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Graphing Quadratic Functions Worksheet Answer Key Algebra 2
Graphing Quadratic Functions Worksheet Answer Key Algebra 2
F worksheet by kuta software llc kuta software infinite algebra 1 name graphing quadratic functions date period. I 2 gm rardced wwpi9t hc vikn xfmikn riyt3eg ha xl kghexbsrza t l1s.
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Graphing quadratic functions worksheet answer key algebra 2. You can graph a quadratic equation using the function grapher but to really understand what is going on you can make the graph yourself.
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Write The Quadratic Function Quadratic Functions Quadratics Functions Algebra | {"url":"https://thekidsworksheet.com/graphing-quadratic-functions-worksheet-answer-key-algebra-2/","timestamp":"2024-11-14T07:10:00Z","content_type":"text/html","content_length":"137069","record_id":"<urn:uuid:b4c7c928-a70b-407a-8faf-9603319e0f0f>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00348.warc.gz"} |
numpy.fft.rfftn(a, s=None, axes=None, norm=None)[source]¶
Compute the N-dimensional discrete Fourier Transform for real input.
This function computes the N-dimensional discrete Fourier Transform over any number of axes in an M-dimensional real array by means of the Fast Fourier Transform (FFT). By default, all axes are
transformed, with the real transform performed over the last axis, while the remaining transforms are complex.
a : array_like
Input array, taken to be real.
s : sequence of ints, optional
Shape (length along each transformed axis) to use from the input. (s[0] refers to axis 0, s[1] to axis 1, etc.). The final element of s corresponds to n for rfft(x, n), while for
the remaining axes, it corresponds to n for fft(x, n). Along any axis, if the given shape is smaller than that of the input, the input is cropped. If it is larger, the input is
padded with zeros. if s is not given, the shape of the input along the axes specified by axes is used.
axes : sequence of ints, optional
Axes over which to compute the FFT. If not given, the last len(s) axes are used, or all axes if s is also not specified.
norm : {None, “ortho”}, optional
New in version 1.10.0.
Normalization mode (see numpy.fft). Default is None.
out : complex ndarray
Returns: The truncated or zero-padded input, transformed along the axes indicated by axes, or by a combination of s and a, as explained in the parameters section above. The length of the
last axis transformed will be s[-1]//2+1, while the remaining transformed axes will have lengths according to s, or unchanged from the input.
If s and axes have different length.
If an element of axes is larger than than the number of axes of a.
See also
The inverse of rfftn, i.e. the inverse of the n-dimensional FFT of real input.
The one-dimensional FFT, with definitions and conventions used.
The one-dimensional FFT of real input.
The n-dimensional FFT.
The two-dimensional FFT of real input.
The transform for real input is performed over the last transformation axis, as by rfft, then the transform over the remaining axes is performed as by fftn. The order of the output is as for rfft
for the final transformation axis, and as for fftn for the remaining transformation axes.
See fft for details, definitions and conventions used.
>>> a = np.ones((2, 2, 2))
>>> np.fft.rfftn(a)
array([[[ 8.+0.j, 0.+0.j],
[ 0.+0.j, 0.+0.j]],
[[ 0.+0.j, 0.+0.j],
[ 0.+0.j, 0.+0.j]]])
>>> np.fft.rfftn(a, axes=(2, 0))
array([[[ 4.+0.j, 0.+0.j],
[ 4.+0.j, 0.+0.j]],
[[ 0.+0.j, 0.+0.j],
[ 0.+0.j, 0.+0.j]]]) | {"url":"https://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.fft.rfftn.html","timestamp":"2024-11-06T11:49:26Z","content_type":"text/html","content_length":"13343","record_id":"<urn:uuid:2e022646-dc77-46d6-b9e1-366925ce7583>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00096.warc.gz"} |
SAT Math Practice 2 - Basic Trigonometry — Steemit
SAT Math Practice 2 - Basic Trigonometry
Solution to SAT Math Practice 1
Graph the function
From last time, we know that there is a shift to the right 2 and a shift down 1. The graph is our answer:
Basic Trigonometric Ratios
These are best known by high schoolers as SOH CAH TOA. It is better to know and understand the equations of course...
Memorize this and know that not all right triangles look the same. It is easy enough to find the correct orientation of sides given the angle. Just remember that the hypotenuse is the longest side
always across from the right angle (90 degrees).
Creating the Unit Circle with GIFs
30, 60, 90 Triangles
Notice that in both of the unit circles we are counting 0, 1, 2, ... from the +x-axis to create the measures with pi. To answer questions, say Sin(30 degrees) or Sin(pi/6), we will look at the
y-value created by the triangle for that angle. Remember that the longer leg of the 30, 60, 90 is sqrt(3)/2 and the shorter leg measures 1/2. Therefore Sin(30) = 1/2.
45, 45, 90 Triangles
The signs, positive or negative, matter as well. For example, Cos(3*pi/4) = - sqrt(2)/2 because it is on the negative side of the x = axis.
1. A triangle has the trigonometric ratio
2. Using the unit circle above, find Tan(120 degrees).
Remember that Tangent is equal to y/x, so the answer is | {"url":"https://steemit.com/steemiteducation/@hansenatortravel/sat-math-practice-2-basic-trigonometry","timestamp":"2024-11-10T05:29:13Z","content_type":"text/html","content_length":"139784","record_id":"<urn:uuid:8a8ce7d5-9fe5-4913-9a81-f4acebc10977>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00814.warc.gz"} |
What is the sum of all four angles of a concave quadrilateral is it true for a convex quadrilateral justify your answer? - Our Planet Today
on April 26, 2022
What is the sum of all four angles of a concave quadrilateral is it true for a convex quadrilateral justify your answer?
Space and Astronomy
Answer. angles of a concave quadrilateral is 360°. It does not matter whether the quadrilateral is concave or convex, the sum of all the four angles present in the concave or the convex quadrilateral
is always 360°.
What is the sum of all 4 angles of a concave quadrilateral?
Hence the required sum of all the angles of a concave quadrilateral is 360∘.
What is the sum of the measures of the angles of a concave quadrilateral will this property?
Let ABCD be a non-convex or a concave quadrilateral. Join BD, which also divides the quadrilateral into two triangles. Therefore, the sum of all the interior angles of this quadrilateral will also be
Does a concave quadrilateral have 4 equal sides?
Video quote: So in summary a convex quadrilateral has all four angles that are less than 180 degrees and a concave quadrilateral has at least one angle greater than 180 degrees.
What is a concave quadrilateral?
Concave quadrilateral: A quadrilateral is called a concave quadrilateral, if at least one line segment joining the vertices is not a part of the same region of the quadrilateral. That is, any line
segment that joins two interior points goes outside the figure.
What is the sum of angles of quadrilaterals?
The sum of interior angles in a quadrilateral is 360°.
What is the sum of angles of a concave polygon?
As with any simple polygon, the sum of the internal angles of a concave polygon is π×(n − 2) radians, equivalently 180×(n − 2) degrees (°), where n is the number of sides.
What is the measure of the angles of a concave quadrilateral?
The measure of angle of concave quadrilateral is more than 180^∘ .
Which of the following represent the angles of a concave quadrilateral?
angles of a concave quadrilateral is 360°.
What are convex and concave quadrilaterals?
A convex quadrilateral has both diagonals completely contained within the figure, while a concave one has at least one diagonal that lies partly or entirely outside of the figure. Those are good
definitions, but the concave shapes having a cavity or cave is probably an easier way to remember it.
How many angles of a concave quadrilateral can be greater than 90 (@)?
Step-by-step explanation:
The sum of the interior angles of a concave quadrilateral must be 360 degrees (you can always divide it into two triangles of 180 degrees each). If three are greater than 90 degrees, the fourth must
be less.
How many angles of a concave quadrilateral are greater than 180 show by figure?
one angle
If an angle of a quadrilateral is greater than 180°, then it is called a concave quadrilateral. In a concave quadrilateral, only one angle can be greater than 180°.
What is the sum of all interior angles for a convex polygon?
The properties of the convex polygon are as follows: The interior angle of a convex polygon is strictly less than 180°. A polygon, with at least one interior angle, is greater than 180° is called a
non-convex polygon or concave polygon. Sum of all the interior angles of a polygon of n sides = (n – 2)180°.
What is the sum of a convex polygon?
Theorem 40: If a polygon is convex, then the sum of the degree measures of the exterior angles, one at each vertex, is 360°.
What is the sum of the interior angles of a convex triangle?
Since we know that the sum of interior angles in a triangle is 180°, and if we subdivide a polygon into triangles, then the sum of the interior angles in a polygon is the number of created triangles
times 180°.
How do you find the sum of interior angles in a polygon?
To find the sum of interior angles of a polygon, multiply the number of triangles in the polygon by 180°. The formula for calculating the sum of interior angles is ( n − 2 ) × 180 ∘ where is the
number of sides.
What is the sum of all interior angle of a regular pentagon?
PENATGON: The sum of the interior angles of a pentagon is always 540°.
How do you find the sum of the interior angles of a pentagon?
Sum of Interior Angles in a Pentagon
We know that the sum of the interior angles of a polygon of n sides = (n – 2) × 180°. = 3× 180°= 540°. Hence, the sum of interior angles of a pentagon is 540°.
What is the polygon whose sum of the interior angles is 1 080?
The polygon with an interior angle sum of 1080° is an octagon, or a polygon with 8 sides.
What is the sum of all angles of a hexagon?
720 degrees
Correct answer:
The sum of the interior angles of a hexagon must equal 720 degrees. Because the hexagon is regular, all of the interior angles will have the same measure. A hexagon has six sides and six interior | {"url":"https://geoscience.blog/what-is-the-sum-of-all-four-angles-of-a-concave-quadrilateral-is-it-true-for-a-convex-quadrilateral-justify-your-answer/","timestamp":"2024-11-13T11:32:19Z","content_type":"text/html","content_length":"190771","record_id":"<urn:uuid:095a3850-f91b-46d7-8052-ac5ccfd2da06>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00296.warc.gz"} |
Regression to the mean (RTM) can be difficult to understand. Oftentimes, this concept is explained in the context of trying to interpret data involving repeated measurements, such as data that have
been collected through time. However, RTM is applicable to tons of scenarios: anytime you have two variables that are imperfectly correlated!
classic example
RTM implies that if you observe an extreme result, the next time you record data, you are more likely to observe results that are closer to the mean value. For example, assume someone’s performance
at a task is not the same every day but has random variability that depends on sleep, mood, weather, and other random events that happened that day. Most of the time their performance is near its
average value, but some days are just really bad (or good!).
RTM implies that if this person performs very poorly on day 1, their performance on day 2 is more likely to be towards its mean value just because of a statistical tendency. That is, this person’s
performance is expected to increase from day 1 to day 2 simply due to RTM. Without knowing about RTM, we could easily come up with some plausible but incorrect explanations as to why this increase
occurred (e.g. say someone yelled at them for performing poorly, so they were ‘motivated’ to do better the next day), but this observation is completely consistent with random fluctuations in
performance each day.
inuitive definition with visualizations
Another definition of this concept, along with the visualizations below, really helped this concept click for me: whenever the correlation between two variables (e.g. performance on day 1 vs day 2)
is less than one, there could be be RTM. That is, if one variable doesn’t perfectly predict another (e.g. when x is 10, y is close to 10 but not exactly), which is almost always the case, then RTM is
a mathematical inevitability.
To visualize this, let’s look at two normally-distributed variables, x and y, that are imperfectly correlated with either a high or low correlation of 0.8 or 0.2, respectively (Figure below). To
visualize RTM, let’s select extreme values of y and look at the conditional sampling distribution of x, given y. This is analagous to the situation above where we condition on observing an extreme
performance on day 1 and then peek at what the distribution of day 2 looks like.
These two variables are plotted below in two scenarios in which they have a high (top row) or low (bottom row) correlation. The sampling distribution of x, conditional on y being extreme (greater
than 2), is highlighted with the red dots in the scatterplots (left column) but also shown explicitly with histograms (middle column). Moreover, RTM may be directly analyzed for these values when y
is extreme by looking at the distribution of x - y, for which negative values indicate x is less than y.
For the case where x and y are highly correlated, x takes on similar values as y, greater than 2, only about half the time (top middle plot), whereas the other half of these x values are less than 2.
Consequently, there is a slight bias towards x generally taking on smaller values than y (top right plot), indicating RTM (here values of 0 indicate no RTM). The effect is a bit subtle here, but if
we do the same exercise when x and y have a lower correlation (bottom row), there is strong RTM as a vast majority of the values x might take on are less than 2 (bottom middle plot) and almost always
less than their corresponding y value (bottom right plot). Here, x almost always regresses to its mean.
I could list some examples here, but there are so many. If you think of two imperfectly correlated variables, you could potentially observe RTM!
Below is the R code used to generate this figure:
Loading in libraries. I will use ‘mvtnorm’ to simulate two noramlly-distributed variables that are imperfectly correlated.
## ── Attaching packages ────────────────────────────────────────────────────────────────────────────── tidyverse 1.3.0 ──
## ✓ ggplot2 3.3.2 ✓ purrr 0.3.4
## ✓ tibble 3.0.3 ✓ dplyr 1.0.0
## ✓ tidyr 1.1.0 ✓ stringr 1.4.0
## ✓ readr 1.3.1 ✓ forcats 0.5.0
## ── Conflicts ───────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## x readr::col_factor() masks scales::col_factor()
## x purrr::discard() masks scales::discard()
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()
Let’s make our two variables have a high correlation of 0.8 and then a low correlation of 0.2 and look at the difference. We will store these values in a vector that we will iterate across.
covars <- c(0.8, 0.2)
For each of these covariances, or correlations between the two variables, we will simulate data and make a plot.
par(bty="n", mfrow=c(2,3)) # make a plot with no box type with 6 panels: 3 rows and 2 columns
for (covar in covars){
covar_mat <- rbind(c(1,covar), c(covar,1)) # create covariance matrix
# create table of draws from a bivariate normal distribution
d <- rmvnorm(n=10000, mean = rep(0,2), sigma=covar_mat) %>%
as_tibble() %>% rename("x" = V1, "y" = V2)
# create another table that selects only extreme values for y
d_red <- d %>%
filter(y >= 2)
# make a plot
plot(d$x, d$y,
col=alpha("black", 0.1),
main=paste("correlation = ", covar, sep=" "),
ylim=c(-4, 4),
xlim=c(-4, 4))
col=alpha("red", 0.3),
axis(side=1, labels=T, at=c(-4,-2,0,2,4))
axis(side=2, labels=T, at=c(-4,-2,0,2,4))
abline(h=2, lty=3, col="blue", lwd=2)
col=alpha("red", 0.5),
main="distribution of x",
xlim=c(-4, 4))
abline(v=2, lty=3, col="blue", lwd=3)
hist(d_red$x - d_red$y,
col=alpha("red", 0.5),
main="distribution of x-y",
xlim=c(-4, 4))
abline(v=0, lty=1, col="black", lwd=3) | {"url":"https://brian-arnold.github.io/resources/datascience/regressionToMean/","timestamp":"2024-11-08T14:44:03Z","content_type":"text/html","content_length":"889014","record_id":"<urn:uuid:a10d20c1-1dc5-4e42-8642-6950da44f58b>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00340.warc.gz"} |
How to convert RGBA color to hex?
Here's a step-by-step guide on how you can convert an RGBA color value to its hexadecimal counterpart:
1. First, convert the RGBA values to their corresponding hexadecimal values. Red = 255 → FF Green = 255 → FF Blue = 0 → 00 Alpha = 0.5 * 255 = 127.5 (rounded to 128) → 80
2. Combine the hexadecimal values in the order Alpha, Red, Green, Blue. So, the resulting hexadecimal color code would be: #80FFFF00
By following these steps, you can accurately convert an RGBA color to its hexadecimal representation for usage in various applications and web development. | {"url":"https://devhubby.com/thread/how-to-convert-rgba-color-to-hex","timestamp":"2024-11-10T00:26:21Z","content_type":"text/html","content_length":"116169","record_id":"<urn:uuid:474ed7d0-9742-4498-ae5e-a1b2ad2b97f9>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00857.warc.gz"} |
Minutes of Arc
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like angle finds its use in
a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the
conversion of different units of measurement like ' to arcmin through multiplicative conversion factors. When you are converting angle, you need a Minute to Minutes of Arc converter that is elaborate
and still easy to use. Converting ' to Minutes of Arc is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Minute to arcmin,
this tool is the answer that gives you the exact conversion of units. You can also get the formula used in ' to arcmin conversion along with a table representing the entire conversion. | {"url":"https://www.unitsconverters.com/en/-To-Arcmin/Utu-3668-7757","timestamp":"2024-11-14T14:30:30Z","content_type":"application/xhtml+xml","content_length":"109116","record_id":"<urn:uuid:ebea5ff7-1773-4198-bd31-91c72ca00984>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00710.warc.gz"} |
How much does a 750ml bottle of liquor weigh?
3 pounds
Consider the weight. A full 750 ml bottle of liquor weighs a little under 3 pounds (1400 grams). The weight limit on most checked luggage is 50 pounds.
How many ml is a unit of alcohol?
One unit equals 10ml or 8g of pure alcohol, which is around the amount of alcohol the average adult can process in an hour.
How many units are in a 1 liter bottle of vodka?
A litre bottle of vodka at 40% contains 40 units of alcohol. To find out ho many units of alcohol are in a drink you need the size of the drink in litres, and the strength of the drink in ABV %. You
multiply these and the answer is the number of units.
How much does a 70cl bottle weigh?
A standard 70cl bottle will typically have a size and weight equivalent to 1.5kg.
How much does 750ml Weigh in ounces?
Is 750 Ml Of Whiskey A Fifth?
Bottle Size, metric Ounces Gallon, quart, or pint “equivalent”
750 milliliters 25.4oz. 4/5 quart, a “fifth” or 1.5 pints
How many units are in a 70cl bottle of gin?
Single measure in a pub is 25ml, usually considered to be one unit. So 28ish units in a bottle.
How many units are in a 70cl bottle of whiskey?
28 units
A 70cl bottle of 40% abv whisky will have around 28 units of alcohol in it.
How many 25ml are in a 70cl bottle?
How Many Shots in a Bottle?
Spirits Size 25ml
Magnum 1.5Ltr 60
Litre 1Ltr 40
70cl 70cl 28
50cl 50cl 20
Is 70cl the same as 700ml?
Yes, 70cl and 700ml are the same size. 21 of 21 found this helpful.
How much does a 70cl bottle of gin weigh?
3.75 Kilograms
General Information
Alcohol Type Gin
Item Weight 3.75 Kilograms
Manufacturer reference AC1112150-3
Volume 2100 Millilitres
Units 3.00 Bottle
How many units of alcohol are in 70cl?
How many measures are in a 70cl bottle?
How Many Shots in a Bottle?
Spirits Size 25ml
Litre 1Ltr 40
75cl 75cl 30
70cl 70cl 28
68cl 68cl 27
How many units are in a 70cl bottle of spirits?
Is drinking 750ml of whiskey a week too much?
Is Drinking A Bottle Of Whiskey Bad? The recommended daily intake of whisky for men is 21 units per week, which is the equivalent to 30-40 units per bottle, depending on the strength. It is therefore
possible to consume a whole bottle of spirits at once — twice the amount of a week’s worth.
What size is 70cl in mL?
What is the difference between 75cL and L the same as 750ml? There are 750 millilitres (ml), 75 centilitres (cl), or 0.75 litres (l) in a standard bottle of wine….How Many Ml Is A 70Cl Bottle Of Gin?
Spirits Size 25ml
70cl 70cl 28
68cl 68cl 27
Is 70cl a full size bottle?
70cl (standard size): the most frequent, almost all distilleries use it as a reference dimension, which is equivalent to 75 cl for wine. | {"url":"https://www.goodgraeff.com/how-much-does-a-750ml-bottle-of-liquor-weigh/","timestamp":"2024-11-05T09:36:34Z","content_type":"text/html","content_length":"54051","record_id":"<urn:uuid:3e2d2401-528a-4239-aa6b-2f054658a66b>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00599.warc.gz"} |
Comparison of methods: Passing and Bablok regression - Biochemia Medica
Method and instrument validation is important issue in clinical laboratory work. Each new method should be validated when introducing in routine laboratory work (1,2). Among different experiments
that should be performed to access method’s performance (precision and accuracy) there is the comparison of methods experiment (3). That experiment compares results obtained using new method to those
obtained using other analytical method. The ideal condition is achieved if analytical method used for comparison is reference or definitive method. Correctness of reference methods is
non-questionable, so new method results should be fitted according to the reference. However, in usual circumstances in routine laboratory, correctness of methods is not well documented so they can
not be considered as reference methods; definitive methods are unavailable, and results can be compared to “comparative method” – one available and used in daily routine work.
The aim of the comparison of methods experiment is to estimate systematic (constant and proportional) difference between two methods e.g. to find out if there is significant difference in their
relative accuracy using real patient samples (3). Results should be interpreted very carefully. If the difference between two methods is small and clinically acceptable than those two methods can be
used simultaneously and interchangeably. If difference is unacceptable it should be investigated further which of two methods is inaccurate (3). The experimental side of method comparison is simple.
It is recommended that at least 40 samples of broad concentration range should be tested with two methods (3). Data analyses and interpretation is complicated issue that has been discussed for
decades and still there is no gold standard for statistical procedure that should be used for method comparison data analyses (3).
Data analyses in the comparison of methods experiment
Standard statistical tests investigated difference between two sets of measured data, are not applicable for method comparison data analyses. Independent sample t-test should never be used because
two sets of data were obtained on same biological samples which defining them as dependent samples. Paired t-test could be used for rough estimation of difference between two sets of data. It
compares means of two samples and results will reveal constant but not proportional difference between two sets of measurement.
The test often used for method comparison data analyses, but does not provide proper conclusions, is Pearson’s correlation coefficient. When the same analyte is measured using two methods it is
expected that correlation coefficient is very high, 0.99 or higher (3). Correlation describes linear relationship between two sets of data but not their agreement (4); it does not detect if there is
a constant or proportional difference between two methods. Linear regression model (the least square regression) is more suitable for method comparison data analyses but it is very sensitive to data
distribution (assumption of normal distribution), linear relationship and especially outliers. Furthermore, it presumes that comparative method results are measured without error (5-7). Considering
all those limitations that model is also not suitable for data analyses.
Thus, several other statistical and graphical methods have been developed and proposed exclusively for method comparison data analyses such as: Passing and Bablok regression, Deming regression,
Mountain plot, Bland and Altman plot (6-8). The aim of this article is to provide an overview of the usage and interpretation of Passing and Bablok regression.
Passing and Bablok regression results interpretation
Considering limitation of the ordinary least square regression model W. Bablok and H. Passing proposed regression model for comparison of methods based on robust, non-parametric model (9). Unlike to
the least square linear regression Passing and Bablok regression is not sensitive towards outliers, assumes that measurement errors in both methods have same distribution, not necessarily normal,
constant ratio of variance, arbitrary sampling distribution and imprecision in both methods. The requirements for Passing and Bablok regression are: continuously distributed measurements (covering
broad concentration range) and linear relationship between two methods (6). Passing and Bablok regression calculates regression line equation from two data sets.
Result of Passing and Bablok regression consists of several parts and each has its role in interpreting method comparison data and concluding on methods agreement. The first result is scatter diagram
with regression line that enables visual inspection of measured data and obvious agreement of fitted regression line and identity line (Figures 1A and 2A). Regression equation (y = a + bx) revealed
constant (regression line’s intercept (a)) and proportional (regression line’s slope (b)) difference with their confidence intervals of 95% (95% CI). Confidence intervals explain if their value
differ from value zero (0) for intercept and value one (1) for slope only by chance. Thus, if 95% CI for intercept includes value zero it can be concluded that there is no significant difference
between obtained intercept value and value zero and there is no constant difference between two methods. Respectively, if 95% CI for slope includes value one, it can be concluded that there is no
significant difference between obtained slope value and value one and there is no proportional difference between two methods. In such case we can assume that x = y and that there is no significant
difference between methods, so both can be used interchangeably. The first example of Passing and Bablok regression analyses on data set obtained by measuring concentration of total bilirubin in
patients’ serums using two different automated analyzers is presented at figure 1. Note that there is small constant difference between two methods (Figure 1). Compensation of that difference can be
made after further investigation of accuracy of both methods. The second example presents data set obtained by measuring direct bilirubin in serums using two methods, revealing small constant but
huge proportional error (Figure 2). Those methods differ seriously and can not be used simultaneously. Note that correlation coefficient in both examples is r = 0.99; proving that method comparison
results can not be assessed using Pearson’s correlation.
Figure 1. Passing and Bablok regression analyses of two methods for total bilirubin, N = 40; concentration range 3-468μmol/L; Pearson correlation coefficient r = 0.99, P < 0.001.
(A) Scatter diagram with regression line and confidence bands for regression line. Identity line is dashed. Regression line equation: y = -3.0 + 1.00 x; 95% CI for intercept -3.8 to -2.1 and for
slope 0.98 to 1.01 indicated good agreement. Cusum test for linearity indicates no significant deviation from linearity (P > 0.10). (B) Residual plot presents distribution of difference around fitted
regression line.
Figure 2. Passing and Bablok regression analyses of two methods for direct bilirubin, N = 70; concentration range 4-357μmol/L; Pearson correlation coefficient r = 0.99, P < 0.001.
(A) Scatter diagram with regression line and confidence bands for regression line. Identity line is dashed. Regression line equation: y = -3.2 + 1.52 x; 95% CI for intercept -4.2 to -1.9 and for
slope 1.47 to 1.58 indicated small constant and huge proportional difference. Cusum test for linearity indicates significant deviation from linearity (P<0.05). (B) Residual plot presents distribution
of difference around fitted regression line.
Besides usual scatter plot Passing and Bablok regression provides the residual plot as well (Figures 1B and 2B). It shows residuals from fitted regression line and clearly revealed outliers,
allover the measurement range and visually identifies non-linearity. Regarding that linear relationship between two measurement data sets is required for obtaining statistically unbiased results,
Passing and Bablok regression analysis calculates cumulative sum linearity test (cusum linearity test) that determinates if residuals are randomly distributed above and below regression line. Cusum
test P value less than 0.05 indicates significant difference from linearity and two compared analytical methods should be further investigated; possibly higher number of samples with better
continuous distribution should be consider.
Passing and Bablok regression is a good and appropriate model for analysis of method comparison results. Constant and proportional bias between two methods can be easily estimated and calculated
parameters allow correction actions. | {"url":"https://www.biochemia-medica.com/en/journal/21/1/10.11613/BM.2011.010/fullArticle","timestamp":"2024-11-02T18:44:55Z","content_type":"text/html","content_length":"79793","record_id":"<urn:uuid:6d75400e-d37d-4911-a2c7-2a5786ba729f>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00042.warc.gz"} |
The Stacks project
Lemma 10.110.9. Let $R \to S$ be a local homomorphism of local Noetherian rings. Assume that $R \to S$ is flat and that $S$ is regular. Then $R$ is regular.
Proof. Let $\mathfrak m \subset R$ be the maximal ideal and let $\kappa = R/\mathfrak m$ be the residue field. Let $d = \dim S$. Choose any resolution $F_\bullet \to \kappa $ with each $F_ i$ a
finite free $R$-module. Set $K_ d = \mathop{\mathrm{Ker}}(F_{d - 1} \to F_{d - 2})$. By flatness of $R \to S$ the complex $0 \to K_ d \otimes _ R S \to F_{d - 1} \otimes _ R S \to \ldots \to F_0 \
otimes _ R S \to \kappa \otimes _ R S \to 0$ is still exact. Because the global dimension of $S$ is $d$, see Proposition 10.110.1, we see that $K_ d \otimes _ R S$ is a finite free $S$-module (see
also Lemma 10.109.3). By Lemma 10.78.6 we see that $K_ d$ is a finite free $R$-module. Hence $\kappa $ has finite projective dimension and $R$ is regular by Proposition 10.110.5. $\square$
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dot-modulationSpectrum: Modulation spectrum per sound in soundgen: Sound Synthesis and Acoustic Analysis
.modulationSpectrum( audio, specSource = c("STFT", "audSpec")[1], windowLength = 15, step = 1, wn = "hanning", zp = 0, audSpec_pars = list(filterType = "butterworth", nFilters = 32, bandwidth = 1/24,
yScale = "bark", dynamicRange = 120), msType = c("1D", "2D")[2], amRes = 5, maxDur = 5, specMethod = c("spec", "meanspec")[2], logSpec = FALSE, logMPS = FALSE, power = 1, normalize = TRUE, roughRange
= c(30, 150), roughMean = NULL, roughSD = NULL, roughMinFreq = 1, amRange = c(10, 200), returnMS = TRUE, returnComplex = FALSE, plot = TRUE, savePlots = NULL, logWarpX = NULL, logWarpY = NULL,
quantiles = c(0.5, 0.8, 0.9), kernelSize = 5, kernelSD = 0.5, colorTheme = c("bw", "seewave", "heat.colors", "...")[1], col = NULL, main = NULL, xlab = "Hz", ylab = "1/kHz", xlim = NULL, ylim = NULL,
width = 900, height = 500, units = "px", res = NA, ... )
audio a list returned by readAudio
specSource 'STFT' = Short-Time Fourier Transform; 'audSpec' = a bank of bandpass filters (see audSpectrogram)
windowLength, parameters for extracting a spectrogram if specType = 'STFT'. Window length and step are specified in ms (see spectrogram). If specType = 'audSpec', these settings have no effect
step, wn, zp
audSpec_pars parameters for extracting an auditory spectrogram if specType = 'audSpec'. If specType = 'STFT', these settings have no effect
msType '2D' = two-dimensional Fourier transform of a spectrogram; '1D' = separately calculated spectrum of each frequency band
target resolution of amplitude modulation, Hz. If NULL, the entire sound is analyzed at once, resulting in a single roughness value (unless it is longer than maxDur, in which case it
amRes is analyzed in chunks maxDur s long). If amRes is set, roughness is calculated for windows ~1000/amRes ms long (but at least 3 STFT frames). amRes also affects the amount of smoothing
when calculating amMsFreq and amMsPurity
maxDur sounds longer than maxDur s are split into fragments, and the modulation spectra of all fragments are averaged
the function to call when calculating the spectrum of each frequency band (only used when msType = '1D'); 'meanspec' is faster and less noisy, whereas 'spec' produces higher
specMethod resolution
logSpec if TRUE, the spectrogram is log-transformed prior to taking 2D FFT
logMPS if TRUE, the modulation spectrum is log-transformed prior to calculating roughness
power raise modulation spectrum to this power (eg power = 2 for ^2, or "power spectrum")
normalize if TRUE, the modulation spectrum of each analyzed fragment maxDur in duration is separately normalized to have max = 1
roughRange the range of temporal modulation frequencies that constitute the "roughness" zone, Hz
the mean (Hz) and standard deviation (semitones) of a lognormal distribution used to weight roughness estimates. If either is null, roughness is calculated simply as the proportion of
roughMean, spectrum within roughRange. If both roughMean and roughRange are defined, weights outside roughRange are set to 0; a very large SD (a flat weighting function) gives the same result as
roughSD just roughRange without any weighting (see examples)
roughMinFreq frequencies below roughMinFreq (Hz) are ignored when calculating roughness (ie the estimated roughness increases if we disregard very low-frequency modulation, which is often strong)
amRange the range of temporal modulation frequencies that we are interested in as "amplitude modulation" (AM), Hz
returnMS if FALSE, only roughness is returned (much faster). Careful with exporting the modulation spectra of a lot of sounds at once as this requires a lot of RAM
returnComplex if TRUE, returns a complex modulation spectrum (without normalization and warping)
plot if TRUE, plots the modulation spectrum of each sound (see plotMS)
savePlots if a valid path is specified, a plot is saved in this folder (defaults to NA)
logWarpX, numeric vector of length 2: c(sigma, base) of pseudolog-warping the modulation spectrum, as in function pseudo_log_trans() from the "scales" package
quantiles labeled contour values, % (e.g., "50" marks regions that contain 50% of the sum total of the entire modulation spectrum)
kernelSize the size of Gaussian kernel used for smoothing (1 = no smoothing)
kernelSD the SD of Gaussian kernel used for smoothing, relative to its size
colorTheme black and white ('bw'), as in seewave package ('seewave'), matlab-type palette ('matlab'), or any palette from palette such as 'heat.colors', 'cm.colors', etc
col actual colors, eg rev(rainbow(100)) - see ?hcl.colors for colors in base R (overrides colorTheme)
xlab, ylab, graphical parameters
main, xlim,
width, height, parameters passed to png if the plot is saved
units, res
... other graphical parameters passed on to filled.contour.mod and contour (see spectrogram)
'STFT' = Short-Time Fourier Transform; 'audSpec' = a bank of bandpass filters (see audSpectrogram)
parameters for extracting a spectrogram if specType = 'STFT'. Window length and step are specified in ms (see spectrogram). If specType = 'audSpec', these settings have no effect
parameters for extracting an auditory spectrogram if specType = 'audSpec'. If specType = 'STFT', these settings have no effect
'2D' = two-dimensional Fourier transform of a spectrogram; '1D' = separately calculated spectrum of each frequency band
target resolution of amplitude modulation, Hz. If NULL, the entire sound is analyzed at once, resulting in a single roughness value (unless it is longer than maxDur, in which case it is analyzed in
chunks maxDur s long). If amRes is set, roughness is calculated for windows ~1000/amRes ms long (but at least 3 STFT frames). amRes also affects the amount of smoothing when calculating amMsFreq and
sounds longer than maxDur s are split into fragments, and the modulation spectra of all fragments are averaged
the function to call when calculating the spectrum of each frequency band (only used when msType = '1D'); 'meanspec' is faster and less noisy, whereas 'spec' produces higher resolution
if TRUE, the spectrogram is log-transformed prior to taking 2D FFT
if TRUE, the modulation spectrum is log-transformed prior to calculating roughness
raise modulation spectrum to this power (eg power = 2 for ^2, or "power spectrum")
if TRUE, the modulation spectrum of each analyzed fragment maxDur in duration is separately normalized to have max = 1
the range of temporal modulation frequencies that constitute the "roughness" zone, Hz
the mean (Hz) and standard deviation (semitones) of a lognormal distribution used to weight roughness estimates. If either is null, roughness is calculated simply as the proportion of spectrum within
roughRange. If both roughMean and roughRange are defined, weights outside roughRange are set to 0; a very large SD (a flat weighting function) gives the same result as just roughRange without any
weighting (see examples)
frequencies below roughMinFreq (Hz) are ignored when calculating roughness (ie the estimated roughness increases if we disregard very low-frequency modulation, which is often strong)
the range of temporal modulation frequencies that we are interested in as "amplitude modulation" (AM), Hz
if FALSE, only roughness is returned (much faster). Careful with exporting the modulation spectra of a lot of sounds at once as this requires a lot of RAM
if TRUE, returns a complex modulation spectrum (without normalization and warping)
if TRUE, plots the modulation spectrum of each sound (see plotMS)
if a valid path is specified, a plot is saved in this folder (defaults to NA)
numeric vector of length 2: c(sigma, base) of pseudolog-warping the modulation spectrum, as in function pseudo_log_trans() from the "scales" package
labeled contour values, % (e.g., "50" marks regions that contain 50% of the sum total of the entire modulation spectrum)
the size of Gaussian kernel used for smoothing (1 = no smoothing)
the SD of Gaussian kernel used for smoothing, relative to its size
black and white ('bw'), as in seewave package ('seewave'), matlab-type palette ('matlab'), or any palette from palette such as 'heat.colors', 'cm.colors', etc
actual colors, eg rev(rainbow(100)) - see ?hcl.colors for colors in base R (overrides colorTheme)
other graphical parameters passed on to filled.contour.mod and contour (see spectrogram)
For more information on customizing the embed code, read Embedding Snippets. | {"url":"https://rdrr.io/cran/soundgen/man/dot-modulationSpectrum.html","timestamp":"2024-11-08T08:15:13Z","content_type":"text/html","content_length":"37550","record_id":"<urn:uuid:ae85133b-d807-4455-a684-18ed90d4b7f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00687.warc.gz"} |
t of
Issue A&A
Volume 564, April 2014
Article Number A107
Number of page(s) 24
Section Interstellar and circumstellar matter
DOI https://doi.org/10.1051/0004-6361/201323332
Published online 16 April 2014
Online material
Appendix A: List of LMXB and HMXB sources used in the calculations
Full list of sources used in the calculations with the corresponding data (also available at http://www.mpa-garching.mpg.de/~molaro).
Table A.1
List of sources used in the calculations, ranked by 2−10 keV luminosity.
Appendix B: Contribution of Sgr A^∗
As discussed in Sect. 2, the past activity of the currently low quiescent source Sgr A^∗ might have significantly contributed to the cumulative X-ray output of the Galaxy, and hence to the diffuse
GRXE component. In Fig. B.1 we show the minimum luminosity required for the flux from this source to outshine the contribution of the entire XBs population at different positions on the Galactic
plane, estimated as (B.1)
Fig. B.1
Minimum Sgr A^∗ luminosity required for the source’s flux to be higher than the total HMXBs (left panel) and LMXBs (right panel) contribution at different positions on the Galactic plane. If we
know the star formation rate and the mass of a galaxy, we can estimate the X-ray luminosity contributed by the X-ray binaries and similarly find when the AGNs contribute more than the X-ray
binaries to the heating of the gas in other galaxies with AGNs.
Open with DEXTER
Appendix C: Scattering cross-sections
In Fig. C.1 we illustrate the energy dependence of the scattering cross-section for the case of H2 scattering. In Figs. C.2 and C.3 we highlight the additional effects caused by bound electrons by
showing the ratio of contributions from H2 and He with respect to the free electrons. The energy dependence of Rayleigh scattering is clearly evident. At high energies, Rayleigh scattering operates
only on a very narrow range of scattering angles and the total scattering cross-section approaches the Klein-Nishina cross-section. At low energies, coherent Rayleigh scattering results in the
enhancement of the cross section for a significant range of scattering angles, resulting in a higher average scattering rate of low-energy photons. The average scattered spectrum is therefore softer
than the average incident spectrum. The contribution of molecular hydrogen as well as helium and heavier elements is compared in Table C.1 in the Rayleigh scattering limit (0° scattering angle). The
elements heavier than helium contribute ≲10% in this extreme case, and the actual contribution when suitably averaged over different scattering angles would be smaller than the values in the table.
Fig. C.1
Cross-section (Rayleigh+ Compton) in polar coordinates, σ(θ)e^iθ, where σ is the amplitude of the cross-section in units of for the scattering angle θ and r[e] is the classical electron radius. The
enhancement of Rayleigh scattering caused by coherence effects is clearly visible at low energies. While the cross section remains constant with energy at angles close to zero, the contribution of
Rayleigh scattering is shown to quickly decrease with increasing energy. Compton scattering is also suppressed by relativistic effects.
Open with DEXTER
Near the Galactic center, a luminosity of ≳10^37 erg/s from Sgr A^∗ would be enough to become comparable with the illumination from X-ray binaries. On the outskirts of the Galaxy, on the other hand,
Sgr A^∗, or some other ultra-luminous source near the Galactic center, can be ignored as long as its luminosity is ≲10^39−10^40 erg/s.
Fig. C.2
Ratio of the Rayleigh + Compton differential cross-section of H2 + He to HI + He (where each element is weighted by relative abundance) as a function of energy for different scattering angles. The
total cross section approaches that of unbound electrons as the importance of coherence effects decreases with energy, because of the suppression of Rayleigh scattering.
Open with DEXTER
Table C.1
Elements according to maximal contribution to Rayleigh scattering if all hydrogen is in atomic form (or molecular form, given in parentheses).
Fig. C.3
Ratio of intensity from the Monte Carlo simulated HMXBs (left column) and LMXBs (right column) scattered along the Galactic plane (b = 0) by H2 (top row) and He (bottom row) to the intensity that
would be scattered if all electrons were unbound. The range of scattering angles over which the Rayleigh scattering dominates the scattering cross section depends on the characteristic size of the
electron distribution in the atom or molecule, which differs for different elements and molecules (see Sect. 4.1). This leads to a nonlinear dependence of the ratio of cross sections for different
elements and molecules on the scattering angle. At each longitude many different scattering angles contribute, corresponding to the relative position of the X-ray sources w.r.t. the gas along the
line of sight, resulting in the apparent longitudinal dependence of the ratio profiles.
Open with DEXTER
© ESO, 2014
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Next: MINIMUM EQUIVALENT DIGRAPH Up: Subgraphs and Supergraphs Previous: MINIMUM EDGE K-SPANNER   Index
• INSTANCE: Graph
• SOLUTION: A subset k-colorable, i.e., there is a coloring for G' of cardinality at most k.
• MEASURE: Cardinality of the vertex set of the induced subgraph, i.e.,
• Good News: As easy to approximate as MAXIMUM INDEPENDENT SET for k independent sets) [221].
• Bad News: As hard to approximate as MAXIMUM INDEPENDENT SET for 389].
• Comment: Transformation from MAXIMUM INDEPENDENT SET. Equivalent to MAXIMUM INDEPENDENT SET for k=1. Admits a PTAS for planar graphs [53]. Admits a PTAS for `263]. The case of degrees bounded by
222] and APX-complete.
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25 Similes for Math -
Mathematics is often seen as a mysterious and daunting realm, filled with complex equations and mind-boggling theorems. But what if we told you that math can be as relatable and easy to understand as
everyday experiences?
That’s where similes come into play. Similes are figures of speech that compare one thing to another using the words “like” or “as.” They can make the abstract world of mathematics feel familiar and
In this article, we’re going to explore various similes for math, giving each one meaning and using them in sentences to make math concepts more engaging and relatable.
Similes for Math
1. As Easy as Pie
Meaning: Extremely easy or effortless.
In a Sentence: Solving that algebraic equation was as easy as pie; I did it in seconds.
2. Like a Walk in the Park
Meaning: A task that is simple and requires little effort.
In a Sentence: Understanding basic arithmetic is like a walk in the park for most people.
3. Clear as Crystal
Meaning: Something that is very easy to understand or see.
In a Sentence: Once the teacher explained the concept, it became clear as crystal to the students.
4. Fast as Lightning
Meaning: Extremely quick or speedy.
In a Sentence: The mathematician calculated the result fast as lightning, surprising everyone with his speed.
5. Smooth as Silk
Meaning: A process or solution that is very well-executed and without any difficulties.
In a Sentence: The way she solved the geometry problem was smooth as silk; she didn’t encounter any obstacles.
6. Sharp as a Tack
Meaning: Someone who is very intelligent and quick-witted.
In a Sentence: Sarah’s math skills are sharp as a tack; she always gets top grades in her math class.
7. Straight as an Arrow
Meaning: Something that is direct and without any deviations.
In a Sentence: The path to solving this calculus problem is straight as an arrow; just follow the steps.
8. Like Clockwork
Meaning: Something that happens with perfect regularity and precision.
In a Sentence: The way the gears of the clock mesh together is like clockwork, just like the principles of trigonometry.
9. Like a Well-Oiled Machine
Meaning: A system or process that runs smoothly and efficiently.
In a Sentence: The teamwork between the students in the math competition was like a well-oiled machine.
10. Like a Puzzle Piece
Meaning: Something that fits perfectly into a larger picture or plan.
In a Sentence: Each mathematical concept we learn is like a puzzle piece that fits into the bigger math puzzle.
11. Neat as a Pin
Meaning: Something that is very orderly and well-organized.
In a Sentence: His math notes were always neat as a pin, making it easy to study from them.
12. Like a Rollercoaster
Meaning: A situation or process with many ups and downs or sudden changes.
In a Sentence: Trying to solve that calculus problem felt like a rollercoaster ride with its twists and turns.
13. Like a Maze
Meaning: Something that is complicated and difficult to navigate.
