3 Solve this
7x +7=12x-3 7x-12x=-3-7 -5x=-10 x=-10/-5 x=+2
• 1
yaar mujhe nahi aa raha tum na google kar do
• -1
I don't know to solve it because I am not clear that which question is it
• 0
What are you looking for?
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https://www.swamonk.com/from-the-following-data-calculate-inventory-turnover-ratio-total-sales-rs-500000-sales-return-rs-50000/
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## From the following data, calculate Inventory Turnover Ratio: Total Sales Rs. 5,00,000; Sales Return Rs. 50,000; Gross Profit Rs. 90,000; Closing Inventory Rs. 1,00,000; Excess of Closing Inventory over Opening Inventory Rs. 20,000.
SOLUTION
Cost of Goods Sold = Net Sales (Sales – Sales Return) – Gross Profit
= 5,00,000 – 50,000 – 90,000
= Rs. 3,60,000
Closing Inventory = 1,00,000
Closing Inventory is Rs. 20,000 more than the Opening Inventory
Therefore, Opening Inventory = 80,000 (1,00,000 – 20,000)
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# chemistry
posted by .
what is the ph of a solution that has [H] = 0.0045
can you show steps pls
• chemistry -
I got - log (0.0045) = 4.75
• chemistry -
pH = -log(H^+)
pH = -log(0.0045)
pH = -(-2.347) = 2.35
## Similar Questions
1. ### chem
are these correct when solving for proper significant figures 2.33 + 4.0045=6.33 .000456 + 1.0023=1.0027 2.33/4.0045=.582
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I GOT THIS WRONG IN AN ASSIGNMENT. I HAVE AN EXAM IN 2 DAYS. CAN YOU SHOW ME HOW TO DO IT PLS. What is the pH of a 2.62 x 10 -1 M NaNO2 solution ?
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If 10.0 mL of 1.0M HCl is added to 90 mL water, what is he concentration of the new solution?
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8x^3-125 X^Y^3-X^2Y^5 Pls show me steps
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pls help me proffesors! I need solution to this question, please show me the steps involve. The mean of the scores of three students in a class is 9. If the modal score is 11, find the lowest score.
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okay so I have two questions that i'm completely stuck on ;-; 1.You want to buy three books that are on sale at 20% off. The original prices of the books are \$2.50, \$4.95, and \$6.00. How much will you save?
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## HW4: Image stitching
### Author: Henrique Weber
The goal of this assignment is to create a panorama from a set of photos. For that the following steps were implemented:
• Manually select some feature points on each image.
• Find the homography matrix that align each pair of neighbor pictures.
• Transform the source image so as to be in the same projective space as the target image.
• Stitch images by taking the target image and placing it in the location given by the multiplication inverse of the homography matrix.
All these steps will be illustrated bellow together with the results.
### Part 0: Warmup
The first part of the assignment was to build the function `"imgTrans = applyTransformation(img, H)"`, which would transform the image "img" by the homography matrix "H". In theory it suffices to apply H to all pixels positions of the image and then copying its intensities to their new position. In practice however such approach leads to three problems: (1) the final image will potentially have holes in places which were there was no correspondency from the original image, (2) there will be many pixels collapsing to the same location if the image needs to be warped (which is almost always the case) and (3) we do not know the size of the transformed image. To overcome problems (1) and (2) we can use the inverse of the matrix H and search on the original image for the pixels intensities for every point in the final image. And to overcome problem (3) we can apply H to the extremes point locations of the original image and from there take the minimum and maximum values to know the final size of the image. The result can be seen bellow.
Target image.
Image after applying the transformation.
### Part 1: Manual matching
The second part of the assignment consisted in stitch images to compose a panorama. For that the images must all be taken with a camera having (aproximately) the same center of projection. In order to know where exactly the pictures collide we must provide an homography matrix H that takes one picture and project it in the same plane as its neighbor picture. The matrix H in turn can be estimated by finding 4 pair of points that describe the same location in both images. Having this 4 (or more) pairs of points gives us the possibility to create a system of linear equations to retrieve the matrix H. In this part of the assignment, these pair of points were chosen manually. Since small errors in the location of those pixels lead to big errors in the homography estimation, it is a better practice to choose more than that. Bellow are the results.
#### Manual results for sequence 1.
Image 1 with points related to image 2.
Image 2 with points related to image 1.
Image 2 with points related to image 3.
Image 3 with points related to image 2.
The result is an interpolation of all images in their right place. This "right place" was for me the most difficult part of the assignment! I had a hard time trying to find a way to keep track of the image places while placing them in the final panorama. Finally my solution was to create a vector containing a translation for each image. As I deform a new image, I calculate its position and update all images that have been already deformed to reflect this new arrangement. In this result I also took the maximum intensity value from both images where they overlap. It was just an experiment to blend both images without visible artifacts in the sky and also to reduce the blur caused by misalignment.
Target image.
#### Manual results for sequence 2.
Image 1 with points related to image 2.
Image 2 with points related to image 1.
Image 2 with points related to image 3.
Image 3 with points related to image 2.
Here the result is similar to the previous one and also to the result presented in the assignment description. No additional points were needed aside from the ones already provided with the assignment.
Target image.
#### Manual results for sequence 3.
Image 1 with points related to image 2.
Image 2 with points related to image 1.
Image 2 with points related to image 3.