In a Sentence: Navigating through the rules of probability theory can be like a maze for beginners.
14. Tight as a Drum
Meaning: Something that is very secure or well-structured.
In a Sentence: The logic in that geometry proof is as tight as a drum; there are no loopholes.
15. Quick as a Flash
Meaning: Extremely fast or rapid.
In a Sentence: When it comes to mental math, John is quick as a flash; he can calculate in no time.
16. Like a Fine-Tuned Instrument
Meaning: Something that is finely calibrated or adjusted for optimal performance.
In a Sentence: His understanding of statistics is like a fine-tuned instrument; he can analyze data with precision.
17. Like a Jigsaw Puzzle
Meaning: Something that requires the assembly of various pieces to make a complete picture.
In a Sentence: Understanding calculus involves putting together various concepts like a jigsaw puzzle.
18. Like a Rubik’s Cube
Meaning: Something that is complex and requires careful manipulation to solve.
In a Sentence: Trying to find the solution to that mathematical problem is like solving a Rubik’s Cube.
19. Like a Mathematical Dance
Meaning: Something that is elegant, coordinated, and harmonious.
In a Sentence: The way the mathematical equations interrelated in that physics problem was like a mathematical dance.
20. Like a Mathematical Equation
Meaning: Something that has a precise and unchanging solution.
In a Sentence: Love can be unpredictable, but math is like a mathematical equation; it always has a solution.
21. Like a Mathematical Masterpiece
Meaning: Something that is beautifully crafted and executed.
In a Sentence: The way she solved the calculus problem was like a mathematical masterpiece; it was a work of art.
22. Like a Geometric Shape
Meaning: Something that has a defined and recognizable structure.
In a Sentence: The patterns in fractal geometry are like a geometric shape, repeating endlessly.
23. Like a Mathematical Formula
Meaning: Something that follows a set pattern or rule.
In a Sentence: The behavior of atoms in a chemical reaction can be predicted like a mathematical formula.
24. Like a Mathematical Theorem
Meaning: Something that is proven to be true based on established principles.
In a Sentence: The Pythagorean Theorem is like a mathematical theorem; its truth is indisputable.
25. Like a Mathematical Pattern
Meaning: Something that repeats in a predictable manner.
In a Sentence: The Fibonacci sequence is like a mathematical pattern; each number is a sum of the two preceding ones.
Simile Meaning Example Sentence
As Easy as Pie Extremely easy or effortless Solving that algebraic equation was as easy as pie.
Like a Walk in the Park A task that is simple and requires little effort Understanding basic arithmetic is like a walk in the park.
Clear as Crystal Very easy to understand or see Once the teacher explained the concept, it became clear as crystal to the students.
Fast as Lightning Extremely quick or speedy The mathematician calculated the result fast as lightning.
Smooth as Silk A process or solution that is well-executed The way she solved the geometry problem was smooth as silk.
Sharp as a Tack Someone who is very intelligent and quick-witted Sarah’s math skills are sharp as a tack.
Straight as an Arrow Direct and without any deviations The path to solving this calculus problem is straight as an arrow.
Like Clockwork Happens with perfect regularity and precision The gears of the clock mesh together like clockwork.
Like a Well-Oiled Machine A system or process that runs smoothly The teamwork in the math competition was like a well-oiled machine.
Like a Puzzle Piece Fits perfectly into a larger picture or plan Each mathematical concept is like a puzzle piece.
Neat as a Pin Very orderly and well-organized His math notes were always neat as a pin.
Like a Rollercoaster Many ups and downs or sudden changes Solving that calculus problem felt like a rollercoaster ride.
Like a Maze Complicated and difficult to navigate Navigating through probability theory can be like a maze.
Tight as a Drum Very secure or well-structured The logic in that geometry proof is tight as a drum.
Quick as a Flash Extremely fast or rapid John is quick as a flash with mental math.
Like a Fine-Tuned Instrument Finely calibrated or adjusted for optimal performance His understanding of statistics is like a fine-tuned instrument.
Like a Jigsaw Puzzle Requires assembly of various pieces Understanding calculus involves putting together various concepts like a jigsaw puzzle.
Like a Rubik’s Cube Complex and requires careful manipulation Finding the solution to that problem is like solving a Rubik’s Cube.
Like a Mathematical Dance Elegant, coordinated, and harmonious The way mathematical equations interrelated was like a mathematical dance.
Like a Mathematical Equation Precise and unchanging solution Math is like a mathematical equation; it always has a solution.
Like a Mathematical Masterpiece Beautifully crafted and executed The way she solved the calculus problem was like a mathematical masterpiece.
Like a Geometric Shape Defined and recognizable structure The patterns in fractal geometry are like a geometric shape.
Like a Mathematical Formula Follows a set pattern or rule The behavior of atoms in a chemical reaction can be predicted like a mathematical formula.
Like a Mathematical Theorem Proven to be true based on established principles The Pythagorean Theorem is like a mathematical theorem.
Like a Mathematical Pattern Repeats in a predictable manner The Fibonacci sequence is like a mathematical pattern.
Mathematics doesn’t have to be a daunting, abstract realm filled with complicated symbols and equations. By using similes, we can make math more engaging, relatable, and, dare we say, enjoyable. | {"url":"https://phrasedictionary.org/similes-for-math/","timestamp":"2024-11-07T00:43:25Z","content_type":"text/html","content_length":"143962","record_id":"<urn:uuid:1df538a3-a8fe-45ab-b2b0-5933fe103aaf>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00180.warc.gz"} |
Mastering Bend Allowance Calculations: Step by Steps
Mastering Bend Allowance Calculations
By meticulously following each step outlined in this guide, you can enhance the accuracy and efficiency of your bending operations. Maybe you also want to check out our Section Bending Machine page.
In the intricate world of sheet metal bending, mastering the concept of the k-factor is crucial for achieving accurate and precise results. The k-factor, representing the ratio of the location of the
neutral axis to the material thickness, plays a pivotal role in determining bend allowances and ensuring optimal bending outcomes. Let’s delve deeper into the factors influencing the k-factor and
explore methods for its calculation and application.
The Influence of Bend Radius and Forming Method
The k-factor is profoundly influenced by the bend radius and the chosen forming method. Altering the inside bend radius relative to the material thickness can lead to significant shifts in the
k-factor. For instance, reducing the inside bend radius may induce cracking on the outer surface of the bend, causing the neutral axis to shift inward and decreasing the k-factor.
Similarly, transitioning between different forming methods, such as air forming, bottoming, or coining, can impact the k-factor. Deformation and thinning of the bend radius during bottoming result in
an increased k-factor, whereas coining, which relieves stress, leads to a decrease in the k-factor as the neutral axis shifts towards the inner surface of the bend.
Effect of Material Thickness and Tooling
Changes in material thickness and tooling configurations also influence the k-factor. Thicker and harder materials tend to decrease the k-factor, whereas adjustments in tooling, such as using
narrower die widths, can increase the k-factor. Moreover, maintaining a constant material thickness while altering tooling setups affects bending force and consequently influences the k-factor.
Levels of Accuracy in K-Factor Determination
Achieving precise bend calculations requires an understanding of the variables influencing the k-factor. While an average k-factor value of 0.4468 suffices for many applications, more accurate
results may necessitate alternative methods of k-factor determination.
One approach involves establishing the k-factor based on the radius-to-material thickness relationship. For instance, if the bend radius is less than double the material thickness, the k-factor is
typically 0.33, while a radius greater than double the material thickness corresponds to a k-factor of 0.5. Additionally, referencing k-factor charts provides further accuracy in selecting
appropriate k-factor values for specific bending scenarios.
Measuring Test Pieces for Precise K-Factor Determination
For utmost accuracy, calculating the k-factor from test bends offers unparalleled precision tailored to specific material grades and bending conditions. This method entails measuring test pieces,
including the bend allowance (BA) and the inside radius (Ir), to derive the k-factor formula.
Accurate measurement techniques, such as employing pin gauges, radius gauges, or optical comparators, ensure precise determination of the Ir. Measuring the BA, which represents the arc length of the
neutral axis, requires careful assessment before and after bending to ascertain accurate results.
Bend Allowance Calculation for 90-Degree Bends
In the case of 90-degree bends, determining the bend allowance involves measuring the total outside dimension of the formed part and subtracting the material thickness (Mt) and the measured inside
radius (Ir) from the outside flange dimension. Adding the two inside leg dimensions together and subtracting the flat dimension yields the bend allowance, providing a clear understanding of the
bending dynamics at this specific angle.
Achieving precision in metal fabrication demands a thorough understanding of bend allowance calculations. In this comprehensive guide, we'll delve into the intricate process of determining bend allowance step by step. Whether you're a seasoned professional or a novice in the field, this article will equip you with the knowledge and skills needed to excel in metal bending operations.
If your bend equals 90 degrees, you can measure the total outside dimension of the formed part, then subtract the Mt and the measured Ir from the outside flange dimension; this gives you the inside
leg dimension. Add your two inside leg dimensions together, then subtract the flat dimension, and you get the BA:Inside leg dimension for 90-degree bend = Outside dimension – Mt – Ir
Measured inside leg dimensions – Measured flat = BA
Again, this equation works only for 90-degree bends, basically because of how the radius and leg dimensions relate at a 90-degree angle. Technically speaking, it’s because the flat leg length meets
the Ir at the tangent point.
Greater or Less Than 90 Degrees: Step-by-Step Guide
To measure the Bend Allowance (BA) for bends with angles greater or less than 90 degrees, trigonometry comes into play. This involves a bit more complexity but is essential for accurate calculations.
It’s important to note that while the trigonometric equations provided here are effective, they are not the only options available. You can consult any trigonometry resource, whether online or in
your library, to explore alternative equations for solving various sides and angles of a right-angle triangle.
Let’s begin by addressing an external angle less than 90 degrees. Let’s take the example of a 60-degree external bend angle as illustrated in Figure 2. The following steps correspond directly to the
referenced figure, and these steps must be repeated for the second inside leg dimension.
Figure 1: The terminology used for this discussion is presented here.
Step 1: Measure Dimension A
The journey begins with precise measurements. Using a calibrated measuring tool, carefully measure dimension A on the test piece. Accurate measurements lay the foundation for successful bend
allowance calculations.
Step 2: Add Material Thickness (Mt) to Dimension A
Dimension A serves as the starting point. By adding the material thickness (Mt) to dimension A, you obtain dimension B, which sets the stage for further calculations.
Figure 2: This shows one way in which you can use right-angle trigonometry to “walk through the triangles” and calculate the inside leg dimension (dimension F) of a bend with an external angle of 60
Step 3: Measure the Inside Bend Radius (Ir)
Utilize specialized instruments such as a pin gauge, radius gauge, or optical comparator to measure the inside bend radius (Ir) accurately. The inside bend radius plays a crucial role in bend
allowance calculations, influencing the final outcome of the bending process.
Step 4: Calculate the Outside Setback (OSSB)
The next step involves computing the outside setback (OSSB), a fundamental parameter in bend allowance calculations. Employ the following formula to determine OSSB:
OSSB = [tangent (external bend angle/2) × (Mt + Ir);
Understanding the geometry of the bending operation is essential for accurate OSSB calculation. Visualize the green triangle formed by OSSB, with angles C and B guiding the determination of side b.
Step-by-Step Guide
To calculate the Outside Setback (OSSB), we’ll utilize trigonometric functions to determine the dimensions of the green triangle. Here’s a step-by-step guide:
• Step 1#: Calculate OSSB:
Use the formula: OSSB = [tangent (external bend angle/2) × (Mt + Ir)].
OSSB represents side ‘a’ of the green triangle.
• Step 2#: Determine Angles of the Green Triangle:
Given that the external bend angle is 60 degrees, angle C of the green triangle is 30 degrees, and angle B is 60 degrees.
• Step 3#: Solve for Side b of the Green Triangle:
Use the formula: b = a × sine B.
Side b corresponds to dimension C, which extends to the tangent point on the material’s outside surface.
• Step 4#: Adjust for True Position of Dimension C:
Note that at this particular bend angle, dimension C may closely match or be very close to the material thickness (Mt). However, dimension C will vary depending on the bend angle.
Therefore, we use the OSSB to accurately calculate dimension C’s true position.
By following these steps, we can accurately determine the Outside Setback (OSSB) and ensure precise measurements for further calculations in the bending process.
Step 5: Determine Dimension D
Dimension D corresponds to side c of the red right-angle triangle. With side a (hypotenuse) representing the material thickness (Mt), angle B of the purple triangle is derived from the external bend
angle. Employing trigonometric principles, calculate dimension D using the cosine of angle B.
Step-by-Step Guide
To determine dimension D, which corresponds to side c of the red right-angle triangle, we’ll follow these steps:
• Identify Angles and Sides:
Side a (the hypotenuse) is equivalent to the material thickness (Mt).
Angle B of the purple triangle represents the external bend angle, which is 60 degrees. This angle ensures that angle C of the purple triangle is 30 degrees, given the sum of angles in a triangle
is 180 degrees.
With the material edge forming a 90-degree angle, angle B of the red triangle is 60 degrees.
• Calculate Side c of the Red Triangle:
Utilize the formula: c = a × cosine B.
Side c, also known as dimension D, represents the length from the material’s edge to the tangent point on its outside surface.
By applying trigonometric principles and the known angles and sides of the triangles involved, we can accurately compute dimension D, facilitating precise measurements in the bending process.
Step 6: Calculate Dimension E
Dimension E serves as a pivotal parameter in bend allowance calculations. By subtracting the sum of dimensions C and D from dimension B, you obtain dimension E, which contributes to the overall
accuracy of the bending process.
Step 7: Solve for Dimension F
Dimension F, the inside leg length, is derived from the purple triangle’s angles and dimensions. Employing cosine C, calculate dimension F to ascertain the precise inside leg length of the bend.
Handling External Bend Angles Greater Than 90 Degrees: For workpieces featuring external bend angles exceeding 90 degrees, a similar procedure is followed. Commence with measured dimensions on the
test piece and navigate through right triangles to determine the inside leg dimension. Replicate this process for the opposite leg to ensure comprehensive accuracy.
Mastering bend allowance calculations is indispensable for achieving precision in metal fabrication. By meticulously following each step outlined in this guide, you can enhance the accuracy and
efficiency of your bending operations. Armed with a thorough understanding of bend allowance principles, you’re well-equipped to tackle complex bending challenges with confidence and expertise.
Works Cited and Img Resources: Analyzing the k-factor in sheet metal bending | {"url":"https://www.angleroller.com/section-bending/other/mastering-bend-allowance-calculations-a-comprehensive-guide.html","timestamp":"2024-11-05T10:44:57Z","content_type":"text/html","content_length":"141747","record_id":"<urn:uuid:202adcb6-fb11-4d08-a82a-aa68a0b3ef75>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00144.warc.gz"} |
What Are Foci of an Ellipse? ⭐ Definition, Formula
Foci of Ellipse – Definition, Formula
Updated on December 31, 2023
At Brighterly, we believe in illuminating young minds with the radiance of knowledge. As we embark on a new exploration today, our compass points towards a fascinating aspect of Geometry – the
Ellipse, specifically its Foci. An ellipse might be a common shape you come across, like an oval racetrack or even the orbit of a planet. It’s one of those simple yet profound concepts that underlie
many of the workings of the universe, from the paths of celestial bodies to the design of whispering galleries. But what gives an ellipse its unique shape? The answer lies in two magical points
called the Foci. So grab your explorer’s hat as we navigate the world of ellipses, focus on its foci, and unearth the mathematical elegance underlying this ubiquitous shape. We will venture from
definitions through properties, dive into formulas, and even solve practice problems together. This voyage, like every journey at Brighterly, promises to enrich your understanding and love for
What Are Foci of an Ellipse?
Welcome to Brighterly, the gateway to magical mathematical journeys. Today, let’s delve into the fascinating world of ellipses. But, what’s an ellipse? Imagine squashing a circle; it forms a
stretched circle or what we call an ‘ellipse.’ In this world of ellipses, there are two special points known as Foci (plural of focus). Picture the scene of two friends holding either end of a piece
of string and drawing an ellipse in the sand; the foci would be where their fingers are holding the string. These two points are the engines that shape the ellipse, giving it its distinctive
elongated shape.
Understanding an Ellipse
An Ellipse is one of the various fundamental shapes in geometry. You may recognize it as the beautifully elongated cousin of a circle. Visually, it looks like a circle that’s been delicately
stretched along one direction. Mathematically speaking, an ellipse is a locus of points in a plane such that the sum of the distances from two fixed points (the foci) is constant. This captivating
definition captures the essence of an ellipse while underscoring the importance of its foci.
Understanding Foci
Now, let’s delve into the Foci. The term ‘Foci’ sounds mystical, and indeed, it plays a vital role in the tale of an ellipse. The foci (pronounced as ‘fo-sai’) are two specific points located along
the major axis of the ellipse. Each point on the ellipse is at a fixed total distance from these two foci, which shapes the ellipse and its properties. The Foci work behind the scenes to define the
shape of the ellipse, much like the puppeteers controlling a puppet show.
Properties of an Ellipse
An Ellipse is no ordinary shape; it is enriched with unique properties. It possesses two axes of symmetry – the major and minor axes. The longest diameter, known as the major axis, passes through the
foci, while the shortest diameter, the minor axis, runs perpendicular to it. Also, the sum of the distances from any point on the ellipse to the two foci is constant, reinforcing the crucial role the
foci play in defining an ellipse.
Properties of Foci
The Foci also possess fascinating properties. They are situated symmetrically on either side of the center of the ellipse along the major axis. The distance between each focus and the center of the
ellipse is given by a magical number c (known as the linear eccentricity). Moreover, the sum of the distances from any point on the ellipse to the two foci is always equal to the length of the major
axis, underscoring the symphony between the ellipse and its foci.
Relationship Between Ellipse and Its Foci
The bond between an Ellipse and its Foci is fundamental. In essence, the foci define the shape and properties of the ellipse. A simple way to envision this is by thinking of a planet orbiting two
stars. The planet follows a path, keeping its total distance from the two stars (the foci) constant. This cosmic dance results in an elliptical orbit, elegantly demonstrating the relationship between
an ellipse and its foci.
Difference Between Foci and Center of an Ellipse
The Foci and Center of an Ellipse may seem similar, but they are distinct. The center is the geometric midpoint of the ellipse, the point from which all distances are measured. In contrast, the foci
are two points situated on either side of the center along the major axis. The interplay between the center and the foci creates the beautiful, unique shape of the ellipse.
Formula for the Foci of an Ellipse
The Formula for the Foci of an Ellipse unearths the hidden relationship between the ellipse’s major axis, minor axis, and its foci. It states that the distance from the center to a focus (c) is given
by c = √(a² – b²), where ‘a’ is the semi-major axis and ‘b’ is the semi-minor axis. This equation is the key to unlocking the elliptical mystery.
Understanding the Formula for Foci of an Ellipse
To Understand the Formula for the Foci of an Ellipse, let’s consider a visual image. Imagine a tightrope walker walking along a rope. The rope’s length is the major axis (2a), and the distance from
the rope to the ground at its highest point is the minor axis (2b). The distances from the center of the rope (the center of the ellipse) to the supporting poles (the foci) are found using the
formula c = √(a² – b²).
Calculating the Foci of an Ellipse Using the Formula
Calculating the Foci of an Ellipse is like unlocking a code. All you need is the lengths of the semi-major axis (a) and semi-minor axis (b), and you can calculate the location of the foci using the
formula c = √(a² – b²). Suppose a = 5 and b = 3, the foci will be situated at c = √(5² – 3²) = √(25 – 9) = √16 = 4 units from the center along the major axis.
Practice Problems on Foci of an Ellipse
We’ve come a long way on our journey. Now, let’s put your newfound knowledge to the test with these Practice Problems on Foci of an Ellipse:
1. If the semi-major axis of an ellipse is 7 units and the semi-minor axis is 5 units, where are the foci located?
2. How does the position of the foci change if we increase the length of the semi-minor axis while keeping the semi-major axis constant?
As we draw our journey to a close, we reflect on the intricate beauty of the ellipse and the pivotal role its foci play. Here at Brighterly, our mission is to light the path of knowledge exploration
for young learners, and we hope this deep dive into the world of ellipses and their foci has fulfilled that promise. You’ve not only uncovered the definitions of an ellipse and its foci but also
delved into their properties, deciphered the formula for foci, and even put your understanding to test with practice problems. Remember, the magic of mathematics lies in understanding and applying,
and we are confident that you’ll carry this knowledge forward in your learning journey. Let the wonder of the ellipse and its mystical foci inspire you to continue exploring the limitless universe of
mathematics. Stay curious, keep learning, and until our next adventure, keep shining brightly with Brighterly.
Frequently Asked Questions on Foci of an Ellipse
What are the foci of an ellipse?
The foci of an ellipse are two special points located on the major axis of the ellipse. These points are unique because the sum of the distances from any point on the ellipse to both foci is always
constant. This constant sum is equal to the length of the major axis of the ellipse. The position of the foci plays a crucial role in shaping the ellipse and determining its eccentricity – the
measure of how ‘stretched’ the ellipse is compared to a circle.
What is the formula for the foci of an ellipse?
The formula for the foci of an ellipse is a fundamental expression that relates the semi-major axis (a), the semi-minor axis (b), and the distance from the center to a focus (c). It’s expressed as c
= √(a² – b²). This formula allows us to calculate the exact location of the foci, which in turn helps us to construct an accurate representation of the ellipse. The formula is derived from the
Pythagorean theorem, illustrating the interconnectedness of various branches of mathematics.
What is the difference between the foci and the center of an ellipse?
The foci and the center of an ellipse, although both are crucial points in the ellipse’s construction, serve distinct roles and have different properties. The center of the ellipse is the geometric
midpoint of the shape, the balance point, and it’s where the major and minor axes intersect. On the other hand, the foci are two points positioned symmetrically on either side of the center along the
major axis. The foci determine the shape and properties of the ellipse because the sum of the distances from any point on the ellipse to both foci remains constant.
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""" Generalized Convex Hull construction for the polymorphs of ROY ============================================================== :Authors: Michele Ceriotti `@ceriottm `_ This notebook analyzes the
structures of 264 polymorphs of ROY, from `Beran et Al, Chemical Science (2022) `__, comparing the conventional density-energy convex hull with a Generalized Convex Hull (GCH) analysis (see `Anelli
et al., Phys. Rev. Materials (2018) `__). It uses features computed with `rascaline `__ and uses the directional convex hull function from `scikit-matter `__ to make the figure. The GCH construction
aims at determining structures, among a collection of candidate configurations, that are stable or have the potential of being stabilized by appropriate thermodynamic boundary conditions (pressure,
doping, external fields, ...). It does so by using microscopic descriptors to determine the diversity of structures, and assumes that configurations that are stable relative to other configurations
with similar descriptors are those that could be made "locally" stable by suitable synthesis conditions. """ # sphinx_gallery_thumbnail_number = 3 import chemiscope import matplotlib.tri import numpy
as np from matplotlib import pyplot as plt from metatensor import mean_over_samples from rascaline import SoapPowerSpectrum from sklearn.decomposition import PCA from skmatter.datasets import
load_roy_dataset from skmatter.sample_selection import DirectionalConvexHull # %% # Loads the structures (that also contain properties in the ``info`` field) roy_data = load_roy_dataset() structures
= roy_data["structures"] density = np.array([s.info["density"] for s in structures]) energy = np.array([s.info["energy"] for s in structures]) structype = np.array([s.info["type"] for s in
structures]) iknown = np.where(structype == "known")[0] iothers = np.where(structype != "known")[0] # %% # Energy-density hull # ------------------- # # The Directional Convex Hull routines can be
used to compute a # conventional density-energy hull (see # `Hautier (2014) # `_ for a pedagogic # introduction to the convex hull construction in the context # of atomistic simulations). dch_builder
= DirectionalConvexHull(low_dim_idx=[0]) dch_builder.fit(density.reshape(-1, 1), energy) # %% # We can get the indices of the selection, and compute the distance from # the hull sel =
dch_builder.selected_idx_ dch_dist = dch_builder.score_samples(density.reshape(-1, 1), energy) # %% # # Hull energies # ^^^^^^^^^^^^^ # # Structures on the hull are stable with respect to synthesis
at constant # molar volume. Any other structure would lower the energy by decomposing # into a mixture of the two nearest structures along the hull. Given that # the lattice energy is an imperfect
proxy for the free energy, and that # synthesis can be performed in other ways than by fixing the density, # structures that are not exactly on the hull might also be stable. One # can compute a
“hull energy” as an indication of how close these # structures are to being stable. fig, ax = plt.subplots(1, 1, figsize=(6, 4)) ax.scatter(density, energy, c=dch_dist, marker=".") ssel = sel
[np.argsort(density[sel])] ax.plot(density[ssel], energy[ssel], "k--") ax.set_xlabel("density / g/cm$^3$") ax.set_ylabel("energy / kJ/mol") plt.show() print( f"Mean hull energy for 'known' stable
structures {dch_dist[iknown].mean()} kJ/mol" ) print(f"Mean hull energy for 'other' structures {dch_dist[iothers].mean()} kJ/mol") # %% # Interactive visualization # ^^^^^^^^^^^^^^^^^^^^^^^^^ # # You
can also visualize the hull with ``chemiscope`` in a juptyer notebook. # cs = chemiscope.show( structures, dict( energy=energy, density=density, hull_energy=dch_dist, structure_type=structype, ),
settings={ "map": { "x": {"property": "density"}, "y": {"property": "energy"}, "color": {"property": "hull_energy"}, "symbol": "structure_type", "size": {"factor": 35}, }, "structure": [{"unitCell":
True, "supercell": {"0": 2, "1": 2, "2": 2}}], }, ) cs # %% # # Save chemiscope file in a format that can be shared and viewed on `chemiscope.org` cs.save("roy_ch.json.gz") # %% # Generalized Convex
Hull # ----------------------- # # A GCH is a similar construction, in which generic structural descriptors # are used in lieu of composition, density or other thermodynamic # constraints. The idea
is that configurations that are found close to the # GCH are locally stable with respect to structurally-similar # configurations. In other terms, one can hope to find a thermodynamic # constraint
(i.e. synthesis conditions) that act differently on these # structures in comparison with the others, and may potentially stabilize # them. # # %% # Compute structural descriptors # ^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^ # # A first step is to computes suitable ML descriptors. Here we have used # ``rascaline`` to evaluate average SOAP features for the structures. # If you don't want to install these
dependencies for this example you # can also use the pre-computed features, but you can use this as a stub # to apply this analysis to other chemical systems hypers = { "cutoff": 4, "max_radial": 6,
"max_angular": 4, "atomic_gaussian_width": 0.7, "cutoff_function": {"ShiftedCosine": {"width": 0.5}}, "radial_basis": {"Gto": {"accuracy": 1e-6}}, "center_atom_weight": 1.0, } calculator =
SoapPowerSpectrum(**hypers) rho2i = calculator.compute(structures) rho2i = rho2i.keys_to_samples(["species_center"]).keys_to_properties( ["species_neighbor_1", "species_neighbor_2"] ) rho2i_structure
= mean_over_samples(rho2i, sample_names=["center", "species_center"]) np.savez("roy_features.npz", feats=rho2i_structure.block(0).values) # features = roy_data["features"] features =
rho2i_structure.block(0).values # %% # PCA projection # ^^^^^^^^^^^^^^ # # Computes PCA projection to generate low-dimensional descriptors that # reflect structural diversity. Any other
dimensionality reduction scheme # could be used in a similar fashion. pca = PCA(n_components=4) pca_features = pca.fit_transform(features) fig, ax = plt.subplots(1, 1, figsize=(6, 4)) scatter =
ax.scatter(pca_features[:, 0], pca_features[:, 1], c=energy) ax.set_xlabel("PCA[1]") ax.set_ylabel("PCA[2]") cbar = fig.colorbar(scatter, ax=ax) cbar.set_label("energy / kJ/mol") plt.show() # %% #
Builds the Generalized Convex Hull # ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ # # Builds a convex hull on the first two PCA features dch_builder = DirectionalConvexHull(low_dim_idx=[0, 1]) dch_builder.fit
(pca_features, energy) sel = dch_builder.selected_idx_ dch_dist = dch_builder.score_samples(pca_features, energy) # %% # Generates a 3D Plot # triang = matplotlib.tri.Triangulation(pca_features[sel,
0], pca_features[sel, 1]) fig = plt.figure(figsize=(7, 5), tight_layout=True) ax = fig.add_subplot(projection="3d") ax.plot_trisurf(triang, energy[sel], color="gray") ax.scatter(pca_features[:, 0],
pca_features[:, 1], energy, c=dch_dist) ax.set_xlabel("PCA[1]") ax.set_ylabel("PCA[2]") ax.set_zlabel("energy / kJ/mol\n \n", labelpad=11) ax.view_init(25, 110) plt.show() # %% # The GCH construction
improves the separation between the hull energies # of “known” and hypothetical polymorphs (compare with the density-energy # values above) print( f"Mean hull energy for 'known' stable structures
{dch_dist[iknown].mean()} kJ/mol" ) print(f"Mean hull energy for 'other' structures {dch_dist[iothers].mean()} kJ/mol") # %% # Visualize in a ``chemiscope`` widget for i, f in enumerate(structures):
for j in range(len(pca_features[i])): f.info["pca_" + str(j + 1)] = pca_features[i, j] structure_properties = chemiscope.extract_properties(structures) structure_properties.update({"per_atom_energy":
energy, "hull_energy": dch_dist}) # You can save a chemiscope file to disk (for viewing on chemiscope.org) chemiscope.write_input( "roy_gch.json.gz", frames=structures, properties=
structure_properties, meta={ "name": "GCH for ROY polymorphs", "description": """ Demonstration of the Generalized Convex Hull construction for polymorphs of the ROY molecule. Molecules that are
closest to the hull built on PCA-based structural descriptors and having the internal energy predicted by electronic-structure calculations as the z axis are the most thermodynamically stable. Indeed
most of the known polymorphs of ROY are on (or very close) to this hull. """, "authors": ["Michele Ceriotti "], "references": [ 'A. Anelli, E. A. Engel, C. J. Pickard, and M. Ceriotti, \ "Generalized
convex hull construction for materials discovery," \ Physical Review Materials 2(10), 103804 (2018).', 'G. J. O. Beran, I. J. Sugden, C. Greenwell, D. H. Bowskill, \ C. C. Pantelides, and C. S.
Adjiman, "How many more polymorphs of \ ROY remain undiscovered," Chem. Sci. 13(5), 1288–1297 (2022).', ], }, settings={ "map": { "x": {"property": "pca_1"}, "y": {"property": "pca_2"}, "z":
{"property": "energy"}, "symbol": "type", "color": {"property": "hull_energy"}, "size": { "factor": 35, "mode": "linear", "property": "", "reverse": True, }, }, "structure": [ { "bonds": True,
"unitCell": True, "keepOrientation": True, } ], }, ) # %% # # ... and also load one as an interactive viewer chemiscope.show_input("roy_gch.json.gz") | {"url":"http://atomistic-cookbook.org/_downloads/372c6744f93b866ccf802a12006619bb/roy-gch.py","timestamp":"2024-11-13T14:28:37Z","content_type":"text/x-python","content_length":"11327","record_id":"<urn:uuid:297a8736-120b-4f64-bb42-1737313e9021>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00455.warc.gz"} |
Worksheet Cells May Contain Text And Numbers
Worksheet Cells May Contain Text And Numbers serve as fundamental devices in the realm of mathematics, offering a structured yet flexible platform for students to check out and understand numerical
ideas. These worksheets use an organized approach to understanding numbers, supporting a strong structure whereupon mathematical effectiveness flourishes. From the simplest counting workouts to the
complexities of advanced calculations, Worksheet Cells May Contain Text And Numbers satisfy students of varied ages and skill degrees.
Unveiling the Essence of Worksheet Cells May Contain Text And Numbers
Worksheet Cells May Contain Text And Numbers
Worksheet Cells May Contain Text And Numbers -
Type the numbers that you want in the formatted cell Numbers that you entered before you applied the Text format to the cells must be entered again in the formatted cells To quickly reenter numbers
as text select each cell press F2 and then press Enter
Match entire cell contents Check this if you want to search for cells that contain only the characters that you typed in the Find what box Replace To replace text or numbers press Ctrl H or go to
Home Editing Find Select Replace
At their core, Worksheet Cells May Contain Text And Numbers are automobiles for conceptual understanding. They envelop a myriad of mathematical concepts, assisting students with the labyrinth of
numbers with a collection of appealing and deliberate exercises. These worksheets go beyond the limits of traditional rote learning, urging energetic engagement and cultivating an instinctive
understanding of numerical connections.
Nurturing Number Sense and Reasoning
Animal And Plant Cells Worksheet
Animal And Plant Cells Worksheet
Three functions ISFORMULA ISTEXT and ISNUMBER let you identify which worksheet cells contain formulas as well as which ones contain numbers or text After watching this video you will be
There are a few different formulas to count cells that contain any text specific characters or only filtered cells All the formulas work in Excel 365 2021 2019 2016 2013 and 2010 Initially Excel
spreadsheets were designed to work with numbers
The heart of Worksheet Cells May Contain Text And Numbers lies in cultivating number sense-- a deep understanding of numbers' meanings and interconnections. They encourage exploration, inviting
students to dissect math procedures, decode patterns, and unlock the secrets of series. Via provocative obstacles and sensible problems, these worksheets end up being gateways to developing reasoning
abilities, supporting the logical minds of budding mathematicians.
From Theory to Real-World Application
Science Cell Worksheet
Science Cell Worksheet
Excel contains two functions designed to check the occurrence of one text string inside another the SEARCH function and the FIND function Both functions return the position of the substring if found
as a number and a VALUE error if the substring is not found
Text Cells can contain text such as letters numbers and dates Formatting attributes Cells can contain formatting attributes that change the way letters numbers and dates are displayed For example
percentages can appear as 0 15 or 15
Worksheet Cells May Contain Text And Numbers serve as avenues bridging theoretical abstractions with the apparent truths of day-to-day life. By instilling useful situations into mathematical
exercises, learners witness the relevance of numbers in their environments. From budgeting and dimension conversions to comprehending analytical data, these worksheets empower students to possess
their mathematical expertise beyond the boundaries of the classroom.
Diverse Tools and Techniques
Flexibility is inherent in Worksheet Cells May Contain Text And Numbers, employing a toolbox of instructional tools to deal with different discovering designs. Aesthetic aids such as number lines,
manipulatives, and digital sources serve as buddies in visualizing abstract ideas. This varied method makes certain inclusivity, fitting students with various choices, staminas, and cognitive
Inclusivity and Cultural Relevance
In an increasingly diverse globe, Worksheet Cells May Contain Text And Numbers embrace inclusivity. They transcend cultural borders, incorporating examples and issues that resonate with students from
diverse backgrounds. By including culturally relevant contexts, these worksheets promote an environment where every learner really feels represented and valued, boosting their connection with
mathematical concepts.
Crafting a Path to Mathematical Mastery
Worksheet Cells May Contain Text And Numbers chart a training course in the direction of mathematical fluency. They infuse perseverance, crucial thinking, and analytic abilities, necessary qualities
not only in mathematics but in numerous facets of life. These worksheets encourage students to navigate the elaborate surface of numbers, supporting a profound admiration for the style and logic
inherent in maths.
Welcoming the Future of Education
In an era marked by technical development, Worksheet Cells May Contain Text And Numbers flawlessly adjust to electronic platforms. Interactive interfaces and digital sources enhance typical learning,
offering immersive experiences that go beyond spatial and temporal limits. This amalgamation of traditional methods with technological technologies declares an appealing era in education, promoting
an extra vibrant and engaging discovering setting.
Final thought: Embracing the Magic of Numbers
Worksheet Cells May Contain Text And Numbers characterize the magic inherent in maths-- a charming journey of expedition, discovery, and mastery. They transcend standard rearing, acting as drivers
for stiring up the fires of inquisitiveness and inquiry. With Worksheet Cells May Contain Text And Numbers, learners embark on an odyssey, unlocking the enigmatic world of numbers-- one problem, one
option, each time.
Cells And Their Organelles Worksheet
Plant And Cell Diagram Worksheet Worksheets For Kindergarten
Check more of Worksheet Cells May Contain Text And Numbers below
Animal And Plant Cells Worksheet
Types Of Cells Worksheet
Introduction To Cells Worksheet Pdf
Excel Formula Count Cells That Contain Text Exceljet Riset
Animal And Plant Cells Worksheet Answers Key Aiminspire
If Cell Contains Text From List 2023
Find Or Replace Text And Numbers On A Worksheet
Match entire cell contents Check this if you want to search for cells that contain only the characters that you typed in the Find what box Replace To replace text or numbers press Ctrl H or go to
Home Editing Find Select Replace
How To Sort Mixed Numbers And Text Hierarchy Numbers In Excel Ablebits
If the target column contains numbers formatted as text all it takes for the numbers to sort normally is to convert text to number The result is in column C in the screenshot below The result is in
column C in the screenshot below
Match entire cell contents Check this if you want to search for cells that contain only the characters that you typed in the Find what box Replace To replace text or numbers press Ctrl H or go to
Home Editing Find Select Replace
If the target column contains numbers formatted as text all it takes for the numbers to sort normally is to convert text to number The result is in column C in the screenshot below The result is in
column C in the screenshot below
Excel Formula Count Cells That Contain Text Exceljet Riset
Animal And Plant Cells Worksheet Answers Key Aiminspire
If Cell Contains Text From List 2023
Parts Of A Cell Worksheets Free Printable Worksheet
Excel Formulas To Sum If Cells Contain Text In Another Cell Images
Excel Formulas To Sum If Cells Contain Text In Another Cell Images
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Algebraic Equations: Single Variable
This post contains word problems from algebraic equations in single variable. All problems are based on relationships in numbers.
1. Thrice the sum of a number and 2 equals 24 added with the same number. Write algebraic equation and find the number.
Let x be the number.
For the given statement, we have following algebraic equation
3* (x+2) =24+x
3 x + 6= 24 +x
Subtracting x on both sides,
2x +6 = 24
Subtracting 6 on both sides,
2x= 18
Dividing by 2,
Therefore, the 9 is the number.
2. Consider three consecutive positive integers. If the third number is subtracted from the sum of first two numbers, the difference is 10. Find the numbers.
Let x, x+1 and x+2 be the three consecutive positive numbers.
X +(x+1)-(x+2) =10
2x +1-x-2=10
x- 1 = 10
Adding 1 on both sides,
Therefore, the three numbers are 11, 12 and 13.
3. A number added with thrice a number is 20. Convert this statement into algebraic equations and find the number.
4. A number divided by 7 is five-fourteenth. Find the number.
5. Sum of a number and 15 is 96. Find the number.
6. The difference of two numbers is 100. If one of the numbers is 91, find the other number.
7. 12 less than twice a number is 20. Find the number.
8. If a number is subtracted from 32, the digits of 32 are interchanged. Write the statement as algebraic equation and also find the number.
9. The difference of 5 times a number and 17 is 23. Find the number and 5 times the number.
10. One-third of a number minus seven gives eight. Find the number.
11. Consider two-third of number and add up six to it. The result is ten. Find the unknown number.
12. I think of a number. If I subtract 6 from the number and divide it by 13, I will get the result as 2. Find the number in my mind.
13. Wilson has a number in mind. If he takes away one-third of the number from it the result is sixteen-third. Find the number.
14. The sum of two consecutive numbers is 19. Find the numbers and its difference.
15. The sum of three consecutive numbers is 18. Find the numbers and check your answer.
16. The sum of two consecutive odd numbers is 20. Find the numbers.
17. The sum of two consecutive even numbers is 198. Find the numbers and its product.
18. The sum of three consecutive odd numbers is 81. Find the numbers and check your answer.
19. The sum of three consecutive even numbers is 42. Choose suitable variable and create single variable algebraic equations and find the difference between the greatest and smallest number.
20. One-third of certain number added with one-fifth of the same number gives eight-fifteenth of the number. Find the number and check your calculation.
21. The difference between one-third of a number and 5 is 10. Find the number.
22. The sum of two numbers is 132, whose ratio is 5:6. Find the numbers.
23. The sum of the digits of a two digit number is 15. If the digits in the tenth place 8, find the two digit number.
24. If twice a number is subtracted from 97, the digits of 97 are interchanged. Find the number.
25. Find the fraction whose numerator is 3 and the denominator is 1 more than twice the numerator.
26. The difference of two numbers is 70. If the larger number is 6 times the smaller number, find the numbers.
27. Two numbers are in the ratio 1:3. If I add 2 to each of the numbers, the ratio becomes 1:2. Find the original numbers.
28. The sum of 7 more than 5 times a number is equal to sum of 6 more than twice the same number. Write the statement into possible algebraic equations and find the unknown number.
29. A number equals 5 times the difference between the number and 4. Find the number.
30. In a fraction, the denominator is 3 more than numerator. The sum of numerator and denominator is 7. Find the fraction.
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​Mr. Kouts. is a professional teacher of Math with Master and Bachelor degrees both on Math and over 14 years of teaching experience on Math, Physics, and Computer Science, including SAT/ACT Math,
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Mr. Perry
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Today in Science History - Quickie Quiz
Who said: “A people without children would face a hopeless future; a country without trees is almost as helpless.”
Category Index for Science Quotations
Category Index A
> Category: Applied Mathematics
Applied Mathematics Quotes (15 quotes)
For all their wealth of content, for all the sum of history and social institution invested in them, music, mathematics, and chess are resplendently useless (applied mathematics is a higher plumbing,
a kind of music for the police band). They are metaphysically trivial, irresponsible. They refuse to relate outward, to take reality for arbiter. This is the source of their witchery.
In 'A Death of Kings', George Steiner at The New Yorker (2009), 209.
From Pythagoras (ca. 550 BC) to Boethius (ca AD 480-524), when pure mathematics consisted of arithmetic and geometry while applied mathematics consisted of music and astronomy, mathematics could be
characterized as the deductive study of “such abstractions as quantities and their consequences, namely figures and so forth” (Aquinas ca. 1260). But since the emergence of abstract algebra it has
become increasingly difficult to formulate a definition to cover the whole of the rich, complex and expanding domain of mathematics.
In 100 Years of Mathematics: a Personal Viewpoint (1981), 2.
I count Maxwell and Einstein, Eddington and Dirac, among “real” mathematicians. The great modern achievements of applied mathematics have been in relativity and quantum mechanics, and these subjects
are at present at any rate, almost as “useless” as the theory of numbers.
In A Mathematician's Apology (1940, 2012), 131.
I do not think the division of the subject into two parts - into applied mathematics and experimental physics a good one, for natural philosophy without experiment is merely mathematical exercise,
while experiment without mathematics will neither sufficiently discipline the mind or sufficiently extend our knowledge in a subject like physics.
to Henry Roscoe, Professor of Chemistry at Owens College (2 Jun 1870), B.C.S Archive Quoted in R.H. Kargon, Science in Victorian Manchester (1977), 215.
I maintain that in every special natural doctrine only so much science proper is to be met with as mathematics; for… science proper, especially [science] of nature, requires a pure portion, lying at
the foundation of the empirical, and based upon a priori knowledge of natural things. … To the possibility of a determinate natural thing, and therefore to cognise it à priori, is further requisite
that the intuition corresponding à priori to the conception should be given; in other words, that the conception should be constructed. But the cognition of the reason through construction of
conceptions is mathematical. A pure philosophy of nature in general, namely, one that only investigates what constitutes a nature in general, may thus be possible without mathematics; but a pure
doctrine of nature respecting determinate natural things (corporeal doctrine and mental doctrine), is only possible by means of mathematics; and as in every natural doctrine only so much science
proper is to be met with therein as there is cognition à priori, a doctrine of nature can only contain so much science proper as there is in it of applied mathematics.