Image 3 with points related to image 2.
This serie of pictures was the most difficult one since there was no much points to match. Also many attempts did not generate a good composition, so I had to try many times until the result bellow was possible.
Target image.
### Part 2: Automatic matching
Now that we are able to stitch images manually, we can go to the next step which is to do it automatically. For this we need to detect and match points on two neighbors images automatically. This apparently simple step involves many smaller procedures. First we need to detect corner features in the image using the algorithm proposed by Harris and provided as a ready-to-use code in the assignment. It creates however a huge number of points, and this is bad for performance and also for matching features points from two different images. To solve this we can use the Adaptive Non-Maximal Suppression. This algorithm works by first defining the radius from each feature point to its closest neighbor given a certain threshold on the neighbor. Points which are bigger than its closer neighbor have its radius set to infinity. After that we can pick a given number of points starting from the one with the biggest radius (which means the one that is more isolated in the image) to the one with smaller radius. In all the images I used 500 feature points. Aside from filtering the points we need to create more information about each one since they must be as much as possible distinct from each other. For that we simply take a window of 40x40 around the pixel and resize it to 8x8 in order to be less susceptible to noise. We also subtract the mean from each point inside this block and divide by its standard deviation, so that the final mean is 0 and standard deviation is 1.
Original points obtained with Harris feature corners detector.
The top 500 points using Adaptive Non-Maximal Suppression.
Once we have the feature points for all images it's time to match pairs of features between neighbor pictures. For that we compare the distance from all points on one image to all points on the other image. Following the paper we use the ratio between the smallest distance and the second smallest distance between one points and its two closer "siblings" on the other picture. We also use the mean of the distance to the second closest point as a threshold to determine if a pair of points is close enough to be considered a good match.
In the images bellow we use the MATLAB function `showMatchedFeatures(I1,I2,matchedPoints1,matchedPoints2);` to see `matchedPoints1` and `matchedPoints2` over images `I1` and `I2`.
In the sequence, we use RANSAC to find the homography matrix that best describes (given a finite number of attempts) the transformation between one set of points and the other. For that we chose 4 pair of points at random and calculate the homography matrix that describes them. Then we count the number of other pairs in the set that are in accordance with this homography given the previously mentioned threshold generated with the mean to the second closest neighbor. After a fixes number of trials (I used 100) we pick the homography that gathered the biggest number of inliers and considered it as the description of the transformation we need to perform to align the images. In the end we reuse the original pairs of inliers to recalculate the homography in order to let it more accurate.
#### Automatic results for sequence 1.
All pair of points that were considered a correct match.
Inliers and their estimated positions given by a pass of RANSAC.
Real location of the final inliers.
The resulting panorama took time to be computed on my computer (around 5min) but the result is mostly good. Since I also used the max between two pixels to get the final value we lost part of the cable in the left of the image. The boat is also blurry but it looks like it was moving while the photos were being taken.
Golden Gate panorama.
#### Automatic results for sequence 2.
Image 1 with points related to image 2.
Image 2 with points related to image 1.
Image 2 with points related to image 3.
Image 3 with points related to image 2.
Above are the points that were found by RANSAC to describe the transformation. The result looks better than the one that was created manually.
Target image.
#### Automatic results for sequence 3.
Image 1 with points related to image 2.
Image 2 with points related to image 1.
Image 2 with points related to image 3.
Image 3 with points related to image 2.
Image 1 with points related to image 4.
Image 4 with points related to image 1.
Image 2 with points related to image 5.
Image 5 with points related to image 2.
Image 3 with points related to image 6.
Image 6 with points related to image 3.
Here is another result with the pictures provided with the assignment. Here I average both images where they overlap, so it's possible to see the blur where they are not perfectly aligned.
Complete panorama.
#### Result over my own pictures.
Image 1 with points related to image 2.
Image 2 with points related to image 1.
Image 2 with points related to image 3.
Image 3 with points related to image 2.
Image 1 with points related to image 4.
Image 4 with points related to image 1.
Here is one result with my own pictures. The alignment looks good and also the exposure of all pictures are similar, which helped to create a nice panorama.
Complete panorama.
#### Another result with my own pictures.
Image 1 with points related to image 2.
Image 2 with points related to image 1.
Image 2 with points related to image 3.
Image 3 with points related to image 2.
Image 1 with points related to image 4.
Image 4 with points related to image 1.
In this serie of photos I slowly turn around the camera, which gives a better result (which can be seen by the borders that are well alligned).
Complete panorama.
#### Bells and whistles: my family on a sign (Source).
To place the picture in the right place it was enough to define the four corners in each image and then compose them.
Photo of my family.
Signs.
Final composition.
#### Bells and whistles: grafitti (Source) on ancient China (Source).
Here the grafitti was composed in a wall. To use only the grafitti the background was set to black, so when we compose them we can ignore the background.
Some place in China.
A grafitti.
Final composition.
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# Question: How Many Basic Shapes Are There?
## What is the 7 sided shape?
heptagonA five-sided shape is called a pentagon.
A six-sided shape is a hexagon, a seven-sided shape a heptagon, while an octagon has eight sides… The names of polygons are derived from the prefixes of ancient Greek numbers..
## What is a diamond shaped sign?
Diamond-shaped signs are used to warn drivers of special conditions or hazards ahead. They are typically yellow or orange in color.