From Preface to The Metaphysical Foundations of Natural Science (1786), as translated by Ernest Belford Boax, in Kant’s Prolegomena: And The Metaphysical Foundations of Natural Science (1883), 140.
I think that the difference between pure and applied mathematics is social rather than scientific. A pure mathematician is paid for making mathematical discoveries. An applied mathematician is paid
for the solution of given problems.
When Columbus set sail, he was like an applied mathematician, paid for the search of the solution of a concrete problem: find a way to India. His discovery of the New World was similar to the work of
a pure mathematician.
In S.H. Lui, 'An Interview with Vladimir Arnol’d', Notices of the AMS (Apr 1997) 44, No. 4, 438. Reprinted from the Hong Kong Mathematics Society (Feb 1996).
It has been asserted … that the power of observation is not developed by mathematical studies; while the truth is, that; from the most elementary mathematical notion that arises in the mind of a
child to the farthest verge to which mathematical investigation has been pushed and applied, this power is in constant exercise. By observation, as here used, can only be meant the fixing of the
attention upon objects (physical or mental) so as to note distinctive peculiarities—to recognize resemblances, differences, and other relations. Now the first mental act of the child recognizing the
distinction between one and more than one, between one and two, two and three, etc., is exactly this. So, again, the first geometrical notions are as pure an exercise of this power as can be given.
To know a straight line, to distinguish it from a curve; to recognize a triangle and distinguish the several forms—what are these, and all perception of form, but a series of observations? Nor is it
alone in securing these fundamental conceptions of number and form that observation plays so important a part. The very genius of the common geometry as a method of reasoning—a system of
investigation—is, that it is but a series of observations. The figure being before the eye in actual representation, or before the mind in conception, is so closely scrutinized, that all its
distinctive features are perceived; auxiliary lines are drawn (the imagination leading in this), and a new series of inspections is made; and thus, by means of direct, simple observations, the
investigation proceeds. So characteristic of common geometry is this method of investigation, that Comte, perhaps the ablest of all writers upon the philosophy of mathematics, is disposed to class
geometry, as to its method, with the natural sciences, being based upon observation. Moreover, when we consider applied mathematics, we need only to notice that the exercise of this faculty is so
essential, that the basis of all such reasoning, the very material with which we build, have received the name observations. Thus we might proceed to consider the whole range of the human faculties,
and find for the most of them ample scope for exercise in mathematical studies. Certainly, the memory will not be found to be neglected. The very first steps in number—counting, the multiplication
table, etc., make heavy demands on this power; while the higher branches require the memorizing of formulas which are simply appalling to the uninitiated. So the imagination, the creative faculty of
the mind, has constant exercise in all original mathematical investigations, from the solution of the simplest problems to the discovery of the most recondite principle; for it is not by sure,
consecutive steps, as many suppose, that we advance from the known to the unknown. The imagination, not the logical faculty, leads in this advance. In fact, practical observation is often in advance
of logical exposition. Thus, in the discovery of truth, the imagination habitually presents hypotheses, and observation supplies facts, which it may require ages for the tardy reason to connect
logically with the known. Of this truth, mathematics, as well as all other sciences, affords abundant illustrations. So remarkably true is this, that today it is seriously questioned by the majority
of thinkers, whether the sublimest branch of mathematics,—the infinitesimal calculus—has anything more than an empirical foundation, mathematicians themselves not being agreed as to its logical
basis. That the imagination, and not the logical faculty, leads in all original investigation, no one who has ever succeeded in producing an original demonstration of one of the simpler propositions
of geometry, can have any doubt. Nor are induction, analogy, the scrutinization of premises or the search for them, or the balancing of probabilities, spheres of mental operations foreign to
mathematics. No one, indeed, can claim preeminence for mathematical studies in all these departments of intellectual culture, but it may, perhaps, be claimed that scarcely any department of science
affords discipline to so great a number of faculties, and that none presents so complete a gradation in the exercise of these faculties, from the first principles of the science to the farthest
extent of its applications, as mathematics.
In 'Mathematics', in Henry Kiddle and Alexander J. Schem, The Cyclopedia of Education, (1877.) As quoted and cited in Robert Édouard Moritz, Memorabilia Mathematica; Or, The Philomath’s
Quotation-book (1914), 27-29.
Mathematics, including not merely Arithmetic, Algebra, Geometry, and the higher Calculus, but also the applied Mathematics of Natural Philosophy, has a marked and peculiar method or character; it is
by preeminence deductive or demonstrative, and exhibits in a nearly perfect form all the machinery belonging to this mode of obtaining truth. Laying down a very small number of first principles,
either self-evident or requiring very little effort to prove them, it evolves a vast number of deductive truths and applications, by a procedure in the highest degree mathematical and systematic.
In Education as a Science (1879), 148.
My decision to leave applied mathematics for economics was in part tied to the widely-held popular belief in the 1960s that macroeconomics had made fundamental inroads into controlling business
cycles and stopping dysfunctional unemployment and inflation.
Nobel Banquet Speech (1995). Collected in Tore Frängsmyr (ed.), Les Prix Nobel/Nobel Lectures/The Nobel Prizes.
Programming is one of the most difficult branches of applied mathematics; the poorer mathematicians had better remain pure mathematicians.
Pure mathematics consists entirely of such asseverations as that, if such and such is a proposition is true of anything, then such and such another propositions is true of that thing. It is essential
not to discuss whether the first proposition is really true, and not to mention what the anything is of which it is supposed to be true. Both these points would belong to applied mathematics. … If
our hypothesis is about anything and not about some one or more particular things, then our deductions constitute mathematics. Thus mathematics may be defined as the the subject in which we never
know what we are talking about, not whether what we are saying is true. People who have been puzzled by the beginnings of mathematics will, I hope, find comfort in this definition, and will probably
agree that it is accurate.
In 'Recent Work on the Principles of Mathematics', International Monthly (1901), 4, 84.
Pure mathematics is much more than an armoury of tools and techniques for the applied mathematician. On the other hand, the pure mathematician has ever been grateful to applied mathematics for
stimulus and inspiration. From the vibrations of the violin string they have drawn enchanting harmonies of Fourier Series, and to study the triode valve they have invented a whole theory of
non-linear oscillations.
In 100 Years of Mathematics: a Personal Viewpoint (1981), 3.
These specimens, which I could easily multiply, may suffice to justify a profound distrust of Auguste Comte, wherever he may venture to speak as a mathematician. But his vast general ability, and
that personal intimacy with the great Fourier, which I most willingly take his own word for having enjoyed, must always give an interest to his views on any subject of pure or applied mathematics.
In R. Graves, Life of W. R. Hamilton (1882-89), Vol. 3, 475.
This trend [emphasizing applied mathematics over pure mathematics] will make the queen of the sciences into the quean of the sciences.
As given, without citation, in Howard W. Eves Mathematical Circles Squared (1972), 158, which attributes it (via Dirk J. Struik) to a memorandum in which Passano wrote of the trend in the Dept. of
Mathematics at M.I.T. Webmaster has as yet been unable to identify a primary source. (Can you help?) [Note: “quean” is an archaic word for: a disreputable woman; specifically :
prostitute.—Merriam-Webster. “Thus the semantic spread between queen and quean could not be greater: from a woman of the highest repute to one of the lowest.” —alphadictionary.com]
Two extreme views have always been held as to the use of mathematics. To some, mathematics is only measuring and calculating instruments, and their interest ceases as soon as discussions arise which
cannot benefit those who use the instruments for the purposes of application in mechanics, astronomy, physics, statistics, and other sciences. At the other extreme we have those who are animated
exclusively by the love of pure science. To them pure mathematics, with the theory of numbers at the head, is the only real and genuine science, and the applications have only an interest in so far
as they contain or suggest problems in pure mathematics.
Of the two greatest mathematicians of modern tunes, Newton and Gauss, the former can be considered as a representative of the first, the latter of the second class; neither of them was exclusively
so, and Newton’s inventions in the science of pure mathematics were probably equal to Gauss’s work in applied mathematics. Newton’s reluctance to publish the method of fluxions invented and used by
him may perhaps be attributed to the fact that he was not satisfied with the logical foundations of the Calculus; and Gauss is known to have abandoned his electro-dynamic speculations, as he could
not find a satisfying physical basis. …
Newton’s greatest work, the Principia, laid the foundation of mathematical physics; Gauss’s greatest work, the Disquisitiones Arithmeticae, that of higher arithmetic as distinguished from algebra.
Both works, written in the synthetic style of the ancients, are difficult, if not deterrent, in their form, neither of them leading the reader by easy steps to the results. It took twenty or more
years before either of these works received due recognition; neither found favour at once before that great tribunal of mathematical thought, the Paris Academy of Sciences. …
The country of Newton is still pre-eminent for its culture of mathematical physics, that of Gauss for the most abstract work in mathematics.
In History of European Thought in the Nineteenth Century (1903), 630.
In science it often happens that scientists say, 'You know that's a really good argument; my position is mistaken,' and then they would actually change their minds and you never hear that old view
from them again. They really do it. It doesn't happen as often as it should, because scientists are human and change is sometimes painful. But it happens every day. I cannot recall the last time
something like that happened in politics or religion. (1987) --
Carl Sagan
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Overview This homework will allow you to demonstrate your
David answered on Nov 30 2019
Secure Software Design
Secure Software Design
Submitted By
As we know that the vulnerability is related with the framework and also this is considered as major threat for framework which result in additional support cost. This is expected for numerous causes
and are not considered during the phases of System Development Life Cycle (SDLC). While having SDLC approach, it might be minimize to least level. In order to have coordination of security in SDLC
stages, the new approach in order to assess the security during its plan stage by applying neural system. This article presents the data for the existing procedures, guidelines, life-cycle models,
structures, as well as its procedures that help or can support secure software development. (Ahmad, Z. & Asif, M., 2015).
As we know that the software that are deployed on commercial level are not of good quality because of various types of flaws which could be hacked by hackers. There are many reasons behind this. At
the time, whenever programming applications are developed without having any type of security as a main priority, the hackers may exploit the security flaws and they enter into the organization's
networking framework by having multiple attacks. For addressing this issue, a new development which is known as software security is emerged. This methodology considers security as a developing
property of the product and various efforts are devoted into meshing security into the product all via using SDLC. Besides, it requires highest skill to learn regardless of whether a product
application has configuration level imperfections which makes it hard to discover as well as to automate. In this way, by considering security at the designing phase of SDLC would help significantly
in creating secured programming applications. (Adebiyi, A. & A
eymbi, J., 2012).
Secure Software Design Principles
Designing an application without considering security architecture are like the
idges that are built without having limited component analysis as well as without having tunnel testing. The developers for building an application are main persons who develop the design to cover
all threats from outrageous assault. The general principles for secure software design are given as following:
1. Minimize the Number of High Consequences Targets - This principle prescribes that records have minimal privileges that are required to play out their business forms. This includes user rights,
resource authorizations like CPU limits, memory, system, as well as file framework permissions. This principle helps to limit the number of performing actors in the framework that are allowed high
privileges, and the measure of time that an actor can have to possess the privileges. (https:
www.owasp.org, 2016). This constrains the extent of the segment's activities that can have two alluring impacts: (1) first is, security effect of failure of debasement of the segment that would be
reduced, and (2) the second one is, the process of analyzing security analysis will be improved. (Levin, T., E. & Irvine, C., E., n.d).
Example: when a middleware server access the network, read access to database table, as well as the capacity for writing a log, this portrays permissions that ought to be allowed. Middleware is not
given any administration privileges. In a customary Web entry application, the end user is just permitted to peruse, post content, as well as enter information into HTML forms, whereas administrator
has all the privileges.
2. Separation of Privileges, Duties, and Roles – According to this, no single element (whether human or programming) ought to have all the privileges that are mainly required to modify, erase, or
to delete the framework, segments as well as assets. This rule expects to deal with i
econcilable circumstance and fraud by directing and approving the measure of the power held by the clients in an association. Administration controls are determined to obstruct the utilization of
power for individual benefit and in addition coordinated effort among people for a similar reason, and to authorize the accurate execution of duties.
Example - This rule is connected in production as well as development environment to ensure that they collaborate as meager as would be prudent and that the installation or the delivery of
utilizations into production ought to be audited by a group of people other than the development group.
Designers ought to approach the development as well as test code/frameworks; they ought to approach the production framework. In the event that designers approach the production framework, they could
make unauthorized access that could prompt a
oken application or include hacking code for their own advantage. The code needs to experience the proper endorsements and testing before being deployed into development. The administrator ought to
have the capacity to deploy the package into production however ought to be able to alter the code.
3. Separation of Domains - Separation of domains makes division of parts as well as privileges less demanding to execute. Few parts have distinctive levels of trust than ordinary clients.
Specifically, the administrators are distinctive to typical clients. The administrator must not be the user of the application. Other example is -
Example - Database executives ought to have control over business rationale and the application administrator ought to have control over the database. If somebody who asks for a system can't login
the system, nor should they specifically get the system access. This keeps the client from asking for some systems, and asserting that they never a
ived. Another example is, when the administrator ought to have the capacity to make the system on or off, set password approach however shouldn't have the capacity to sign on to the user as a super
privileged user like having the capacity to "purchase" products in the interest of other users.
4. Keep Program Data, Executables, and Configuration data separated - This key practice is the part of General Principle 2. It is used to minimize the probability that an assailant who is able to
gain the access to the information will effectively find as well as access program executables or potentially control the configuration information. These procedures are used to isolate project
information from the control information, and from the defined executables at every level of the framework to various execution environments like level strategies. It is not necessarily the case that
keeping insider facts is not a good idea, it essentially implies that the security of key frameworks ought to be dependent after keeping subtle elements hidden.
It will help to reduce the probability that the attacker who accesses program information will effectively find and access program executables or control information.
Example - In case of Unix or Linux frameworks, the chroot "jail" characteristic of the standard operating system controls could be configured to make isolated execution points for programming,
consequently filling an indistinguishable need from Java or Perl "sandbox." Every directory to which programs developed ought to be designed to be writable just by those projects. By considering Web
server application, there is an express requirement for clients in order to view information which is utilized by a script, every single such data ought to be put outside the Web server's record
This key practice will help to prohibit projects and contents from composing documents to world-writable catalogs, for example, the Unix or tmp directory. Every directory where the programs compose
ought to be configured should be writable just by those projects.
5. Segregate Trusted Entities from Untrusted Entities - The untrusted element (segment, operator, and process) can be viewed as untrustworthy which based on the failure to fulfill some predefined
paradigm or criteria for deciding dependability. Assurance of reliability might be made in developing a software, or it might be resolved during its operation. It helps to minimize the exposure of
the product's high-result functions could be susceptible to assaults.
Example - COTS segment whose source code couldn't be analyzed the framework's improvement may be regarded as "untrusted" in light of the fact that its reliability couldn't be enough evaluated before
its implementation. A mobile operator that can't be validated during the framework's operation might be assigned "untrusted", like Java applet which isn't signed or whose code signature can't be
Java as well as Perl's sandboxing and Code Access Security system in .NET Code Access Security system allots a level privilege to be executable which is inside it. This level ought to be the
insignificant required by the function(s) to play out its typical expected operation. When the anomalies happen, the sandbox/CLR will produce a special case and an exemption handler will keep the
executable from playing out the unexpected operation.
6. Assume Environment Data is Not Trustworthy - This key practice minimizes the introduction of the product to possibly malevolent execution environment segments or the hackers caught as well as
modified information. The designer ought to expect that all parts of the execution condition are not one or the other tried and true nor dependable unless and until the point that this suspicion is
demonstrated off-base. Some application systems are confirmed to give reliable condition information to the applications facilitated inside those systems.
Example - Java Platform, Enterprise Edition (JEE) segments keep running inside "context" like Framework Context, Login Context, Session Context, Naming and Directory Context, and so forth. These
could be depended on to give reliable environment information at runtime to Java programs.
7. Use Only Safe Interfaces to Environment Resources - This key practice decreases the exposure of the information that is passed between the product as well as its environment. Almost every
programming and script language permits application level projects for issuing the system calls that give the command or information to the fundamental OS. Because of these types of calls, the OS
executes these commands that are shown by the framework call, and it also returns the outcomes to the product alongside different return codes that mainly shows that the requested order was executed
effectively or not.
As we know that the framework commands can appear as the most proficient approach to actualize the interface to the fundamental OS, which is a safe application that will never issue an immediate call
to the fundamental OS, or to framework level system projects like send mail unless these controls are forced which are sufficient to keep the client, assailant, or malicious program from gaining the
control of the calling program and misusing its immediate calling mechanism. Every application call to a framework level capacity make a potential focus for this attack, at whatever point the product
issues a framework call, the homogeneity of the framework's plan is minimized, also, its unwavering quality decreases. (McGi
on, T., 2008).
Example - Application-level projects can call other application-layer programs, middleware, or explicit APIs to framework assets. These applications ought to utilize APIs planned for human clients as
opposed to programming nor depend on a framework level tool versus the application-level device to channel or change the output.
Implementation Details
The implementation process is the methodology to convert a framework determination into an executable framework. It might also include refinement of the product detail. The software configuration is
a depiction of the structure of the product to be implemented, information models, interfaces between framework parts, and possibly the algorithms utilized. The implementation process requires the
representation of the code the security rules that are defined for the application. Since these guidelines are communicated as classes, affiliations, and requirements, they can be actualized as extra
The organization must choose particular security package like firewall, a cryptographic package. Also after implementation process, auditing must be performed in order to perform reviews to check
that the main policies for implementation process are followed. The security constraints can be made more exact by utilizing (Object Constraint Language (OCL) rather than any textual requirements.
Various examples for security models characterize the highest level. At lower level the organization can apply the model examples to particular components that uphold these models. Patterns for file
systems can be defined. Also, we can likewise assess an existing framework utilizing designs. Patterns will help to comprehend the security structure of every part to permit their part as well as
characterize secure interfaces. If the framework doesn't contain a suitable example then it can't support the relating secure model or component. (Fernandez, E., B., 2004).
Ahmad, Z. & Asif, M. (2015). Implementation of Secure Software Design and their impact on Application. International Journal of Computer Applications (0975 – 8887) Volume 120 – No.10, June 2015.
Retrieved from - http:
Adebiyi, A. & A
eymbi, J. (2012). Security Assessment of Software Design using Neural Network. (IJARAI) International Journal of Advanced Research in Artificial Intelligence, Vol. 1, No. 4, 2012. Retrieved from -
Security by Design Principles. Retrieved from - https:
Fernandez, E., B. (2004). A Methodology for Secure Software Design. Retrieved from - https:
Levin, T., E. & Irvine, C., E. (n.d). Design Principles and Guidelines for Security. Retrieved from - ftp:
on, T. (2008). Enhanced the development life cycle to produce secure software. A reference guidebook on assurance. Retrieved from - http: | {"url":"https://www.topassignmentexperts.com/questions/overview-this-homework-will-allow-you-to-demonstrate-your-understanding-of-guid-113845.html","timestamp":"2024-11-06T07:36:09Z","content_type":"text/html","content_length":"251269","record_id":"<urn:uuid:49c2fba7-166c-4388-bb2f-3c7f2e82223b>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00134.warc.gz"} |
Implementation of the Thornthwaite-Mather procedure to map groundwater recharge | Google Earth Engine | Google for Developers
Groundwater recharge represents the amount of water coming from precipitation reaching the groundwater table. Its determination helps to better understand the available/renewable groundwater in
watersheds and the shape of groundwater flow systems.
One of the simplest methods to estimate groundwater recharge is the Thornthwaite-Mather procedure (Steenhuis and Van Der Molen, 1986). This procedure was published by Thornthwaite and Mather (1955,
1957). The idea of this procedure is to calculate the water balance in the root zone of the soil where water can be (1) evaporated into the atmosphere under the effect of heat, (2) transpired by
vegetation, (3) stored by the soil, and eventually (4) infiltrated when stored water exceeds the field capacity.
This procedures relies on several parameters and variables described as follows:
• information about soil texture (e.g. sand and clay content) to describe the hydraulic properties of the soil and its capacity to store/infiltrate,
• meteorological records: precipitation and potential evapotranspiration.
Of course groundwater recharge can be influenced by many other factors such as the slope of the terrain, the snow cover, the variability of the crop/land cover and the irrigation. In the following
these aspects are not taken into account.
In the first part of the tutorial, the Earth Engine python API will be initialized, some useful libraries will be imported, and the location/period of interest will be defined.
In the second part, OpenLandMap datasets related to soil properties will be explored. The wilting point and field capacity of the soil will be calculated by applying some mathematical expressions to
multiple images.
In the third part, evapotranspiration and precipitation datasets will be imported. A function will be defined to resample the time resolution of an ee.ImageCollection and to homogenize time index of
both datasets. Both datasets will then be combined into one.
In the fourth and final part, the Thornthwaite-Mather(TM) procedure will be implemented by iterating over the meteorological ee.ImageCollection. Finally, a comparison between groundwater recharge in
two places will be described and the resulting mean annual groundwater recharge will be displayed over France.
Run me first
Earth Engine API
First of all, run the following cell to initialize the API. The output will contain instructions on how to grant this notebook access to Earth Engine using your account.
import ee
# Trigger the authentication flow.
# Initialize the library.
*** Earth Engine *** Share your feedback by taking our Annual Developer Satisfaction Survey: https://google.qualtrics.com/jfe/form/SV_0JLhFqfSY1uiEaW?source=Init
Other libraries
Import other libraries/modules used in this notebook.
import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
import folium
import pprint
import branca.colormap as cm
Some input parameters
We additionally define some parameters used to evaluate the results of our implementation. Particularly:
• the period of interest to get meteorological records,
• a location of interest based on longitude and latitude coordinates. In the following, the point of interest is located in a productive agricultural region which is about 30 kilometers outside of
the city of Lyon (France). This point is used to evaluate and illustrate the progress of the described procedure.
# Initial date of interest (inclusive).
i_date = "2015-01-01"
# Final date of interest (exclusive).
f_date = "2020-01-01"
# Define the location of interest with a point.
lon = 5.145041
lat = 45.772439
poi = ee.Geometry.Point(lon, lat)
# A nominal scale in meters of the projection to work in [in meters].
scale = 1000
From soil texture to hydraulic properties
Two hydraulic properties of soil are commonly used in the TM procedure:
• the wilting point represents the point below what water cannot be extracted by plant roots,
• the field capacity represents the point after which water cannot be stored by soil any more. After that point, gravitational forces become too high and water starts to infiltrate the lower
Some equations given by Saxton & Rawls (2006) are used to link both parameters to the texture of the soil. The calculation of water content at wilting point \(θ_{WP}\) can be done as follows:
$$ \theta_{WP}= \theta_{1500t} + (0.14 \theta_{1500t} - 0.002) $$
$$ \theta_{1500t} = -0.024 S + 0.487 C + 0.006 OM + 0.005(S \times OM) - 0.013 (C \times OM) + 0.068 (S \times C) + 0.031 $$
• \(S\): represents the sand content of the soil (mass percentage),
• \(C\): represents the clay content of the soil (mass percentage),
• \(OM\): represents the organic matter content of the soil (mass percentage).
Similarly, the calculation of the water content at field capacity \(θ_{FC}\) can be done as follows:
$$ \theta_{FC} = \theta_{33t} + (1.283 \theta_{33t}^{2} - 0.374 \theta_{33t}-0.15) $$
$$ \theta_{33t} = -0.251 S + 0.195 C + 0.011 OM + 0.006 (S \times OM) - 0.027 (C \times OM) + 0.452 (S \times C) + 0.299 $$
Determination of soil texture and properties
In the following, OpenLandMap datasets are used to describe clay, sand and organic carbon content of soil. A global dataset of soil water content at the field capacity with a resolution of 250 m has
been made available by Hengl & Gupta (2019). However, up to now, there is no dataset dedicated to the water content of soil at the wilting point. Consequently, in the following, both parameters will
be determined considering the previous equations and using the global datasets giving the sand, clay and organic matter contents of the soil. According to the description, these datasets are based on
machine learning predictions from global compilation of soil profiles and samples. Processing steps are described in detail here. The information (clay, sand content, etc.) is given at 6 standard
depths (0, 10, 30, 60, 100 and 200 cm) at 250 m resolution.
These standard depths and associated bands are defined into a list as follows:
# Soil depths [in cm] where we have data.
olm_depths = [0, 10, 30, 60, 100, 200]
# Names of bands associated with reference depths.
olm_bands = ["b" + str(sd) for sd in olm_depths]
We now define a function to get the ee.Image associated to the parameter we are interested in (e.g. sand, clay, organic carbon content, etc.).
def get_soil_prop(param):
This function returns soil properties image
param (str): must be one of:
"sand" - Sand fraction
"clay" - Clay fraction
"orgc" - Organic Carbon fraction
if param == "sand": # Sand fraction [%w]
snippet = "OpenLandMap/SOL/SOL_SAND-WFRACTION_USDA-3A1A1A_M/v02"
# Define the scale factor in accordance with the dataset description.
scale_factor = 1 * 0.01
elif param == "clay": # Clay fraction [%w]
snippet = "OpenLandMap/SOL/SOL_CLAY-WFRACTION_USDA-3A1A1A_M/v02"
# Define the scale factor in accordance with the dataset description.
scale_factor = 1 * 0.01
elif param == "orgc": # Organic Carbon fraction [g/kg]
snippet = "OpenLandMap/SOL/SOL_ORGANIC-CARBON_USDA-6A1C_M/v02"
# Define the scale factor in accordance with the dataset description.
scale_factor = 5 * 0.001 # to get kg/kg
return print("error")
# Apply the scale factor to the ee.Image.
dataset = ee.Image(snippet).multiply(scale_factor)
return dataset
We apply this function to import soil properties:
# Image associated with the sand content.
sand = get_soil_prop("sand")
# Image associated with the clay content.
clay = get_soil_prop("clay")
# Image associated with the organic carbon content.
orgc = get_soil_prop("orgc")
To illustrate the result, we define a new method for handing Earth Engine tiles and using it to display the clay content of the soil at a given reference depth, to a Leaflet map.
def add_ee_layer(self, ee_image_object, vis_params, name):
"""Adds a method for displaying Earth Engine image tiles to folium map."""
map_id_dict = ee.Image(ee_image_object).getMapId(vis_params)
attr="Map Data © <a href='https://earthengine.google.com/'>Google Earth Engine</a>",
# Add Earth Engine drawing method to folium.
folium.Map.add_ee_layer = add_ee_layer
my_map = folium.Map(location=[lat, lon], zoom_start=3)
# Set visualization parameters.
vis_params = {
"bands": ["b0"],
"min": 0.01,
"max": 1,
"opacity": 1,
"palette": ["white", "#464646"],
# Add the sand content data to the map object.
my_map.add_ee_layer(sand, vis_params, "Sand Content")
# Add a marker at the location of interest.
folium.Marker([lat, lon], popup="point of interest").add_to(my_map)
# Add a layer control panel to the map.
# Display the map.
Now, a function is defined to get soil properties at a given location. The following function returns a dictionary indicating the value of the parameter of interest for each standard depth (in
centimeter). This function uses the ee.Image.sample method to evaluate the ee.Image properties on the region of interest. The result is then transferred client-side using the ee.Image.getInfo method.
In the example below, we are asking for the sand content.
def local_profile(dataset, poi, buffer):
# Get properties at the location of interest and transfer to client-side.
prop = dataset.sample(poi, buffer).select(olm_bands).getInfo()
# Selection of the features/properties of interest.
profile = prop["features"][0]["properties"]
# Re-shaping of the dict.
profile = {key: round(val, 3) for key, val in profile.items()}
return profile
# Apply the function to get the sand profile.
profile_sand = local_profile(sand, poi, scale)
# Print the result.
print("Sand content profile at the location of interest:\n", profile_sand)
Sand content profile at the location of interest:
{'b0': 0.36, 'b10': 0.35, 'b100': 0.37, 'b200': 0.39, 'b30': 0.34, 'b60': 0.35}
We now apply the function to plot the profile of the soil regarding sand and clay and organic carbon content at the location of interest:
# Clay and organic content profiles.
profile_clay = local_profile(clay, poi, scale)
profile_orgc = local_profile(orgc, poi, scale)
# Data visualization in the form of a bar plot.
fig, ax = plt.subplots(figsize=(15, 6))
# Definition of label locations.
x = np.arange(len(olm_bands))
# Definition of the bar width.
width = 0.25
# Bar plot representing the sand content profile.
rect1 = ax.bar(
x - width,
[round(100 * profile_sand[b], 2) for b in olm_bands],
# Bar plot representing the clay content profile.
rect2 = ax.bar(
[round(100 * profile_clay[b], 2) for b in olm_bands],
# Bar plot representing the organic carbon content profile.
rect3 = ax.bar(
x + width,
[round(100 * profile_orgc[b], 2) for b in olm_bands],
label="Organic Carbon",
# Definition of a function to attach a label to each bar.
def autolabel_soil_prop(rects):
"""Attach a text label above each bar in *rects*, displaying its height."""
for rect in rects:
height = rect.get_height()
"{}".format(height) + "%",
xy=(rect.get_x() + rect.get_width() / 2, height),
xytext=(0, 3), # 3 points vertical offset.
textcoords="offset points",
# Application of the function to each barplot.
# Title of the plot.
ax.set_title("Properties of the soil at different depths (mass content)", fontsize=14)
# Properties of x/y labels and ticks.
x_labels = [str(d) + " cm" for d in olm_depths]
ax.set_xticklabels(x_labels, rotation=45, fontsize=10)
# Shrink current axis's height by 10% on the bottom.
box = ax.get_position()
ax.set_position([box.x0, box.y0 + box.height * 0.1, box.width, box.height * 0.9])
# Add a legend below current axis.
loc="upper center", bbox_to_anchor=(0.5, -0.15), fancybox=True, shadow=True, ncol=3
Expression to calculate hydraulic properties
Now that soil properties are described, the water content at the field capacity and at the wilting point can be calculated according to the equation defined at the beginning of this section. Please
note that in the equation of Saxton & Rawls (2006), the wilting point and field capacity are calculated using the Organic Matter content (\(OM\)) and not the Organic Carbon content (\(OC\)). In the
following, we convert \(OC\) into \(OM\) using the corrective factor known as the Van Bemmelen factor:
$$ 0M = 1.724 \times OC $$
Several operators are available to perform basic mathematical operations on image bands: add(), subtract(), multiply() and divide(). Here, we multiply the organic content by the Van Bemmelen factor.
It is done using the ee.Image.multiply method on the organic carbon content ee.Image.
# Conversion of organic carbon content into organic matter content.
orgm = orgc.multiply(1.724)
# Organic matter content profile.
profile_orgm = local_profile(orgm, poi, scale)
print("Organic Matter content profile at the location of interest:\n", profile_orgm)
Organic Matter content profile at the location of interest:
{'b0': 0.043, 'b10': 0.043, 'b100': 0.009, 'b200': 0.009, 'b30': 0.017, 'b60': 0.009}
When the mathematical operation to apply to the ee.Image becomes too complex, the ee.Image.expression is a good alternative. We use it in the following code block since the calculation of wilting
point and field capacity relies on multiple parameters and images. This method takes two arguments:
• a string formalizing the arithmetic expression we want to evaluate,
• a dict associating images to each parameter of the arithmetic expression.
The mathematical expression is applied as follows to determine wilting point and field capacity:
# Initialization of two constant images for wilting point and field capacity.
wilting_point = ee.Image(0)
field_capacity = ee.Image(0)
# Calculation for each standard depth using a loop.
for key in olm_bands:
# Getting sand, clay and organic matter at the appropriate depth.
si = sand.select(key)
ci = clay.select(key)
oi = orgm.select(key)
# Calculation of the wilting point.
# The theta_1500t parameter is needed for the given depth.
theta_1500ti = (
"-0.024 * S + 0.487 * C + 0.006 * OM + 0.005 * (S * OM)\
- 0.013 * (C * OM) + 0.068 * (S * C) + 0.031",
"S": si,
"C": ci,
"OM": oi,
# Final expression for the wilting point.
wpi = theta_1500ti.expression(
"T1500ti + ( 0.14 * T1500ti - 0.002)", {"T1500ti": theta_1500ti}
# Add as a new band of the global wilting point ee.Image.
# Do not forget to cast the type with float().
wilting_point = wilting_point.addBands(wpi.rename(key).float())
# Same process for the calculation of the field capacity.
# The parameter theta_33t is needed for the given depth.
theta_33ti = (
"-0.251 * S + 0.195 * C + 0.011 * OM +\
0.006 * (S * OM) - 0.027 * (C * OM)+\
0.452 * (S * C) + 0.299",
"S": si,
"C": ci,
"OM": oi,
# Final expression for the field capacity of the soil.
fci = theta_33ti.expression(
"T33ti + (1.283 * T33ti * T33ti - 0.374 * T33ti - 0.015)",
{"T33ti": theta_33ti.select("T33ti")},
# Add a new band of the global field capacity ee.Image.
field_capacity = field_capacity.addBands(fci.rename(key).float())
Let's see the result around our location of interest:
profile_wp = local_profile(wilting_point, poi, scale)
profile_fc = local_profile(field_capacity, poi, scale)
print("Wilting point profile:\n", profile_wp)
print("Field capacity profile:\n", profile_fc)
Wilting point profile:
{'b0': 0.152, 'b10': 0.158, 'b100': 0.175, 'b200': 0.169, 'b30': 0.175, 'b60': 0.181}
Field capacity profile:
{'b0': 0.271, 'b10': 0.278, 'b100': 0.289, 'b200': 0.28, 'b30': 0.294, 'b60': 0.297}
The result is displayed using barplots as follows:
fig, ax = plt.subplots(figsize=(15, 6))
# Definition of the label locations.
x = np.arange(len(olm_bands))
# Width of the bar of the barplot.
width = 0.25
# Barplot associated with the water content at the wilting point.
rect1 = ax.bar(
x - width / 2,
[round(profile_wp[b] * 100, 2) for b in olm_bands],
label="Water content at wilting point",
# Barplot associated with the water content at the field capacity.
rect2 = ax.bar(
x + width / 2,
[round(profile_fc[b] * 100, 2) for b in olm_bands],
label="Water content at field capacity",
# Add Labels on top of bars.
# Title of the plot.
ax.set_title("Hydraulic properties of the soil at different depths", fontsize=14)
# Properties of x/y labels and ticks.
x_labels = [str(d) + " cm" for d in olm_depths]
ax.set_xticklabels(x_labels, rotation=45, fontsize=10)
# Shrink current axis's height by 10% on the bottom.
box = ax.get_position()
ax.set_position([box.x0, box.y0 + box.height * 0.1, box.width, box.height * 0.9])
# Put a legend below current axis.
loc="upper center", bbox_to_anchor=(0.5, -0.15), fancybox=True, shadow=True, ncol=2
Getting meteorological datasets
Datasets exploration
The meteorological data used in our implementation of the TM procedure relies on the following datasets:
Both datasets are imported as follows, specifying the bands of interest using .select() and the period of interest using .filterDate().
# Import precipitation.
pr = (
.filterDate(i_date, f_date)
# Import potential evaporation PET and its quality indicator ET_QC.
pet = (
.select(["PET", "ET_QC"])
.filterDate(i_date, f_date)
/tmpfs/src/tf_docs_env/lib/python3.9/site-packages/ee/deprecation.py:207: DeprecationWarning:
Attention required for MODIS/006/MOD16A2! You are using a deprecated asset.
To ensure continued functionality, please update it.
Learn more: https://developers.google.com/earth-engine/datasets/catalog/MODIS_006_MOD16A2
warnings.warn(warning, category=DeprecationWarning)
Now we can have a closer look around our location of interest. To evaluate the properties of an ee.ImageCollection, the ee.ImageCollection.getRegion method is used and combined with
ee.ImageCollection.getInfo method for a client-side visualization.
# Evaluate local precipitation conditions.
local_pr = pr.getRegion(poi, scale).getInfo()
[['id', 'longitude', 'latitude', 'time', 'precipitation'],
['20150101', 5.142855001584261, 45.77365530231022, 1420070400000, 0],
['20150102', 5.142855001584261, 45.77365530231022, 1420156800000, 0],
['20150103', 5.142855001584261, 45.77365530231022, 1420243200000, 0],
['20150104', 5.142855001584261, 45.77365530231022, 1420329600000, 0]]
We now establish a procedure to get meteorological data around a given location in form of a pandas.DataFrame:
def ee_array_to_df(arr, list_of_bands):
"""Transforms client-side ee.Image.getRegion array to pandas.DataFrame."""
df = pd.DataFrame(arr)
# Rearrange the header.
headers = df.iloc[0]
df = pd.DataFrame(df.values[1:], columns=headers)
# Convert the data to numeric values.
for band in list_of_bands:
df[band] = pd.to_numeric(df[band], errors="coerce")
# Convert the time field into a datetime.
df["datetime"] = pd.to_datetime(df["time"], unit="ms")
# Keep the columns of interest.
df = df[["time", "datetime", *list_of_bands]]
# The datetime column is defined as index.
df = df.set_index("datetime")
return df
We apply the function and see the head of the resulting pandas.DataFrame:
pr_df = ee_array_to_df(local_pr, ["precipitation"])
We do the same for potential evaporation:
# Evaluate local potential evapotranspiration.
local_pet = pet.getRegion(poi, scale).getInfo()
# Transform the result into a pandas dataframe.
pet_df = ee_array_to_df(local_pet, ["PET", "ET_QC"])
Looking at both pandas.DataFrame shows the following points:
• the time resolution between both datasets is not the same,
• for some reasons, potential evapotranspiration cannot be calculated at some dates. It corresponds to lines where the quality indicator ET_QC is higher than 1.
Both issues must be handled before implementing the iterative process: we want to work on a similar timeline with potential evapotranspiration and precipitation, and we want to avoid missing values.
Resampling the time resolution of an ee.ImageCollection
To address these issues (homogeneous time index and missing values), we make a sum resampling of both datasets by month. When PET cannot be calculated, the monthly averaged value is considered. The
key steps and functions used to resample are described below:
• A new date index is defined as a sequence using the ee.List.sequence method.
• A function representing the resampling operation is defined. This function consists of grouping images of the desired time interval and calculating the sum. The sum is calculated by taking the
mean between available images and multiplying it by the duration of the interval.
• The user-supplied function is then mapped over the new time index using .map().
Finally, the resampling procedure reads as follows:
def sum_resampler(coll, freq, unit, scale_factor, band_name):
This function aims to resample the time scale of an ee.ImageCollection.
The function returns an ee.ImageCollection with the averaged sum of the
band on the selected frequency.
coll: (ee.ImageCollection) only one band can be handled
freq: (int) corresponds to the resampling frequency
unit: (str) corresponds to the resampling time unit.
must be 'day', 'month' or 'year'
scale_factor (float): scaling factor used to get our value in the good unit
band_name (str) name of the output band
# Define initial and final dates of the collection.
firstdate = ee.Date(
coll.sort("system:time_start", True).first().get("system:time_start")
lastdate = ee.Date(
coll.sort("system:time_start", False).first().get("system:time_start")
# Calculate the time difference between both dates.
# https://developers.google.com/earth-engine/apidocs/ee-date-difference
diff_dates = lastdate.difference(firstdate, unit)
# Define a new time index (for output).
new_index = ee.List.sequence(0, ee.Number(diff_dates), freq)
# Define the function that will be applied to our new time index.
def apply_resampling(date_index):
# Define the starting date to take into account.
startdate = firstdate.advance(ee.Number(date_index), unit)
# Define the ending date to take into account according
# to the desired frequency.
enddate = firstdate.advance(ee.Number(date_index).add(freq), unit)
# Calculate the number of days between starting and ending days.
diff_days = enddate.difference(startdate, "day")
# Calculate the composite image.
image = (
coll.filterDate(startdate, enddate)
# Return the final image with the appropriate time index.
return image.set("system:time_start", startdate.millis())
# Map the function to the new time index.
res = new_index.map(apply_resampling)
# Transform the result into an ee.ImageCollection.
res = ee.ImageCollection(res)
return res
The precipitation dataset is now resampled by month as follows:
• the collection to resample is defined as pr,
• we want a collection on a monthly basis then freq = 1 and unit = "month",
• there is no correction factor to apply according to the dataset description then scale_factor = 1,
• "pr" is the name of the output band.
# Apply the resampling function to the precipitation dataset.
pr_m = sum_resampler(pr, 1, "month", 1, "pr")
# Evaluate the result at the location of interest.
pprint.pprint(pr_m.getRegion(poi, scale).getInfo()[:5])
[['id', 'longitude', 'latitude', 'time', 'pr'],
['0', 5.142855001584261, 45.77365530231022, 1420070400000, 83.12794733047485],
['1', 5.142855001584261, 45.77365530231022, 1422748800000, 61.235233306884766],
['2', 5.142855001584261, 45.77365530231022, 1425168000000, 60.416391015052795],
['3', 5.142855001584261, 45.77365530231022, 1427846400000, 48.404173851013184]]
For evapotranspiration, we have to be careful with the unit. The dataset gives us an 8-day sum and a scale factor of 10 is applied. Then, to get a homogeneous unit, we need to rescale by dividing by
8 and 10: \(\frac{1}{10 \times 8} = 0.0125\).
# Apply the resampling function to the PET dataset.
pet_m = sum_resampler(pet.select("PET"), 1, "month", 0.0125, "pet")
# Evaluate the result at the location of interest.
pprint.pprint(pet_m.getRegion(poi, scale).getInfo()[:5])
[['id', 'longitude', 'latitude', 'time', 'pet'],
['0', 5.142855001584261, 45.77365530231022, 1420070400000, 23.637500000000003],
['1', 5.142855001584261, 45.77365530231022, 1422748800000, 39.2],
['2', 5.142855001584261, 45.77365530231022, 1425168000000, 76.434375],
['3', 5.142855001584261, 45.77365530231022, 1427846400000, 137]]
We now combine both ee.ImageCollection objects (pet_m and pr_m) using the ee.ImageCollection.combine method. Note that corresponding images in both ee.ImageCollection objects need to have the same
time index before combining.
# Combine precipitation and evapotranspiration.
meteo = pr_m.combine(pet_m).sort("system:time_start")
# Import meteorological data as an array at the location of interest.
meteo_arr = meteo.getRegion(poi, scale).getInfo()
# Print the result.
[['id', 'longitude', 'latitude', 'time', 'pr', 'pet'],
We evaluate the result on our location of interest:
# Transform the array into a pandas dataframe and sort the index.
meteo_df = ee_array_to_df(meteo_arr, ["pr", "pet"]).sort_index()
# Data visualization
fig, ax = plt.subplots(figsize=(15, 6))
# Barplot associated with precipitations.
meteo_df["pr"].plot(kind="bar", ax=ax, label="precipitation")
# Barplot associated with potential evapotranspiration.
kind="bar", ax=ax, label="potential evapotranspiration", color="orange", alpha=0.5
# Add a legend.
# Add some x/y-labels properties.
ax.set_ylabel("Intensity [mm]")
# Define the date format and shape of x-labels.
x_labels = meteo_df.index.strftime("%m-%Y")
ax.set_xticklabels(x_labels, rotation=90, fontsize=10)
Implementation of the TM procedure
Some additional definitions are needed to formalize the Thornthwaite-Mather procedure. The following definitions are given in accordance with Allen et al. (1998) (the document can be downloaded here
$$ TAW = 1000 \times (\theta_{FC} − \theta_{WP})\times Z{r} $$
• \(TAW\): the total available soil water in the root zone [\(mm\)],
• \(\theta_{FC}\): the water content at the field capacity [\(m^{3} m^{-3}\)],
• \(\theta_{WP}\): the water content at wilting point [\(m^{3} m^{-3}\)],
• \(Z_{r}\): the rooting depth [\(m\)],
Typical values of \(\theta_{FC}\) and \(\theta_{WP}\) for different soil types are given in Table 19 of Allen et al. (1998).