## What is a one sided shape called?
List of n-gons by Greek numerical prefixesSidesNames1henagonmonogon2digonbigon3trigontriangle4tetragonquadrilateral7 more rows
## What are the 3 categories of shape?
There are 3 basic types of shapes: geometric, natural/organic and abstract.
## What are the 16 basic shapes?
Terms in this set (16)equilateral triangle. A triangle with all sides of equal length.isosceles triangle. A triangle with two sides of equal length.scalene triangle. A triangle with no sides of equal length. … scalene right triangle. … isosceles right triangle. … square. … rectangle. … parallelogram.More items…
## What is a 1000000000000000 sided shape called?
In geometry, a chiliagon (/ˈkɪliəɡɒn/) or 1000-gon is a polygon with 1,000 sides. Philosophers commonly refer to chiliagons to illustrate ideas about the nature and workings of thought, meaning, and mental representation.
## What are the 10 basic shapes?
Basic shapes Learninging charts introduce 10 basic shapes are circle, oval, triangle, rhombus, square, rectangle, trapezoid, pentagon, hexagon and octagon.
hexagonThe hexagon – a shape with 6 sides – is one of the most common shapes in nature. From honeycombs to snowflakes and patterns found on fruit skins, the hexagon is present everywhere!
## What shapes have less than 4 sides?
2D ShapesTriangle – 3 SidesSquare – 4 SidesPentagon – 5 SidesHexagon – 6 sidesHeptagon – 7 SidesOctagon – 8 SidesNonagon – 9 SidesDecagon – 10 SidesMore …
## What are the 5 basic shapes in art?
And that’s really all we’re going to do here, except we use a pencil and simplify a complex figure to just five basic geometric shapes – the triangle, oval, oblong, circle and square.
## What is a 20 sided shape called?
IcosagonIn geometry, an icosagon or 20-gon is a twenty-sided polygon. The sum of any icosagon’s interior angles is 3240 degrees.
## Is there a shape with 1 sides?
Indicate to dCode the number of sides and it will find the name….What is the name of a polygon with…?#Name of the Polygon + Geometric Drawing1 sidemonogon (impossible figure in Euclidean geometry)2 sidesdigon3 sidestriangle or trigon4 sidesquadrilateral or quadrangle or tetragon54 more rows
## What are 10 geometric concepts?
Geometry is the fourth math course in high school and will guide you through among other things points, lines, planes, angles, parallel lines, triangles, similarity, trigonometry, quadrilaterals, transformations, circles and area.
## What is a 2 sided shape?
In geometry, a digon is a polygon with two sides (edges) and two vertices. … A regular digon has both angles equal and both sides equal and is represented by Schläfli symbol {2}. It may be constructed on a sphere as a pair of 180 degree arcs connecting antipodal points, when it forms a lune.
## What is a 9999 sided shape called?
nonanonacontanonactanonaliagonWhat do you call a 9999-sided polygon? A nonanonacontanonactanonaliagon.
## What are the 8 basic shapes?
Basic Geometric Shapes Mini Book Worksheet. The Basic Geometric Shapes Mini Book is fun and simple for children in preschool to practice recognizing the eight basic shapes: square, circle, triangle, diamond, oval, rectangle, star, and heart.
## What is a 100 sided shape called?
hectogonIn geometry, a hectogon or hecatontagon or 100-gon is a hundred-sided polygon. The sum of all hectogon’s interior angles are 17640 degrees.
## What is the most simple shape?
The square, circle, and triangle are the most basic shapes on Earth, supporting structures both synthetic and natural.
## What are the 2 types of shapes?
All objects are either shapes or forms. A shape is a two-dimensional area that is defined in some way. There are two types of shapes: geometric and free-form. Geometric shapes are precise shapes that can be described using mathematical formulas.
## How many shapes are there?
In solid geometry, the three-dimensional shapes are cube, cuboid, cone, sphere and cylinder. We can observe all these shapes in our daily existence also. For example books (cuboid shape), glasses (cylindrical shape), traffic cones (conical shape) and so on.
## What are the basic shapes?
The three basic shapes are a square, a triangle and a circle. All other shapes are derived from these.