The readily available water (\(RAW\)) is given by \(RAW = p \times TAW\), where \(p\) is the average fraction of \(TAW\) that can be depleted from the root zone before moisture stress (ranging
between 0 to 1). This quantity is also noted \(ST_{FC}\) which is the available water stored at field capacity in the root zone.
Ranges of maximum effective rooting depth \(Z_{r}\), and soil water depletion fraction for no stress \(p\), for common crops are given in the Table 22 of Allen et al. (1998). In addition, a global
effective plant rooting depth dataset is provided by Yang et al. (2016) with a resolution of 0.5° by 0.5° (see the paper here and the dataset here).
According to this global dataset, the effective rooting depth around our region of interest (France) can reasonably assumed to \(Z_{r} = 0.5\). Additionally, the parameter \(p\) is also assumed
constant and equal to and \(p = 0.5\) which is in line with common values described in Table 22 of Allen et al. (1998).
zr = ee.Image(0.5)
p = ee.Image(0.5)
In the following, we also consider an averaged value between reference depths of the water content at wilting point and field capacity:
def olm_prop_mean(olm_image, band_output_name):
This function calculates an averaged value of
soil properties between reference depths.
mean_image = olm_image.expression(
"(b0 + b10 + b30 + b60 + b100 + b200) / 6",
"b0": olm_image.select("b0"),
"b10": olm_image.select("b10"),
"b30": olm_image.select("b30"),
"b60": olm_image.select("b60"),
"b100": olm_image.select("b100"),
"b200": olm_image.select("b200"),
return mean_image
# Apply the function to field capacity and wilting point.
fcm = olm_prop_mean(field_capacity, "fc_mean")
wpm = olm_prop_mean(wilting_point, "wp_mean")
# Calculate the theoretical available water.
taw = (
# Calculate the stored water at the field capacity.
stfc = taw.multiply(p)
The Thornthwaite-Mather procedure used to estimate groundwater recharge is explicitly described by Steenhuis and Van der Molen (1985). This procedure uses monthly sums of potential evaporation,
cumulative precipitation, and the moisture status of the soil which is calculated iteratively. The moisture status of the soils depends on the accumulated potential water loss (\(APWL\)). This
parameter is calculated depending on whether the potential evaporation is greater than or less than the cumulative precipitation. The procedure reads as follow:
Case 1: potential evapotranspiration is higher than precipitation.
In that case, \(PET>P\) and \(APWL_{m}\) is incremented as follows: \(APWL_{m} = APWL_{m - 1} + (PET_{m} - P_{m})\) where:
• \(APWL_{m}\) (respectively \(APWL_{m - 1}\)) represents the accumulated potential water loss for the month \(m\) (respectively at the previous month \(m - 1\))
• \(PET_{m}\) the cumulative potential evapotranspiration at month \(m\),
• \(P_{m}\) the cumulative precipitation at month \(m\),
and the relationship between \(APWL\) and the amount of water stored in the root zone for the month \(m\) is expressed as: \(ST_{m} = ST_{FC} \times [\textrm{exp}(-APWL_{m}/ST_{FC})]\) where \(ST_{m}
\) is the available water stored in the root zone for the month \(m\).
Case 2: potential evapotranspiration is lower than precipitation.
In that case, \(PET<P\) and \(ST_{m}\) is incremented as follows: \(ST_{m} = ST_{m-1} + (P_{m} - PET_{m})\).
Case 2.1: the storage \(ST_{m}\) is higher than the water stored at the field capacity.
If \(ST_{m} > ST_{FC}\) the recharge is calculated as: $$R_{m} = ST_{m} - ST_{FC} + P_{m} - PET_{m}$$
In addition, the water stored at the end of the month \(m\) becomes equal to \(ST_{FC}\) and \(APWL_{m}\) is set equal to zero.
Case 2.2: the storage \(ST_{m}\) is less than or equal to the water stored at the field capacity.
If \(ST_{m} <= ST_{FC}\), \(APWL_{m}\) is implemented as follows: \(APWL_{m} = - ST_{FC} \times \textrm{ln}(ST_{m}/ST_{FC})\), and no percolation occurs.
The initial time of the calculation is defined according to the first date of the meteorological collection:
# Define the initial time (time0) according to the start of the collection.
time0 = meteo.first().get("system:time_start")
Then, we initialize the calculation with an ee.Image where all bands associated to the hydric state of the soil are set equal to ee.Image(0), except for the initial storage which is considered to be
equal to the water content at field capacity, meaning that \(ST_{0} = ST_{FC}\).
# Initialize all bands describing the hydric state of the soil.
# Do not forget to cast the type of the data with a .float().
# Initial recharge.
initial_rech = ee.Image(0).set("system:time_start", time0).select([0], ["rech"]).float()
# Initialization of APWL.
initial_apwl = ee.Image(0).set("system:time_start", time0).select([0], ["apwl"]).float()
# Initialization of ST.
initial_st = stfc.set("system:time_start", time0).select([0], ["st"]).float()
# Initialization of precipitation.
initial_pr = ee.Image(0).set("system:time_start", time0).select([0], ["pr"]).float()
# Initialization of potential evapotranspiration.
initial_pet = ee.Image(0).set("system:time_start", time0).select([0], ["pet"]).float()
We combine all these bands into one ee.Image adding new bands to the first using the ee.Image.addBands method:
initial_image = initial_rech.addBands(
ee.Image([initial_apwl, initial_st, initial_pr, initial_pet])
We also initialize a list in which new images will be added after each iteration. We create this server-side list using the ee.List method.
image_list = ee.List([initial_image])
Iteration over an ee.ImageCollection
The procedure is implemented by means of the ee.ImageCollection.iterate method, which applies a user-supplied function to each element of a collection. For each time step, groundwater recharge is
calculated using the recharge_calculator considering the previous hydric state of the soil and current meteorological conditions.
Of course, considering the TM description, several cases must be distinguished to calculate groundwater recharge. The distinction is made by the definition of binary layers with different logical
operations. It allows specific calculations to be applied in areas where a given condition is true using the ee.Image.where method.
The function we apply to each element of the meteorological dataset to calculate groundwater recharge is defined as follows.
def recharge_calculator(image, image_list):
Contains operations made at each iteration.
# Determine the date of the current ee.Image of the collection.
localdate = image.date().millis()
# Import previous image stored in the list.
prev_im = ee.Image(ee.List(image_list).get(-1))
# Import previous APWL and ST.
prev_apwl = prev_im.select("apwl")
prev_st = prev_im.select("st")
# Import current precipitation and evapotranspiration.
pr_im = image.select("pr")
pet_im = image.select("pet")
# Initialize the new bands associated with recharge, apwl and st.
# DO NOT FORGET TO CAST THE TYPE WITH .float().
new_rech = (
.set("system:time_start", localdate)
.select([0], ["rech"])
new_apwl = (
.set("system:time_start", localdate)
.select([0], ["apwl"])
new_st = (
prev_st.set("system:time_start", localdate).select([0], ["st"]).float()
# Calculate bands depending on the situation using binary layers with
# logical operations.
# CASE 1.
# Define zone1: the area where PET > P.
zone1 = pet_im.gt(pr_im)
# Calculation of APWL in zone 1.
zone1_apwl = prev_apwl.add(pet_im.subtract(pr_im)).rename("apwl")
# Implementation of zone 1 values for APWL.
new_apwl = new_apwl.where(zone1, zone1_apwl)
# Calculate ST in zone 1.
zone1_st = stfc.multiply(
# Implement ST in zone 1.
new_st = new_st.where(zone1, zone1_st)
# CASE 2.
# Define zone2: the area where PET <= P.
zone2 = pet_im.lte(pr_im)
# Calculate ST in zone 2.
zone2_st = prev_st.add(pr_im).subtract(pet_im).rename("st")
# Implement ST in zone 2.
new_st = new_st.where(zone2, zone2_st)
# CASE 2.1.
# Define zone21: the area where PET <= P and ST >= STfc.
zone21 = zone2.And(zone2_st.gte(stfc))
# Calculate recharge in zone 21.
zone21_re = zone2_st.subtract(stfc).rename("rech")
# Implement recharge in zone 21.
new_rech = new_rech.where(zone21, zone21_re)
# Implement ST in zone 21.
new_st = new_st.where(zone21, stfc)
# CASE 2.2.
# Define zone 22: the area where PET <= P and ST < STfc.
zone22 = zone2.And(zone2_st.lt(stfc))
# Calculate APWL in zone 22.
zone22_apwl = (
# Implement APWL in zone 22.
new_apwl = new_apwl.where(zone22, zone22_apwl)
# Create a mask around area where recharge can effectively be calculated.
# Where we have have PET, P, FCm, WPm (except urban areas, etc.).
mask = pet_im.gte(0).And(pr_im.gte(0)).And(fcm.gte(0)).And(wpm.gte(0))
# Apply the mask.
new_rech = new_rech.updateMask(mask)
# Add all Bands to our ee.Image.
new_image = new_rech.addBands(ee.Image([new_apwl, new_st, pr_im, pet_im]))
# Add the new ee.Image to the ee.List.
return ee.List(image_list).add(new_image)
The TM procedure can now be applied to the meteorological ee.ImageCollection:
# Iterate the user-supplied function to the meteo collection.
rech_list = meteo.iterate(recharge_calculator, image_list)
# Remove the initial image from our list.
rech_list = ee.List(rech_list).remove(initial_image)
# Transform the list into an ee.ImageCollection.
rech_coll = ee.ImageCollection(rech_list)
Let's have a look at the result around the location of interest:
arr = rech_coll.getRegion(poi, scale).getInfo()
rdf = ee_array_to_df(arr, ["pr", "pet", "apwl", "st", "rech"]).sort_index()
The result can be displayed in the form of a barplot as follows:
# Data visualization in the form of barplots.
fig, ax = plt.subplots(figsize=(15, 6))
# Barplot associated with precipitation.
rdf["pr"].plot(kind="bar", ax=ax, label="precipitation", alpha=0.5)
# Barplot associated with potential evapotranspiration.
kind="bar", ax=ax, label="potential evapotranspiration", color="orange", alpha=0.2
# Barplot associated with groundwater recharge
rdf["rech"].plot(kind="bar", ax=ax, label="recharge", color="green", alpha=1)
# Add a legend.
# Define x/y-labels properties.
ax.set_ylabel("Intensity [mm]")
# Define the date format and shape of x-labels.
x_labels = rdf.index.strftime("%m-%Y")
ax.set_xticklabels(x_labels, rotation=90, fontsize=10)
The result shows the distribution of precipitation, potential evapotranspiration, and groundwater recharge along the year. It shows that in our area of interest, groundwater recharge generally occurs
from October to March. Even though a significant amount of precipitation occurs from April to September, evapotranspiration largely dominates because of high temperatures and sun exposure during
these months. The result is that no percolation into aquifers occurs during this period.
Now the annual average recharge over the period of interest can be calculated. To do that, we resample the DataFrame we've just created:
# Resample the pandas dataframe on a yearly basis making the sum by year.
rdfy = rdf.resample("Y").sum()
# Calculate the mean value.
mean_recharge = rdfy["rech"].mean()
# Print the result.
"The mean annual recharge at our point of interest is", int(mean_recharge), "mm/an"
The mean annual recharge at our point of interest is 154 mm/an
/tmpfs/tmp/ipykernel_4144/1140476268.py:2: FutureWarning: 'Y' is deprecated and will be removed in a future version, please use 'YE' instead.
rdfy = rdf.resample("Y").sum()
Groundwater recharge comparison between multiple places
We now may want to get local information about groundwater recharge and/or map this variable on an area of interest.
Let's define a function to get the local information based on the ee.ImageCollection we've just built:
def get_local_recharge(i_date, f_date, lon, lat, scale):
Returns a pandas df describing the cumulative groundwater
recharge by month
# Define the location of interest with a point.
poi = ee.Geometry.Point(lon, lat)
# Evaluate the recharge around the location of interest.
rarr = rech_coll.filterDate(i_date, f_date).getRegion(poi, scale).getInfo()
# Transform the result into a pandas dataframe.
rdf = ee_array_to_df(rarr, ["pr", "pet", "apwl", "st", "rech"]).sort_index()
return rdf
We now use this function on a second point of interest located near the city of Montpellier (France). This city is located in the south of France, and precipitation and groundwater recharge are
expected to be much lower than in the previous case.
# Define the second location of interest by longitude/latitude.
lon2 = 4.137152
lat2 = 43.626945
# Calculate the local recharge condition at this location.
rdf2 = get_local_recharge(i_date, f_date, lon2, lat2, scale)
# Resample the resulting pandas dataframe on a yearly basis (sum by year).
rdf2y = rdf2.resample("Y").sum()
/tmpfs/tmp/ipykernel_4144/891754906.py:9: FutureWarning: 'Y' is deprecated and will be removed in a future version, please use 'YE' instead.
rdf2y = rdf2.resample("Y").sum()
# Data Visualization
fig, ax = plt.subplots(figsize=(15, 6))
# Define the x-label locations.
x = np.arange(len(rdfy))
# Define the bar width.
width = 0.25
# Bar plot associated to groundwater recharge at the 1st location of interest.
rect1 = ax.bar(
x - width / 2, rdfy.rech, width, label="Lyon (France)", color="blue", alpha=0.5
# Bar plot associated to groundwater recharge at the 2nd location of interest.
rect2 = ax.bar(
x + width / 2,
label="Montpellier (France)",
# Define a function to attach a label to each bar.
def autolabel_recharge(rects):
"""Attach a text label above each bar in *rects*, displaying its height."""
for rect in rects:
height = rect.get_height()
"{}".format(int(height)) + " mm",
xy=(rect.get_x() + rect.get_width() / 2, height),
xytext=(0, 3), # 3 points vertical offset
textcoords="offset points",
# Calculate the averaged annual recharge at both locations of interest.
place1mean = int(rdfy["rech"].mean())
place2mean = int(rdf2y["rech"].mean())
# Add an horizontal line associated with averaged annual values (location 1).
xmin=min(x) - width,
xmax=max(x) + width,
label="average " + str(place1mean) + " mm/y",
# Add an horizontal line associated with averaged annual values (location 2).
xmin=min(x) - width,
xmax=max(x) + width,
label="average " + str(place2mean) + " mm/y",
# Add a title.
ax.set_title("Groundwater recharge comparison between two places", fontsize=12)
# Define some x/y-axis properties.
x_labels = rdfy.index.year.tolist()
ax.set_xticklabels(x_labels, rotation=45, fontsize=10)
# Shrink current axis's height by 10% on the bottom.
box = ax.get_position()
ax.set_position([box.x0, box.y0 + box.height * 0.1, box.width, box.height * 0.9])
# Add a legend below current axis.
loc="upper center", bbox_to_anchor=(0.5, -0.15), fancybox=True, shadow=True, ncol=2
The result shows that the annual recharge in Lyon is almost twice as high as in the area of Montpellier. The result also shows a great variability of the annual recharge ranging from 98 mm/y to 258
mm/y in Lyon and from 16 mm/y to 147 mm/y in Montpellier.
Groundwater recharge map of France
To get a map of groundwater recharge around our region of interest, let's create a mean composite ee.Image based on our resulting ee.ImageCollection.
# Calculate the averaged annual recharge.
annual_rech = rech_coll.select("rech").mean().multiply(12)
# Calculate the average annual precipitation.
annual_pr = rech_coll.select("pr").mean().multiply(12)
# Get a feature collection of administrative boundaries.
countries = ee.FeatureCollection("FAO/GAUL/2015/level0").select("ADM0_NAME")
# Filter the feature collection to subset France.
france = countries.filter(ee.Filter.eq("ADM0_NAME", "France"))
# Clip the composite ee.Images around the region of interest.
rech_france = annual_rech.clip(france)
pr_france = annual_pr.clip(france)
And finally the map can be drawn.
# Create a folium map.
my_map = folium.Map(location=[lat, lon], zoom_start=6, zoom_control=False)
# Set visualization parameters for recharge.
rech_vis_params = {
"bands": "rech",
"min": 0,
"max": 500,
"opacity": 1,
"palette": ["red", "orange", "yellow", "green", "blue", "purple"],
# Set visualization parameters for precipitation.
pr_vis_params = {
"bands": "pr",
"min": 500,
"max": 1500,
"opacity": 1,
"palette": ["white", "blue"],
# Define a recharge colormap.
rech_colormap = cm.LinearColormap(
# Define a precipitation colormap.
pr_colormap = cm.LinearColormap(
# Caption of the recharge colormap.
rech_colormap.caption = "Average annual recharge rate (mm/year)"
# Caption of the precipitation colormap.
pr_colormap.caption = "Average annual precipitation rate (mm/year)"
# Add the precipitation composite to the map object.
my_map.add_ee_layer(pr_france, pr_vis_params, "Precipitation")
# Add the recharge composite to the map object.
my_map.add_ee_layer(rech_france, rech_vis_params, "Recharge")
# Add a marker at both locations of interest.
folium.Marker([lat, lon], popup="Area of Lyon").add_to(my_map)
folium.Marker([lat2, lon2], popup="Area of Montpellier").add_to(my_map)
# Add the colormaps to the map.
# Add a layer control panel to the map.
# Display the map.
The resulting map represents the averaged annual precipitation and groundwater recharge rate in France over the period 2015-2019. Groundwater recharge is represented between 0 (red) to 500 mm/year
(purple). Precipitation is represented between 500 (white) to 1500 mm/year (blue).
Please note the following points:
• The precipitation dataset used in this tutorial is not available over +/- 50 deg latitudes. That is why (1) the northern part of France is not covered by the precipitation/recharge, and (2) some
anomalies/discontinuities can be observed near the boundary of the dataset.
• The procedure described in this tutorial does not consider the terrain slope which has a strong influence on runoff, especially in areas with significant relief. In these areas, groundwater
recharge is overestimated because runoff is not considered.
• Snow cover has a more complex contribution to groundwater recharge than precipitation. This process is not taken into account in this tutorial. Hence, groundwater recharge is overestimated in
areas where snow cover period is significant.
• The recharge calculation can be highly influenced by missing precipitation or potential evapotranspiration values. This issue is partially addressed by the dataset resampling procedure, but
missing value can still exist and this point must be checked before considering that the calculated recharge is representative and reliable.
Allen RG, Pereira LS, Raes D, Smith M (1998). Crop evapotranspiration: guidelines for computing crop water requirements. Irrigation and Drainage Paper 56, FAO, Rome.
Saxton, K. E., & Rawls, W. J. (2006). Soil water characteristic estimates by texture and organic matter for hydrologic solutions. Soil science society of America Journal, 70(5), 1569-1578.
Steenhuis, T. S., & Van der Molen, W. H. (1986). The Thornthwaite-Mather procedure as a simple engineering method to predict recharge. Journal of Hydrology, 84(3-4), 221-229.
Thornthwaite, C. W., & Mather, J. R. (1957). Instructions and tables for computing potential evapotranspiration and the water balance. Publ. Climatol., 10(3).
Yang, Y., Donohue, R. J., & McVicar, T. R. (2016). Global estimation of effective plant rooting depth: Implications for hydrological modeling. Water Resources Research, 52(10), 8260-8276.
Thanks to Susanne Benz for precious advice. | {"url":"https://developers.google.com/earth-engine/tutorials/community/groundwater-recharge-estimation","timestamp":"2024-11-15T05:00:06Z","content_type":"text/html","content_length":"162037","record_id":"<urn:uuid:092abcad-dc4e-4a49-b840-cacf882341bb>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00844.warc.gz"} |
Microsoft Dynamics 365 Business Central – Sorting Algorithms: Bubble, Merge, and Quick
Jul 24 2023
Microsoft Dynamics 365 Business Central – Sorting Algorithms: Bubble, Merge, and Quick
The other day I felt my computer files were a little disorganized, so I went through some old files to do some hard-core cleanup. I wanted to purge and archive old programs, documents, and repos.
While reviewing the files, I found some old C# code (from the days of when I tried to stay sharp through CodeWars), I wrote to test different sorting algorithms. Sorting algorithms are used to
organize data in a specific order, and in my code, there was the good ole Bubble Sort, the Merge Sort, and their friend, the Quick Sort.
There are several options when sorting data, and depending on the data set, some are better than others. Once I came across the sorting code, For no other reason than wanting to, I strayed from my
task and converted them to AL for Microsoft Dynamics 365 Business Central.
Bubble Sort
One such sorting algorithm is called Bubble Sort. It’s a basic comparison-based method that gets its name from how smaller or larger elements “bubble” to the top of the list. With a Bubble Sort,
start at the beginning of the list and compare the first element to the second. If the first element is greater than the second, swap their positions. If not, leave them as they are. Move one
position to the right and compare the second and third elements. Continue this process for each pair of adjacent numbers until the end of the list is reached. It’s important to note that Bubble Sort
is not very efficient, especially for large lists.
procedure BubbleSort(List: List of [Integer])
i, j : Integer;
ListItem: Integer;
for i := 1 to List.Count do begin
for j := 1 to List.Count - 1 do begin
if List.Get(i) < List.Get(j) then begin
ListItem := List.Get(i);
List.Set(i, List.Get(j));
List.Set(j, ListItem);
Merge Sort
Another option, if you need to sort data, is Merge Sort. This type of comparison-based sorting algorithm is effective for most practical data sets, especially on larger lists. The strategy is to
divide and conquer. The data set is split in half continuously until each set contains only one element, which is considered a sorted list. Then, the adjacent lists are merged back together while
maintaining sorted order. This is done by comparing the first elements of each list and adding the smaller one to the new list, then moving on to the next element in the list from which the element
was taken. The process is repeated until all the elements have been merged back into a single sorted list. Compared to Bubble Sort, Merge Sort is more efficient.
procedure MergeSort(List: List of [Integer])
MergeSort(List, 1, List.Count);
local procedure Merge(List: List of [Integer]; Left: Integer; Middle: Integer; Right: Integer)
i, j, k : Integer;
LeftList: List of [Integer];
RightList: List of [Integer];
for i := Left to Middle do begin
for i := Middle + 1 to Right do begin
i := 1;
j := 1;
k := Left;
while (i <= LeftList.Count) and (j <= RightList.Count) do begin
if LeftList.Get(i) <= RightList.Get(j) then begin
List.Set(k, LeftList.Get(i));
i := i + 1;
end else begin
List.Set(k, RightList.Get(j));
j := j + 1;
k := k + 1;
while i <= LeftList.Count do begin
List.Set(k, LeftList.Get(i));
i := i + 1;
k := k + 1;
while j <= RightList.Count do begin
List.Set(k, RightList.Get(j));
j := j + 1;
k := k + 1;
local procedure MergeSort(List: List of [Integer]; Left: Integer; Right: Integer)
Middle: Integer;
if Left < Right then begin
Middle := (Left + Right) div 2;
MergeSort(List, Left, Middle);
MergeSort(List, Middle + 1, Right);
Merge(List, Left, Middle, Right);
Quick Sort
Quicksort is another efficient and commonly used sorting algorithm that uses a divide-and-conquer approach. It’s widely used due to its efficiency and ease of implementation. The first step in the
Quicksort algorithm is to choose a pivot. The pivot can be any element from the array, but one commonly chooses the first, last, or middle elements. The role of the pivot is to assist in splitting
the array. Next, rearrange the array elements so that all elements less than or equal to the pivot come before (to the left of) the pivot and all elements greater than the pivot come after (to the
right of) the pivot. At this point, the pivot is in its final position in the sorted array.
procedure QuickSort(List: List of [Integer])
i: Integer;
QuickSort(List, 1, List.Count);
local procedure Partition(List: List of [Integer]; Left: Integer; Right: Integer): Integer
i, j : Integer;
Pivot: Integer;
Pivot := List.Get(Right);
i := Left - 1;
for j := Left to Right - 1 do begin
if List.Get(j) <= Pivot then begin
i := i + 1;
Swap(List, i, j);
Swap(List, i + 1, Right);
exit(i + 1);
local procedure QuickSort(List: List of [Integer]; Left: Integer; Right: Integer)
Pivot: Integer;
if Left < Right then begin
Pivot := Partition(List, Left, Right);
QuickSort(List, Left, Pivot - 1);
QuickSort(List, Pivot + 1, Right);
local procedure Swap(List: List of [Integer]; i: Integer; j: Integer)
ListItem: Integer;
ListItem := List.Get(i);
List.Set(i, List.Get(j));
List.Set(j, ListItem);
Note: The code and information discussed in this article are for informational and demonstration purposes only. This content was created referencing Microsoft Dynamics 365 Business Central 2023 Wave
1 online.
1 ping
1. […] Source : DvlprLife.com Read more… […] | {"url":"https://www.dvlprlife.com/2023/07/microsoft-dynamics-365-business-central-sorting-algorithms-bubble-merge-and-quick/","timestamp":"2024-11-09T09:28:17Z","content_type":"text/html","content_length":"69056","record_id":"<urn:uuid:7e74efa7-015e-40ba-987f-e68d2b928c5b>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00741.warc.gz"} |
The Thinking Like a Mathematician Series
In an age when apparently human beings use only ten percent of their brain, the public hasn't been spared of the Think Like a Genius syndrome. We've titles like Think like Da Vinci, Think like
Einstein, and the like. Interestingly, but not surprisingly, we've yet to see a Thinking Like a Mathematician Series.
Probably we won't see such self-help series for some time, if the rate of innumeracy among writers and journalists doesn't dip in future. It's unlikely that untenured professional mathematicians
(whose rice bowls depend on the number of published journal papers) would be lured into writing this type of pop math titles. However, non-fiction writers with post-graduate qualifications in science
and mathematics are potential candidates to popularize such math-lite titles to the lay public.
Those who are fluent with abstract ideas would undoubtedly want to read about their mathematical idols of yesteryear – the habits of those highly successful mathematical minds. Monkeying the language
of positive thinking, our chances of becoming a mathematician or mathematics educator would presumably be higher if we started acting or thinking like one. It is the intention and going through the
motions of being or thinking like a mathematician that counts.
Self-help books and circuit speakers tell us that if we want to think like a mathematician, and actually go through the motions of being one, we will become at least an adequate [probably,
third-class] mathematician. We may not become another Archimedes, Gauss or Newton, but we'll be much more of a pseudo-mathematician than someone who doesn't practice the yoga of applied positive
Meanwhile, let's look forward to some of these feel-good math titles, which may give the mathematical brethren some psychological boost to their often-unappreciated vocations.
Chicken Soup for the Mathematician Series
Chicken Soup for the Pure Mathematician
Chicken Soup for the Applied Mathematician
Chicken Soup for the Math Professor
Chicken Soup for the Math Assistant Professor
Chicken Soup for the Math Associate Professor
Chicken Soup for the Math Adjunct Professor
Chicken Soup for the Algebraist
Chicken Soup for the Geometer
Chicken Soup for the Group Theorist
Chicken Soup for the Number Theorist
Chicken Soup for the Probabilist
Chicken Soup for the Statistician
Chicken Soup for the Topologist ...
© Yan Kow Cheong, March 8, 2010
1 comment:
Singapore Math said...
Anyone interested to co-write some "Maths Series for Goondus & Suakus" (the Asian equivalents of dummies, idiots or morons) with me? Looking forward to co-authoring these pop maths titles. | {"url":"https://www.singaporemathplus.com/2010/03/thinking-like-mathematician-series.html","timestamp":"2024-11-14T14:29:03Z","content_type":"application/xhtml+xml","content_length":"80588","record_id":"<urn:uuid:cdfbb189-c60a-47f5-b26c-a458e19e80fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00074.warc.gz"} |
This module contains classes and procedures for computing various statistical quantities related to the GenGamma distribution. More...
This module contains classes and procedures for computing various statistical quantities related to the GenGamma distribution.
Specifically, this module contains routines for computing the following quantities of the GenGamma distribution:
1. the Probability Density Function (PDF)
2. the Cumulative Distribution Function (CDF)
3. the Random Number Generation from the distribution (RNG)
4. the Inverse Cumulative Distribution Function (ICDF) or the Quantile Function
A variable \(X\) is said to be Generalized Gamma (GenGamma) distributed if its PDF with the scale \(0 < \sigma < +\infty\), shape \(0 < \omega < +\infty\), and shape \(0 < \kappa < +\infty\)
parameters is described by the following equation,
$$\large \pi(x | \kappa, \omega, \sigma) = \frac{1}{\sigma \omega \Gamma(\kappa)} ~ \bigg( \frac{x}{\sigma} \bigg)^{\frac{\kappa}{\omega} - 1} \exp\Bigg( -\bigg(\frac{x}{\sigma}\bigg)^{\frac{1}{\
omega}} \Bigg) ~,~ 0 < x < \infty$$
where \(\eta = \frac{1}{\sigma \omega \Gamma(\kappa)}\) is the normalization factor of the PDF.
When \(\sigma = 1\), the GenGamma PDF simplifies to the form,
$$\large \pi(x) = \frac{1}{\omega \Gamma(\kappa)} ~ x^{\frac{\kappa}{\omega} - 1} \exp\Bigg( -x^{\frac{1}{\omega}} \Bigg) ~,~ 0 < x < \infty$$
If \((\sigma, \omega) = (1, 1)\), the GenGamma PDF further simplifies to the form,
$$\large \pi(x) = \frac{1}{\Gamma(\kappa)} ~ x^{\kappa - 1} \exp(-x) ~,~ 0 < x < \infty$$
Setting the shape parameter to \(\kappa = 1\) further simplifies the PDF to the Exponential distribution PDF with the scale parameter \(\sigma = 1\),
$$\large \pi(x) = \exp(x) ~,~ 0 < x < \infty$$
1. The parameter \(\sigma\) determines the scale of the GenGamma PDF.
2. When \(\omega = 1\), the GenGamma PDF reduces to the PDF of the Gamma distribution.
3. When \(\kappa = 1, \omega = 1\), the GenGamma PDF reduces to the PDF of the Exponential distribution.
The CDF of the Generalized Gamma distribution over a strictly-positive support \(x \in (0, +\infty)\) with the three (shape, shape, scale) parameters \((\kappa > 0, \omega > 0, \sigma > 0)\) is
defined by the regularized Lower Incomplete Gamma function as,
\begin{eqnarray} \large \mathrm{CDF}(x | \kappa, \omega, \sigma) & = & P\bigg(\kappa, \big(\frac{x}{\sigma}\big)^{\frac{1}{\omega}} \bigg) \\ & = & \frac{1}{\Gamma(\kappa)} \int_0^{\big(\frac{x}{\
sigma}\big)^{\frac{1}{\omega}}} ~ t^{\kappa - 1}{\mathrm e}^{-t} ~ dt ~, \end{eqnarray}
where \(\Gamma(\kappa)\) represents the Gamma function.
The distribution mean is given by,
$$\large \overline{x} = \frac{\Gamma\left(\kappa + \omega\right)}{\Gamma(\kappa)} \sigma ~.$$
The distribution mode is given by,
$$\large \widehat{x} = \begin{cases} \sigma \left( \kappa - \omega \right)^\omega ~~~ , ~~~ \omega < \kappa ~, onumber \\ 0 ~~~ , ~~~ \kappa \leq \omega ~. \end{cases}$$
The distribution variance is given by,
$$\large \mathrm{VAR}(x) = \sigma^2 \left[ \frac{\Gamma(\kappa + 2\omega)}{\Gamma(\kappa)} - \left( \frac{\Gamma(\kappa + \omega)}{\Gamma(\kappa)} \right)^2 \right] ~.$$
The relationship between the GenExpGamma and GenGamma distributions is similar to that of the Normal and LogNormal distributions.
In other words, a better more consistent naming for the GenExpGamma and GenGamma distributions could have been GenGamma and GenLogGamma distributions, respectively, similar to Normal and
LogNormal distributions.
See also
Stacy, E. W. (1962). A generalization of the gamma distribution. The Annals of mathematical statistics, 33(3), 1187-1192.
Wolfram Research (2010), GenGammaDistribution, Wolfram Language function, https://reference.wolfram.com/language/ref/GenGammaDistribution.html (updated 2016)
Final Remarks ⛓
If you believe this algorithm or its documentation can be improved, we appreciate your contribution and help to edit this page's documentation and source file on GitHub.
For details on the naming abbreviations, see this page.
For details on the naming conventions, see this page.
This software is distributed under the MIT license with additional terms outlined below.
1. If you use any parts or concepts from this library to any extent, please acknowledge the usage by citing the relevant publications of the ParaMonte library.
2. If you regenerate any parts/ideas from this library in a programming environment other than those currently supported by this ParaMonte library (i.e., other than C, C++, Fortran, MATLAB, Python,
R), please also ask the end users to cite this original ParaMonte library.
This software is available to the public under a highly permissive license.
Help us justify its continued development and maintenance by acknowledging its benefit to society, distributing it, and contributing to it.
Amir Shahmoradi, Oct 16, 2009, 11:14 AM, Michigan | {"url":"https://www.cdslab.org/paramonte/fortran/latest/namespacepm__distGenGamma.html","timestamp":"2024-11-12T02:33:26Z","content_type":"application/xhtml+xml","content_length":"22587","record_id":"<urn:uuid:5d370923-a8bf-4666-8dbc-0ef0b1126a18>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00052.warc.gz"} |
Thermodynamic Processes | Types and Equations | ChemTalk
Core Concepts
This article will cover the four thermodynamic processes: Adiabatic, Isochoric, Isothermal, and Isobaric processes along with their Pressure-Volume curves. After this article, you will be able to
describe and understand how a system is able to interact with its surrounding through work and heat.
Topics Covered in Other Articles
Systems and Surroundings
In thermodynamics, a system is the entity under study. Everything else is, therefore, the surrounding of this system. A system and its surrounding can interact by exchanging matter and energy. There
are three types of systems:
• Isolated Systems: are systems that are unable to exchange neither matter nor energy with their surroundings. One example of this is the thermos flask. Because the flask is sealed, no matter is
exchanged. The thermos flask also prevents any form of heat transmission by preventing heat convection, conduction, and radiation. This allows the flask to keep the liquid inside of it hot for an
extended period of time.
• Closed Systems: are systems that can only exchange energy with their surroundings. For example, heating pads and ice packs are closed systems. They allow for exchange of energy, which is how heat
is transferred from and to them, respectively. However, the liquids inside of them do not leak out, hence, no matter is exchanged.
• Open Systems: are systems that can exchange both energy and matter with their surroundings. Such is the case of a boiling pot with no lid atop it. Energy is exchanged since the area surrounding
the pot gets warmer. Moreover, some of the water vaporizes and leaves the pot, this is matter exchange.
Notions to be Familiar with
Thermodynamic processes characterize the transfer of energy (as heat) between and within systems. They help us understand the properties of the system that change or remain constant throughout the
process. But before going into the processes, we must be familiar with two things:
• The First law of thermodynamics: which states that energy is always conserved, even as changes from one form to another. This is illustrated in the equation : to the system, and by the system.
• Pressure-Volume curves: when a gas expands or is compressed, work is done by or on the system, respectively. We represent this by the equation pressure (y-axis) with respect to changes in volume
(x-axis), the area under the resulting curve would be equal to the value of work.
Thermodynamic Processes
We have four processes: Isobaric, Isochoric, Isothermal, and Adiabatic
Isobaric Process
An isobaric process is a process that occurs when pressure is constant. Since the pressure is constant irrespective of the changes in volume, then the
Isochoric Process (Isovolumetric)
Isochoric processes occur under constant volume. Since the volume is fixed,
Substituting this in the equation of the first law of thermodynamics yields
Isothermal Process
In an isothermal process, the temperature is constant. Thus, there is no change in the internal energy of the system, ideal gas equation.
For More Help, Watch Our Interactive Video Explaining Isothermal Processes
Adiabatic Process
Adiabatic processes occur when the system and its surroundings are unable to exchange heat. Therefore, on the system.
Practice Problems
Using the graph below, solve the following problems :
Problem 1
Identify the thermodynamic process depicted in the graph above
Problem 2
Calculate the work done using the information in the graph
Practice Problems Solutions
Problem 1
As we can see in the graph, the pressure stays constant throughout the process. Therefore, it is an isobaric process.
Problem 2
For More Help, Watch Our Interactive Video Explaining Adiabatic Processes and Calculations
And Another Video Explaining Reversible Processes!
Further Reading | {"url":"https://chemistrytalk.org/thermodynamic-processes-types-and-equations/","timestamp":"2024-11-14T01:09:43Z","content_type":"text/html","content_length":"253140","record_id":"<urn:uuid:a061fe85-90b0-4815-a253-846a2f9eff4b>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00152.warc.gz"} |
Kilowatt-hour to Megawatt-hour Converter
Enter Kilowatt-hour
⇅ Switch toMegawatt-hour to Kilowatt-hour Converter
How to use this Kilowatt-hour to Megawatt-hour Converter 🤔
Follow these steps to convert given energy from the units of Kilowatt-hour to the units of Megawatt-hour.
1. Enter the input Kilowatt-hour value in the text field.
2. The calculator converts the given Kilowatt-hour into Megawatt-hour in realtime ⌚ using the conversion formula, and displays under the Megawatt-hour label. You do not need to click any button. If
the input changes, Megawatt-hour value is re-calculated, just like that.
3. You may copy the resulting Megawatt-hour value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Kilowatt-hour to Megawatt-hour?
The formula to convert given energy from Kilowatt-hour to Megawatt-hour is:
Energy[(Megawatt-hour)] = Energy[(Kilowatt-hour)]/1e3
Substitute the given value of energy in kilowatt-hour, i.e., Energy[(Kilowatt-hour)] in the above formula and simplify the right-hand side value. The resulting value is the energy in megawatt-hour,
i.e., Energy[(Megawatt-hour)].
Calculation will be done after you enter a valid input.
Consider that you have an electric car that consumes 5 kilowatt-hours (kWh) of energy for a full charge.
Convert this energy consumption from Kilowatt-hours to Megawatt-hour.
The energy in kilowatt-hour is:
Energy[(Kilowatt-hour)] = 5
The formula to convert energy from kilowatt-hour to megawatt-hour is:
Energy[(Megawatt-hour)] = Energy[(Kilowatt-hour)]/1e3
Substitute given weight Energy[(Kilowatt-hour)] = 5 in the above formula.
Energy[(Megawatt-hour)] = 5/1e3
Energy[(Megawatt-hour)] = 0.005
Final Answer:
Therefore, 5 kWh is equal to 0.005 MWh.
The energy is 0.005 MWh, in megawatt-hour.
Consider that a residential solar panel system generates 2 kilowatt-hours (kWh) of energy in a day.
Convert this energy generation from kilowatt-hours to Megawatt-hour.
The energy in kilowatt-hour is:
Energy[(Kilowatt-hour)] = 2
The formula to convert energy from kilowatt-hour to megawatt-hour is:
Energy[(Megawatt-hour)] = Energy[(Kilowatt-hour)]/1e3
Substitute given weight Energy[(Kilowatt-hour)] = 2 in the above formula.
Energy[(Megawatt-hour)] = 2/1e3
Energy[(Megawatt-hour)] = 0.002
Final Answer:
Therefore, 2 kWh is equal to 0.002 MWh.
The energy is 0.002 MWh, in megawatt-hour.
Kilowatt-hour to Megawatt-hour Conversion Table
The following table gives some of the most used conversions from Kilowatt-hour to Megawatt-hour.
Kilowatt-hour (kWh) Megawatt-hour (MWh)
0.01 kWh 0.00001 MWh
0.1 kWh 0.0001 MWh
1 kWh 0.001 MWh
2 kWh 0.002 MWh
3 kWh 0.003 MWh
4 kWh 0.004 MWh
5 kWh 0.005 MWh
6 kWh 0.006 MWh
7 kWh 0.007 MWh
8 kWh 0.008 MWh
9 kWh 0.009 MWh
10 kWh 0.01 MWh
20 kWh 0.02 MWh
50 kWh 0.05 MWh
100 kWh 0.1 MWh
1000 kWh 1 MWh
A Kilowatt-hour (kWh) is a unit of energy that measures the amount of electrical energy consumed or generated over time. One kilowatt-hour is equivalent to one kilowatt (1,000 watts) of power used or
produced for one hour. This unit is commonly used to quantify energy usage in households, industries, and various devices. For example, if a 1,000-watt appliance runs for one hour, it consumes 1 kWh
of energy. Kilowatt-hours are essential for understanding energy consumption, billing in electric utilities, and managing energy efficiency.
A Megawatt-hour (MWh) is a unit of energy that measures the amount of electrical energy consumed or generated over time. One megawatt-hour is equivalent to one megawatt (1,000,000 watts) of power
used or produced for one hour. This unit is commonly used to quantify large-scale energy usage, such as that of power plants, industrial facilities, or in the context of national and regional energy
consumption. For example, if a power plant operates at 1 megawatt of output for one hour, it produces 1 MWh of energy. Megawatt-hours are crucial for understanding and managing large-scale energy
production and consumption.
Frequently Asked Questions (FAQs)
1. How do I convert kilowatt-hours to megawatt-hours?
To convert kilowatt-hours to megawatt-hours, divide the number of kilowatt-hours by 1,000, since there are 1,000 kilowatt-hours in a megawatt-hour. For example, 5,000 kilowatt-hours divided by 1,000
equals 5 megawatt-hours.
2. What is the formula for converting kilowatt-hours to megawatt-hours?
The formula is: \( \text{megawatt-hours} = \dfrac{\text{kilowatt-hours}}{1,000} \).
3. How many megawatt-hours are in a kilowatt-hour?
There are 0.001 megawatt-hours in 1 kilowatt-hour.
4. Is 1,000 kilowatt-hours equal to 1 megawatt-hour?
Yes, 1,000 kilowatt-hours is equal to 1 megawatt-hour.
5. How do I convert megawatt-hours to kilowatt-hours?
To convert megawatt-hours to kilowatt-hours, multiply the number of megawatt-hours by 1,000. For example, 3 megawatt-hours multiplied by 1,000 equals 3,000 kilowatt-hours.
6. Why do we divide by 1,000 to convert kilowatt-hours to megawatt-hours?
Because there are 1,000 kilowatt-hours in one megawatt-hour, dividing by 1,000 converts kilowatt-hours to megawatt-hours.
7. How many megawatt-hours are in 2,500 kilowatt-hours?
2,500 kilowatt-hours divided by 1,000 equals 2.5 megawatt-hours. | {"url":"https://convertonline.org/unit/?convert=kilowatt_hour-megawatt_hour","timestamp":"2024-11-09T20:00:58Z","content_type":"text/html","content_length":"87746","record_id":"<urn:uuid:9d07e15f-da7b-4e5a-a981-fa0015320c29>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00509.warc.gz"} |
Hi everyone,
If you’ve had trouble accessing the first WeBWorK assignment, take another look – it should be available now. If you have any trouble, please let me know.
Prof. Reitz
WeBWorK is accessible from on and off campus (anywhere you have access to the internet). Your first WeBWorK assignment is due on Tuesday, September 3rd, at midnight, and will cover the material from
the second day of class. Here’s what you have to do:
Assignment. To get started , you must complete the following three steps.
Step 1. Log in to WeBWorK here: http://mathww.citytech.cuny.edu/webwork2/MAT2071-F19-Reitz. I have created Usernames and Passwords for each student registered for my class.
Username. Your username for WeBWorK consists of your first initial plus your last name, all lowercase (for example, John Smith would have username ‘jsmith’).
Password. Your password is your Student ID (EmplID in CUNYFirst).
Step 2. Update your email address if you wish. To do this, select “Password/Email” from the main menu on the left. Use whatever email address you like (I suggest using one that you check often).
Step 3. Complete the first assignment, titled Assignment1-Sec1.2-1.3. Click on an assignment on the main screen to get started.
If you have any trouble – either with logging in, or with completing the assignment, post a comment here or send me an email and I will get back to you.