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## Gravity Simulator
This simulation … Applied a geocentric coordinate system. Air friction is ignored. Covers fall and orbital motion due to Earth’s gravity. (However, the same can be applied to other celestial bodies.) Gravity Objects with mass exert a pulling force on … more
## Free Fall Experiment
This simulation allows you to measure speed as a function of fall height. Free fall movement All objects on Earth are affected by gravity. If there were no friction with air, the speed of any object would increase by 9.8 … more
## General Relativity of Gravity
* After setting the small object’s mass, position, initial velocity, and initial direction of movement, turn on the ‘Run’ checkbox to start the simulation. * The gravitational constant of the celestial body were assumed to be ‘1’. General relativity General … more
## Three Body Problem
x y Vx Vx * The mass and gravitational constant of the celestial body were each assumed to be ‘1’. * You can edit the position and speed of each celestial body while not running. * As the operation is … more
## Movement of an object suspended from a pulley
Movement on the pulley Two objects are connected by a string on a fixed, frictionless pulley and are moving together. In this case, the mass of the string and pulley is ignored. Since tension $$T$$ and gravity $$g$$ act in … more
## Free Fall Motion Analysis
Why do light and heavy objects fall at the same time? The change in speed due to gravity is the same for all objects because the force and inertia acting on an object are proportional to the object’s mass. The … more
## Graph of Constant Velocity Motion
The distance covered by an object moving at a constant speed increases proportionally with time. If the distance is plotted against time on a graph, it will result in a straight line with a slope that indicates the speed. A … more
## “Free falling” or “Thrown to the horizontal direction”
How to use the simulation Drag the ball with your mouse to free fall it in the air. Drag your mouse on the cliff to observe the parabolic motion. You can change the horizontal speed by adjusting the drag speed. … more
## Webcam Movement Analysis (on the table)
Try pushing an imaginary ball placed on the table with your finger. If your PC doesn’t have a camera, it won’t work. When the web browser asks for access to the camera, please allow it. (Personal data never stored on … more
## Analysis of Free Fall Motion
* Please refer to the text below for how to make a graph. Free fall motion If you hold the ball in your hand and release it, it will fall. In the absence of air resistance, the only force acting … more
## Average velocity and instantaneous velocity
Record of movements The motion of an object can be more easily understood by graphing it. In the time-distance graph, the slope is equal to the velocity. The average velocity that an object has moved over a certain period of … more
## Ticker Timer (Movement on the slope)
Ticker Timer It is a device to record an object’s movement by taking a spot on a paper tape at regular time intervals. A typical ticker timer is plugged into a household outlet. So the ticker cycle of the ticker … more
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# Divisible by i Solution Codechef
## Divisible by i Solution Codechef
You are given an integer N. Construct a permutation P of length N such that For all i (1≤iN-1), i divides abs(Pi+1−Pi). Recall that a permutation of length N is an array where every integer from 1 to N occurs exactly once. It can be proven that for the given constraints at least one such P always exists.
Input Format
• The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follow.
• The only line of each test case contains an integer N – the length of the array to be constructed.
Output Format
• For each test case, output a single line containing N space-separated integers P1,P2,…,PN, denoting the elements of the array P.
• If there exist multiple such arrays, print any.
Constraints
• 1≤T≤5⋅104
• 2≤N≤105
• The sum of N over all test cases does not exceed 105.
Sample 1:
Input
```2
2
3
```
Output
```1 2
2 1 3
```
Explanation:
Test case 1: A possible array satisfying all the conditions is [1,2]:
• For i=1: abs(A2−A1)=abs(2-1)=1 is divisible by 1.
Test case 2: A possible array satisfying all the conditions is [2,1,3]:
• For i=1: abs(A2A1)=abs(1-2)=1 is divisible by 1.
• For i=2: abs(A3A2)=abs(3-1)=2 is divisible by 2.
### SOLUTION
Program: Divisible by i Solution in Python
``````for _ in range(int(input())):
n = int(input())
p = [n]
m = n
for i in range(n-1,0,-1):
el = p[0]-i
if 1<=el and el<=m and el not in p:
p.insert(0,el)
else:
p.insert(0,p[0]+i)
for i in p:
print(i, end=' ')
print()``````
Program: Divisible by i Solution in C++
``````#include <iostream>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int arr[n];
int k=0;
int e =n;
int s =1;
for(int i=n-1;i>=0;i--){
if(k==0){
arr[i]= e;
e--;
k=1;
}
else if(k==1){
arr[i]=s;
s++;
k=0;
}
}
for(int i=0;i<n;i++){
cout<<arr[i]<<" ";
}
cout<<endl;
}
return 0;
}``````
Program: Divisible by i Solution in Java
``````import java.util.*;
import java.lang.*;
import java.io.*;
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc= new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int num=sc.nextInt();
int n=num;
int a[]= new int[n];
a[n-1]=n;
int counter=0;
int k=n-1;
for(int i=n-2;i>=0;i--)
{
if(counter%2==0)
{
a[i]=n-k;
n-=k;
}
else
{
a[i]=n+k;
n+=k;
}
k-=1;
counter+=1;
}
for(int i=0;i<num;i++)
System.out.print(a[i]+" ");
System.out.println();
}
}
}``````
Related:
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# C++ float number to nan
I want to know what makes a float number nan in c++. I am using a large dataset and it is really hard to trace. I want to know the ways of changing a float number to nan to reduce bug possibilities.
I found the code that causes the nan problem. I found that s/m is nan in some cases. But I don't know how to solve it.
``````float gp(float x){
float e = 2.71828183;
x *= -1;
float s = pow(e,x);
float m = (1 + pow(e,x)) * (1 + pow(e,x));
return s / m;}
``````
-
Do you have any unchecked casts in your code? I’m almost certain that you cannot get a NaN if you only use +, – and * on well-defined (= properly initialised, non-NaN) values. – Konrad Rudolph Mar 26 '11 at 13:20
I posted some of my codes here. Do you think that it will cause a problem? – AliBZ Mar 26 '11 at 13:22
infinity times zero (as stated in my answer) is possible – Ronny Brendel Mar 26 '11 at 13:27
I edited the question. Could you plz take a look at it? – AliBZ Mar 26 '11 at 13:51
I updated my answer. – Ronny Brendel Mar 26 '11 at 17:31
Taken from wikipedia -> special values -> nan
• 0/0
• ∞×0
• sqrt(−1)
• in general "invalid operations" (I am not sure wether there are not more than the three above)
Looking at you code: infinity times 0 is possible, is it?
## edit:
• 0 <= s <= +inf
• 1 <= m <= +inf
s / m:
• +inf / +inf does indeed make minus NaN (I tested it)
I think that's the only thing that makes a NaN.
-
I edited the question. Could you plz take a look at it? – AliBZ Mar 26 '11 at 13:52
by the way that function could be more efficient, if you used s instead of calculating the same power 3 times. – Ronny Brendel Mar 26 '11 at 17:28
I think that if you divede by zero a finite number you also get a NaN (not only 0/0) – sergico Mar 27 '11 at 21:45
did you test it? – Ronny Brendel Mar 27 '11 at 21:47
If you can keep `x` between 0 and `FLT_MAX` (3.40E+38 in my case), your gp function will not return NaN.
-
You say in a comment that you only use `*`, `+`, `-`.
[Edit: you've since said that you also use `pow` and division, which introduce some extra ways to get NaN. For example if the parameter `x` is a large negative value then `pow(e,-x)` is infinity, so you can easily end up computing infinity/infinity, which is another NaN]
So, if you have IEEE floating-point then assuming this summary is correct, the only ways you can generate NaN are:
1. Generate a positive or negative infinity by going out of range,
2. Multiply it by zero.
or:
1. Generate a positive and a negative infinity,
2. Add them (or equivalently, subtract two infinities of the same sign).
So if you check for and catch infinities, you don't have to worry about NaNs as well. That said, the usual way is to let such values propagate as quiet NaNs, and check at the end.
For C++ implementations using non-IEEE arithmetic, I'm not sure what the rules are when a NaN is permitted. I could look them up in the standard, but then again so could you ;-)
-
``````sqrt(-1)
``````
give you NaN, for example. http://www.gnu.org/s/libc/manual/html_node/Infinity-and-NaN.html
-
I don't have sqrt() in my code. I only have *, + and -. – AliBZ Mar 26 '11 at 13:19
you can check when float became NaN by using function isnan() and post operation here. – Mark.Ablov Mar 26 '11 at 13:24
EDIT Try use `double` instead of `float`.
Probably it depends to compiler you using but general option is:
• variable is too small
• variable is too big
• divide by 0 (zero)
-
Doesn't the variable overflow if it gets too big !? – AliBZ Mar 26 '11 at 13:20
Yes and it could be marked as NaN – ProblemFactory Mar 26 '11 at 13:22
I used float for my variables. Can I use bigger floats? like double float !? – AliBZ Mar 26 '11 at 13:23
yes, just use `double` ;) – ProblemFactory Mar 26 '11 at 13:25
"too big" or "too small" should lead to infinities or zeroes, not NaN – Michael Borgwardt Mar 26 '11 at 13:26
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https://harbourfronts.com/formula-for-amortizing-loan-payment/
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# Formula for Amortizing Loan Payment
### What is an Amortizing Loan?
An amortizing loan (or amortized loan) is a type of loan that comes with periodic, scheduled payments of both the principal and interest amount. With amortized loans, borrowers need to repay the interest expense for the period. After that, they must pay an additional amount which will reduce the principal amount of the loan. There are various types of amortizing loans, such as auto loans, home loans, personal loans, etc.
Amortizing loans can be beneficial for both lenders and borrowers. For the borrower, these come with lower costs. Since the principal amount decreases with each payment, they have to pay lower interests on further installments. Similarly, with each repayment, lenders get a portion of their principal amount back. Therefore, they can reduce their credit risk through these loans.
### How to calculate Amortizing Loan payments?
Usually, amortizing loans come with amortization schedules. The schedule contains details of the payments that borrowers need to make every period. For each period, the interest payment depends on the principal amount for the last period. Therefore, the interest payments at the start of the loan will be the highest. On the other hand, interest payments during the end will be the lowest.
Similarly, the principal repayments at the start of the period will be the lowest while in the end, it will be the highest. It is because amortizing loans come with fixed payments each period for borrowers. These amortizing loan payments include a mix of both interest and principal amounts. However, for each repayment, the mix between both will be different.
As mentioned, the first payments will include a higher mix of interest payments and lower principal amounts. The final payments will consist of lower interest amounts and higher principal amounts. It is because the total repayment stays the same. Therefore, as the interest payment decreases, the principal repayments increase.
Borrowers and lenders can use the following formula for amortizing loan payments.
Amortizing Loan Payment = (PV x R) / [1 – (1 + R)^(-n)]
In the above formula, ‘PV’ represents the present value of the loan. ‘R’ denotes the per period interest rate, usually calculated by dividing the annual interest rate by the number of interest payments per year. Lastly, ‘n’ represents the number of payments in a year. Using the above formula for amortizing loan payments, lenders can calculate their total amount.
The following formula is also helpful in calculating the interest portion of the amortizing loan payment.
Interest on Amortizing Loan Payment = PV x (R / n)
In the above formula, ‘PV’ represents the present value of the loan. ‘R’ is the periodic interest rate on it. Lastly, ‘n’ represents the total number of payments.
### Example
A lender pays a \$100,000 loan to a borrower at an interest rate of 9%. The loan period is five years, and the borrower has to make monthly repayments. Therefore, they can calculate the amortizing loan payment using the following formula.