WeBWorK Tips:
1. Click on a problem to see the details (the list of problems appears in the menu on the left). Enter an answer and hit “Submit Answers”. Don’t worry, if you get it wrong you can try it again.
2. You can work on the problems in any order you wish. You can do some problems now, and come back and do the rest another day (your work will be saved, as long as you submit your answers).
3. If you want to print out a copy of the assignment, click on the assignment name in the main menu on the left, and then click the link in the main screen area that reads “Download a hardcopy of
this homework set.”
For the past several years I have taught this same course in the Fall semester. At the end of each course, I give my students the following assignment:
Imagine that you are invited to speak on the first day of MAT 2071, to give advice to entering students. Write at least three sentences … describing what you would tell them.
To see the assignment and the students’ responses, follow this link for Fall 2018 and this link for Fall 2017.
Your assignment, due at the beginning of class on Tuesday, September 3th, is to:
1. Read through ALL the responses (there are 25 of them altogether).
2. Write a reply to this post (1 paragraph) responding to all of the following:
1. What advice seemed most relevant to you personally? Why? (you can copy/paste a short statement, or put it in your own words)
2. Based on this advice, what changes can you make right now to help you succeed in this course?
Extra Credit. For extra credit, write a response to one of your classmates’ comments. Do you have any advice? Be kind.
Hi everyone,
Your first homework assignment consists of some problems from the book, as well as a WeBWorK assignment – these are due on Tuesday (written work must be handed in in class, WeBWorK must be completed
online by the end of the day). Your first OpenLab Assignment is due next Tuesday, September 3 (future OpenLab assignments will generally be due on Thursday – the due date for the first assignment is
unusual as Thursday, September 5th, runs on a Monday schedule).
Welcome back,
Prof. Reitz
Week 1 Assignments
Written work due Tuesday September 3th – Sec 1.1 p.7: 1, 12, 19, 26, 29, 35
NOTE: On this assignment, odd problems are worth 2 points and even problems worth 4 points.
WeBWorK – WeBWorK 1, due Tuesday September 3th at midnight
OpenLab – Register for the OpenLab and join this course (instructions provided in a separate post). “OpenLab #1: Advice from the Past” due Tuesday, September 3.
This course is MAT 2071, Introduction to Proofs and Logic, taking place in the Fall 2019 semester with Professor Reitz. We will be using this website in a variety of ways this semester – as a
central location for information about the course (assignments, review sheets, policies, and so on), a place to write about the work we are doing, to ask and answer questions, to post examples of our
work, and to talk about logic, proofs, mathematics, reality and so on.
Getting Started
Anyone on the internet can look around the site and see what we are doing, and even leave a comment on one of the pages. However, only registered users can create new posts and participate in the
discussion boards.
How do I register?
You will need to do two things:
1. If you have not used the OpenLab before, you must first create an account. You will need access to your citytech email address for this. Detailed instructions for signing up on the OpenLab can
be found here.
2. Once you have created an account on the OpenLab, log in and then join this particular course, 2019 Fall – MAT 2071 Proofs and Logic – Reitz. To do this, first click the “Course Profile” link at
the top left of this page (just under the picture). Then click the “Join Now” button, which should appear just underneath the picture there.
Problems with the OpenLab or with your CityTech email:
Please let me know if you run into any problems registering or joining our course (send me an email, jreitz@citytech.cuny.edu). I also wanted to give you two resources to help out in the process:
1. For problems with your citytech email account, contact the Student Computing Helpdesk, either in person, by phone, or by email:
Student Computing Helpdesk
Location: Library Building, First Floor (L-114)
Hours: TBD
Phone: 718.260.4900
E-mail: Studenthelpdesk@citytech.cuny.edu
Their website also contains tutorials and FAQ on common problems
2. For problems registering for the OpenLab, contact the OpenLab support team, either by email at openlab@citytech.cuny.edu, or by following this link. | {"url":"https://openlab.citytech.cuny.edu/2019-fall-mat-2071-reitz/?m=201908","timestamp":"2024-11-06T00:07:15Z","content_type":"text/html","content_length":"130811","record_id":"<urn:uuid:d88d5c69-ee03-4c5d-800f-41c83889e493>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00269.warc.gz"} |
How Do You Simplify Logarithms Using the Product Property?
If you're adding logarithms that have the same base, you can use the Product Property of Logarithms to make that job easier! Follow along with this tutorial to see how the Product Property of
Logarithms is used to add logarithms with the same base. | {"url":"https://virtualnerd.com/common-core/hsf-functions/HSF-BF-building-functions/B/5/simplify-log-product-property","timestamp":"2024-11-13T13:08:36Z","content_type":"text/html","content_length":"16725","record_id":"<urn:uuid:4b208373-367e-433c-9131-aca10f326ca3>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00515.warc.gz"} |
Toroidal Chess Queens’ Problem
The toroidal M × N chess board is obtained from the usual M × N one in the same way as one obtains a torus from rectangle: by gluing together the upper side with the lower one and the left side with
the right one. The arrow directions after the gluing should coincide. The figure below represents the toroidal chess desk 4 × 3: So, shifting to the left from the very left column you get to the very
right one. Shifting to the right from the very right column, you find yourself on the first one. In the same way, moving up from the top row, you arrive on the bottom one and moving down from the
bottom row, you get to the top one. Toroidal chess figures move in the same way, as the plain ones. In particular, the queen attacks any figure which is situated in the same row, column or diagonal.
The problem is to place K queens on the toroidal chess board so that no one will attack another. Input The input will consist of several lines. Each line contains three integers M, N and K, separated
by space. Here M denotes number of board columns, N - number of board rows, K - number of queens to be placed. (1 ≤ M,N,K ≤ 14) The input is terminated by . Output For each input line the output
should consist of solution representation. The position of every queen is placed in the separate line as a pair of two integers separated by space. The first integer denotes the column number (from 1
to M), the second - the row number (from 1 to N). If there is no solution for given M, N, K, you should output two zeroes in one line. If there are several solutions, you should output only one of
them. Sample Input 323 632 Sample Output 00 11 43 | {"url":"https://ohbug.com/uva/10265/","timestamp":"2024-11-03T16:10:43Z","content_type":"text/html","content_length":"2923","record_id":"<urn:uuid:39c0723a-030a-4220-abe3-ef4992dc0446>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00493.warc.gz"} |
Question #2b71d | Socratic
Question #2b71d
2 Answers
The idea here is that when you're diluting a solution, its concentration decreases and its volume increases by the same factor called the dilution factor.
$\text{DF" = "concentration stock"/"concentration diluted" = V_"diluted"/V_"stock}$
In your case, you know that the concentration of the solution must decrease by a factor of
#"DF" = (1000 color(red)(cancel(color(black)("ppm"))))/(50color(red)(cancel(color(black)("ppm")))) = color(blue)(20)#
This means that the volume of the stock solution, i.e. of the concentrated solution, must be #color(blue)(20#times smaller than the volume of the diluted solution.
You will thus have
${V}_{\text{stock" = V_"diluted}} / \textcolor{b l u e}{20}$
which, in your case, is equal to
#V_"stock" = "100 mL"/color(blue)(20) = color(darkgreen)(ul(color(black)("5 mL")))#
The answer is rounded to one significant figure.
So, in order to prepare your target solution, use $\text{5 mL}$ of $\text{1000-ppm}$ stock solution and add enough water to get its volume to $\text{100 mL}$.
100mL, if that is the smallest volume accurate measurement available.
The volumes and concentrations are ratios.
#?mL_(Conc) * 1000"ppm" = 100mL_(Dil) * 50"ppm"#
#?mL_(Conc) = (100mL_(Dil) * 50"ppm")/1000#
#?mL_(Conc) = 5mL#
Thus, you would need to put exactly 5mL of the 1000ppm stock solution into a solution diluted to 50mL final volume.
If all you have is a 100mL volumetric pipet, you would need to scale that up to use the minimum 100mL volume. Then, starting with 100mL of stock solution we can use the calculation to see how much
final solution we will need.
$100 m {L}_{C o n c} \cdot 1000 \text{ppm" = ?mL_(Dil) * 50"ppm}$
# ?mL_(Dil) = (100mL_(Conc) * 1000"ppm")/50#
# ?mL_(Dil) = 2000mL#
So, taking 100mL of the stock solution with the pipet and putting it in a 2L+ size flask or beaker, you would then pipet diluent (water?) until the final solution is 2000mL – add 1900mL of diluent
with 19 additional loads and discharges of the pipet.
THEN you can use the pipet to take exactly 100mL of your final 50ppm solution.
Impact of this question
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Does Your Strategy Generate Positive Expected Value - Total Alpha Trading
No strategy stands the test of time without producing positive expected value…even if you’re Warren Buffett. The problem is traders confuse getting lucky or a hot streak with a repeatable strategy.
Trust me – they aren’t the same thing. But can you tell the difference?
I fell into this trap when I first started trading during the dot.com bubble. The market was so buoyant that even bad strategies turned a profit. There was even an experiment where monkeys picked
stocks…and beat the market!
Unfortunately, it lulled me into a false sense of security. After the market crashed, I stuck with a strategy that didn’t produce a positive expected value.
Expected value tells you what your trade should return if you repeat it over and over. The critical component here – repeating the trade. That’s why you need to trade small and often. Otherwise, one
single trade can demolish your account, even if it had a 90% chance of success.
I trade multiple strategies in my Total Alpha service. But no matter what, ever single trade must yield a positive expected value.
Don’t worry. It’s simple to calculate and easy to incorporate.
This simple formula provides invaluable insight into your trading strategy…probably more than any other metric.
As I mentioned previously, the expected value tells you the return for that trade on average. That doesn’t mean you will get that exact return every time. In fact, it’s likely you never get the exact
amount. But if you repeat that same trade over and over…then average the outcomes…you get near the expected value.
You need three components to calculate expected value: win rate, winning outcome, and losing outcome.
Expected value weights the values of the outcomes based on the likelihood of that outcome (IE the win/lose rate).
• A value below 0 means your strategy will lose money over time.
• A value equal to 0 means your strategy will breakeven over time.
• A value greater than 0 means your strategy will make money over time.
The formula is as follows.
Expected Value = (% odds of a win x potential profit) + (% odds of a loss x potential loss)
Note: The chances of losing is 100% – Win Rate. Also, losses are always negative numbers.
Let’s use an example from my Total Alpha portfolio last week.
In this trade, I sold a credit spread in Netflix (NFLX).
• I received a $2.30 premium for the sale, the maximum amount I could win on the trade.
• The maximum potential loss would be the difference between the strike prices less the premium I received.
□ In this case, I sold the $310/$315 strikes
□ That makes my maximum loss $315 – $310 – $2.30 = $5.00 – $2.30 = $2.70
If you want to learn more about option credit spreads, check out this free article from Jason Bond.
Now I need my win rate. There are a few ways to figure this out. The most common is to look at your results from your trade journal on similar trades.
Another choice is to look at the option probabilities provided by your trading platform. You look at the probability for the lower strike price (in a call spread-high strike price in a put spread)
landing out-of-the-money at expiration.
Here’s a snapshot of an option chain in a trading platform. I’ve pointed to the probabilities I’m referring to.
I went with my typical win-rate on these trades, which is around 65% when I let them expire at full profit.
Now, let’s plug that into the equation and see what we get.
Expected Value = (65% x +$2.30) + (35% x –$2.70) = $1.495 – $0.945 = $0.55
That means I should make $0.55 on this trade over time if I repeat it over and over on average. However, each individual trade will produce results of $2.30 profit or $2.70 loss.
Professional notes about expected value
I provided a very basic explanation of how the expected value works. It gets you started in the right direction. But, if you’re a quick study, then let me give you some additional insights.
• Multiple winning or losing outcomes are possible – I simplified things down to a binary win or loss. The truth is that many outcomes are possible. You achieve partial profit by expiration or
partial loss. Each outcome has its own percentage. The trick is to make sure you don’t overlap in your outcomes.
• Don’t figure out your win rate with real money – New strategies should be thoroughly tested. You can program backtests, look through historical charts, or take simulated trades.
• Keep your trades small – The key to making this all work is to trade often. That means you need to trade small. Otherwise, even a 90% chance of success trade can blow up your account.
More trades and more excitement
I’m really excited for what’s to come. I took member feedback and started incorporating different strategies into my Total Alpha service. Now there are trades to suit each person’s tastes.
Total Alpha teaches you how to trade multiple option strategies from beginner to advanced. It’s a great way to learn how to trade profitably.
Amazon (AMZN) can’t seem to get a foothold these days. Why? After sifting through the data, I determined it comes down to three key reasons: Borrowing costs Increased competition Government action/
inaction I can monitor these if I know where to look. And it’s crucial...
As long as the Fed backstops the market, stocks can and will continue to trade higher. And based on what they’re saying...they have no intention of putting their foot off the gas. It’s one of the
main reasons why some market pundits believe that some stocks and...
Crude oil futures, and energy related stocks have benefited from positive vaccine news. And its creating a slew of trading opportunities…. So much so it makes me dance just thinking about it... I
use my favorite chart pattern to narrow down which plays I’ll take. ...
The S&P 500 has been on a tear, and it seems as if every week it makes new all-time highs. To me, it just doesn’t feel as if these gains will stick right now. Why? THE MARKET IS HOLDING BACK! You
see, we could be in another record bull run instead of this...
The best performing index year-to-date — Nasdaq 100 — is up over 40%. When most traders see those returns, they stare in awe and think they should just buy an ETF and hold. While that may be a good
strategy for some… It’s not a strategy I love, when it comes to...
These are serious obstacles for AMZN
Watch for these signals to find a top in WKHS
Here’s How To Play The Surge In Oil
These Indicators Signal A Short-Term Pullback
One strategy + One sector = All winners in November | {"url":"https://totalalphatrading.com/2019/12/14/does-your-strategy-generate-positive-expected-value/","timestamp":"2024-11-13T22:25:53Z","content_type":"text/html","content_length":"292762","record_id":"<urn:uuid:4f9a0f03-4eb1-439f-aafb-90441bfe5d70>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00077.warc.gz"} |
7 Statistical Analysis Methods Beginners Should Know
Nearly every social or scientific discipline uses statistics to inform decisions and improve outcomes. They do this through statistical analysis methods, which make sense of data by giving analytical
insights into it. Statistical analysis drives informed approaches with business analytics. The insights gained from statistical analysis allow you to see patterns in data that have the potential to
make future predictions, informing your business decision-making process.
This article explores some basic statistical analysis methods to help you get started using statistics to improve your decision-making. It also examines how statistical analysis compares to data
analysis when to use descriptive or inferential analysis and some jobs that use statistical analysis.
Statistical analysis vs. data analysis
Statistical and data analysis do similar things and often work together to discover similar outcomes, such as behavior predictions. The main difference is the discipline's tactics to find patterns
and predictions. Let’s examine some differences between statistical analysis and data analysis:
Statistical analysis Data analysis
Data analyzed is from smaller sample sizes Data analyzed is from large or massive amounts of data
Analysis focuses on the use of mathematical techniques, including probability, calculus, and Analysis focuses on data science techniques, including machine learning and computer programming, to
linear algebra identify patterns
Uses descriptive and inferential statistics to analyze data Uses descriptive, diagnostic, predictive, and prescriptive data analysis to inform decisions
Looks to understand a particular aspect of a data set Draws conclusions and finds patterns from the entire data set
Descriptive statistical analysis methods
Descriptive statistical analysis describes aspects of a set of data. These quantitative statistical methods show representations of what a set of data represents. Graphs and charts help visualize the
findings of these methods. Some important beginner descriptive statistical analysis methods to know are:
Let’s take a closer look at each method and its application.
The mean is a central tendency that calculates the average value in a data set. The formula is the sum of all data points divided by the quantity of data points in the set. For example, if you want
to find the average grade from this series of tests: 89, 99, 100, 75, 86, 95, 86, 73, and 86, you would start by adding them together, getting the sum of the series, which is 789. Then, divide that
by the number of data points (nine), which equals 87.67—the mean or average test score.
The median is another central tendency that finds the data set’s middle value. To find the median, order data from the lowest to highest value. Using the test scores from above, the data set should
look as follows: 73, 75, 86, 86, 86, 89, 95, 99, 100. Since this data set contains odd numbers, the median becomes 86.
However, if it had one more number, it might look as follows: 73, 75, 86, 86, 86, 88, 89, 95, 99, and 100. Then, you would calculate the mean value of the two middle numbers. In this example, you
would add 86 and 88, which sum to 174. Divide that by the two numbers and arrive at the new median of 87. In this case, the mean and median are similar. However, the median is sometimes a more
accurate indicator of the average if the mean contains large outliers that weigh the average.
The mode is the last central tendency of a data set and is simply data set’s most common number. With our original data set, put in order 73, 75, 86, 86, 86, 89, 95, 99, and 100, the mode reveals
itself as 86—the most frequently repeated number in the data set. Mode is a valuable method for finding data patterns when predicting a common occurrence. In this case, while the median is also 86,
the mode indicates there could be something about the test that makes 86 a common score.
Standard deviation
The standard deviation is a test of variability you use to measure the average distance data points vary from the mean. This method explains how far data points spread out from the mean value. Low
values indicate a closeness to the mean, while high values indicate the values are more spread out. Standard deviation uses this formula:
s = √ ( Σ (x - x̄ )2 / n -1 )
Here are the steps to find the standard deviation using the data set from above 73, 75, 86, 86, 86, 89, 95, 99, 100:
1. Find the mean of the data set. In this example, it would be 87.6667.
2. Subtract the value of each data point from the mean to find the deviation, then square each value.
3. Sum the squared deviations. In this case, it is 720.
4. Using the formula, you get √720/8 = 9.49
Using this calculation, 9.49 is the standard deviation from the mean.
Inferential statistical analysis methods
Inferential statistical analysis methods work to draw general conclusions and make predictions about populations through smaller data sets. These methods examine the quality of samples and findings
of descriptive statistical findings to ensure their inferences to the larger population are valid. Many methods test the quality of the results. Some of these essential inferential methods include:
• Hypothesis testing
• Confidence intervals
• Regression analysis
Let’s take a closer look at each method and its application.
Hypothesis testing
In hypothesis testing, you formulate two hypotheses to discover which statement about a data sample is valid. These two hypotheses are:
A typical test to reject the null hypothesis, which is assumed correct until you reject it, is analyzing a p-value. You can reject the null hypothesis if the p-value is less than or equal to the
chosen significance level. The smaller the p-value, the more the evidence supports the alternative hypothesis.
Using the data on test scores above, let’s calculate a p-value with a significance value of a .05 level to perform a hypothesis test. This example is for the more common two-tailed p-value. Let’s say
you think the mean of the test scores is 90 instead of 87.67.
1. Make your null and alternative hypotheses known.
μ = hypothesis mean
The two hypotheses for this problem become:
H0: μ = 90
H1: μ ≠ 90
2. After you state the hypothesis, use a t-test to calculate the value of the test concerning the data set.
The formula for “t” is t = x-μs÷n
x = 87.67 = data set mean
μ = 90 = hypothesis mean
s = 9.49 = standard deviation
n = 9 = the size of the data set
Plug in your numbers from the sample problem and calculate the t. Once calculated, use the absolute value of t to keep the number positive |t| = 0.7366.
3. Once you have your t value, consult a t-table to find a p-value.
In this case, the p-value = 0.482425. Because this value is greater than the significance value of 0.05, you would not reject the null hypothesis H0: μ = 90 because you lack sufficient evidence.
Confidence interval
This test determines how accurate a mean is from data set to data set. In the example of test scores above, the confidence interval determines a degree of confidence that the mean of the test scores
will fall into a specific percentage of the time. The confidence interval is the sample mean margin of error.
In the test score example, you want a confidence interval = 95 percent.
1. Calculate the margin of error.
The formula for the margin of error is ME = z*sn
In the margin of error formula, z* represents a level of confidence consulted to the confidence table. For a 95 percent level of confidence, z* = 1.96
Using the standard deviation above 9.49 and the number of data points 9, the ME = 6.2
2. Calculate the confidence interval using the margin of error
Use the sample mean of 87.67 calculated earlier.
C = 87.67±6.2, or from 81.47 to 93.87.
With 95 percent confidence, you can say that the mean of the test scores in a different class falls between 81.47 and 93.87.
Regression analysis
Simple regression analysis uses a line of best fit drawn through a graph of data points, showing how many data points the line hits. This is the regression line. A regression analysis gives you the
slope of the line, the correlation, and how well the line fits the data based on variation. Simple linear regression uses two variables, while multi-variable regressions use three or more variables.
Simple regression analysis primarily aims to find the relationship between the dependent and independent variables. The formula for regression analysis is Y = a + b(x). In the formula:
• Y = the independent variable
• x = the dependent variable
• a = the y-intercept
• b = the slope of the graph
What jobs use statistical analysis methods?
The government, marketing, business, and engineering industries rely on statistical analysis methods. Additionally, you can get jobs in data analytics with a background in statistics as well. Many
jobs require a master’s degree, but some entry-level positions accept a bachelor’s degree if your math background is strong enough. US Bureau of Labor Statistics projects the current job outlook from
2023 to 2033 for statisticians to grow 11 percent [1].
To get an idea of the types of roles you might pursue, consult the following list of statistics jobs and their average annual base salaries:
(All salary data is average annual base pay from Glassdoor as of October 2024)
• Statistician: $99,937
• Statistical analyst: $89,552
• Data analyst: $89,552
• Business analyst: $92,569
• Financial analyst: $79,434
• Market researcher: $61,053
• Actuarial analyst: $116,501
• Investment analyst: $115,718
• Data scientist: $115,691
Getting started with Coursera
If you want to understand statistical analysis methods more deeply, consider an online course or degree to gain in-demand skills. For example, you can try the Introduction to Statistics course from
Stanford University to gain beginner skills in statistics. If you want a background in statistical analysis tied to data science, you can try Statistics with Python Specialization from the University
of Michigan on Coursera. | {"url":"https://www.coursera.org/articles/statistical-analysis-methods","timestamp":"2024-11-07T03:11:43Z","content_type":"text/html","content_length":"694380","record_id":"<urn:uuid:9031a072-c316-4a19-9ae9-2002284eed4d>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00397.warc.gz"} |
Euler line
In any triangle $\triangle ABC$, the Euler line is a line which passes through the orthocenter $H$, centroid $G$, circumcenter $O$, nine-point center $N$ and de Longchamps point $L$. It is named
after Leonhard Euler. Its existence is a non-trivial fact of Euclidean geometry. Certain fixed orders and distance ratios hold among these points. In particular, $\overline{OGNH}$ and $OG:GN:NH =
Euler line is the central line $L_{647}$.
Given the orthic triangle $\triangle H_AH_BH_C$ of $\triangle ABC$, the Euler lines of $\triangle AH_BH_C$,$\triangle BH_CH_A$, and $\triangle CH_AH_B$ concur at $N$, the nine-point circle of $\
triangle ABC$.
Proof Centroid Lies on Euler Line
This proof utilizes the concept of spiral similarity, which in this case is a rotation followed homothety. Consider the medial triangle $\triangle O_AO_BO_C$. It is similar to $\triangle ABC$.
Specifically, a rotation of $180^\circ$ about the midpoint of $O_BO_C$ followed by a homothety with scale factor $2$ centered at $A$ brings $\triangle ABC \to \triangle O_AO_BO_C$. Let us examine
what else this transformation, which we denote as $\mathcal{S}$, will do.
It turns out $O$ is the orthocenter, and $G$ is the centroid of $\triangle O_AO_BO_C$. Thus, $\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}$. As a homothety preserves angles, it follows that $\
measuredangle O_AOG = \measuredangle AHG$. Finally, as $\overline{AH} || \overline{O_AO}$ it follows that $\[\triangle AHG = \triangle O_AOG\]$ Thus, $O, G, H$ are collinear, and $\frac{OG}{HG} = \
Another Proof
Let $M$ be the midpoint of $BC$. Extend $CG$ past $G$ to point $H'$ such that $CG = \frac{1}{2} GH$. We will show $H'$ is the orthocenter. Consider triangles $MGO$ and $AGH'$. Since $\frac{MG}{GA}=\
frac{H'G}{GC} = \frac{1}{2}$, and they both share a vertical angle, they are similar by SAS similarity. Thus, $AH' \parallel OM \perp BC$, so $H'$ lies on the $A$ altitude of $\triangle ABC$. We can
analogously show that $H'$ also lies on the $B$ and $C$ altitudes, so $H'$ is the orthocenter.
Proof Nine-Point Center Lies on Euler Line
Assuming that the nine point circle exists and that $N$ is the center, note that a homothety centered at $H$ with factor $2$ brings the Euler points $\{E_A, E_B, E_C\}$ onto the circumcircle of $\
triangle ABC$. Thus, it brings the nine-point circle to the circumcircle. Additionally, $N$ should be sent to $O$, thus $N \in \overline{HO}$ and $\frac{HN}{ON} = 1$.
Analytic Proof of Existence
Let the circumcenter be represented by the vector $O = (0, 0)$, and let vectors $A,B,C$ correspond to the vertices of the triangle. It is well known the that the orthocenter is $H = A+B+C$ and the
centroid is $G = \frac{A+B+C}{3}$. Thus, $O, G, H$ are collinear and $\frac{OG}{HG} = \frac{1}{2}$
The points of intersection of the Euler line with the sides of the triangle
Acute triangle
Let $\triangle ABC$ be the acute triangle where $AC > BC > AB.$ Denote $\angle A = \alpha, \angle B = \beta, \angle C = \gamma$$\[\implies \beta > \alpha > \gamma.\]$
Let Euler line cross lines $AB, AC,$ and $BC$ in points $D, E,$ and $F,$ respectively.
Then point $D$ lyes on segment $AB, \frac {\vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.$
Point $E$ lyes on segment $AC, \frac {\vec AE}{\vec EC} = \frac {\tan \beta – \tan \gamma}{\tan \beta – \tan \alpha} > 0.$
Point $F$ lyes on ray $BC, \frac {\vec {BF}}{\vec {CF}} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} > 0.$
Denote $n = \frac {\vec BD}{\vec DA}, m = \frac {\vec CE}{\vec EA}, p = \frac {\vec CX}{\vec XB}, X \in BC, q = \frac {AY}{YX}, Y \in AX.$
We use the formulae $m + pn = \frac {p+1}{q}$ (see Claim “Segments crossing inside triangle” in “Schiffler point” in “Euler line”).
Centroid $G$ lyes on median $AA'' \implies X = A'' , Y = G, p = 1, q = 2 \implies m+n=1.$
Orthocenter $H$ lyes on altitude $AA' \implies X = A', Y = H, p = \frac {\vec {CX}}{\vec {XB}} = \frac {\tan \beta}{\tan \gamma}, q = \frac {AY}{YX}.$$\[q = \frac {\vec {AH}}{\vec {HA'}} = \frac {\
cos \alpha}{\cos \beta \cdot \cos \gamma} = \tan \beta \cdot \tan \gamma – 1 \implies\]$$\[m \tan \gamma + n \tan \beta = \frac {\tan \beta + \tan \gamma}{1 – \tan \beta \cdot \tan \gamma} = – \tan \
alpha.\]$ Therefore $\[\frac {\vec {BD}}{\vec {DA}} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0,\]$$\[\frac {\vec {CE}}{\vec {EA}} = m =\frac {\tan \beta – \tan \alpha}{\tan
\beta – \tan \gamma} > 0.\]$ We use the signed version of Menelaus's theorem and get $\[\frac {\vec {BF}}{\vec {FC}} = \frac {\tan \alpha – \tan \gamma}{ \tan \alpha –\tan \beta} < 0.\]$
Obtuse triangle
Let $\triangle ABC$ be the obtuse triangle where $BC > AC > AB \implies \alpha > \beta > \gamma.$
Let Euler line cross lines $AB, AC,$ and $BC$ in points $D, E,$ and $F,$ respectively.
Similarly we get $F \in BC, \frac {\vec {BF}}{\vec {FC}} = \frac {\tan \gamma – \tan \alpha}{ \tan \beta –\tan \alpha} > 0.$
$\[E \in AC, \frac {\vec {CE}}{\vec {EA}} = \frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma} > 0.\]$$D \in$ ray $BA, \frac {\vec {BD}}{\vec {AD}} = \frac {\tan \gamma – \tan \alpha}{\tan \
beta – \tan \gamma} > 0.$
Right triangle
Let $\triangle ABC$ be the right triangle where $\angle BAC = 90^\circ.$ Then Euler line contain median from vertex $A.$
Isosceles triangle
Let $\triangle ABC$ be the isosceles triangle where $AC = AB.$ Then Euler line contain median from vertex $A.$
Corollary: Euler line is parallel to side
Euler line $DE$ is parallel to side $BC$ iff $\tan \beta \cdot \tan \gamma = 3.$
$DE||BC \implies \frac {BD}{DA} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma}= \frac {CE}{EA} = \frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma}.$
After simplification in the case $\beta e \gamma$ we get $2 \tan \alpha = \tan \beta + \tan \gamma.$$\[180^\circ – \alpha = \beta + \gamma \implies \tan \alpha = \frac{\tan \beta + \tan \gamma}{\tan
\beta \cdot \tan \gamma – 1} \implies \tan \beta \cdot \tan \gamma = 3.\]$
vladimir.shelomovskii@gmail.com, vvsss
Angles between Euler line and the sides of the triangle
Let Euler line of the $\triangle ABC$ cross lines $AB, AC,$ and $BC$ in points $D, E,$ and $F,$ respectively. Denote $\angle A = \alpha, \angle B = \beta, \angle C = \gamma,$ smaller angles between
the Euler line and lines $BC, AC,$ and $AB$ as $\theta_A, \theta_B,$ and $\theta_C,$ respectively.
Prove that $\tan \theta_A = \vert \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} \vert,$$\[\tan \theta_B = |\frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}\
vert, \tan \theta_C = |\frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}\vert.\]$
WLOG, $AC > BC > AB \implies \frac {\vec {BF}}{\vec {CF}} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} > 0.$
Let $|BC| = 2a, M$ be the midpoint $BC, O$ be the circumcenter of $\triangle ABC \implies OM = \frac {a}{\tan \alpha}.$
$\[MF = MC + CF, \frac {\vec {BF}}{\vec {CF}} = \frac {\vec {BC + CF}}{\vec {CF}} = \frac {2a}{|CF|}+ 1 =\frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} \implies\]$$\[\frac {|MF|}{a} = \
frac {\tan \beta – \tan \gamma}{2 \tan \alpha – \tan \beta – \tan \gamma} \implies\]$$\[\tan \theta_A = \frac {|OM|}{|MF|} = |\frac {3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}|.\]$
Symilarly, for other angles.
vladimir.shelomovskii@gmail.com, vvsss
Distances along Euler line
Let $H, G, O,$ and $R$ be orthocenter, centroid, circumcenter, and circumradius of the $\triangle ABC,$ respectively. $\[a = BC, b = AC, c = AB,\]$$\[\alpha = \angle A,\beta = \angle B,\gamma = \
angle C.\]$
Prove that $HO^2 = R^2 (1 - 8 \cos A \cos B \cos C),$$\[GO^2 = R^2 - \frac {a^2 + b^2 + c^2}{9}.\]$
WLOG, $ABC$ is an acute triangle, $\beta \ge \gamma.$
$\[OA = R, AH = 2 R \cos \alpha, \angle BAD = \angle OAC = 90^\circ - \beta \implies\]$$\[\angle OAH = \alpha - 2\cdot (90^\circ - \beta) = \alpha + \beta + \gamma - 180^\circ + \beta - \gamma = \
beta - \gamma .\]$$\[HO^2 = AO^2 + AH^2 - 2 AH \cdot AO \cos \angle OAC = R^2 + (2 R \cos \alpha)^2 - 2 R \cdot 2R \cos \alpha \cdot \cos (\beta - \gamma).\]$$\[\frac {HO^2}{R^2} = 1 + 4 \cos \alpha
(\cos \alpha - \cos (\beta - \gamma) = 1 - 4 \cos \alpha (\cos (\beta + \gamma) + \cos (\beta – \gamma)) = 1 - 8 \cos \alpha \cos \beta \cos \gamma.\]$$\[\frac {HO^2}{R^2} = 1 + 4 \cos^2 \alpha - 4 \
cos \alpha \cos (\beta - \gamma) = 5 - 4 \sin^2 \alpha + 4 \cos (\beta + \gamma) \cos (\beta - \gamma)\]$$\[HO^2 = 5R^2 - 4R^2 \sin^2 \alpha + 2R^2 \cos 2\beta + 2R^2 \cos 2 \gamma = 9R^2 - 4R^2 \sin
^2 \alpha - 4R^2 \sin^2 \beta - 4 R^2\sin^2 \gamma\]$$\[HO^2 = 9R^2 - a^2 - b^2 - c^2, GO^2 = \frac {HO^2}{9} = R^2 - \frac {a^2 + b^2 + c^2}{9}.\]$vladimir.shelomovskii@gmail.com, vvsss
Position of Kimberling centers on the Euler line
Let triangle ABC be given. Let $H = X(4), O = X(3), R$ and $r$ are orthocenter, circumcenter, circumradius and inradius, respectively.
We use point $\vec O = \vec X(3)$ as origin and $\vec {HO}$ as a unit vector.
We find Kimberling center X(I) on Euler line in the form of $\[\vec X(I) = \vec O + k_i \cdot \vec {OH}.\]$ For a lot of Kimberling centers the coefficient $k_i$ is a function of only two parameters
$J = \frac {|OH|}{R}$ and $t = \frac {r}{R}.$
Centroid $X(2)$$\[X(2) = X(3) + \frac {1}{3} (X(4) - X(3)) \implies k_2 = \frac {1}{3}.\]$ Nine-point center $X(5)$$\[X(5) = X(3) + \frac {1}{2} (X(4) - X(3)) \implies k_5 = \frac {1}{2}.\]$ de
Longchamps point $X(20)$$\[X(20) = X(3) - (X(4) - X(3)) \implies k_{20} = - 1.\]$ Schiffler point $X(21)$$\[X(21) = X(3) + \frac {1}{3 + 2r/R} (X(4) + X(3)) \implies k_{21} = \frac {1}{3 + 2t}.\]$
Exeter point $X(22)$$\[X(22) = X(3) + \frac {2}{J^2 - 3} (X(4) - X(3)) \implies k_{22} = \frac {2}{J^2 - 3}.\]$ Far-out point $X(23)$$\[X(23) = X(3) + \frac {3}{J^2} (X(4) - X(3)) \implies k_{23} = \
frac {3}{J^2}.\]$ Perspector of ABC and orthic-of-orthic triangle $X(24)$$\[X(24) = X(3) + \frac {2}{J^2+1} (X(4) - X(3)) \implies k_{24} = \frac {2}{J^2 + 1}.\]$ Homothetic center of orthic and
tangential triangles $X(25)$$\[X(25) = X(3) + \frac {4}{J^2+3} (X(4) - X(3)) \implies k_{25} = \frac {4}{J^2 + 3}.\]$ Circumcenter of the tangential triangle $X(26)$$\[X(26) = X(3) + \frac {2}{J^2 -
1}(X(4) - X(3)) \implies k_{26} = \frac {2}{J^2 - 1}.\]$
Midpoint of X(3) and $X(5)$$\[X(140) = X(3) + \frac {1}{4} (X(4) - X(3)) \implies k_{140} = \frac {1}{4}.\]$vladimir.shelomovskii@gmail.com, vvsss
Triangles with angles of $60^\circ$ or $120^\circ$
Claim 1
Let the $\angle C$ in triangle $ABC$ be $120^\circ.$ Then the Euler line of the $\triangle ABC$ is parallel to the bisector of $\angle C.$
Let $\omega$ be circumcircle of $\triangle ABC.$
Let $O$ be circumcenter of $\triangle ABC.$
Let $\omega'$ be the circle symmetric to $\omega$ with respect to $AB.$
Let $E$ be the point symmetric to $O$ with respect to $AB.$
The $\angle C = 120^\circ \implies O$ lies on $\omega', E$ lies on $\omega.$
$EO$ is the radius of $\omega$ and $\omega' \implies$ translation vector $\omega'$ to $\omega$ is $\vec {EO}.$
Let $H'$ be the point symmetric to $H$ with respect to $AB.$ Well known that $H'$ lies on $\omega.$ Therefore point $H$ lies on $\omega'.$
Point $C$ lies on $\omega, CH || OE \implies CH = OE.$
Let $CD$ be the bisector of $\angle C \implies E,O,D$ are concurrent. $OD = HC, OD||HC \implies CD || HO \implies$
Euler line $HO$ of the $\triangle ABC$ is parallel to the bisector $CD$ of $\angle C$ as desired.
Claim 2
Let the $\angle C$ in triangle $ABC$ be $60^\circ.$ Then the Euler line of the $\triangle ABC$ is perpendicular to the bisector of $\angle C.$
Let $\omega, O, H, I$ be circumcircle, circumcenter, orthocenter and incenter of the $\triangle ABC.$$\[\angle AHB = 180^\circ – \angle ACB = 180^\circ – 60^\circ = 120^\circ.\]$$\[\angle AOB = 2 \
angle ACB = 120^\circ.\]$$\angle AIB = 90^\circ + \frac {\angle ACB}{2} = 120^\circ \implies$ points $A, H, I, O, B$ are concyclic.
The circle $AIB$ centered at midpoint of small arc $AB \implies$
$EH = EO = CO, EO||CH \implies COEH$ is rhomb.
Therefore the Euler line $HO$ is perpendicular to $CI$ as desired.
Claim 3
Let $ABCD$ be a quadrilateral whose diagonals $AC$ and $BD$ intersect at $P$ and form an angle of $60^\circ.$ If the triangles PAB, PBC, PCD, PDA are all not equilateral, then their Euler lines are
pairwise parallel or coincident.
Let $l$ and $l'$ be internal and external bisectors of the angle $\angle BPC, l \perp l'$.
Then Euler lines of $\triangle ABP$ and $\triangle CDP$ are parallel to $l'$ and Euler lines of $\triangle BCP$ and $\triangle ADP$ are perpendicular to $l$ as desired.
vladimir.shelomovskii@gmail.com, vvsss
Euler lines of cyclic quadrilateral (Vittas’s theorem)
Claim 1
Let $ABCD$ be a cyclic quadrilateral with diagonals intersecting at $P (\angle APB e 60^\circ).$ The Euler lines of triangles $\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA$ are
Let $O_1, O_2, O_3, O_4 (H_1, H_2, H_3, H_4)$ be the circumcenters (orthocenters) of triangles $\triangle APD, \triangle CPD, \triangle BPC, \triangle APB.$ Let $I_2I_4$ be the common bisector of $\
angle APB$ and $\angle CPD.$$\[O_1O_4 \perp AC, H_3H_4 \perp AC, O_2O_3 \perp AC, H_1H_2 \perp AC \implies\]$$\[O_1O_4 || H_3H_4 || O_2O_3 || H_1H_2.\]$$\[O_1O_2 \perp BD, H_2H_3 \perp BD, O_3O_4 \
perp BD, H_1H_4 \perp BD \implies\]$$\[O_1O_2 || H_2H_3 || O_3O_4 || H_1H_4 .\]$ Therefore $O_1O_2O_3O_4$ and $H_1H_2H_3H_4$ are parallelograms with parallel sides.
$\triangle APB \sim \triangle DPC \implies \angle O_2PH_2 = \angle O_4PH_4 \implies I_2I_4$ bisect these angles. So points $O_2, P, H_4$ are collinear and lies on one straight line which is side of
the pare vertical angles $\angle O_2PH_2$ and $\angle O_4PH_4.$ Similarly, points $O_4, P, H_2$ are collinear and lies on another side of these angles. Similarly obtuse $\triangle APD \sim \triangle
BPC$ so points $H_1, P$ and $O_3$ are collinear and lies on one side and points $H_3, P$ and $O_1$ are collinear and lies on another side of the same vertical angles.
We use Claim and get that lines $O_1H_1, O_2H_2, O_3H_3, O_4H_4$ are concurrent (or parallel if $\angle APD = 60^\circ$ or $\angle APD = 120^\circ$).
Claim 2 (Property of vertex of two parallelograms)
Let $ABCD$ and $EFGH$ be parallelograms, $AB||EF, AD||EH.$ Let lines $AE, BH,$ and $CG$ be concurrent at point $O.$ Then points $D, O,$ and $F$ are collinear and lines $AG, BF, CE,$ and $DH$ are
We consider only the case $AB \perp AD.$ Shift transformation allows to generalize the obtained results.
We use the coordinate system with the origin at the point $O,$ and axes $Ox||AD, Oy||BA.$
We use $x_A, y_A, y_B, x_E, y_F$ and get $x_B = x_A, y_E = \frac {x_E \cdot y_A}{x_A}, x_F = x_E, y_H = y_E, x_H=\frac {x_E \cdot y_A}{y_B}, y_D = y_A,$$\[x_D=\frac {x_E \cdot y_A}{y_F}, x_C=x_D, y_C
= y_B, x_G=\frac {x_E \cdot y_A}{y_B}, y_G = y_F \implies\]$$\frac {y_C}{x_C}=\frac {y_G}{x_G} \implies$ points $C, O,$ and $G$ are colinear.
We calculate point of crossing $AG$ and $BF, AG$ and $DH, AG$ and $CE$ and get the same result: $\[x_I = \frac {y_A+ y_B – y_F – {\frac {x_E \cdot y_A}{x_A}}} {\frac{y_B}{x_E}– \frac {y_F}{x_A}}, y_I
= \frac {\frac{y_A \cdot y_B}{x_A}+ \frac{y_B \cdot y_F}{x_E}– \frac{y_A \cdot y_F}{x_A} – \frac {y_B \cdot y_F}{x_A}} {\frac{y_B}{x_E} – \frac {y_F}{x_A}}\]$ as desired (if $\frac{y_B}{x_E}= \frac
{y_F}{x_A}$ then point $I$ moves to infinity and lines are parallel, angles $\angle APD = 60^\circ$ or $\angle APD = 120^\circ).$
vladimir.shelomovskii@gmail.com, vvsss ~minor edit by Yiyj1
Concurrent Euler lines and Fermat points
Consider a triangle $ABC$ with Fermat–Torricelli points $F$ and $F'.$ The Euler lines of the $10$ triangles with vertices chosen from $A, B, C, F,$ and $F'$ are concurrent at the centroid $G$ of
triangle $ABC.$ We denote centroids by $g$, circumcenters by $o.$ We use red color for points and lines of triangles $F**,$ green color for triangles $F'**,$ and blue color for triangles $FF'*.$
Case 1
Let $F$ be the first Fermat point of $\triangle ABC$ maximum angle of which smaller then $120^\circ.$ Then the centroid of triangle $ABC$ lies on Euler line of the $\triangle ABF.$ The pairwise
angles between these Euler lines are equal $60^\circ.$
Let $G', O,$ and $\omega$ be centroid, circumcenter, and circumcircle of $\triangle ABF,$ respectevely.
Let $\triangle ABD$ be external for $\triangle ABC$ equilateral triangle $\implies F = CD \cap \omega.$$\angle AFB = 120^\circ \implies AFBD$ is cyclic.
Point $O$ is centroid of $\triangle ABD \implies \vec O = \frac {\vec A + \vec B + \vec D}{3}.$$\[\vec G' = \frac {A + B + F}{3}, G = \frac {A + B + C}{3} \implies\]$$\[\vec {OG} = \frac {\vec C – \
vec D}{3} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec C – \vec F}{3} = \frac {\vec FC}{3} \implies\]$$OG||G'G \implies$ Points $O, G',$ and $G$ are colinear, so point $G$ lies on Euler line $OG'$
of $\triangle ABF.$
$\vec {GG_0} = \frac {A – F}{3} \implies GG_0||AF, \vec {GG_1} = \frac {B – F}{3} \implies GG_1||BF.$
Case 2
Let $F$ be the first Fermat point of $\triangle ABC, \angle BAC > 120^\circ.$
Then the centroid $G$ of triangle $ABC$ lies on Euler lines of the triangles $\triangle ABF,\triangle ACF,$ and $\triangle BCF.$ The pairwise angles between these Euler lines are equal $60^\circ.$
Let $\triangle ABD$ be external for $\triangle ABC$ equilateral triangle, $\omega$ be circumcircle of $\triangle ABD \implies F = CD \cap \omega.$$\angle ABD = 60^\circ, \angle AFD = 120^\circ \
implies ABDF$ is cyclic.