Amortizing Loan Payment = (PV x R) / [1 – (1 + R)^(-n)]
In the above formula, ‘PV’ will be the \$100,000 value of the loan. ‘R’ will be the periodic interest rate, which will be 0.75% (9%/12 payments). ‘n’ will be the number of total payments. Since the loan lasts for five years and the borrower has to make a payment every month, the total number of payments will be 60 (5 years x 12 months). Therefore, the amortizing loan payment will be.
Amortizing Loan Payment = (\$100,000 x 0.75%) / [1 – (1 + 0.75%)^(-60)]
Amortizing Loan Payment = \$2,076
### Conclusion
Amortizing loans are debt obligations where borrowers have to repay a portion of the principal amount along with periodic interest payments. Borrowers can use the amortizing loan payment formula above to calculate the repayments on their amortizing loans.
## Further questions
Have an answer to the questions below? Post it here or in the forum
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Purchase Solution
# Finding the derivative of the inverse cosine of x
Not what you're looking for?
The inverse cosine function has domain [-1,1] and range [0, pi]. Prove that (cos^-1)'(x) = -1 / sqrt(1-x^2)
##### Solution Summary
The step by step classical answer with proper explanations
##### Solution Preview
Let y = Cos^-1 (x)
<br>
<br>then x = Cos(y)
<br>
<br>Now we will use the implicit ...
##### Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
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Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.
##### Probability Quiz
Some questions on probability
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https://computergraphics.stackexchange.com/questions/4651/additive-blending-with-weighted-blended-order-independent-transparency/5937
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# Additive blending with weighted-blended order independent transparency
I am trying to retrofit weighted blended OIT to my rendering pipeline and while it works well, producing convincing results, for normal alpha blending (based on the OVER) operator, I am struggling to make it support additive blending (for example Colour1 + Colour2 + Background Colour) correctly using the blending formula:
$$\frac{C_1 w_1 + C_2 w_2}{\alpha_1 w_1 + \alpha_2 w_2} \cdot \bigl(1 - (1 - \alpha_1)(1 - \alpha_2)\bigr) + \text{Bg} \cdot (1 - \alpha_1)(1 - \alpha_2)$$
I could hack it outputting weight values of 1 and very low alphas (eg 0.01), which would make it to sort of converge to additive blending, trouble is, this does not weight the colours at all and does not blend nicely with normal alpha blending.
I was wondering how people have tackled this problem.
I have worked with this specific formula for the OVER operator but not with additive blending. I'll use the paper's nomenclature in the following discussion:
$$C_f = \frac{\sum_{i=1}^{n}C_i \cdot w(z_i, \alpha_i)}{\sum_{i=1}^{n}\alpha_i \cdot w(z_i, \alpha_i)}(1 - \prod_{i=1}^{n}(1 - \alpha_i)) + C_0\prod_{i=1}^{n}(1 - \alpha_i)$$
This is not explicitly stated in the paper, but the term $$C_i$$ is the premultiplied-alpha color (i.e. color.rgb * color.a)
As described in the paper, the term $$C_0\prod_{i=1}^{n}(1 - \alpha_i)$$ provides the "revealage" of the background color. If all the transparent surfaces are transparent, the product will be 1 and the background will be fully visible. The rest of the equation provides an approximation of the result of sorting the transparent surfaces by distance and using the OVER operator with pre-multiplied alpha colors.
However, the equation for additive blending (using glBlendFunc(GL_SRC_ALPHA, GL_ONE) and glBlendEquation(GL_FUNC_ADD)) without weights is: $$C_f = \sum_{i=1}^{n}\alpha_iRGB_i + C_0$$
This equation is already order independent! In order to add weights to the transparent surfaces with a normalization step in the end, the equation can then be simplified to:
$$C_f = \frac{\sum_{i=1}^{n}C_i \cdot w(z_i, \alpha_i)}{\sum_{i=1}^{n}w(z_i, \alpha_i)} + C_0$$
If you'd rather keep the same equation as before, you can achieve the same result by changing the shader outputs in listing 3 to:
gl_FragData[0] = vec4(Ci, 1) * w(zi, ai);
gl_FragData[1] = vec4(0);
Use the following equation for blending $$\frac{\alpha_1 w_1 C_1 + \alpha_2 w_2 C_2}{\alpha_1 w_1 + \alpha_2 w_2}.(1-(1-\alpha_1)(1-\alpha_2)) + Bg.(1-\alpha_1)(1-\alpha_2)$$ Also note that $\alpha$ value must always be between $0$ and $1$
• This is not much different to the one above, is it? :-) The point is that you need to output really small alphas to keep as much of Bg as possible and rely on large scales of output Colours to compensate for the suppression that those alphas will cause to the accumulated colour. Making sure that the denominator is not zero as well. Commented Feb 3, 2017 at 9:02
• There is a difference. Note that the alpha is multiplied by the color in the numerator and that fixes the problem Commented Feb 3, 2017 at 20:03
• In order to help people assess whether to use this approach, could you add an explanation of why this solves the problem with the approach in the question? Commented Mar 5, 2017 at 13:38
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http://stackoverflow.com/questions/5050250/fast-way-of-getting-the-dominant-color-of-an-image/5728702
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# Fast way of getting the dominant color of an image
I have a question about how to get the dominant color of an image (a photo). I thought of this algorithm: loop through all pixels and get their color, either red, green, yellow, orange, blue, magenta, cyan, white, grey or black (with some margin of course) and it's darkness (light, dark or normal) and afterwards check which colors occurred the most. I think this is slow and not very precise. Is there a better way?