Point $O$ is centroid of $\triangle ABD \implies$$\vec {OG} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec FC}{3} \implies OG||G'G \implies$
Points $O, G',$ and $G$ are colinear, so point $G$ lies on Euler line $OG'$ of $\triangle ABF$ as desired.
Case 3
Let $F'$ be the second Fermat point of $\triangle ABC.$ Then the centroid $G$ of triangle $ABC$ lies on Euler lines of the triangles $\triangle ABF',\triangle ACF',$ and $\triangle BCF'.$
The pairwise angles between these Euler lines are equal $60^\circ.$
Let $\triangle ABD$ be internal for $\triangle ABC$ equilateral triangle, $\omega$ be circumcircle of $\triangle ABD \implies F' = CD \cap \omega.$
Let $O_1, O_0,$ and $O'$ be circumcenters of the triangles $\triangle ABF',\triangle ACF',$ and $\triangle BCF'.$ Point $O_1$ is centroid of the $\triangle ABD \implies GO_1G_1$ is the Euler line of
the $\triangle ABF'$ parallel to $CD.$
$O_1 O_0$ is bisector of $BF', O'O_1$ is bisector of $AF', O'O_0$ is bisector of $CF' \implies \triangle O'O_1O_0$ is regular triangle.
$\triangle O'O_0 O_1$ is the inner Napoleon triangle of the $\triangle ABC \implies G$ is centroid of this regular triangle. $\[\angle GO_1O_0 = 30^\circ, O_1O_0 \perp F'B, \angle AF'B = 120^\circ \
implies GO_0||F'A.\]$
$\vec {GG_0} = \frac {\vec {F'A}}{3} \implies GG_0||F'A \implies$ points $O_0,G,$ and $G_0$ are collinear as desired.
Similarly, points $O',G,$ and $G'$ are collinear.
Case 4
Let $F$ and $F'$ be the Fermat points of $\triangle ABC.$ Then the centroid of $\triangle ABC$ point $G$ lies on Euler line $OG' (O$ is circumcenter, $G'$ is centroid) of the $\triangle AFF'.$
Step 1 We will find line $F'D$ which is parallel to $GG'.$
Let $M$ be midpoint of $BC.$ Let $M'$ be the midpoint of $FF'.$
Let $D$ be point symmetrical to $F$ with respect to $M.$
$M'M||F'D$ as midline of $\triangle FF'D.$$\[\vec {G'G} = \frac {\vec A + \vec B + \vec C}{3} – \frac {\vec A + \vec F + \vec F'}{3} = \frac {2}{3} \cdot (\frac {\vec B + \vec C}{2} – \frac {\vec F +
\vec F'}{3})\]$$\[\vec {G'G} = \frac {2}{3} (\vec M – \vec M') = \frac {2}{3} \vec {M'M} \implies M'M||F'D||G'G.\]$
Step 2 We will prove that line $F'D$ is parallel to $OG.$
Let $\triangle xyz$ be the inner Napoleon triangle. Let $\triangle XYZ$ be the outer Napoleon triangle. These triangles are regular centered at $G.$
Points $O, z,$ and $x$ are collinear (they lies on bisector $AF').$
Points $O, Z,$ and $Y$ are collinear (they lies on bisector $AF).$
Points $M, X,$ and $y$ are collinear (they lies on bisector $BC).$$\[E = YZ \cap BF, E' = Zx \cap BC.\]$$\[BF \perp XZ \implies \angle BEZ = 30^\circ.\]$$BC \perp Xy,$ angle between $Zx$ and $Xy$ is
$60^\circ \implies \angle BE'Z = 30^\circ.$
$\[\angle AF'B = \angle AZB = 120^\circ, AZ = BZ \implies \overset{\Large\frown} {BZ} = 60^\circ \implies\]$
Points $A, Z, F', B, E',$ and $E$ are concyclic $\implies \angle OZx = \angle CBF.$$\[FM = MD, BM = MC \implies \angle CBF = \angle BCD.\]$ Points $C, D, X, B,$ and $F'$ are concyclic $\implies \
angle BCD = \angle BF'D.$
$\angle GZO = \angle GxO = 30^\circ \implies$ points $Z, O, G,$ and $x$ are concyclic
$\[\implies \angle GOx = \angle OZx – 30^\circ = \angle DF'B – 30^\circ.\]$$\[\angle AF'B = 120^\circ, Ox \perp AF' \implies OG||F'D.\]$ Therefore $OG||G'G \implies O, G',$ and $G$ are collinear or
point $G$ lies on Euler line $OG'.$
vladimir.shelomovskii@gmail.com, vvsss
Euler line of Gergonne triangle
Prove that the Euler line of Gergonne triangle of $\triangle ABC$ passes through the circumcenter of triangle $ABC.$
Gergonne triangle is also known as the contact triangle or intouch triangle. If the inscribed circle touches the sides of $\triangle ABC$ at points $D, E,$ and $F,$ then $\triangle DEF$ is Gergonne
triangle of $\triangle ABC$.
Other wording: Tangents to circumcircle of $\triangle ABC$ are drawn at the vertices of the triangle. Prove that the circumcenter of the triangle formed by these three tangents lies on the Euler line
of the original triangle.
Let $H$ and $I$ be orthocenter and circumcenter of $\triangle DEF,$ respectively. Let $A'B'C'$ be Orthic Triangle of $\triangle DEF.$
Then $IH$ is Euler line of $\triangle DEF,$$I$ is the incenter of $\triangle ABC,$$H$ is the incenter of $\triangle A'B'C'.$
$\angle DEF = \angle DB'C' = \angle BDF = \frac { \overset{\Large\frown} {DF}}{2} \implies B'C' || BC.$
Similarly, $A'C' || AC, A'B' || AB \implies A'B'C'\sim ABC \implies$
$AA' \cap BB' \cap CC' = P,$ where $P$ is the perspector of triangles $ABC$ and $A'B'C'.$
Under homothety with center P and coefficient $\frac {B'C'}{BC}$ the incenter $I$ of $\triangle ABC$ maps into incenter $H$ of $\triangle A'B'C'$, circumcenter $O$ of $\triangle ABC$ maps into
circumcenter $I$ of $\triangle A'B'C' \implies P,H,I,O$ are collinear as desired.
vladimir.shelomovskii@gmail.com, vvsss
Thebault point
Let $AD, BE,$ and $CF$ be the altitudes of the $\triangle ABC,$ where $BC> AC > AB, \angle BAC e 90^\circ.$
a) Prove that the Euler lines of triangles $\triangle AEF, \triangle BFD, \triangle CDE$ are concurrent on the nine-point circle at a point T (Thebault point of $\triangle ABC.$)
b) Prove that if $\angle BAC < 90^\circ$ then $TE = TF + TD,$ else $TF = TE + TD.$
Case 1 Acute triangle
a) It is known, that Euler line of acute triangle $\triangle ABC$ cross AB and BC (shortest and longest sides) in inner points.
Let $O_0, O, O'$ be circumcenters of $\triangle AEF, \triangle BFD, \triangle CDE.$
Let $G_0, G,$ and $G'$ be centroids of $\triangle AEF, \triangle BFD, \triangle CDE.$
Denote $\angle ABC = \beta, K = DE \cap G'O', L = EF \cap G_0 O_0, M = EC \cap G'O', \omega$ is the circle $DEF$ (the nine-points circle).
$\angle CEH = \angle CDH = 90^\circ \implies O'$ is the midpoint $CH,$ where $H$ is the orthocenter of $\triangle ABC \implies O' \in \omega.$
Similarly $O_0 \in \omega, O \in \omega.$
$CO' = HO', BO = OH \implies OO'$ is the midline of $\triangle BHC \implies \triangle O_0OO' \sim \triangle ABC.$
Let $O'G'$ cross $\omega$ at point $T$ different from $O'.$
$\triangle ABC \sim \triangle AEF \sim \triangle DBF \sim \triangle DEC \implies$ spiral similarity centered at $E$ maps $\triangle AEF$ onto $\triangle DEC.$
This similarity has the rotation angle $180^\circ – \beta \implies$ acute angle between Euler lines of these triangles is $\beta.$
Let these lines crossed at point $T'.$ Therefore $\angle O_0T'O' = \angle O_0OO' \implies$ points $O, O_0, O',T$ and $T'$ are concyclic $\implies T = T'.$
Similarly, $OG \cap O'G' = T$ as desired.
b) $\triangle AEF \sim \triangle DEC \implies \frac {FL}{LE} = \frac {CM}{ME}.$ Point $G'$ lies on median of $\triangle DEC$ and divide it in ratio 2 : 1.
Point $G'$ lies on Euler line of $\triangle DEC.$
According the Claim, $\frac {DK}{KE}+ \frac {CM}{ME} = 1 \implies \frac {DK}{KE}+ \frac {FL}{LE} = 1.$$FO_0 = EO_0 \implies \overset{\Large\frown} {FO_0} = \overset{\Large\frown} {EO_0} \implies \
angle FTO_0 = \angle ETO_0 \implies \frac {TF}{TE} = \frac {FL}{LE}.$
Similarly $\frac {TD}{TE} = \frac {DK}{KE} \implies \frac {TF}{TE} + \frac {TD}{TE} = 1 \implies TE = TD + TF.$
Case 2 Obtuse triangle
a) It is known, that Euler line of obtuse $\triangle ABC$ cross AC and BC (middle and longest sides) in inner points.
Let $O_0, O, O'$ be circumcenters of $\triangle AEF, \triangle BFD, \triangle CDE.$
Let $G_0, G,$ and $G'$ be centroids of $\triangle AEF, \triangle BFD, \triangle CDE.$
Denote $\angle ABC = \beta, K = DF \cap GO, K' = DC \cap G'O',$$L = EF \cap G_0 O_0, L' = EC \cap G'O', \omega$ is the circle $DEF$ (the nine-points circle).
$\angle AEH = \angle AFH = 90^\circ \implies O_0$ is the midpoint $AH,$ where $H$ is the orthocenter of $\triangle ABC \implies O_0 \in \omega.$
Similarly $O' \in \omega, O \in \omega.$
$CO' = HO', BO = OH \implies OO'$ is the midline of $\triangle BHC \implies \triangle O_0OO' \sim \triangle ABC.$
Let $O'G'$ cross $\omega$ at point $T$ different from $O'.$
$\triangle ABC \sim \triangle AEF \sim \triangle DBF \sim \triangle DEC \implies$ spiral similarity centered at $E$ maps $\triangle AEF$ onto $\triangle DEC.$
This similarity has the rotation angle $\beta \implies$ acute angle between Euler lines of these triangles is $\beta.$
Let these lines crossed at point $T'.$ Therefore $\angle O_0T'O' = \angle O_0OO' \implies$ points $O, O_0, O',T$ and $T'$ are concyclic $\implies T = T'.$
Similarly, $OG \cap O'G' = T$ as desired.
b) $\triangle AEF \sim \triangle DEC \implies \frac {EL}{LF} = \frac {EL'}{L'C}.$$\triangle AEF \sim \triangle DBF \implies \frac {DK}{KF} = \frac {DK'}{K'C}.$
Point $G'$ lies on median of $\triangle DEC$ and divide it in ratio $2 : 1.$
Point $G'$ lies on Euler line of $\triangle DEC.$ According the Claim, $\frac {DK'}{K'C} + \frac {EL'}{L'C} = 1 \implies \frac {DK}{KF}+ \frac {EL}{LF} = 1.$$FO_0 = EO_0 \implies \overset{\Large\
frown} {FO_0} = \overset{\Large\frown} {EO_0} \implies \angle FTO_0 = \angle ETO_0 \implies \frac {TE}{TF} = \frac {EL}{LF}.$
Similarly $\frac {TD}{TF} = \frac {DK}{KF} \implies \frac {TE}{TF} + \frac {TD}{TF} = 1 \implies TF = TD + TE.$
Claim (Segment crossing the median)
Let $M$ be the midpoint of side $AB$ of the $\triangle ABC, D \in AC,$$\[E \in BC, G = DE \cap CM.\]$$\[\frac {BE}{CE} = m, \frac {AD}{CD} = n.\]$
Then $\frac {DG}{GE} = \frac{1+m}{1+n}, \frac {MG}{GC} = \frac{n+m}{2}.$
Let $[ABC]$ be $1$ (We use sign $[t]$ to denote the area of $t).$
Denote $[CDG] = x, [CEG] = y, [DGM] = z.$$\[[ACM] = [ BCM] = \frac {1}{2}.\]$$\[x = \frac {CD \cdot CG}{2 \cdot CA \cdot CM}, y = \frac {CE \cdot CG}{2 \cdot CB \cdot CM} \implies\]$$\[\frac {DG}{GE}
= \frac {x}{y} = \frac {CD \cdot CB}{CA \cdot CE} = \frac {1+m}{1+n}.\]$$\[x + y = \frac {CD \cdot CE}{AC \cdot BC} = \frac {1}{(1+m)(1+n)} \implies\]$$\[x + y = x(1+\frac {y}{x}) = x (1 + \frac
{1+n}{1+m} )= \frac {1}{(1+n)(1+m)} \implies x = \frac{1}{(1+n)(2+m+n)}.\]$$\[z+x = \frac {[CDM]}{2[CAM]} = \frac {1}{2(1 + n)}.\]$$\[\frac {MG}{GC} = \frac {z}{x} = \frac {z + x}{x} - 1 =\frac {1}{2
(1 + n)} (1 + n)(2 + m + n) - 1 = \frac {n + m}{2}.\]$vladimir.shelomovskii@gmail.com, vvsss
Schiffler point
Let $I, O, G, R, \alpha,$ and $r$ be the incenter, circumcenter, centroid, circumradius, $\angle A,$ and inradius of $\triangle ABC,$ respectively. Then the Euler lines of the four triangles $\
triangle BCI, \triangle CAI, \triangle ABI,$ and $\triangle ABC$ are concurrent at Schiffler point $S = X(21), \frac {OS}{SG} = \frac {3R}{2r}$.
We will prove that the Euler line $O'G'$ of $\triangle BCI$ cross the Euler line $OG$ of $\triangle ABC$ at such point $S,$ that $\frac {OS}{SG} = \frac {3R}{2r}$.
Let $O'$ and $G'$ be the circumcenter and centroid of $\triangle IBC,$ respectively.
It is known that $O'$ lies on circumcircle of $\triangle ABC, \overset{\Large\frown} {BO'} = \overset{\Large\frown} {CO'}.$
Denote $E = OO' \cap BC, X = AE \cap G'O', Y = GG' \cap OO'.$
It is known that $E$ is midpoint $BC,$ point $G$ lies on median $AE,$ points $A, I, O'$ belong the bisector of $\angle A, \frac {AE}{GE} = \frac {IE}{IG'} = 3 \implies GY||AO', \frac {O'E}{YE} = 3, \
frac {GG'}{G'Y} = \frac {AI}{IO'}.$
Easy to find that $AI = \frac {r} {\sin {\frac {\alpha}{2}}}$, $IO' = BO' = 2 R {\sin {\frac {\alpha}{2}}} \implies \frac {AI}{IO'} = \frac {r}{R \cdot (1 – \cos \alpha)}.$
We use sigh [t] for area of t. We get $\[n = \frac {GG'}{G'Y} = \frac {[GXO']}{[YXO'} = \frac {[GXO']}{[EXO']} \cdot \frac {[EXO']}{[YXO'} = \frac {GX}{XE} \cdot \frac {3}{2} \implies\]$
$\[m = \frac {GX}{XE} = \frac {2n}{3}.\]$$\[p = \frac {OE}{EY} = \frac {\cos \alpha}{(1 - \cos \alpha)/3} = \frac {3 \cos \alpha}{1 - \cos \alpha}\]$ Using Claim we get $\[\frac {OS}{SG} = \frac {p +
1}{m} - \frac {p}{n} = \frac {3(p + 1)}{2n} - \frac {p}{n} = \frac {p + 3}{2n} = \frac {3R}{2r}.\]$ Therefore each Euler line of triangles $\triangle BCI, \triangle CAI, \triangle ABI,$ cross Euler
line of $\triangle ABC$ in the same point, as desired.
Claim (Segments crossing inside triangle)
Given triangle GOY. Point $S$ lies on $GO, k = \frac {OS}{SG}.$
Point $E$ lies on $YO, p = \frac {OE}{EY}.$
Point $G'$ lies on $GY, n = \frac {GG'}{G'Y}.$
Point $X$ lies on $GE, m = \frac {GX}{XE}.$ Then $k = \frac {p + 1}{m} - \frac {p}{n}.$
Let $[OGY]$ be $1$ (We use sigh $[t]$ for area of $t).$$\[[GSG'] = \frac{n}{(n + 1)(k + 1)}, [YEG'] = \frac{1}{(n + 1)(p + 1)},\]$$\[[SOE] = \frac{kp}{(k + 1)(p + 1)}, [ESG'] = \frac {[GSG']}{m},\]$$
\[[OGY] = [GSG'] + [YEG'] + [SOE] + [ESG'] = 1 \implies\]$$\[\frac{n(p + 1)}{m}=nk + p \implies k = \frac {p + 1}{m} - \frac {p}{n}.\]$vladimir.shelomovskii@gmail.com, vvsss
Euler line as radical axis
Let $\triangle ABC$ with altitudes $AA_1, BB_1,$ and $CC_1$ be given.
Let $\Omega, O, H$ and $R$ be circumcircle, circumcenter, orthocenter and circumradius of $\triangle ABC,$ respectively.
Circle $\omega_1$ centered at $Q_1$ passes through $A, A_1$ and is tangent to the radius AO. Similarly define circles $\omega_2$ and $\omega_3.$
Then Euler line of $\triangle ABC$ is the radical axis of these circles.
If $\triangle ABC$ is acute, then these three circles intersect at two points located on the Euler line of the $\triangle ABC.$
The power of point $O$ with respect to $\omega_1, \omega_2,$ and $\omega_3$ is $R^2.$
The power of point $H$ with respect to $\omega_1$ is $AH \cdot HA_1.$
The power of point $H$ with respect to $\omega_2$ is $BH \cdot HB_1.$
The power of point $H$ with respect to $\omega_3$ is $CH \cdot HC_1.$
It is known that $AH \cdot HA_1 = BH \cdot HB_1 = CH \cdot HC_1.$
Therefore points $H$ and $O$ lies on radical axis of these three circles as desired.
vladimir.shelomovskii@gmail.com, vvsss
De Longchamps point X(20)
Definition 1
The De Longchamps’ point of a triangle is the radical center of the power circles of the triangle. Prove that De Longchamps point lies on Euler line.
We call A-power circle of a $\triangle ABC$ the circle centered at the midpoint $BC$ point $A'$ with radius $R_A = AA'.$ The other two circles are defined symmetrically.
Let $H, O,$ and $L$ be orthocenter, circumcenter, and De Longchamps point, respectively.
Denote $B-$power circle by $\omega_B, C-$power circle by $\omega_C, D = \omega_B \cap \omega_C,$$a = BC, b = AC, c = AB.$ WLOG, $a \ge b \ge c.$
Denote $X_t$ the projection of point $X$ on $B'C', E = D_t.$
We will prove that radical axes of $B-$power and $C-$power cicles is symmetric to altitude $AH$ with respect $O.$ Further, we will conclude that the point of intersection of the radical axes,
symmetrical to the heights with respect to O, is symmetrical to the point of intersection of the heights $H$ with respect to $O.$
Point $E$ is the crosspoint of the center line of the $B-$power and $C-$power circles and there radical axis. $B'C' = \frac {a}{2}.$ We use claim and get:
$\[C'E = \frac {a}{4} + \frac {R_C^2 - R_B^2}{a}.\]$$R_B$ and $R_C$ are the medians, so $\[R_B^2 = \frac {a^2}{2}+ \frac {c^2}{2} - \frac {b^2}{4}, R_C^2 = \frac {a^2}{2}+ \frac {b^2}{2} - \frac {c^
2}{4} \implies C'E = \frac {a}{4} + \frac {3(b^2 - c^2)}{4a}.\]$
We use Claim some times and get: $\[C'A_t = \frac {a}{4} - \frac {b^2 - c^2}{4a}, A_tO_t = \frac {a}{2} - 2 C'A_t = \frac {b^2 - c^2}{2a} \implies\]$$\[O_t L_t = C'E - C'A_t - A_t O_t = \frac {b^2 -
c^2}{2a} = A_t O_t = H_t O_t \implies\]$ radical axes of $B-$power and $C-$power cicles is symmetric to altitude $AH$ with respect $O.$
Similarly radical axes of $A-$power and $B-$power cicles is symmetric to altitude $CH,$ radical axes of $A-$power and $C-$power cicles is symmetric to altitude $BH$ with respect $O.$ Therefore the
point $L$ of intersection of the radical axes, symmetrical to the heights with respect to $O,$ is symmetrical to the point $H$ of intersection of the heights with respect to $O \implies \vec {HO} = \
vec {OL} \implies L$ lies on Euler line of $\triangle ABC.$
Claim (Distance between projections)
$\[x + y = a, c^2 - x^2 = h^2 = b^2 - y^2,\]$$\[y^2 - x^2 = b^2 - c^2 \implies y - x = \frac {b^2 - c^2}{a},\]$$\[x = \frac {a}{2} - \frac {b^2 - c^2}{2a}, y = \frac {a}{2} + \frac {b^2 - c^2}{2a}.\]
Definition 2
We call $\omega_A = A-$circle of a $\triangle ABC$ the circle centered at $A$ with radius $BC.$ The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical
center of $A-$circle, $B-$circle, and $C-$circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under Definition 1.
Let $H, G,$ and $L_o$ be orthocenter, centroid, and De Longchamps point, respectively. Let $\omega_B$ cross $\omega_C$ at points $A'$ and $E.$ The other points $(D, F, B', C')$ are defined
symmetrically. $\[AB' = BC, B'C = AB \implies \triangle ABC = \triangle CB'A \implies\]$$\[AB||B'C \implies CH \perp B'C.\]$ Similarly $CH \perp A'C \implies A'B'$ is diameter $\omega_C \implies$$\[\
angle A'EB' = 90^\circ, 2\vec {BG} = \vec {GB'}.\]$
Therefore $\triangle A'B'C'$ is anticomplementary triangle of $\triangle ABC, \triangle DEF$ is orthic triangle of $\triangle A'B'C'.$ So $L_o$ is orthocenter of $\triangle A'B'C'.$
$2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L$ as desired.
vladimir.shelomovskii@gmail.com, vvsss
De Longchamps line
The de Longchamps line $l$ of $\triangle ABC$ is defined as the radical axes of the de Longchamps circle $\omega$ and of the circumscribed circle $\Omega$ of $\triangle ABC.$
Let $\Omega'$ be the circumcircle of $\triangle DEF$ (the anticomplementary triangle of $\triangle ABC).$
Let $\omega'$ be the circle centered at $G$ (centroid of $\triangle ABC$) with radius $\rho = \frac {\sqrt{2}}{3} \sqrt {a^2 + b^2 + c^2},$ where $a = BC, b = AC, c = AB.$
Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of $\Omega, \Omega', \omega,$ and $\omega'.$
Center of $\Omega$ is $O$, center of $\omega$ is $L \implies OL \perp l,$ where $OL$ is Euler line. The homothety with center $G$ and ratio $-2$ maps $\triangle ABC$ into $\triangle DEF.$ This
homothety maps $\Omega$ into $\Omega'.$$R_{\Omega} e R_{\Omega'}$ and $\Omega \cap \Omega' = K \implies$ there is two inversion which swap $\Omega$ and $\Omega'.$
First inversion $I_{\omega'}$ centered at point $G = \frac {\vec O \cdot 2R + \vec H \cdot R}{2R + R} = \frac {2\vec O + \vec H}{3}.$ Let $K$ be the point of crossing $\Omega$ and $\Omega'.$
The radius of $\omega'$ we can find using $\triangle HKO:$
$\[OK = R, HK = 2R, HG = 2GO \implies GK^2 = 2(R^2 – GO^2), GO^2 = \frac {HO^2}{9} \implies\]$$\[R_G = GK = \frac {\sqrt {2(a^2 + b^2 + c^2)}}{3}.\]$
Second inversion $I_{\omega}$ centered at point $L = \frac {\vec O \cdot 2R – \vec H \cdot R}{2R – R} = 2 \vec O – \vec H.$ We can make the same calculations and get $R_L = 4R \sqrt{– \cos A \cos B \
cos C}$ as desired.
vladimir.shelomovskii@gmail.com, vvsss
Prove that the circumcenter of the tangential triangle $\triangle A'B'C'$ of $\triangle ABC$ (Kimberling’s point $X(26))$ lies on the Euler line of $\triangle ABC.$
Let $A_0, B_0,$ and $C_0$ be midpoints of $BC, AC,$ and $AB,$ respectively.
Let $\omega$ be circumcircle of $\triangle A_0B_0C_0.$ It is nine-points circle of the $\triangle ABC.$
Let $\Omega$ be circumcircle of $\triangle ABC.$ Let $\Omega'$ be circumcircle of $\triangle A'B'C'.$
$A'B$ and $A'C$ are tangents to $\Omega \implies$ inversion with respect $\Omega$ swap $B'$ and $B_0.$ Similarly, this inversion swap $A'$ and $A_0, C'$ and $C_0.$ Therefore this inversion swap $\
omega$ and $\Omega'.$
The center $N$ of $\omega$ and the center $O$ of $\Omega$ lies on Euler line, so the center $O'$ of $\Omega'$ lies on this line, as desired.
After some calculations one can find position of point $X(26)$ on Euler line (see Kimberling's point $X(26)).$
vladimir.shelomovskii@gmail.com, vvsss
Let $\triangle A_1B_1C_1$ be the orthic triangle of $\triangle ABC.$ Let $N$ be the circumcenter of $\triangle A_1B_1C_1.$ Let $\triangle A'B'C'$ be the tangencial triangle of $\triangle ABC.$ Let
$O'$ be the circumcenter of $\triangle A'B'C'.$
Prove that lines $A_1A', B_1B',$ and $C_1C'$ are concurrent at point, lies on Euler line of $\triangle ABC.$
$B'C'$ and $B_1C_1$ are antiparallel to BC with respect $\angle BAC \implies B'C' ||B_1C_1.$
Similarly, $A'C' ||A_1C_1, A'B' ||A_1B_1.$
Therefore $\triangle A_1B_1C_1 \sim \triangle A'B'C' \implies$ homothetic center of $\triangle A_1B_1C_1$ and $\triangle A'B'C'$ is the point of concurrence of lines $A_1A', B_1B',$ and $C_1C'.$
Denote this point as $K.$
The points $N$ and $O'$ are the corresponding points (circumcenters) of $\triangle A_1B_1C_1$ and $\triangle A'B'C',$ so point $K$ lies on line $NO'.$
Points $N$ and $O' = X(26)$ lies on Euler line, so $K$ lies on Euler line of $\triangle ABC.$
vladimir.shelomovskii@gmail.com, vvsss
Exeter point X(22)
Exeter point is the perspector of the circummedial triangle $A_0B_0C_0$ and the tangential triangle $A'B'C'.$ By another words, let $\triangle ABC$ be the reference triangle (other than a right
triangle). Let the medians through the vertices $A, B, C$ meet the circumcircle $\Omega$ of triangle $ABC$ at $A_0, B_0,$ and $C_0$ respectively. Let $A'B'C'$ be the triangle formed by the tangents
at $A, B,$ and $C$ to $\Omega.$ (Let $A'$ be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through $A_0A', B_0B',$ and $C_0C'$ are concurrent, the point
of concurrence lies on Euler line of triangle $ABC,$ the point of concurrence $X_{22}$ lies on Euler line of triangle $ABC, \vec {X_{22}} = \vec O + \frac {2}{J^2 - 3} (\vec H - \vec O), J = \frac {|
OH|}{R},$ where $O$ - circumcenter, $H$ - orthocenter, $R$ - circumradius.
At first we prove that lines $A_0A', B_0B',$ and $C_0C'$ are concurrent. This follows from the fact that lines $AA_0, BB_0,$ and $CC_0$ are concurrent at point $G$ and Mapping theorem.
Let $A_1, B_1,$ and $C_1$ be the midpoints of $BC, AC,$ and $AB,$ respectively. The points $A, G, A_1,$ and $A_0$ are collinear. Similarly the points $B, G, B_1,$ and $B_0$ are collinear.
Denote $I_{\Omega}$ the inversion with respect $\Omega.$ It is evident that $I_{\Omega}(A_0) = A_0, I_{\Omega}(A') = A_1, I_{\Omega}(B_0) = B_0, I_{\Omega}(B') = B_1.$
Denote $\omega_A = I_{\Omega}(A'A_0), \omega_B = I_{\Omega}(B'B_0) \implies$$\[A_0 \in \omega_A, A_1 \in \omega_A, O \in \omega_A, B_0 \in \omega_B, B_1 \in \omega_B, O \in \omega_B \implies O = \
omega_A \cap \omega_B.\]$
The power of point $G$ with respect $\omega_A$ is $GA_1 \cdot GA_0 = \frac {1}{2} AG \cdot GA_0.$
Similarly the power of point $G$ with respect $\omega_B$ is $GB_1 \cdot GB_0 = \frac {1}{2} BG \cdot GB_0.$
$G = BB_0 \cap AA_0 \implies AG \cdot GA_0 = BG \cdot GB_0 \implies G$ lies on radical axis of $\omega_A$ and $\omega_B.$
Therefore second crosspoint of $\omega_A$ and $\omega_B$ point $D$ lies on line $OG$ which is the Euler line of $\triangle ABC.$ Point $X_{22} = I_{\Omega}(D)$ lies on the same Euler line as desired.
Last we will find the length of $OX_{22}.$$\[A_1 = BC \cap AA_0 \implies AA_1 \cdot A_1A_0 = BA_1 \cdot CA_1 = \frac {BC^2}{4}.\]$$\[GO \cdot GD =GO \cdot (GO + OD) = GA_1 \cdot GA_0\]$$\[GA_1 \cdot
GA_0 = \frac {AA_1}{3} \cdot ( \frac {AA_1}{3} + A_1A_0) = \frac {AA_1^2}{9} + \frac {BC^2}{3 \cdot 4} = \frac {AB^2 + BC^2 + AC^2}{18}= \frac {R^2 - GO^2} {2}.\]$$\[2GO^2 + 2 GO \cdot OD = R^2 - GO^
2 \implies 2 GO \cdot OD = R^2 - 3GO^2.\]$$\[I_{\Omega}(D) = X_{22} \implies OX_{22} = \frac {R^2} {OD} = \frac {R^2 \cdot 2 GO}{R^2 - 3 GO^2} = \frac {2 HO}{3 - \frac {HO^2}{R^2}} = \frac {2}{3 - J^
2} HO\]$ as desired.
Mapping theorem
Let triangle $ABC$ and incircle $\omega$ be given. $\[D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega.\]$ Let $P$ be the point in the plane $ABC.$ Let lines $DP, EP,$ and $FP$ crossing $\
omega$ second time at points $D_0, E_0,$ and $F_0,$ respectively.
Prove that lines $AD_0, BE_0,$ and $CF_0$ are concurrent.
$\[k_A = \frac {\sin {D_0AE'}}{\sin {D_0AF'}} = \frac {D_0E'}{D_0A} \cdot \frac {D_0A}{D_0F'} = \frac {D_0E'}{D_0F'}.\]$ We use Claim and get: $k_A = \frac {D_0E^2}{D_0F^2}.$$\[k_D = \frac {\sin
{D_0DE}}{\sin {D_0DF}} = \frac {D_0E}{2R} \cdot \frac {2R}{D_0F} = \frac {D_0E}{D_0F} \implies k_A = k_D^2.\]$ Similarly, $k_B = k_E^2, k_C = k_F^2.$
We use the trigonometric form of Ceva's Theorem for point $P$ and triangle $\triangle DEF$ and get $\[k_D \cdot k_E \cdot k_F = 1 \implies k_A \cdot k_B \cdot k_C = 1^2 = 1.\]$ We use the
trigonometric form of Ceva's Theorem for triangle $\triangle ABC$ and finish proof that lines $AD_0, BE_0,$ and $CF_0$ are concurrent.
Claim (Point on incircle)
Let triangle $ABC$ and incircle $\omega$ be given. $\[D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega, P \in \omega, F' \in AB,\]$$\[PF' \perp AB, E' \in AC, PE' \perp AC, A' \in EF, PA' \
perp EF.\]$ Prove that $\frac {PF'}{PE'} = \frac {PF^2}{PE^2}, PA'^2 = PF' \cdot PE'.$
$\[AF = AE \implies \angle AFE = \angle AEF = \angle A'EE'.\]$$\[\angle EFP = \angle PEE' \implies \angle PFF' = \angle PEE' \implies\]$$\[\triangle PFF' \sim \triangle PEA' \implies \frac {PF}{PF'}
= \frac {PE}{PA'}.\]$
Similarly $\triangle PEE' \sim \triangle PFA' \implies \frac {PE}{PE'} = \frac {PF}{PA'}.$
We multiply and divide these equations and get: $\[PA'^2 = PF' \cdot PE', \frac {PF'}{PE'} = \frac {PF^2}{PE^2}.\]$
vladimir.shelomovskii@gmail.com, vvsss
Far-out point X(23)
Let $\triangle A'B'C'$ be the tangential triangle of $\triangle ABC.$
Let $G, \Omega, O, R,$ and $H$ be the centroid, circumcircle, circumcenter, circumradius and orthocenter of $\triangle ABC.$
Prove that the second crosspoint of circumcircles of $\triangle AA'O, \triangle BB'O,$ and $\triangle CC'O$ is point $X_{23}.$ Point $X_{23}$ lies on Euler line of $\triangle ABC, X_{23} = O + \frac
{3}{J^2} (H – O), J = \frac {OH}{R}.$
Denote $I_{\Omega}$ the inversion with respect $\Omega, A_1, B_1, C_1$ midpoints of $BC, AC, AB.$
It is evident that $I_{\Omega}(A') = A_1, I_{\Omega}(B') = B_1, I_{\Omega}(C') = C_1.$
The inversion of circles $AA'O, BB'O, CC'O$ are lines $AA_1, BB_1,CC_1$ which crosses at point $G \implies X_{23} = I_{\Omega}(G).$
Therefore point $X_{23}$ lies on Euler line $OG$ of $\triangle ABC, OG \cdot OX_{23} = R^2 \implies \frac {OX_{23}} {OH} = \frac {R^2}{OG \cdot OH} = \frac {3}{J^2},$ as desired.
vladimir.shelomovskii@gmail.com, vvsss
Symmetric lines
Let triangle $ABC$ having the circumcircle $\omega$ be given.
Prove that the lines symmetric to the Euler line with respect $BC, AC,$ and $AB$ are concurrent and the point of concurrence lies on $\omega.$
The orthocenter $H$ lies on the Euler line therefore the Euler line is $H-line.$ We use H-line Clime and finish the proof.
H–line Claim
Let triangle $ABC$ having the orthocenter $H$ and circumcircle $\omega$ be given. Denote $H–line$ any line containing point $H.$
Let $l_A, l_B,$ and $l_C$ be the lines symmetric to $H-line$ with respect $BC, AC,$ and $AB,$ respectively.
Prove that $l_A, l_B,$ and $l_C$ are concurrent and the point of concurrence lies on $\omega.$
Let $D, E,$ and $F$ be the crosspoints of $H–line$ with $AB, AC,$ and $BC,$ respectively.
WLOG $D \in AB, E \in AC.$ Let $H_A, H_B,$ and $H_C$ be the points symmetric to $H$ with respect $BC, AC,$ and $AB,$ respectively.
Therefore $H_A \in l_A, H_B \in l_B, H_C \in l_C, AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies$$\[\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.\]$
Let $P$ be the crosspoint of $l_B$ and $l_C \implies BH_CH_BP$ is cyclic $\implies P \in \omega.$
Similarly $\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP$ is cyclic $\implies P \in \omega \implies$ the crosspoint of $l_B$ and $l_A$ is point $P.$
vladimir.shelomovskii@gmail.com, vvsss
See also
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Power Analysis for One-sample t-test | SAS Data Analysis Examples
Example 1. A company that manufactures light bulbs claims that a particular type of light bulb will last 850 hours on average with standard deviation of 50. A consumer protection group thinks that
the manufacturer has overestimated the lifespan of their light bulbs by about 40 hours. How many light bulbs does the consumer protection group have to test in order to prove their point with
reasonable confidence?
Example 2. It has been estimated that the average height of American white male adults is 70 inches. It has also been postulated that there is a positive correlation between height and intelligence.
If this is true, then the average height of a white male graduate students on campus should be greater than the average height of American white male adults in general. You want to test this theory
out by random sampling a small group of white male graduate students. But you need to know how small the group can be or how few people that you need to measure such that you can still prove your
Prelude to The Power Analysis
For the power analysis below, we are going to focus on Example 1 testing the average lifespan of a light bulb. Our first goal is to figure out the number of light bulbs that need to be tested. That
is, we will determine the sample size for a given a significance level and power. Next, we will reverse the process and determine the power, given the sample size and the significance level.
We know so far that the manufacturer claims that the average lifespan of the light bulb is 850 with the standard deviation of 50, and the consumer protection group believes that the manufactory has
overestimated by about 40 hours. So in terms of hypotheses, our null hypothesis is H[0] = 850 and our alternative hypothesis is H[a]= 810.
The significance level is the probability of a Type I error, that is the probability of rejecting H[0] when it is actually true. We will set it at the .05 level. The power of the test against H[a] is
the probability of that the test rejects H[0]. We will set it at .90 level.
We are almost ready for our power analysis. But let’s talk about the standard deviation a little bit. Intuitively, the number of light bulbs we need to test depends on the variability of the lifespan
of these light bulbs. Take an extreme case where all the light bulbs have exactly the same lifespan. Then we just need to check a single light bulb to prove our point. Of course, this will never
happen. On the other hand, suppose that some light bulbs last for 1000 hours and some only last 500 hours. We will have to select quite a few of light bulbs to cover all the ground. Therefore, the
standard deviation for the distribution of the lifespan of the light bulbs will play an important role in determining the sample size.
Power Analysis
In SAS, it is fairly straightforward to perform a power analysis for comparing means. For example, we can use proc power of SAS for our calculation as shown below. First, we specify the two means,
the mean for the null hypothesis and the mean for the alternative hypothesis. Then we specify the standard deviation for the population. The default significance level (alpha level) is at .05 so we
are not going to specify it. The power is set to be .9. Last, we tell SAS that we are performing a one-sample t-test.
proc power;
onesamplemeans test=t
nullmean = 850
mean = 810
stddev = 50
power = .9
ntotal = . ;
One-sample t Test for Mean
Fixed Scenario Elements
Distribution Normal
Method Exact
Null Mean 850
Mean 810
Standard Deviation 50
Nominal Power 0.9
Number of Sides 2
Alpha 0.05
Computed N Total
Actual N
Power Total
0.909 19
The result tells us that we need a sample size at least 19 light bulbs to reject H[0] under the alternative hypothesis H[a] to have a power of 0.9.
Next, suppose we have a sample of size 10, how much power do we have keeping all of the other numbers the same? We can use the same program to calculate it.
proc power;
onesamplemeans test=t
nullmean = 850
mean = 810
stddev = 50
power = .
ntotal = 10 ;
One-sample t Test for Mean
Fixed Scenario Elements
Distribution Normal
Method Exact
Null Mean 850
Mean 810
Standard Deviation 50
Total Sample Size 10
Number of Sides 2
Alpha 0.05
Computed Power
You can see that the power is about .616 for a sample size of 10. What then is the power for sample size of 15 or 20? We can use a list of sample sizes as input to proc power.
proc power;
onesamplemeans test=t
nullmean = 850
mean = 810
stddev = 50
power = .
ntotal = 10 to 45 by 5 ;
One-sample t Test for Mean
Fixed Scenario Elements
Distribution Normal
Method Exact
Null Mean 850
Mean 810
Standard Deviation 50
Number of Sides 2
Alpha 0.05
Computed Power
Index Total Power
1 10 0.616
2 15 0.821
3 20 0.924
4 25 0.970
5 30 0.988
6 35 0.996
7 40 0.999
8 45 >.999
We can also expect that if we actually know that the standard deviation is smaller, then the sample size could be also be smaller. We can experiment with different values of standard deviation as
shown below.
proc power;
onesamplemeans test=t
nullmean = 850
mean = 810
stddev = 30 to 100 by 10
power = .8
ntotal = . ;
One-sample t Test for Mean
Fixed Scenario Elements
Distribution Normal
Method Exact
Null Mean 850
Mean 810
Nominal Power 0.8
Number of Sides 2
Alpha 0.05
Computed N Total
Std Actual N
Index Dev Power Total
1 30 0.834 7
2 40 0.803 10
3 50 0.821 15
4 60 0.807 20
5 70 0.815 27
6 80 0.808 34
7 90 0.803 42
8 100 0.808 52
There is another technical assumption, the normality assumption. If the variable is not normally distributed, a small sample size usually will not have the power indicated in the results, because
those results are calculated using the common method based on the normality assumption. It might not even be a good idea to do a t-test on such a small sample to begin with if the normality
assumption is in question.
Here is another technical point. What we really need to know is the difference between the two means, not the individual values. In fact, what really matters is the difference of the means over the
standard deviation. We call this the effect size. For example, we would get the same power if we subtracted 800 from each mean, changing 850 to 50 and 810 to 10.
proc power;
onesamplemeans test=t
nullmean = 50
mean = 10
stddev = 50
power = .9
ntotal = . ;
One-sample t Test for Mean
Fixed Scenario Elements
Distribution Normal
Method Exact
Null Mean 50
Mean 10
Standard Deviation 50
Nominal Power 0.9
Number of Sides 2
Alpha 0.05
Computed N Total
Actual N
Power Total
0.909 19
If we standardize our variable, we can calculate the means in terms of change in standard deviations.
proc power;
onesamplemeans test=t
nullmean = 1
mean = .2
stddev = 1
power = .9
ntotal = . ;
One-sample t Test for Mean
Fixed Scenario Elements
Distribution Normal
Method Exact
Null Mean 1
Mean 0.2
Standard Deviation 1
Nominal Power 0.9
Number of Sides 2
Alpha 0.05
Computed N Total
Actual N
Power Total
0.909 19
It is usually not an easy task to determine the “true” effect size. We make our best guess based upon the existing literature or a pilot study. A good estimate of the effect size is the key to a
successful power analysis.
See Also
• References
D. Moore and G. McCabe, Introduction to the Practice of Statistics, Third Edition, Section 6.4. | {"url":"https://stats.oarc.ucla.edu/sas/dae/power-analysis-for-one-sample-t-test/","timestamp":"2024-11-03T02:54:48Z","content_type":"text/html","content_length":"45420","record_id":"<urn:uuid:ae7bdca9-e84e-43b3-98c7-35a97e0043c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00360.warc.gz"} |
Raw, unnormalized outputs of the last layer in a neural network before applying the softmax function in classification tasks.
Logits represent the vector of raw predictions that a classification model generates, which are then transformed through the softmax function to produce probabilities. The term is significant because
logits directly reflect the model's internal decision values for each class, before any normalization to make these outputs interpretable as probabilities. In neural networks, especially those
designed for classification, understanding and analyzing logits can provide insights into the model's confidence and decision-making process. They are crucial for loss calculations in training, where
functions like the softmax cross-entropy loss directly work with logits to compute gradients for model optimization.
The concept of logits has its roots in logistic regression, predating deep learning, where it referred to the log-odds of probabilities. Its application in neural networks, particularly in the
context of deep learning, gained prominence with the rise of these models for classification tasks in the 2010s.
While the concept of logits is foundational and not attributed to a single contributor, the development and popularization of neural networks and deep learning techniques by researchers such as
Geoffrey Hinton, Yoshua Bengio, and Yann LeCun have significantly influenced the widespread use and understanding of logits in AI. | {"url":"https://www.envisioning.io/vocab/logits","timestamp":"2024-11-11T23:27:27Z","content_type":"text/html","content_length":"438043","record_id":"<urn:uuid:41ba27c3-60cf-4cfc-8e35-728e44f6d2e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00661.warc.gz"} |
Performance limits of greedy maximal matching in multi-hop wireless networks
In this paper, we characterize the performance limits of an important class of scheduling schemes, called Greedy Maximal Matching (GMM), for multi-hop wireless networks. For simplicity, we focus on
the well-established node-exclusive interference model, although many of the stated results can be readily extended to more general interference models. The study of the performance of GMM is
intriguing because although a lower bound on its performance is well known, empirical observations suggest that this bound is quite loose, and that the performance of GMM is often close to optimal.
In fact, recent results have shown that GMM achieves optimal performance under certain conditions. In this paper, we provide new analytic results that characterize the performance of GMM through the
topological properties of the underlying graphs. To that end, we generalize a recently developed topological notion called the local pooling condition to a far weaker condition called the σ-local
pooling. We then define the local-pooling factor on a graph, as the supremum of all σ such that the graph satisfies σ-local pooling. We show that for a given graph, the efficiency ratio of GMM (i.e.,
the worst-case ratio of the throughput of GMM to that of the optimal) is equal to its local-pooling factor. Further, we provide results on how to estimate the local-pooling factor for arbitrary
graphs and show that the efficiency ratio of GMM is no smaller than d* /(2d* - 1) in a network topology of maximum node-degree d*. We also identify a specific network topology for which the
efficiency ratio of GMM is strictly less than 1.
Publication series
Name Proceedings of the IEEE Conference on Decision and Control
ISSN (Print) 0743-1546
ISSN (Electronic) 2576-2370
Conference 46th IEEE Conference on Decision and Control 2007, CDC
Country/Territory United States
City New Orleans, LA
Period 07/12/12 → 07/12/14
ASJC Scopus subject areas
• Control and Systems Engineering
• Modelling and Simulation
• Control and Optimization
Dive into the research topics of 'Performance limits of greedy maximal matching in multi-hop wireless networks'. Together they form a unique fingerprint. | {"url":"https://pure.korea.ac.kr/en/publications/performance-limits-of-greedy-maximal-matching-in-multi-hop-wirele","timestamp":"2024-11-11T01:54:39Z","content_type":"text/html","content_length":"57125","record_id":"<urn:uuid:3d752897-2492-4eb8-9a74-e687b9b606d7>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00203.warc.gz"} |
Blog Series: Building a transformer model – Part 3 – Training the model
In the previous two posts, we built the basic and scalable versions of the Transformer model. Now, it’s time to move on to the next critical step of training the model. In this post, we’ll focus on:
• Preparing the dataset.
• Defining the training loop.
• Using loss functions and optimizers.
• Monitoring performance.
• Understanding the model output.
By the end of this post, you’ll have a Transformer model that can be trained on real data, ready to make predictions.
Preparing the Dataset
The first step in training any machine learning model is to prepare a dataset. In the case of Transformers, the dataset usually consists of sequences of text.
Sample Dataset: Text Tokenization
We’ll use a toy dataset of sentences to demonstrate. In practice, you might use larger datasets like Wikitext, OpenWebText, or other public datasets.
Here’s how we can tokenize a sample dataset using torchtext or the transformers library for tokenization.
from transformers import BertTokenizer
import torch
# Initialize the tokenizer (using BERT's tokenizer for example)
tokenizer = BertTokenizer.from_pretrained('bert-base-uncased')
# Sample dataset (toy example)
sentences = ["The quick brown fox jumps over the lazy dog.",
"The Transformers architecture is very powerful."]
# Tokenize sentences
inputs = tokenizer(sentences, return_tensors="pt", padding=True, truncation=True)
# Extract input ids and attention masks for the Transformer
input_ids = inputs['input_ids']
attention_mask = inputs['attention_mask']
# Print tokenized input
What’s happening here:
• We use BERT’s tokenizer to convert text into sequences of integers (token IDs).
• Attention masks tell the model which parts of the input are padding and which are actual tokens.
Defining the Training Loop
Once we have the dataset ready, we need to define the training loop. The loop is responsible for:
• Passing inputs through the Transformer model.
• Calculating loss between predicted and actual outputs.
• Backpropagating the error and updating model parameters.
Here’s a basic training loop:
import torch
import torch.optim as optim
# Assuming the Transformer model from Post 2
model = Transformer(embed_size=256, heads=8, depth=4, forward_expansion=4, max_len=50, dropout=0.1, vocab_size=30522)
# Define the optimizer and loss function
optimizer = optim.Adam(model.parameters(), lr=0.001)
loss_fn = torch.nn.CrossEntropyLoss()
# Dummy target (random example for simplicity)
target_ids = torch.tensor([[101, 2023, 2003, 1037, 3944, 102], [101, 2057, 2024, 1037, 2037, 102]])
# Training loop
def train(model, input_ids, target_ids, attention_mask, epochs=10):
model.train() # Set the model in training mode
for epoch in range(epochs):
optimizer.zero_grad() # Clear previous gradients
outputs = model(input_ids) # Forward pass through the model
# Reshape outputs and target to match the loss function expectations
outputs = outputs.view(-1, outputs.size(-1)) # Flatten the output to match the target
target_ids = target_ids.view(-1)
# Calculate loss and backpropagate
loss = loss_fn(outputs, target_ids)
loss.backward() # Backpropagate the error
optimizer.step() # Update the model weights
print(f"Epoch {epoch + 1}, Loss: {loss.item()}")
# Example of training
train(model, input_ids, target_ids, attention_mask)
Loss Function and Optimizer
Loss Function: Cross-Entropy Loss
• Since we are dealing with text, the cross-entropy loss function is appropriate. It helps compare the predicted word distribution (output) against the actual word distribution (target).
Optimizer: Adam
• We use Adam optimizer for training. It’s widely used for neural networks because it adapts the learning rate for each parameter.
Monitoring Performance
During training, it’s essential to monitor the loss to ensure that the model is learning effectively. If the loss is not decreasing, the model might need adjustments, such as:
• Learning rate tweaks.
• More epochs.
• Regularization techniques like dropout or weight decay.
To visualize training progress, we can plot the loss over time using matplotlib:
import matplotlib.pyplot as plt
def train_with_monitoring(model, input_ids, target_ids, attention_mask, epochs=10):
loss_values = []
for epoch in range(epochs):
outputs = model(input_ids)
outputs = outputs.view(-1, outputs.size(-1))
target_ids = target_ids.view(-1)
loss = loss_fn(outputs, target_ids)
print(f"Epoch {epoch + 1}, Loss: {loss.item()}")
# Plot loss over epochs
plt.plot(range(1, epochs + 1), loss_values, label="Training Loss")
# Train and monitor
train_with_monitoring(model, input_ids, target_ids, attention_mask)
This will give you a real-time look at how the model is performing across different epochs.
Explanation of the Output
Here’s what happens when you train the Transformer model:
• Input Sentence: The input sentences are tokenized into numbers (IDs), representing the words in a vocabulary. Example input:
tensor([[ 101, 1996, 4248, 2829, 4419, 2058, 1996, 13971, 3899, 102], [ 101, 1996, 17288, 19085, 2324, 2003, 2200, 3787, 102, 0]])
• Target: The target sequence is also tokenized in a similar way. This is used to calculate the loss.
• Output: During each epoch, the model outputs a set of probabilities (logits) for each word, and the loss function compares these predictions to the actual words (target). The loss value is
printed at each epoch. Example output:
Epoch 1, Loss: 5.673 Epoch 2, Loss: 4.532 ...
• Loss Values: As training progresses, the loss decreases, which means the model is improving at predicting the next word in the sequence.
Visual Output:
After running the train_with_monitoring function, you’ll see a loss curve similar to this:
• The x-axis represents the number of epochs.
• The y-axis represents the loss value.
• The curve will ideally go down as the model gets better at predicting the correct words.
Evaluation (Optional)
After training, you may want to evaluate the model on a validation dataset to check its performance before using it for actual predictions.
Example of Model Evaluation:
def evaluate(model, input_ids, target_ids, attention_mask):
model.eval() # Set the model to evaluation mode (no backpropagation)
with torch.no_grad(): # Disable gradient calculation for efficiency
outputs = model(input_ids)
outputs = outputs.view(-1, outputs.size(-1))
target_ids = target_ids.view(-1)
loss = loss_fn(outputs, target_ids)
print(f"Evaluation Loss: {loss.item()}")
# Evaluate the model on validation data
evaluate(model, input_ids, target_ids, attention_mask)
In this post, we walked through the process of training a Transformer model. Here’s a quick recap:
1. We prepared a dataset and tokenized it for the Transformer.
2. We defined a training loop with a loss function and optimizer.
3. We monitored the model’s performance during training using loss values.
4. Finally, we explained the model’s output during training and how to visualize the loss.
With this foundation, you can now train your own Transformer models on any dataset. In the next post, we’ll dive deeper into evaluating model performance and fine-tuning.
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2. What is Inductive Logic?
2. What is Inductive Logic?
In this lecture I want to revisit the first point we raised, which is about inductive logic. I want to lay out some terms here so that it’s clear what we’re talking about, and the role that
probability concepts play in inductive reasoning.
Deductive vs Inductive Logic
We distinguish deductive logic from inductive logic. Deductive logic deals with deductive arguments, inductive logic deals with inductive arguments. So what’s the difference between a deductive
argument and an inductive argument?
The difference has to do with the logical relationship between the premises and the conclusion. Here we’ve got a schematic representation of an argument, a set of premises from we infer some
1. Premise
2. Premise
n. Premise
∴ Conclusion
That three-point triangle shape is the mathematicians symbol for “therefore”, so when you see that just read it as “premise 1, premise 2, and so on, THEREFORE, conclusion.
Now, in a deductive argument, the intention is for the conclusion to follow from the premises with CERTAINTY. And by that we mean that IF the premises are all true, the conclusion could not possibly
be false. So the inference isn’t a risky one at all — if we assume the premises are true, we’re guaranteed that the conclusion will also be true.
For those who’ve worked through the course on “Basic Concepts in Logic and Argumentation”, you’ll recognize this as the definition of a logically VALID argument. A deductive argument is one that is
intended to be valid. Here’s a simple example, well-worn example.
1. All humans are mortal.
2. Socrates is human.
∴ Socrates is mortal.
If we grant both of these premises, it follows with absolute deductive certainty that Socrates must be mortal.
Now, by contrast, with inductive arguments, we don’t expect the conclusion to follow with certainty. With an inductive argument, the conclusion only follows with some probability, some likelihood.
This makes it a “risky” inference in the sense that, even if the premises are all true, and we’re 100 % convinced of their truth, the conclusion that we infer from them could still be false. So
there’s always a bit of a gamble involved in accepting the conclusion of an inductive argument.
Here’s a simple example.
1. 90% of humans are right-handed.
2. John is human.
∴ John is right-handed.
This conclusion obviously doesn’t follow with certainty. If we assume these two premises are true, the conclusion could still be false, John could be one of the 10% of people who are left-handed. In
this case it’s highly likely that John is right-handed, so we’d say that, while the inference isn’t logically valid, it is a logically STRONG inference. On the other hand, an argument like this …
1. Roughly 50% of humans are female.
2. Julie has a new baby.
∴ Julie’s baby is female.
... is not STRONG. In this case the odds of this conclusion being correct are only about 50%, no better than a coin toss. Simply knowing that the baby is human doesn’t give us good reasons to infer
that the baby is a girl; the logical connection is TOO WEAK to justify this inference.
These two examples show how probability concepts play a role in helping us distinguish between logically strong and logically weak arguments.
Now, I want to draw attention to two different aspects of inductive reasoning.
Two Different Questions to Consider
When you’re given an inductive argument there are two questions that have to be answered before you can properly evaluate the reasoning.
The first question is this: How strong is the inference from premises to conclusion? In other words, what is the probability that the conclusion is true, given the premises?
This was easy to figure out with the previous examples, because the proportions in the population were made explicit, and we all have at least some experience with reasoning with percentages — if 90%
of people are right-handed, and you don’t know anything else about John, we just assume there’s a 90% chance that John is right-handed, and 10% chance that he’s left-handed. We're actually doing a
little probability calculation in our head when we draw this inference.
This is where probability theory can play a useful role in inductive reasoning. For more complicated inferences the answers aren’t so obvious. For example, if I shuffle a deck of cards and I ask you
what are the odds that the first two cards I draw off the top of the deck will both be ACES, you’ll probably be stumped. But you actually do have enough information to answer this question, assuming
you’re familiar with the layout of a normal deck of cards. It’s just a matter of using your background knowledge and applying some simple RULES for reasoning with probabilities.
Now, the other question we need to ask about inductive arguments isn’t so easy to answer.
The question is, how high does the probability have to be before it’s rational to accept the conclusion?.
This is a very different question. This is a question about thresholds for rational acceptance, how high the probability should be before we can say “okay, it’s reasonable for me to accept this
conclusion — even though I know there’s still a chance it’s wrong”. In inductive logic, this is the threshold between STRONG and WEAK arguments — strong arguments are those where the probability is
high enough to warrant accepting the conclusion, weak arguments are those where the probability isn’t high enough.
Now, I’m just going to say this up front. THIS is an unresolved problem in the philosophy of inductive reasoning. Why? Because it gets into what is known as the “problem of induction”.
This is a famous problem in philosophy, and it’s about how you justify inductive reasoning in the first place. The Scottish philosopher David Hume first formulated the problem and there’s no
consensus on exactly how it should be answered. And for those who do think there’s an answer and are confident that we are justified in distinguishing between strong and weak inductive arguments, the
best that we can say is that it’s at least partly a conventional choice where we set the threshold.
To refine our reasoning on this question we need to get into rational choice theory where we start comparing the costs and benefits of setting the bar too low versus the costs and benefits of setting
it high, and to make a long story short, that’s an area that I’m not planning on going into in this course.
In this course we’re going to stick with the first question, and look at how probability theory, and different interpretations of the probability concept, can be used to assign probabilities to
individual claims AND to logical inferences between claims.
With this under our belt we’ll then be in a good position to understand the material on probabilistic fallacies and probability blindness, which is really, really important from a critical thinking | {"url":"https://criticalthinkeracademy.com/courses/what-is-probability/lectures/51587","timestamp":"2024-11-02T03:21:36Z","content_type":"text/html","content_length":"104797","record_id":"<urn:uuid:cb28cef4-bac6-459c-8ae7-d4232b35248e>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00429.warc.gz"} |
Important Questions for CBSE Class 9 Maths, Chapter wise Solutions - Free PDF Download - CoolGyan
CBSE Class 9 Maths Chapter wise Important Questions – Free PDF Download
Math’s requires a strategic approach and the important questions for class 9 Maths provided on our website for every chapter form a cornerstone for thorough preparation. Math is one of the most
important subjects in your board exams. This subject can help to get a high overall score and hence requires dedicated preparation.
In the run-up to board exams of class 10, it is important to build the base in class 9 itself. Getting a good score in Maths in class 9 exams will not only boost your confidence but will also give
you the motivation to study for class 10. In addition, our website also has important questions for class 9 Math’s chapter-wise solutions which will help students to refer to the answers once they
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You can also Download NCERT Solutions for Class 9 Maths to help you to revise the complete Syllabus and score more marks in your examinations. CoolGyan provides students with a Free PDF download
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Chapter wise Important Questions for CBSE Class 9 Maths
Important Questions for Class 9 Maths CBSE Board
Our aim is to make this question bank available to every student. This will enable you to practice the subject better and score high marks in the exam. Practising these important questions is a good
way to ensure that you have not missed out on any topic and that you have complete coverage of the syllabus. Also, students can use these questions to take a mock test and use it as a tool to gauge
their preparation. Once done answering these questions, students can access important questions class 9 Math’s solution in free pdf download option on our website. These solutions can be used as a
reference to understand how answers are written in the exam.
The entire section of NCERT CBSE board important questions for class 9 Math’s with answers is curated by our team of highly experienced teachers who understand the pulse of the examiners and have
gained experience over the years to understand what are the important topics and how they need to be presented in the exam.
The questions and answers are developed by referring to previous year question papers and question bank given in the NCERT textbooks. The format of the answers is in accordance with the latest
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A half-plane is one part of a plane divided by a line, referred to as the "half-plane axis" or "origin of the half-plane."
This concept comes from the two-dimensional plane.
For example, if I draw a line on a piece of paper, I create two half-planes: one on the left side of the line and one on the right.
Each half-plane consists of an infinite number of points.
Note: Generally, the half-plane includes both the points within one of the regions created by the division and the points on the half-plane axis itself.
Half-Plane Postulates
A half-plane must satisfy the properties of the postulate of plane partition by a line.
Given a line that splits the plane into two sets of points not on the line:
• If we consider any two points P and Q in the same region, the segment connecting them does not intersect the line.
• If we consider any two points P and Q in different regions, the segment connecting them intersects the line.
If a set of points meets this postulate, then it is considered a half-plane.
Types of Half-Planes
There are two types of half-planes:
• Open Half-Planes
An open half-plane does not include the dividing line. It consists only of the points inside the half-plane.
• Closed Half-Planes
A closed half-plane includes the dividing line. It consists of both the internal points and the points on the dividing line.
Thus, in the case of a plane divided into two closed half-planes, the points on the half-plane axis are considered to belong to both half-planes.
And so on. | {"url":"https://www.andreaminini.net/math/half-plane","timestamp":"2024-11-01T21:56:48Z","content_type":"text/html","content_length":"12114","record_id":"<urn:uuid:3266e709-fef8-4462-9bd1-b5faf05313ee>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00737.warc.gz"} |
Semi-Supervised LDA: Thoughts
My current problem domain is characterized by large amounts of unlabeled data and a smaller amount of labeled data. Recently I
experienced some success
by applying LDA on the unlabeled data to create a representation of the social graph, and using the resulting features in my supervised classifier.
Several papers have pointed out that simultaneously extracting topics and estimating a classifier can yield better performance than the above procedure of 1) extracting topics without regard to the
classification problem and then 2) using the topics in a supervised classifier. Unfortunately the papers I've read assume that the document class is always observed. My case is more classic
semi-supervised: I want to leverage all the data, unlabeled and labeled, to build the best classifier possible. Surprisingly I don't see this case being explicitly treated in the literature, although
I feel it is typical.
LDA is a generative model, so intuitively it feels like it should be easy to adapt it to the case where associated document information is only partially observed: in practice there are some
pitfalls. I'm going to dig into the inference strategies for two extensions of LDA designed for use with (fully observed) associated document labels to see if I can find a way to adapt these to the
case of semi-supervised data.
Supervised LDA
First up is the
supervised LDA approach
from ``the Godfather" himself (Blei) and Jon McAuliffe. Before I get into the details, I'll repeat the overall result from the paper: jointly estimating topics and the classifier leads to better
overall performance than estimating topics first and then a classifier separately. Indulge me another reminder that this result was demonstrated when supervised information was associated with every
Here is the slate representation of the model from Figure 1 of the paper.
The model is similar to the original LDA, but with an extra label emission step.
1. Draw topic proportions $\theta | \alpha \sim \mbox{Dir} (\alpha)$.
2. For each word
1. Draw topic assignment $z_n | \theta \sim \mbox{Mult} (\theta)$.
2. Draw word $w_n | z_n, \beta_{1:K} \sim \mbox{Mult} (\beta_{z_n})$.
3. Draw response variable $y | z_{1:N}, \eta, \delta \sim \mbox{GLM} (\tilde z, \eta, \delta)$.
where $\tilde z = \frac{1}{N} \sum_n z_n$ is the empirical topic frequency in the document, and
is a generalized linear model.
In the supervised LDA paper, inference proceeds by variational EM. The auxiliary function $\mathcal{L}$ is derived using a variational distribution $q$ where each topic vector is drawn from a
per-document $\gamma$ parametrized Dirichlet distribution and each word is drawn from a per-document-position $\phi_n$ parametrized Multinomial distribution. For a single document this looks like, \[
\log p (w_{1:N}, y | \alpha, \beta_{1:K}, \eta, \delta)
&\geq \mathcal{L} (\gamma, \phi_{1:N}; \alpha, \beta_{1:K}, \eta, \delta) \\
&= E_q[ \log p (\theta | \alpha)] + \sum_{n=1}^N E_q[\log p (Z_n | \theta)] \\
&\;\;\;\; + E_q[\log p(y | Z_{1:N}, \eta, \delta)] + H (q).
\] Ok so no problem adapting for the semi-supervised case, right? In those cases $y$ is not observed for the document in question, so the $E_q[\log p(y | Z_{1:N}, \eta, \delta)]$ term goes away in
the auxiliary function for that document, and basically the variational parameters for unlabeled documents follow the update rules from the original LDA. So basically I need to take the
public implementation of sLDA
and modify it to accept a ``nothing observed'' class label which elides the corresponding portions of the objective function.
Well as Matt Hoffman pointed out to me, this might not behave as I hope. Since the unlabeled data vastly outnumbers the labeled data, most of the $\phi_n$ will not be under any pressure to be
explanatory with respect to the known labels, and these will dominate the ultimate estimation of the $\beta_{1:K}$. This is because the M-step for $\beta_{1:K}$ is given by \[
\hat \beta_{k,w} \propto \sum_{d=1}^D \sum_{n=1}^{N_d} 1_{w_{d,n} = w} \phi^k_{d,n}.
\] Therefore the prediction is that as I throw more unlabeled data at this technique, it will degenerate into something equivalent to running a topic estimator first and then a classifier second.
Badness 10000?
Perhaps not. Given a set of $D_s = \{ (d, l) \}$ of labeled documents and a larger set of $D_u = \{ d \}$ of unlabeled documents, which is better?
1. Run supervised LDA on $D_s$ and ignore $D_u$.
2. Run unsupervised LDA on $D_s \cup D_u$ and then classify $D_s$.
Intuition suggests the second approach is better: after all, the first is ignoring most of the data. Perhaps the generative model is saying that when the amount of unsupervised data is massive,
separate feature extraction and classification steps are close to optimal.
Still the situation is unsatisfactory. One idea would be to importance weight the documents with labels on them so that they don't get overwhelmed by the unlabeled data. This has the practical virtue
that it would be a hopefully straightforward modification of the publicly available sLDA implementation (basically, treat each labeled document as if it occurs multiple times; therefore in the above
M-step for $\beta_{1:K}$, overweight the $\phi$ associated with labeled documents). However it feels dirty.
Another idea is to take inspiration from the
transductive SVM
. In the binary case, the basic idea is that although the labels on the unlabeled data are unknown, they are believed to be either 0 or 1; therefore in practice the decision boundary should avoid
areas of high density in the unlabeled distribution. Additionally, one can also use the empirical distribution of labels to prefer decision boundaries that create a similar distribution of labels in
the unlabeled data. Analogously for ``transductive LDA'' in the binary case I'll need a prior on $y$ which prefers extreme values for unlabeled points, possibly biased to promote a certain label
distribution on the unlabeled data. Ultimately this would imply the $E_q[\log p(y | Z_{1:N}, \eta, \delta)]$ term in the variation bound becomes, for unlabeled points, an $E_q[E_\zeta (\log p (y | Z_
{1:N}, \eta, \delta)]$ term where $\zeta$ is the extreme value preferring prior on $y$. Since an easy way to shatter the unlabeled data would be to set $||\eta|| \to \infty$ or $\delta \to 0$ some
prior distributions on those parameters will be necessary to keep things under control.
While I'm very excited about ``transductive LDA'', in practice I think it would take me a long time to get it working.
Next up is
from Julien, Sha, and the ``Michael Jordan of statistics''. Here is the slate representation of the model from Figures 1 through 3 of the paper.
The model modifies the original LDA such that a document class label $y_d$ can remap the original document topics $z_d$ into transformed topics $u_d$ which then control the word emission
1. Draw topic proportions $\theta | \alpha \sim \mbox{Dir} (\alpha)$.
2. Draw word emission vectors $\phi | \beta \sim \mbox{Dir} (\beta)$.
3. Draw class label $y | \pi \sim p (y | \pi)$ from some prior distribution.
4. For each word
1. Draw topic assignment $z_n | \theta \sim \mbox{Mult} (\theta)$.
2. Draw transformed topic assignment $u_n | z_n, T, y \sim \mbox{Mult} (T^y_{u_n, z_n})$.
3. Draw word $w_n | z_n, \phi_{1:K} \sim \mbox{Mult} (\phi_{z_n})$.
In practice the $T$ matrices are fixed to be a mixture of block zeroes and block diagonals which basically arranges for 1) some number of topics $K_1$ shared across all class labels and 2) some
number $|Y| K_0$ of topics of which each class label gets $K_0$ topics reserved for use with that particular class label, \[
T^y = \begin{pmatrix}
0 & I_{K_0} 1_{y = 1} \\
\vdots & \vdots \\
0 & I_{K_0} 1_{y = |Y|} \\
I_{K_1} & 0
\] Used in this manner, the resulting model is very similar to
Labeled LDA
; and in fact I recommend reading the Labeled LDA paper to get an intuition of what happens in this case (quick summary: the collapsed Gibbs sampler is the same as vanilla LDA, except that
transitions are only allowed between feasible topics according to the class label.)
For the semi-supervised case we can start by saying that $y$ is not always observed. However it becomes clear in the DiscLDA generative model that we need to marginalize over $y$. In other words,
unlabeled documents would not be restricted to using only the $K_1$ topics shared across all class labels. Instead, they would also be allowed to use some mixture of the per-class-label topics,
relating to the posterior distribution $p (y_d | \cdot)$ of the unobserved class label.
So what happens when the amount of unlabeled data vastly outnumbers the amount of labeled data? I could end up degenerating into something equivalent to vanilla LDA, e.g., in the binary case most of
the unobserved labels will end up with $p (y_d = 1 | \cdot) \approx 1/2$ which means that in practice all the per-class topics are being ``washed out''. So again I might need a ``transductive
prior'', which is really a prior on distributions of $y$, which prefers low entropy distributions over the class label.
In practice I'll use DiscLDA (Labeled LDA) as a starting point. This is basically because the collapsed Gibbs sampler is straightforward to implement in case of restricted transformation matrices.
Getting this to work in the semi-supervised case will presumably be challenging because the model might have the propensity to let the invented class labels on the unlabeled data dominate the
actually observed class labels on the labeled data. | {"url":"http://www.machinedlearnings.com/2011/04/semi-supervised-lda-thoughts.html","timestamp":"2024-11-09T04:27:30Z","content_type":"application/xhtml+xml","content_length":"89814","record_id":"<urn:uuid:f12d98db-d2d7-43c0-8816-434b8dc10055>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00731.warc.gz"} |
The potential at a point xmeasured in mu m due to some class 12 physics JEE_Main
Hint: We have the relationship between electric field and potential difference which is given by:
$E = \dfrac{V}{d}$. Where E is the electric field of the charge present, V is the potential difference of the cell, d is the length of balance for E.
Where electric field is given when change in potential takes place by:
$E = - \dfrac{{dV}}{{dx}}$(Minus sign shows the opposite direction from potential difference)
Using the above relations we will solve the problem.
Complete step by step solution:
Let’s define the electric field and potential difference first.
Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is
radially outwards from the positive charge and radially inwards from the negative charge. Electric field is generated around positive or negative charge, field may be attracting or repelling.
Potential difference: Potential difference: potential difference is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test
charge between the two points.
Now comes the calculation part:
$V(x) = \dfrac{{20}}{{({x^2} - 4)}}$ (Potential difference)
Where electric field is given as,
$E = - \dfrac{{dV}}{{dx}}$ (negative sign indicates direction of electric field is opposite to the direction of potential difference)
Let’s do the differentiation of potential differences.
$ \Rightarrow E = - \dfrac{{d\dfrac{{20}}{{({x^2} - 4)}}}}{{dx}}$
$ \Rightarrow E = - \dfrac{{ - 40x}}{{{{({x^2} - 4)}^2}}}$
$ \Rightarrow E = \dfrac{{40x}}{{{{\left( {{x^2} - 4} \right)}^2}}}$ ...............(1)
Value of x is given as $x = 4\mu m$
$ \Rightarrow E = \dfrac{{40 \times 4}}{{{{\left( {{4^2} - 4} \right)}^2}}} \\
\Rightarrow E = \dfrac{{10}}{9}volt/\mu m \\$ (We have substituted the value of x in equation 1)
Electric field is positive which means the positive direction is in the positive direction of x.
Thus option (A) is correct.
Note: Electric field has many applications which we observe in our daily life such used in Van de graaff generators, Xerography is a dry copying process based on electrostatics, laser printers,
inkjet printers and many other uses in medical fields such as MRI machines. | {"url":"https://www.vedantu.com/jee-main/the-potential-at-a-point-xmeasured-in-mu-m-due-physics-question-answer","timestamp":"2024-11-13T04:48:12Z","content_type":"text/html","content_length":"155225","record_id":"<urn:uuid:2d205dae-cc5b-47b5-b1d3-058b8c64a682>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00693.warc.gz"} |
seminars - An introduction to geometric representation theory and 3d mirror symmetry
※ 2월 21일(화), 23일(목), 28일(화), 10:30-12:00
The Beilinson-Bernstein theorem, which identifies representations of semi-simple Lie algebra \mathfrak{g} with D-modules on the flag variety G/B, makes it possible to use powerful techniques from
algebraic geometry, especially Hodge theory, to attack problems in representation theory. Some successes of this program are the proofs of the Kazhdan-Lusztig and Jantzen conjectures as well as
discovery that the Bernstein-Gelfand-Gelfand categories O for Langlands dual Lie algebras are Koszul dual.
The modern perspective on these results places them in the context of deformation quantizations of holomorphic symplectic manifolds: The universal enveloping algebra U(\mathfrak{g}) is isomorphic to
the ring of differential operators on G/B which is a non-commutative deformation of the ring of functions on the cotangent bundle T^*G/B. Thanks to work of Braden-Licata-Proudfoot-Webster it is known
that an analogue of BGG category O can be defined for any associative algebra which quantizes a conical symplectic resolution. Examples include finite W-algebras, rational Cherednik algebras, and
hypertoric enveloping algebras.
Moreover BLPW collected a list of pairs of conical symplectic resolutions whose categories O are Koszul dual. Incredibly, these “symplectic dual” pairs had already appeared in physics as Higgs and
Coulomb branches of the moduli spaces of vacua in 3d N=4 gauge theories. Moreover, there is a duality of these field theories known as 3d mirror symmetry which exchanges the Higgs and Coulomb branch.
Based on this observation Bullimore-Dimofte-Gaiotto-Hilburn showed that the Koszul duality of categories O is a shadow of 3d mirror symmetry.
In this series of lectures I will give an introduction to these ideas assuming only representation theory of semi-simple Lie algebras and a small amount of algebraic geometry. | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&sort_index=speaker&order_type=desc&page=76&document_srl=1036799","timestamp":"2024-11-08T06:26:52Z","content_type":"text/html","content_length":"47061","record_id":"<urn:uuid:b7ff83b8-c38f-449a-87df-e42a0d162f4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00275.warc.gz"} |
Antonio Garcia-Martinez - MATLAB Central
Antonio Garcia-Martinez
Last seen: Today |  Active since 2012
Followers: 0 Following: 0
Expert Matlab programing, conventional and advanced control designer. State estimator designer/filters. Work with Matlab, Simulink, Arduino, LabVIEW and their linked. Theorical and experimental
Programming Languages:
Python, MATLAB, Arduino, Visual Basic
Spoken Languages:
English, Spanish
Professional Interests:
Simulink, Control Design, Modeling and Prediction, Sciences, Robotics
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Tutorials Archives | Data + People
I first heard this term used at the Tableau Conference a few years ago. In one of their first HR tracks, I attended every presentation remotely close to the intersection of HR and Tableau. The
disclaimer from every presenter at the start included a blurb that we weren’t looking at real data. I would have cringed had I not heard this. One presenter though introduced the term ‘fata’ – fake
data. I’ve adopted it since.
In the pursuit of sharing ideas in the space of People Analytics, one hurdle is the extremely sensitive nature of the actual data we’re working with. Names, emails, gender, age, social security
numbers, and much more are all often part of an employee data file and useful in analysis. However, sharing this information is not proper and could result in you losing your job in people analytics.
Especially outside your organization, but even within, the information must be protected from those who shouldn’t have access.
This tutorial will show a few ways in which you can create this data. Useful for development, sharing internally, and presenting your work to the work – without losing your job.
Using one of these options you’ll have the ability to create a complete data set, quickly, and without exposing any real user or employee data.
view on github
Photo by Mohsin Khusro on Unsplash
Beginner’s Guide: Python for Analytics | Seaborn
Beginner’s Guide to Using Python with HR Data | Exploration Series
Part Three – Seaborn
In this first tutorial series, I’m exploring the IBM HR Attrition and Performance data set. This is a great data set used to demonstrate the possibilities from using machine learning and other data
science techniques.
Now we’ll move on to using Seaborn for our visualizations. The benefit of Seaborn is it continues to abstract the complex, underlying calls to visualize your data – further allowing you to focus on
your analysis task and not having to think about how to implement what you want to do. It goes even further and provides built-in functionality that would be incredibly complex to implement without
the benefit of Seaborn.
Series Outline
0: basic operations & summary statistics
1: matplotlib
2: pandas visualization
3: seaborn
4: plotly
5: series summary
3: Seaborn
view on github
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Photo: Photo by Randall Ruiz on Unsplash
Beginner’s Guide: Python for Analytics | Pandas
Beginner’s Guide to Using Python with HR Data | Exploration Series
Part Two – Pandas
In this first tutorial series, I’m exploring the IBM HR Attrition and Performance data set. This is a great data set used to demonstrate the possibilities from using machine learning and other data
science techniques.
Next, we’ll take a look at the power of Pandas to plot our data. As a budding data [analyst/scientist/enthusiast], Pandas has become my most common import and tool. Plotting directly from pandas
objects makes it very easy to stay in the flow of analyzing data. Let’s get going.
Series Outline
0: basic operations & summary statistics
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3: seaborn
4: plotly
5: series summary
2: Pandas
view on github
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Beginner’s Guide: Python for Analytics | Matplotlib
Beginner’s Guide to Using Python with HR Data | Exploration Series
Part One – Matplotlib
In this first tutorial series, I’m exploring the IBM HR Attrition and Performance data set. This is a great data set used to demonstrate the possibilities from using machine learning and other data
science techniques.
In this next walkthrough, we’ll begin to ‘see’ our data through the use of visualization packages. In R there are 3 commons plotting tools, and other packages extend these main items. In Python,
there is Matplotlib, and most other packages build on this foundation. So, the decision of where to start with Python plotting is an easy one – let’s get going.
Series Outline
0: basic operations & summary statistics
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4: plotly
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view on github
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Beginner’s Guide: Python for Analytics | The Basics
Beginner’s Guide to Using Python with HR Data | Exploration Series
Part Zero – The Basics
In this first tutorial series, I’m exploring the IBM HR Attrition and Performance data set. This is a great data set used to demonstrate the possibilities from using machine learning and other data
science techniques.
I’ll be back with tutorial posts that walk through how to apply more advanced techniques to generate predictive and prescriptive insights from the data. But that’d be jumping ahead. First, the
basics. Exploratory Data Analysis, or EDA.
It’s often tempting to jump right in and try to find the most advanced insight possible. When I’m in the process of learning something new, it’s my first instinct to begin applying it straight away,
skipping the basics. Eventually, I’ll stumble; and it’s always something I could have avoided by simply spending a little bit of time really understanding the data I have.
To properly analyze data, you must understand it. Is it complete (missing values), are the errors (values out of normal bounds – is this correct), and generally what information is contained within
the data? Depending on where the request is coming from in a work-context, you may not control the data, so what you get is what you have; it’s often much easier when you’ve pulled your own data –
it’s just not always possible, or even smart to do so.
Always begin with an exploration of your data. In this tutorial, I’m digging out my current favorite tool – Python. If you’ve never programmed, if Excel still frightens you a bit, or you’re firmly in
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view on github | {"url":"https://www.datapluspeople.com/category/tutorials/","timestamp":"2024-11-13T13:16:08Z","content_type":"text/html","content_length":"68889","record_id":"<urn:uuid:9d21347d-ea7e-4074-9b69-a77051b1acb7>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00109.warc.gz"} |
Area of Hendecagon Calculators | List of Area of Hendecagon Calculators
List of Area of Hendecagon Calculators
Area of Hendecagon calculators give you a list of online Area of Hendecagon calculators. A tool perform calculations on the concepts and applications for Area of Hendecagon calculations.
These calculators will be useful for everyone and save time with the complex procedure involved to obtain the calculation results. You can also download, share as well as print the list of Area of
Hendecagon calculators with all the formulas. | {"url":"https://www.calculatoratoz.com/en/area-of-hendecagon-Calculators/CalcList-2540","timestamp":"2024-11-03T21:48:05Z","content_type":"application/xhtml+xml","content_length":"118757","record_id":"<urn:uuid:1ce83888-d3fe-4310-8af0-3ec90fd4a805>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00659.warc.gz"} |
Very Short Introductions – Book Review 1
Very Short Introductions – Book Review 1
I love the “Very Short Introductions” series by Oxford University Press as generally they are very well written and accessible to a general audience but not patronising.
Number 260 in this series, “Numbers” by Peter M. Higgins is no exception to this rule. The text flows nicely, and (considering the brevity of the text) is a comprehensive introduction to the types
of number – starting from the counting numbers, moving through rationals, reals and transcendentals before looking at different types of infinity and countable versus uncountable sets. The book ends
with an introduction to complex numbers and operations on complex numbers.
Here are a few highlights from the book:
• When discussing primes there is the following aid to memory given “inequality signs always point to the smaller quantity”. It may seem stupid, but I don’t think I have ever said that.
• He often mentions things but doesn’t discuss them in full – this is really effective at encouraging people to go away and find out more – for example the Cattle Problem attributed to Archimedes.
• There is a great exploration of Cantor’s Middle Third Set where there is a suggestion that using a ternary representation of numbers instead of decimal makes it easier to see that there are
infinitely many numbers in the Middle Third Set. Recall that to construct the Middle Third Set you start with the unit interval \([0,1] \) and delete the middle third (i.e remove all the numbers
between \(\frac{1}{3}\) and \(\frac{2}{3}\).) This process is then repeated recursively on the remaining intervals. If you represent numbers in ternary then working in base 3 a third is given by
0.1 and 0.2 represents two thirds. So removing the first middle portion removes all the numbers whose ternary expansion starts with a 1, the next iteration removes all those numbers remaining
which have a 1 in the second place in their ternary expansion and so on. So in the end, all numbers with a 1 anywhere in their ternary expansion have been removed. I really like this approach, as
it makes it a lot clearer that Cantor’s Middle Third Set is uncountable.
• Introducing matrices by relating them to complex numbers is really nice. The number \(z = a+ i b \) can be represented by the matrix \( \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \). If I do
ever write an A-Level textbook I think I could be tempted to introduce matrices after complex numbers to make the link with matrix multiplication etc.
I’d encourage anyone to get this book and they are also all £4.99 instead of £7.99 at the moment if you buy them through Blackwells. | {"url":"http://blog.ifem.co.uk/blog/2015/08/15/very-short-introductions-book-review-1/","timestamp":"2024-11-04T01:04:49Z","content_type":"text/html","content_length":"77027","record_id":"<urn:uuid:ca21d58c-a81b-41d0-8712-3b88e7115f8c>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00722.warc.gz"} |
BRep Edges
Home Page > Models > Entities > Geometry > BRep > Edges Go to DGKC docs Search Documentation
BRep Edges
BRep Edges are part of the BRep structure, which represent the curved sides of Faces. Edges are added to a shape as part of wires. Edges are shared by wires of adjacent faces.
Edges always have finite length. Its ends are represented as BRep vertices (IBRepVertex_DG). Edge can be closed, in which case its ends is the same vertex.
Edges are accessed via IBRepEdge_DG obtained with IBRepWire_DG.GetEdge(), IBRepShape_DG.GetSubShapes() and other ways.
The actual geometry of an edge is defined by its 3D curve, accessed via IBRepEdge_DG.GetCurve(). The first vertex (vertex 0) geometrically coincides with the first end of the curve (and the second
vertex coincides with the second end of the curve). When a curve is attached to an edge, its natural (standalone/unattached) parameter range is reduced to comply with this restriction.
An edge has internal direction from the first vertex to the second along the curve as defined by the 3D curve parameter. ICurve_DG.GetTangent() returns the direction at any point.
An edge in a valid surface has references to attached faces. These are returned by IBRepEdge_DG.GetFace(int side). The side parameter 0 specifies the face on the felft hand side looking from positive
side of the face (from end of a normal). An external edge of an open surface (like a rectangle) references a single face.
An initialised edge in a BRep shape also has zero or more 2D curves, one per each adjacent face, in u,v plane of parameters of surface of the face. They are called P-curves. Normally P-curves are
created automatically as result of initialisation. So, it is often enough to define only the 3D curve. All curves (3D, and P-curves) have the same range and parametrisation: For a parameter u, points
returned by all curves either coincide or map to the same 3D point on the surface.
An edge is positively (forward) oriented in a wire (and hence in a face) if its internal direction (see above) is counter-clock-wise while looking from the positive side of the face. Otherwise the
edge is called reversed or negative.
As vertices of a wire are always indexed in CCW direction, an edge is positively (forward) oriented in a wire if its second vertex, as returned by IBRepEdge_DG.GetEndVertex(), is next after the first
in the wire. More precisely, i-th positively oriented edge in a wire has ends, which are i-th and i+1-th (modulo count) vertices of the wire. Negative (reversed) i-th edge has ends in the opposite
order: (i+1)%count is the first end and i-th is the second.
If an edge is positively oriented in one wire, its orientation is negative is the adjacent wire, which shares the edge. And vice versa. Orientation can be queried via IBRepWire_DG.IsEdgeReversed().
A box, for an example, is a solid with a single shell, six faces, six wires, twelve edges, and eight vertices | {"url":"https://dynoinsight.com/Help/V7_1/AX/Models/Entities/Geometry/BRep/Edges.aspx","timestamp":"2024-11-14T15:08:41Z","content_type":"application/xhtml+xml","content_length":"100639","record_id":"<urn:uuid:0709e59a-9364-4355-aef5-3e4da550a354>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00478.warc.gz"} |
George Dantzig
by Andrew Boyd
Today, a Nobel Prize lost. The University of Houston's College of Engineering presents this series about the machines that make our civilization run, and the people whose ingenuity created them.
George B. Dantzig was born in 1914, named after Irish playwright George Bernard Shaw. His Russian father, Tobias, briefly studied with French mathematician Henri Poincaré before emigrating to the
U.S. When his son showed proclivity in geometry, Tobias fed the boy's interest with geometry problems. Dantzig would later comment that "the mental exercise required to solve them was the great gift
from my father."
In 1939, Dantzig began his doctorate in statistics at Berkeley. But, with WW-II underway, he soon put his studies on hold and went to work as Head of the Combat Analysis Branch at the Air Force
Headquarters for Statistical Control. The armed forces had to get trucks, planes, people, food -- you name it -- into the field. The detailed plans for doing so were called "programs." These programs
would serve as the foundation of Dantzig's legacy.