If it matters, it's a UIImage taken from an iPhone or iPod touch camera which is at most 5 Mpx. The reason it has to be fast is that simply showing a progress indicator doesn't make very much sense as this is for an app for people with bad sight, or no sight at all. Because it's for a mobile device, it may not take very much memory (at most 50 MB).
-
Your general approach should work, but I'd highlight some details.
Instead of your given list of colors, generate a number of color "bins" in the color spectrum to count pixels. Here's another question that has some algorithms for that: How to generate spectrum color palettes Make the number of bins configurable, so you can experiment to get the results you want.
Next, for each pixel you're considering, you need to find the "nearest" color bin to increment. You'll need to define "nearest"; see this article on "color difference": http://en.wikipedia.org/wiki/Color_difference
For performance, you don't need to look at every pixel. Since image elements usually cover large areas (e.g. the sky, grass, etc.), you can get the result you want by only sampling a few pixels. I'd guess that you could get good results sampling every 10th pixel, or even every 100th. You can experiment with that factor as well.
Finally, your question title says "average", but I'd resist the urge to actually average pixels. That can give you a surprising result. For example, see the resulting average color in this demo: http://jsfiddle.net/xLF38/
-
I just released a small javascript library (pieroxy.net/blog/pages/color-finder/index.html ) which does pretty much what you are describing. You can also pass a feedback method to skew the algo toward your color preference. Here is a live demo: pieroxy.net/blog/pages/color-finder/demo.html . Hope this helps. – pieroxy Jun 19 at 7:39
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Tensor Operations Are NP Hard
### New Book Reviews!
Tensor Operations Are NP Hard
Written by Mike James
Thursday, 01 August 2013
Most non-mathematicians might think that tensor operations are pretty hard without any formal proof, but new results prove that they are NP Hard which is not good news if you are trying to work something out using them.
A tensor is the n-dimensional generalization of a matrix. A matrix is a 2D tensor in that it has rows and columns. A 3D tensor is a cube of numbers with rows, columns and slices. In a matrix the numbers are indexed by two variables, e.g. something like aij; in a tensor the number of indexing variables can be more than two, e.g. aijk for a tensor of dimension, or more accurately rank 3.
We use matrix algebra and matrix numerical methods very often in almost any numerical program - it is the foundation of just about all the number crunching we do. We solve linear systems of equations, find eigenvalues, perform singular value decompositions and so on. These might be regarded as difficult math by some programmers, but for the majority of these tasks we have polynomial time algorithms. That is, vector and matrix computations are in P.
Tensor algebra generalizes the the linear algebra that we are all familiar with to higher dimensions - multi-linear algebra. It turns up in advanced geometry, the theory of gravity - the general theory of relativity that is, numerical mechanics, and so on. It is also being used in AI and computer vision and a number of cutting edge approaches to all sorts of topics. The good thing about tensor algebra is that from an algorithmic point of view it isn't really anything new - in most cases just a few more indexes to cope with.
Surely the algorithms that we need to work with say rank-3 tensors is no more difficult than for rank-2 tensors, i.e. matrices? It has to be in P - right?
Well no.
According to a recent paper, things are much worse for rank-3 tensors. It seems that they mark a dividing line between the linear convex problems that are tractable and the more difficult non-linear non-convex class. In the paper a whole list of generalizations of tractable matrix problems are shown to be NP-hard in their rank 3 formulations. These include finding eigenvalues, finding approximate eigenvalues, symmetric eigenvalues, singular values, proving positive definiteness, approximating a tensor by rank 1 tensors and so on. There is even a mention of a generalized question about finding a matrix function which is shown to be undecidable for any tensor 20x20x2 or greater. This is a big surprise because the equivalent matrix function can be easily found and another hint that throwing in just one extra index to a matrix problem makes it very much more difficult. As the title of the paper puts it - Most tensor problems are NP-hard.
Should we abandon tensor methods because most of them are not in P?
You need to keep in mind that these sorts of analysis are only valid asymptotically when the size of the tensor n in an n x n x n tensor gets very large. It could well be that for a range of small n we can find algorithms that get the job done in reasonable time. The fact that a problem is NP-hard doesn't mean its impossible!
The paper concludes with some open questions and, if you are reasonable at math you should find it an interesting read.
Most tensor problems are NP-hard
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Why Article 13 Must Be Stopped14/06/2018With only days to go before a crucial vote on EU copyright legislation could have a drastic impact on the internet and, as many others have pointed out, spell the end for memes, Internet luminaries ha [ ... ] + Full Story Are Perl Programmers Different?30/05/2018The results of the 2018 Perl Developer Survey are now available. It looks at users of Perl 5 and includes questions on their attitude to Perl 6 and to other languages. + Full Story More News
Last Updated ( Thursday, 01 August 2013 )
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# Reasoning Quiz For SBI Clerk Prelims 2023 -01st November
Directions (1-5): Study the following information carefully and answer the given questions:
There are eight persons P, R, Q, U, W, S, T, and V are seated in a row (but not necessarily in the same direction). Some of them are facing towards north direction and some of them are facing towards south direction.
Q sits 2nd to the left of the one who is 3rd to the right of T. T is not facing towards the north. V and Q faces opposite direction. P does not face the same direction as W faces. Q sits 4th to the left of R. W is an immediate neighbor of U who faces the same direction as R faces. Neither Q nor T sits any of the extremes ends of the row. S sits 2nd right of U. Only one person sits to the left of the T. R does not sit any of the extremes ends of the row. P who sits at one of the extreme ends of the row, sits 2nd left of U. S faces towards south direction. The one who sits 3rd to the right of the one who sits immediate right of W, doesn’t face to the north direction.
Q1. How many persons in a row are facing towards north direction?
(a) Two
(b) Four
(c) Three
(d) More than four
(e) None of these
Q2. Who among the following sits immediate left of the one who is 3rd to the left of R?
(a) Q
(b) P
(c) U
(d) S
(e) None of these
Q3. The one who sits immediate right of W, is facing in which direction?
(a) North
(b) South.
(c) Both (b) & (d).
(d) Same direction in which T faces.
(e) None of these.
Q4. Who sit at extreme ends of the row?
(a) PT
(b) VR
(c) None of these
(d) PV
(e) WS
Q5. How many persons sit to the left of R?
(a) One
(b) Two
(c) Four
(d) More than four
(e) None of these
Directions (6-10): The following questions are based on the five words in each of three letters given below.
TRA PTQ YTV WEN CBN
Q6. If all the letters in each of the words are arranged in alphabetical order within the word, then how many meaningful words will be formed?
(a) One
(b) Two
(c) Three
(d) More than three
(e) None
Q7. If the first letter of each word replaces by succeeding letter in English alphabet, then how many words have more than one vowel?
(a) Three
(b) Two
(c) One
(d) More than three
(e) None
Q8. If all the words are arranged in alphabetical order, then how many words remains at the same position?
(a) None
(b) One
(c) Two
(d) Three
(e) More than three
Q9. If first and last letter of each word are interchanged their position, then how many meaningful words are formed?
(a) One
(b) Two
(c) Three
(d) More than three
(e) None
Q10. If all words are arranged in alphabetical order from left to right, after interchanging first and second letter of each word, then which of the following word is 3rd from right?
(a) YTV
(b) TRA
(c) WEN
(d) CBN
(e) None of these
Solutions
. .
## FAQs
### What are sections included in the SBI Clerk Prelims Exam?
Reasoning Ability, Numerical Ability and English sections are included in the SBI Clerk Prelims Exam
#### Congratulations!
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## Fundamentals of Physics Extended (10th Edition)
1 almude= 2 medio 7 almude = 2$\times$ 7 medio 7 almude =14 medio
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Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C Bib Ind
### 12 Exterior Algebra and Koszul Complex
What follows are several operations related to the exterior algebra of a free module:
• A constructor for the graded parts of the exterior algebra ("exterior powers")
• Several Operations on elements of these exterior powers
• A constructor for the "Koszul complex"
• An implementation of the "Cayley determinant" as defined in [CQ11], which allows calculating greatest common divisors from finite free resolutions.
#### 12.1 Exterior Algebra: Constructor
##### 12.1-1 ExteriorPower
‣ ExteriorPower( k, M ) ( operation )
Returns: a homalg module
Construct the k-th exterior power of module M.
#### 12.2 Exterior Algebra: Properties and Attributes
##### 12.2-1 IsExteriorPower
‣ IsExteriorPower( M ) ( property )
Returns: true or false
Marks a module as an exterior power of another module.
##### 12.2-2 ExteriorPowerExponent
‣ ExteriorPowerExponent( M ) ( attribute )
Returns: an integer
The exponent of the exterior power.
##### 12.2-3 ExteriorPowerBaseModule
‣ ExteriorPowerBaseModule( M ) ( attribute )
Returns: a homalg module
The module that M is an exterior power of.
#### 12.3 Exterior Algebra: Element Properties
##### 12.3-1 IsExteriorPowerElement
‣ IsExteriorPowerElement( x ) ( property )
Returns: true or false
Checks if the element x is from an exterior power.
#### 12.4 Exterior Algebra: Element Operations
##### 12.4-1 Wedge
‣ Wedge( x, y ) ( operation )
Returns: an element of an exterior power
Calculate xy.
##### 12.4-2 ExteriorPowerElementDual
‣ ExteriorPowerElementDual( x ) ( operation )
Returns: an element of an exterior power
For x in a q-th exterior power of a free module of rank n, return x* in the (n-q)-th exterior power, as defined in [CQ11].
##### 12.4-3 SingleValueOfExteriorPowerElement
‣ SingleValueOfExteriorPowerElement( x ) ( operation )
Returns: a ring element
For x in a highest exterior power, returns its single coordinate in the canonical basis; i.e. [x] as defined in [CQ11].
#### 12.5 Koszul complex and Cayley determinant
##### 12.5-1 KoszulCocomplex
‣ KoszulCocomplex( a, E ) ( operation )
Returns: a homalg cocomplex
Calculate the E-valued Koszul complex of a.
##### 12.5-2 CayleyDeterminant
‣ CayleyDeterminant( C ) ( operation )
Returns: a ring element
Calculate the Cayley determinant of the complex C, as defined in [CQ11].
##### 12.5-3 Gcd_UsingCayleyDeterminant
‣ Gcd_UsingCayleyDeterminant( x, y[, ...] ) ( function )
Returns: a ring element
Returns the greatest common divisor of the given ring elements, calculated using the Cayley determinant.
Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C Bib Ind
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