Just as there are many ways to get from New York to Los Angeles, there are many programs that will solve a military planning problem. Some are better than others. We need to know the possible
solutions; then we need to know how to find the best one. It isn't easy. There are usually more solutions than we could ever evaluate one by one in the next billion years.
Dantzig's efforts gave birth to one of the most important mathematical tools ever used for solving complex decision problems. That tool is linear programming. The word "programming" does not refer to
computer programs. It refers to those old military problems. His algorithm is called the simplex method.
Dantzig is known throughout the world as the father of linear programming. He received countless honors and awards in his life, including the National Medal of Science. But he was passed over by the
Nobel Prize committee, even though linear programming was not. That decision shocked and dismayed many. Tjalling Koopmans was deeply distressed. He shared the 1975 Nobel Prize with Leonid Kantorovich
for their contributions to linear programming. Koopmans even approached Kantorovich about refusing the prize. Nothing came of it, but he did donate what would've been Dantzig's share of the prize
money to the institute where they'd once worked together.
We can only speculate about the decision of the prize committee, but there's an important distinction between Dantzig's work and the work of Koopmans and Kantorovich. The two prize winners emphasized
economic theory. Dantzig emphasized the mathematical engineering needed to solve linear programs. There's no Nobel Prize in engineering or mathematics. Maybe there should be.
I'm Andy Boyd, at the University of Houston, where we're interested in the way inventive minds work.
(Theme music)
See: George Dantzig. Retrieved January 31, 2008.
Press release for the 1975 Nobel Prize in Economics. Retrieved January 31, 2008.
Simplex algorithm. Retrieved January 31, 2008, from Wikipedia. | {"url":"https://engines.egr.uh.edu/episode/2327","timestamp":"2024-11-04T17:03:20Z","content_type":"text/html","content_length":"31847","record_id":"<urn:uuid:0c158129-efb1-4e2b-9c59-aeb89011205c>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00575.warc.gz"} |
Accounting Rate of Return Definition, Formula Calculate ARR - Xây Nhà Trọn Gói Tại Buôn Ma Thuột - Đắk Lắk
Someone on our team will connect you with a financial professional in our network holding the correct designation and expertise. Our mission is to empower readers with the most factual and reliable
financial information possible to help them make informed decisions for their individual needs. Our writing and editorial staff are a team of experts holding advanced financial designations and have
written for most major financial media publications. Our work has been directly cited by organizations including Entrepreneur, Business Insider, Investopedia, Forbes, CNBC, and many others.
1. Accounting Rate of Return formula is used in capital budgeting projects and can be used to filter out when there are multiple projects, and only one or a few can be selected.
2. For example, if the minimum required return of a project is 12% and ARR is 9%, a manager will know not to proceed with the project.
3. Accounting Rate of Return, shortly referred to as ARR, is the percentage of average accounting profit earned from an investment in comparison with the average accounting value of investment over
the period.
4. The Accounting Rate of Return (ARR) provides firms with a straight-forward way to evaluate an investment’s profitability over time.
5. While it can be used to swiftly determine an investment’s profitability, ARR has certain limitations.
6. Calculate the denominator Look in the question to see which definition of investment is to be used.
Every investment one makes is generally expected to bring some kind of return, and the accounting rate of return can be defined as the measure to ascertain the profits we make on our investments. If
the ARR is positive (equals or is more than the required rate of return) for a certain project it indicates profitability, if it’s less, you can reject a project for it may attract loss on
investment. To find this, the profit for the whole project needs to be calculated, which is then divided by the number of years for which the project is running (in this case five years). An ARR of
10% for example means that the investment would generate an average of 10% annual accounting profit over the investment period based on the average investment. In today’s fast-paced corporate world,
using technology leave management for xero to expedite financial procedures and make better decisions is critical.
Investment evaluation, capital budgeting, and financial analysis are all areas where ARR has a strong foundation. Its adaptability makes it useful for a wide range of applications, including
assessing the economic profitability of projects, benchmarking performance, and improving resource allocation. Kings & Queens started a new project where they expect incremental annual revenue of
50,000 for the next ten years, and the estimated incremental cost for earning that revenue is 20,000. Based on this information, you are required to calculate the accounting rate of return.
The Accounting Rate of Return is also sometimes referred to as the “Internal Rate of Return” (IRR). Candidates should note that accounting rate of return can not only be examined within the FFM
syllabus, but also the F9 syllabus. Of course, that doesn’t mean too much on its own, so here’s how to put that into practice and actually work out the profitability of your investments. Note that
the value of investment assets at the end of 5th year (i.e. $50m) is the sum of scrap value ($10 m) and working capital ($40 m). HighRadius Autonomous Accounting Application consists of End-to-end
Financial Close Automation, AI-powered Anomaly Detection and Account Reconciliation, and Connected Workspaces. Delivered as SaaS, our solutions seamlessly integrate bi-directionally with multiple
systems including ERPs, HR, CRM, Payroll, and banks.
Great! The Financial Professional Will Get Back To You Soon.
If you’re not comfortable working this out for yourself, you can use an ARR calculator online to be extra sure that your figures are correct. EasyCalculation offers a simple tool for working out your
ARR, although there are many different ARR calculators online to explore. Read on as we take a look at the formula, what it is useful for, and give you an example of an ARR calculation in action.
Finance Strategists is a leading financial education organization that connects people with financial professionals, priding itself on providing accurate and reliable financial information to
millions of readers each year. Some limitations include the Accounting Rate of Returns not taking into account dividends or other sources of finance. Calculating the denominator Now we have the
numerator, we need to consider the denominator, i.e. the investment figure.
Finance Calculators
The accounting rate of return (ARR) is a simple formula that allows investors and managers to determine the profitability of an asset or project. Because of its ease of use and determination of
profitability, it is a handy tool to compare the profitability of various projects. However, the formula does not consider the cash flows of an investment or project or the overall timeline of
return, which determines the entire value of an investment or project. The Accounting Rate of Return (ARR) provides financial reporting small business firms with a straight-forward way to evaluate an
investment’s profitability over time. A firm understanding of ARR is critical for financial decision-makers as it demonstrates the potential return on investment and is instrumental in strategic
Impact on accounting profit
It can help a business define if it has enough cash, loans or assets to keep the day to day operations going or to improve/add facilities to eventually become more profitable. ARR takes into account
any potential yearly costs for the project, including depreciation. Depreciation is a practical accounting practice that allows the cost of a fixed asset to be dispersed or expensed. This enables the
business to make money off the asset right away, even in the asset’s first year of operation. Accounting Rate of Return helps companies see how well a project is going in terms of profitability while
taking into account returns on investments over a certain period. The accounting rate of return is one of the most common tools used to determine an investment’s profitability.
However, in the general sense, what would constitute a “good” rate of return varies between investors, may differ according to individual circumstances, and may also differ according to investment
goals. It is important that you have confidence if the financial calculations made so that your decision based on the financial data is appropriate. ICalculator helps you make an informed financial
decision with the ARR online calculator. The main difference is that IRR is a discounted cash flow formula, while ARR is a non-discounted cash flow formula.
This is a solid tool for evaluating financial performance and it can be applied across multiple industries and businesses that take on projects with varying degrees of risk. Based on the below
information, you are required to calculate the accounting rate of return, assuming a 20% tax rate. For example, if your business needs to decide whether to continue with a particular investment,
whether it’s a project or an acquisition, an ARR calculation can help to determine whether going ahead is the right move. Accounting Rate of Return (ARR) is a formula used to calculate the net income
expected from an investment or asset compared to the initial cost of investment.
The accounting rate of return (ARR) is a formula that shows the percentage rate of return that is expected on an asset or investment. This is when it is compared to the initial average capital cost
of the investment. ARR for projections will give you an idea of how well your project has done or is going to do. Calculating the accounting rate of return conventionally is a tiring task so using a
calculator is preferred to manual estimation. | {"url":"https://nhadep47.com/accounting-rate-of-return-definition-formula.html","timestamp":"2024-11-08T09:34:06Z","content_type":"text/html","content_length":"264677","record_id":"<urn:uuid:2afa95f3-84ec-44fd-a8cd-869c2d99cad8>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00276.warc.gz"} |
Excel Formula: INDEX Function
In this tutorial, we will learn how to use the INDEX function in Excel to retrieve specific values from a range of cells. The INDEX function is a powerful tool that allows you to extract data based
on its position within a given range.
To use the INDEX function, you need to provide several arguments. The first argument is the range of cells from which you want to retrieve the value. In our example, the range is A3:A111, which means
it includes cells from A3 to A111.
The second argument is the row number within the range from which you want to retrieve the value. In our example, we use 5 as the row number.
The third argument is the column number within the range from which you want to retrieve the value. In our example, we use 1 as the column number.
The fourth argument is the height of the range from which you want to retrieve the value. In our example, we use 1, indicating that we only want to consider a single row.
The fifth argument is the width of the range from which you want to retrieve the value. In our example, we use the COUNT function to count the number of cells in the range A3:A111.
Let's take a look at an example to better understand how the formula works. Suppose we have the following data in the range A3:A111:
If we use the formula =INDEX(A3:A111, 5, 1, 1, COUNT(A3:A111)), it will return the value 5. This is because it retrieves the value from the 5th row and 1st column within the range A3:A111,
considering a range height of 1 and a range width equal to the number of cells in the range (111 in this case).
In conclusion, the INDEX function is a useful tool in Excel that allows you to retrieve specific values from a range of cells. By providing the appropriate arguments, you can extract data based on
its position within the range.
An Excel formula
=INDEX(A3:A111, 5, 1, 1, COUNT(A3:A111))
Formula Explanation
This formula uses the INDEX function to retrieve a specific value from a range of cells. The formula has multiple arguments that determine the range and position of the value to be returned.
• A3:A111: This is the range of cells from which the value will be retrieved. In this example, it is assumed that the range contains 111 cells, starting from A3.
• 5: This argument specifies the row number within the range from which the value will be retrieved. In this case, it is the 5th row.
• 1: This argument specifies the column number within the range from which the value will be retrieved. In this case, it is the 1st column.
• 1: This argument specifies the height of the range from which the value will be retrieved. In this case, it is 1, indicating that only a single row will be considered.
• COUNT(A3:A111): This argument specifies the width of the range from which the value will be retrieved. It uses the COUNT function to count the number of cells in the range A3:A111. This ensures
that the entire range is considered for the retrieval.
Assuming the following data is present in the range A3:A111:
| A |
| |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
| 10 |
| ... |
| 110 |
| 111 |
The formula =INDEX(A3:A111, 5, 1, 1, COUNT(A3:A111)) would return the value 5. This is because it retrieves the value from the 5th row and 1st column within the range A3:A111, considering a range
height of 1 and a range width equal to the number of cells in the range (111 in this case). | {"url":"https://codepal.ai/excel-formula-generator/query/74B3hLpe/excel-formula-index-function","timestamp":"2024-11-10T11:43:34Z","content_type":"text/html","content_length":"96392","record_id":"<urn:uuid:38020d17-312d-4867-92b2-d066037dbdee>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00109.warc.gz"} |
Floating point specials and shader emulation
¶Floating point specials and shader emulation
I decided to work a bit more on my shader compiler and emulator last night, and ran into some unexpectedly ugly problems with floating point specials.
The first problem had to do with the following innocent looking expression:
This returns the magnitude of a vector. The conventional way to compute this is to compute the squared magnitude by taking the dot product of the vector with itself -- the vector analog of squaring a
scalar -- and then taking the square root of the result. It turns out that shader hardware doesn't actually support a direct square root. The reciprocal square root, 1/sqrt(x) is easier to compute,
and it's also useful in some other cases, most notably normalizing a vector. In this particular case, though, we need the normal square root, and therefore we need to invert it in the assembly:
dp3 r0.x, r_vec, r_vec ;compute dot(vec, vec)
rsq r0.x, r0.x ;compute reciprocal square root of squared magnitude
rcp r0.x, r0.x ;invert reciprocal square root
This little code fragment has a major surprise lurking in it, which may not be apparent until you try optimizing it. Both rsq and rcp are scalar instructions, so in cases where multiple magnitudes
are being computed, the temptation is high to replace 1/rsqrt(x) with x*rsqrt(x) instead to take advantage of the much more available mul instruction:
dp3 r0.x, r_vec, r_vec
rsq r0.y, r0.x
mul r0.x, r0.x, r0.y
This works just fine... until you have the misfortune of trying to compute the length of a zero vector. In that case, the reciprocal square root operation returns +Infinity, and then the next thing
that happens is the computation 0*+Infinity, which then returns a NaN (invalid result). Suck. Therefore, for the general case, that rcp has to stay there.
The real gotcha comes when you try implementing the rsq and rcp instructions in software. Reciprocal is a very slow instruction on most FPUs, being done with the divide unit and usually taking dozens
of nonpipelineable clocks. Reciprocal square root may not even exist in full precision form, and 1/sqrt(x) is a horribly painful way to implement it. If you want to implement this fragment quickly in
SSE, you need to take advantage of the rcpps and rsqrtps instructions, which are very fast and work in parallel on four values. They only provide limited precision up to about 2^-12, though. The WPF
shader engine just goes ahead and uses the approximation result directly, which works and is accurate enough for half precision (10 bit mantissa), but technically it's not Direct3D compliant as 22
bits of precision are needed.
The usual way to improve the accuracy of the reciprocal and reciprocal square root operations is by iteration through Newton's Method. For the reciprocal, it looks like this:
x = reciprocal_approx(c);
x' = x * (2 - x * c);
...and for the reciprocal square root, it looks like this:
x = rsqrt_approx(c);
x' = 0.5 * x * (3 - x*x*c);
Assuming that you have a good estimate, these will tend to double the number of significant digits per iteration, which means that just one iteration will give us pretty good precision, and quickly,
too. And unfortunately wrong, as I discovered when I implemented it. The problem is once again zero. In order for the zero case to work, we need rsq to transform 0 -> Inf and rcp to transform Inf ->
0, but thanks to the x*c expression in both of these iterations, you once again get 0 * Inf = NaN. The way I fixed it was to insert a couple of carefully placed min/max operations in the iteration,
although I'm not quite sure they're 100% correct.
The specials struck again when I was trying to optimize the code generated for a gamma correction shader. Gamma correction is primarily a power operation, which expands as follows:
pow(x, y) = exp(y * log(x))
If you actually try gamma correcting an image in this manner, you'll be waiting a long time for the result. For limited precision (8-bit), you can get away with a lookup table, but that doesn't scale
well to higher precision or vector computations and definitely not to a shader where floating point is involved. Therefore, in order to get a faster version working, I had to implement the log and
exp instructions, which compute the base 2 logarithm and exponential functions. SSE doesn't provide you any help to do this, so you're stuck implementing this from the ground floor. It's a bit like
an old BASIC interpreter, except at least you start with add and multiply. This triggered the following conversation with a friend:
"What are you doing?"
"I'm implementing the log() and exp() functions."
"Doesn't the runtime provide those?"
"This is the runtime."
Anyway, I ended up implementing exp2(x) = c*floor(x) + f(x - floor(x)) and log2(x) = c*exponent(x) + g(mantissa(x)). For the most part they're not too hard, as long as you find good polynomial
expansions and make sure it's exact at the right values, i.e. you don't want exp(0) = 1.004. In the end, I used a fifth-order polynomial for exp2() and a first-order Padé approximation with a change
of variables for log2()... but I digress.
As you have probably already guessed, zero rears its ugly head again here, because 0^y becomes exp2(y * log2(0)), and for this to work you need log2(0) = -Infinity and exp2(-Infinity) = 0. A couple
of well-placed min/max operations in the expansions once again fixed the SSE version, but I unexpectedly ran into problems with the x87 version. I hadn't bothered to optimize the x87 version, so it
simply called into expf() and scaled the result. Since I compile with /fp:fast /Oi, expf() ended up getting expanded in intrinsic form like this:
fld st(0)
fxch st(1)
fsub st, st(1)
If you look closely, you'll see that this computes (x - round(x)), which in this case is -Inf - (-Inf) = NaN. There are two basic ways to fix this. One is to force the runtime library version of the
function, which is faster for SSE2 but unfortunately quite slow for x87. The other way, which is what I did, is to just check for infinity and special-case the result.
Ordinarily I don't think much about floating point specials, and the last place I would have expected to find them is in a pixel shader. I have to say it's been a bit of a humbling (and frustrating) | {"url":"https://www.virtualdub.org/blog2/entry_229.html","timestamp":"2024-11-14T02:12:58Z","content_type":"text/html","content_length":"9965","record_id":"<urn:uuid:ab9d1065-3cfc-40f8-9a82-d00d34c566fb>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00888.warc.gz"} |
Does the observational evidence in AR5 support its/the CMIP5 models’ TCR ranges?
Does the observational evidence in AR5 support its/the CMIP5 models’ TCR ranges?
Originally posted on Dec 9, 2013 – 12:11 PM at Climate Audit
Steve McIntyre pointed out some time ago, here, that almost all the global climate models around which much of the IPCC’s AR5 WGI report was centred had been warming faster than the real climate
system over the last 35-odd years, in terms of the key metric of global mean surface temperature. The relevant figure from Steve’s post is reproduced as Figure 1 below.
Figure 1 Modelled versus observed decadal global surface temperature trend 1979–2013
Temperature trends in °C/decade. Models with multiple runs have separate boxplots; models with single runs are grouped together in the boxplot marked ‘singleton’. The orange boxplot at the right
combines all model runs together. The default settings in the R boxplot function have been used. The red dotted line shows the actual increase in global surface temperature over the same period per
the HadCRUT4 observational dataset.
Transient climate response
Virtually all the projections of future climate change in AR5 are based on the mean and range of outcomes simulated by this latest CMIP5 generation of climate models (AOGCMs). Changes in other
variables largely scale with changes in global surface temperature. The key determinant of the range and mean level of projected increases in global temperature over the rest of this century is the
transient climate response (TCR) exhibited by each CMIP5 model, and their mean TCR. Model equilibrium climate sensitivity (ECS) values, although important for other purposes, provide little
information regarding surface warming to the last quarter of this century beyond that given by TCR values.
TCR represents the increase in 20-year mean global temperature over a 70 year timeframe during which CO[2] concentrations, rising throughout at 1% p.a. compound, double. More generally, paraphrasing
from Section 10.8.1 of AR5 WG1,TCR can be thought of as a generic property of the climate system that determines the global temperature response ΔT to any gradual increase in (effective) radiative
forcing (ERF – see AR5 WGI glossary, here ) ΔF taking place over a ~70-year timescale, normalised by the ratio of the forcing change to the forcing due to doubling CO[2], F[2xCO2]: TCR = F[2xCO2] ΔT/
ΔF. This equation permits warming resulting from a gradual change in ERF over a 60–80 year timescale, at least, to be estimated from the change in ERF and TCR. Equally, it permits TCR to be estimated
from such changes in global temperature and in ERF.
The TCRs of the 30 AR5 CMIP5 models featured in WGI Table 9.5 vary from 1.1°C to 2.6°C, with a mean of slightly over 1.8°C. Many projections in AR5 are for changes up to 2081–2100. Applying the CMIP5
TCRs to the changes in CO[2] concentration and other drivers of climate change from the first part of this century up to 2081–2100, expressed as the increase in total ERF, explains most of the
projected rises in global temperature on the business-as-usual RCP8.5 scenario, although the relationship varies from model to model. Overall the models project about 10–20% faster warming than would
be expected from their TCR values, allowing for warming ‘in-the-pipeline’. That discrepancy, which will not be investigated in this article, implies that the mean ‘effective’ TCR of the AR5 CMIP5
models for warming towards the end of this century under RCP8.5 is in the region of 2.0–2.2°C.
Observational evidence in AR5 about TCR
AR5 gives a ‘likely’ (17–83% probability) range for TCR of 1.0–2.5°C, pretty much in line with the 5–95% CMIP5 model TCR range (from fitting a Normal distribution) but with a downgraded certainty
level. How does that compare with the observational evidence in AR5? Figure 10.20a thereof, reproduced as Figure 2 here, shows various observationally based TCR estimates.
Figure 2. Reproduction of Figure 10.20a from AR5
Bars show 5–95% uncertainty ranges for TCR.[1]
On the face of it, the observational study TCR estimates in Figure 2 offer reasonable support to the AR5 1.0–2.5°C range, leaving aside the Tung et al. (2008) study, which uses a method that AR5 WGI
discounts as unreliable. However, I have undertaken a critical analysis of all these TCR studies, here. I find serious fault with all the studies other than Gillett et al. (2013), Otto et al. (2013)
and Schwartz (2012). Examples of the faults that I find with other studies are:
Harris et al. (2013): This perturbed physics/parameter ensemble (PPE) study’s TCR range, like its ECS range, almost entirely reflects the characteristics of the UK Met Office HadCM3 model. Despite
the HadCM3 PPE (as extended by emulation) sampling a wide range of values for 31 key model atmospheric parameters, the model’s structural rigidities are so strong that none of the cases results in
the combination of low-to-moderate climate sensitivity and low-to-moderate aerosol forcing that the observational data best supports – nor could perturbing aerosol model parameters achieve this.
Knutti and Tomassini (2008): This study used initial estimates of aerosol forcing totalling −1.3 W/m² in 2000, in line with AR4 but far higher than the best estimate in AR5. Although it attempted to
observationally-constrain these initial estimates, the study’s use of only global temperature data makes it impossible to separate properly greenhouse gas and aerosol forcing, the evolution of which
are very highly (negatively) correlated at a global scale. The resulting final estimates of aerosol forcing are still significantly stronger than the AR5 estimates, biasing up TCR estimation. The use
of inappropriate uniform and expert priors for ECS in the Bayesian statistical analysis further biases TCR estimation.
Rogelj et al. (2012): This study does not actually provide an observationally-based estimate for TCR. It explicitly sets out to generate a PDF for ECS that simply reflects the AR4 ‘likely’ range and
best estimate; in fact it reflects a slightly higher range. Moreover, the paper and its Supplementary Information do not even mention estimation of TCR or provide any estimated PDF for TCR.
Stott and Forest (2007): This TCR estimate is based on the analysis in Stott et al. (2006), an AR4 study from which all four of the unlabelled grey dashed-line PDFs in Figure 10.20a are sourced. It
used a detection-and-attribution regression method applied to 20th century temperature observations to scale TCR values, and 20th century warming attributable to greenhouse gases, for three AOGCMs.
Gillett et al. (2012) found that just using 20th century data for this purpose biased TCR estimation up by almost 40% compared with when 1851–2010 data was used. Moreover, the 20th century greenhouse
gas forcing increase used in Stott and Forest (2007) to derive TCR (from the Stott et al. (2006) attributable warming estimate) is 11% below that per AR5, biasing up its TCR estimation by a further
In relation to the three studies that I do not find any serious fault with, some relevant details from my analysis are:
Gillett et al. (2013): This study uses temperature observations over 1851–2010 and a detection-and-attribution regression method to scale AOGCM TCR values. The individual CMIP5 model regression-based
observationally-constrained TCRs shown in a figure in the Gillett et al. (2013) study imply a best (median[2]) estimate for TCR of 1.4°C, with a 5–95% range of 0.8–2.0°C.[3] That compares with a
range of 0.9–2.3°C given in the study based on a single regression incorporating all models at once, which it is unclear is as suitable a method.
Otto et al. (2013): There are two TCR estimates from this energy budget study included in Figure 10.20a. One estimate uses 2000–2009 data and has a median of 1.3°C, with a 5–95% range of 0.9–2.0°C.
The other estimate uses 1970–2009 data and has a median of slightly over 1.35°C, with a 5–95% range of 0.7–2.5°C. Since mean forcing was substantially higher over 2000–2009 than over 1970–2009, and
was also less affected by volcanic activity, the TCR estimate based on 2000–2009 data is less uncertain, and arguably more reliable, than that based on 1970–2009 data.
Schwartz (2012): This study derived TCR by zero-intercept regressions of changes, from the 1896–1901 mean, in observed global surface temperature on corresponding changes in forcing, up to 2009,
based on forcing histories used in historical model simulations. The mean change in forcing up to 1990 (pre the Mount Pinatubu eruption) per the five datasets used to derive the TCR range is close to
the best estimate of the forcing change per AR5. The study’s TCR range is 0.85–1.9°C, with a median estimate of 1.3°C.
So the three unimpeached studies in Figure 10.20a support a median TCR estimate of about 1.35°C, and a top of the ‘likely’ range for TCR of about 2.0°C based on downgrading 5–95% ranges, following
The implication for TCR of the substantial revision in AR5 to aerosol forcing estimates
There has been a 43% increase in the best estimate of total anthropogenic radiative forcing between that for 2005 per AR4, and that for 2011 per AR5. Yet global surface temperatures remain almost
unchanged: 2012 was marginally cooler than 2007, whilst the trailing decadal mean temperature was marginally higher. The same 0.8°C warming now has to be spread over a 43% greater change in total
forcing, natural forcing being small in 2005 and little different in 2012. The warming per unit of forcing is a measure of climate sensitivity, in this case a measure close to TCR, since most of the
increase in forcing has occurred over the last 60–70 years. It follows that TCR estimates that reflect the best estimates of forcing in AR5 should be of the order of 30% lower than those that
reflected AR4 forcing estimates.
Two thirds of the 43% increase in estimated total anthropogenic forcing between AR4 and AR5 is accounted for by revisions to the 2005 estimate, reflecting improved understanding, with the increase in
greenhouse gas concentrations between 2005 and 2011 accounting for almost all of the remainder. Almost all of the revision to the 2005 estimate relates to aerosol forcing. The AR5 best (median)
estimate of recent total aerosol forcing is −0.9 W/m^2, a large reduction from −1.3 W/m^2 (for a more limited measure of aerosol forcing) in AR4. This reduction has major implications for TCR and ECS
Moreover, the best estimate the IPCC gives in AR5 for total aerosol forcing is not fully based on observations. It is an expert judgement based on a composite of estimates derived from simulations by
global climate models and from satellite observations. The nine satellite-observation-derived aerosol forcing estimates featured in Figure 7.19 of AR5 WGI range from −0.09 W/m^2 to −0.95 W/m^2, with
a mean of −0.65 W/m^2. Of these, six satellite studies with a mean best estimate of −0.78 W/m^2 were taken into account in deciding on the −0.9 W/m^2 AR5 composite best estimate of total aerosol
TCR calculation based on AR5 forcing estimates
Arguably the most important question is: what do the new ERF estimates in AR5 imply about TCR? Over the last century or more we have had a period of gradually increasing ERF, with some 80% of the
decadal mean increase occurring fairly smoothly, volcanic eruptions apart, over the last ~70 years. We can therefore use the TCR = F[2xCO2] ΔT/ΔF equation to estimate TCR from ΔT and ΔF, taking the
change in each between the means for two periods, each long enough for internal variability to be small.
That is exactly the method used, with a base period of 1860–1879, by the ‘energy budget’ study Otto et al. (2013), of which I was a co-author. That study used estimates of radiative forcing that are
approximately consistent with estimates from Chapters 7 and 8 of AR5, but since AR5 had not at that time been published the forcings were actually diagnosed from CMIP5 models, with an adjustment
being made to reflect satellite-observation-derived estimates of aerosol forcing. However, in a blog-published study, here, I did use the same method but with forcing estimates (satellite-based for
aerosols) taken from the second draft of AR5. That study estimated only ECS, based on changes between 1871–1880 and 2002–2011, but a TCR estimate of 1.30°C is readily derived from information in it.
We can now use the robust method of the Otto et al. (2013) paper in conjunction with the published AR5 forcing best (median) estimates up to 2011, the most recent year given. The best periods to
compare appear to be 1859–1882 and 1995–2011. These two periods are the longest ones in respectively the earliest and latest parts of the instrumental period that were largely unaffected by major
volcanic eruptions. Volcanic forcing appears to have substantially less effect on global temperature than other forcings, and so can distort TCR estimation. Using a final period that ends as recently
as possible is important for obtaining a well-constrained TCR estimate, since total forcing (and the signal-to-noise ratio) declines as one goes back in time. Measuring the change from early in the
instrumental period maximises the ratio of temperature change to internal variability, and since non-volcanic forcings were small then it matters little that they are known less accurately than in
recent decades. Moreover, these two periods are both near the peak of the quasi-periodic ~65 year AMO cycle. Using a base period extending before 1880 limits one to using the HadCRUT surface
temperature dataset. However, that is of little consequence since the HadCRUT4 v2 change in global temperature from 1880–1900 to 1995–2011 is identical to that per NCDC MLOST and only marginally
below that per GISS.
In order to obtain a TCR estimate that is as independent of global climate models as possible, one should scale the aerosol component of the AR5 total forcing estimates to match the AR5 recent
satellite-observation-derived mean of −0.78 W/m^2. Putting this all together gives ΔF = 2.03 W/m^2 and ΔT = 0.71, which, since AR5 uses F[2xCO2] = 3.71 W/m, gives a best estimate of 1.30°C for TCR.
The best estimate for TCR would be 1.36°C without scaling aerosol forcing to match the satellite-observation derived mean.
So, based on the most up to date numbers from the IPCC AR5 report itself and using the most robust methodology on the data with the best signal-to-noise ratio, one arrives at an observationally based
best estimate for TCR of 1.30°C, or 1.36°C based on the unadjusted AR5 aerosol forcing estimate.
I selected 1859–1882 and 1995–2011 as they seem to me to be the best periods for estimating TCR. But it is worth looking at longer periods as well, even though the signal-to-noise ratio is lower.
Using 1850–1900 and 1985–2011, two periods with mean volcanic forcing levels that, although significant, are well matched, gives a TCR best estimate of 1.24°C, or 1.30°C based on the unadjusted AR5
aerosol forcing estimate. The TCR estimates are even lower using 1850–1900 to 1972–2011, periods that are also well-matched volcanically.
What about estimating TCR over a shorter timescale? If one took ~65 rather than ~130 years between the middles of the base and end periods, and compared 1923–1946 with 1995–2011, the TCR estimates
would be almost unchanged. But there is some sensitivity to the exact periods used. An alternative approach is to use information in the AR5 Summary for Policymakers (SPM) about anthropogenic-only
changes over 1951–2010, a well-observed period. The mid-range estimated contributions to global mean surface temperature change over 1951–2010 per Section D.3 of the SPM are 0.9°C for greenhouse
gases and ‑0.25°C for other anthropogenic forcings, total 0.65°C. The estimated change in total anthropogenic radiative forcing between 1950 and 2011 of 1.72 Wm^-2 per Figure SPM.5, reduced by 0.04
Wm^-2 to adjust to 1951–2010, implies a TCR of 1.4°C after multiplying by an F[2xCO2] of 3.71 Wm^-2. When instead basing the estimate on the linear trend increase in observed total warming of 0.64°C
over 1951–2010 per Jones et al. (2013) – the study cited in the section to which the SPM refers – (the estimated contribution from internal variability being zero) and the linear trend increase in
total forcing per AR5 of 1.73 Wm^-2, the implied TCR is also 1.4°C. Scaling the AR5 aerosol forcing estimates to match the mean satellite observation derived aerosol forcing estimate would reduce the
mean of these two TCR estimates to 1.3°C.
So does the observational evidence in AR5 support its/the CMIP5 models’ TCR ranges?
The evidence from AR5 best estimates of forcing, combined with that in solid observational studies cited in AR5, points to a best (median) estimate for TCR of 1.3°C if the AR5 aerosol forcing best
estimate is scaled to match the satellite-observation-derived best estimate thereof, or 1.4°C if not (giving a somewhat less observationally-based TCR estimate). We can compare this with model TCRs.
The distribution of CMIP5 model TCRs is shown in Figure 3 below, with a maximally observationally-based TCR estimate of 1.3°C for comparison.
Figure 3. Transient climate response distribution for CMIP5 models in AR5 Table 9.5
The bar heights show how many models in Table 9.5 exhibit each level of TCR
Figure 3 shows an evident mismatch between the observational best estimate and the model range. Nevertheless, AR5 states (Box 12.2) that:
“the ranges of TCR estimated from the observed warming and from AOGCMs agree well, increasing our confidence in the assessment of uncertainties in projections over the 21st century.”
How can this be right, when the median model TCR is 40% higher than an observationally-based best estimate of 1.3°C, and almost half the models have TCRs 50% or more above that? Moreover, the fact
that effective model TCRs for warming to 2081–2100 are the 10%–20% higher than their nominal TCRs means that over half the models project future warming on the RCP8.5 scenario that is over 50% higher
than what an observational TCR estimate of 1.3°C implies.
Interestingly, the final draft of AR5 WG1 dropped the statement in the second draft that TCR had a most likely value near 1.8°C, in line with CMIP5 models, and marginally reduced the ‘likely’ range
from 1.2–2.6°C to 1.0–2.5°C, at the same time as making the above claim.
So, in their capacity as authors of Otto et al. (2013), we have fourteen lead or coordinating lead authors of the WG1 chapters relevant to climate sensitivity stating that the most reliable data and
methodology give ‘likely’ and 5–95% ranges for TCR of 1.1–1.7°C and 0.9–2.0°C, respectively. They go on to suggest that some CMIP5 models have TCRs that are too high to be consistent with recent
observations. On the other hand, we have Chapter 12, Box 12.2, stating that the ranges of TCR estimated from the observed warming and from AOGCMs agree well. Were the Chapter 10 and 12 authors misled
by the flawed TCR estimates included in Figure 10.20a? Or, given the key role of the CMIP5 models in AR5, did the IPCC process offer the authors little choice but to endorse the CMIP5 models’ range
of TCR values?
Nicholas Lewis
[1] Note that the PDFs and ranges given for Otto et al. (2013) are slightly too high in the current version of Figure 10.20a. It is understood that those in the final version of AR5 will agree to the
ranges in the published study.
[2] All best estimates given are medians (50% probability points for continuous distributions), unless otherwise stated.
[3] This range for Gillett et al. (2013) excludes an outlier at either end; doing so does not affect the median.
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The Possibility of Time Travel through General Relativity
Philosophers strive for conceptual clarity, making the physical possibility of time travel on Einstein’s General Theory of Relativity (hereafter, ‘GR’) a topic of particular philosophical interest
(Arntzenius and Maudlin, 2023). There are indeed universes on GR with closed time-like curves (Effingham, 2020). Time travel—along such structures—is therefore a physical possibility according to GR
in the kind of spacetime that permits it. This article will explain how time travel is physically possible and, through a philosophical lens, offer a conceptual study of GR and the time travel
possibility. To this end, this essay evaluates how one must very carefully address the concept of 'physical possibility'. The analysis starts by providing some brief background on Einstein’s field
equations and the locality of GR, thus going on to explain how GR allows for universes with time travel. A major conceptual threat to this is also considered here, namely the Grandfather Paradox
(Effingham, 2020). Then, it is finally discussed what physically possible really means according to GR. Via the Global Constraint concept (Lewis, 1976), this feature overall aims to show that the
so-called Grandfather Paradox does not pose a serious threat to the physical possibility of time travel permitted by Einstein’s theory of GR.
Einstein’s Theory of General Relativity
Einstein's Theory of GR reflects his conviction that the theoretical physicist must trust simplicity, thus moving steadily into domains even further removed from direct contact with observation and
experiment (Norton, 2000). Having published the Special Theory of Relativity in 1905, the physicist sought to incorporate gravity into his new relativistic framework (Matthews, 2013), and notably,
via his field equations, subsequently developed GR in 1915 by providing the key mathematical framework on which he fit his physical ideas of gravity (Carmeli, 2008). His proposed idea of a single
continuum where space and time are interwoven is what is now known as spacetime (Kennedy, 2014). In GR, spacetime is notably intrinsically curved. For example, energy bends spacetime, and spacetime
is not embedded in a higher dimensional space but is instead intrinsic. Hence, there is no extrinsic curvature since there is no exterior space to look at it from – maps of large regions may not be
drawn without distortion of the map (Collas, 1997). Broadly speaking, though, GR detects curvature in space-space sheets (i.e., the ordinary two-dimensional slices of three-dimensional space). The
summed curvature (of all sheets) is then equal to the total matter-energy density (Hobson, Efstathiou, and Lasenby, 2006). Einstein’s famous gravitational field equations, therefore, relate the
geometry of spacetime to the distribution of matter within spacetime. More specifically, the Einstein tensor equals the stress-energy tensor (Norton, 2007), meaning that GR describes spacetime
curvature in a manner that is consistent with the conservation of energy and momentum (Lovelock, 1971):
Importantly, the field equations make up the core of Einstein’s theory and embrace the entirety of gravitational phenomena as well as the geometry of space (Norton, 2007). This explains the locality
of GR since Gμν = Tμν can explain every local region (Hobson, Efstathiou, and Lasenby, 2006). Gμν and Tμν are equal then to decide which spacetimes (i.e., which universes) are admissible. If the
spacetime curvature and matter density are related appropriately, then the universe is admissible (Norton, 2007). Hence, if a universe is physically possible according to GR, then Gμν = Tμν holds at
every region of spacetime in that universe (Norton, 2007), roughly speaking.
Figure 1. NASA's Gravity Probe B corroborates Einstein's prediction of massive objects warping their surrounding spacetime (Wall, 2015)
Time Travel and The Grandfather Paradox
Gμν = Tμν explains all the possible ways a universe can be. Remarkably, in such a universe where space is curved back onto itself, GR discovers that there is a physically possible (admissible)
universe where the prospect of time travel arises. This is the Cylinder Universe, which allows for time travel where every single spacetime region is a Minkowski spacetime, that is, with zero
curvature and zero-sum curvature, named after German mathematician Hermann Minkowski and allowing for spacetime to curve with the resulting effect of gravity (Norton, 2007). One spatial dimension is
wrapped onto itself (hence, ‘cylinder’ universe). What is more, this universe involves only one dimension of time. The cylinder is wrapped onto itself in the direction of time: a closed time-like
curve (Weingard, 1979). A time traveller, travelling along the curve, will thus always come back to the same moment in time since the closed time-like curve forms a loop (Weingard, 1979). Every
region of spacetime is Minkowski spacetime and is therefore permissible according to GR. As such, time travel is a physical possibility because this universe is a physical possibility (it is
important to note that this is all according to GR).
Gμν = Tμν is the solution, and each region of spacetime (Minkowski spacetime) connects to the next. Remarkably, this is all that is needed for spacetime to count as a solution to Einstein’s
gravitational field equations (Norton, 2007). So, a time traveller on this closed time-like curve would—technically, at some point—wrap all the way around the curve and come back to the same moment
in time. Put differently, the time traveller would reconnect with their former self (see figure 2 for a diagrammatical representation of this universe). This extraordinary possibility is somewhat
complicated, however. The very possibility (and problem) of time travel seems to yield a paradox (Maudlin, 1990). Said logical problem is famously known as the Grandfather Paradox, and conceptually
presents a real challenge to the physical possibility of time travel in GR. In a somewhat simple fashion, it says that if a time traveller goes back to a time before their grandfather had children,
and kills him, the grandfather’s death would make their own birth impossible. Indeed, the killing of the grandfather also entails no time traveller to go back in time to kill him (Norton, 2007). The
closed loop, therefore, appears contradictory – the time traveller both does and does not exist at some and the same time. The time traveller goes back in time if and only if the time traveller does
not go back in time (both P and ~ P, the kind of logical contradiction that must be taken seriously in any physical theory). Perhaps, one wonders, time travel is not physically possible after all.
But what does ‘physically possible’ mean in the sense of ‘permitted-by-GR’, though?
Figure 2. The 'grandfather paradox' resulting from a closed time-like curve (Norton, 2007).
As discussed throughout the article, the argument set out for the physical possibility of time travel is as follows (Norton, 2007):
Premise 1. A universe is physically possible according to GR if it satisfies Einstein’s gravitational field equations in every region of spacetime.
Premise 2. There are time travel universes with closed time-like curves (i.e., the cylinder universe) that satisfy Einstein’s field equations and are physically possible as a result.
Premise 3. If a universe that is physically possible can allow for time travel to arise, then time travel is a physical possibility too.
Premise 4. Einstein’s theory of GR describes universes that allow for time travel to arise.
Conclusion. Therefore, time travel is a physical possibility according to GR.
The Grandfather Paradox objects to the above by asserting that time travel is inherently paradoxical (Maudlin, 1990). It is worth noting, however, that there is no contradiction if the bullet is
never fired or if the attempt to kill the grandfather fails. Despite stating the obvious, the point to take away is that a global constraint can suffice as a response to the paradox. As this article
discusses hereafter, both a global constraint in this (cylinder) universe and an adequate understanding of what it means for time travel to be physically possible—according to GR—may overcome the
Grandfather Paradox. Essentially, this is to say that the paradox is only a serious issue when physically possible is conceptually misunderstood and since the paradox assumes that the attempt to kill
the grandfather would always be successful. The global constraint says that the grandson cannot actually kill his grandfather (Lewis, 1976). Hence, the global structure of the time travel universe is
quite different to our own. In the time travel universe, all processes must conform to these additional ‘global’ constraints so that the distant future and distant past ‘mesh’ when they meet (Norton,
2007). Unlike our own universe, the cosmos, (or any time-travel-free universe for that matter), global constraints ensure that any attempt to kill the grandfather fails. This event would never happen
since processes must conform to the global constraints of the universe (Norton, 2007). The past and the future have to connect or ‘mesh’ to make sense in a time-travel universe as such.
Figure 3. Stephen Hawking's 'M theory' offers the possibility of travelling back in time (Kaiser, 2018).
The physical possibility of time travel according to GR is consistent with the said global constraints. This is therefore how physically possible time travel may be understood. The kind of time
travel universe that is physically possible according to GR is indeed both physically possible and wholly different: time travel is a physical possibility in a universe that is vastly different to
the cosmos. Crucially, this then must play a role in one’s understanding of physical possibility (according to GR). It is a possible universe since Einstein’s field equations are satisfied (Norton,
2007). As for the Grandfather Paradox, one must remember that there are global constraints and other dissimilar structural features. According to GR, the cylinder universe is physically possible thus
making time travel achievable. The Grandfather Paradox is problematic when one assigns features of one’s own universe (i.e., the cosmos) to a time-travel possible universe. With a more adequate
description of what it means for time travel to be physically possible according to GR (i.e., in an utterly different universe that permits time travel and contains closed time-like curves where a
traveller simply by being travels in time), the Grandfather Paradox does not pose nearly as a serious threat.
The very possibility of a time travel universe enables time travel itself to become a physical possibility, as permitted by GR (where every region of spacetime connects to the next and Gμν = Tμν
holds) (Maudlin, 1990). After providing some background on the locality of GR—albeit briefly—this article then went on to describe what time travel really refers to. Unlike what might be portrayed in
Hollywood movies, a time traveller (merely by being) will travel in time via a closed time-like curve which forms a loop. By travelling along such a loop, the traveller will eventually end up at the
same moment in time. Indeed, this is all physically possible in the Cylinder Universe according to GR. But what does it mean for time travel to be 'physically possible'? The analysis next considered
the Grandfather Paradox, which arguably calls the very possibility of time travel into question by deriving a contradiction. The paradox, however, forgets that the time-travel universe is quite
different to the cosmos. An adequate understanding of time travel being physically possible, this piece has argued, involves some recognition of the difference between our cosmos universe and one
that permits time travel (i.e., the cylinder universe). The latter is entirely different, and this is part of what makes it physically possible. The very physical possibility of time travel acts like
a kind of global constraint. Hence, the feature has drawn on the global constraint concept here, where all processes in a time travel universe conform to those constraints. Such a universe is
entirely different to our own because of the physical possibility of time travel.
Bibliographical References
Arntzenius, F., & Maudlin, T. (2002). Time travel and modern physics. Royal Institute of Philosophy Supplements, 50, 169-200.
Carmeli, M. (2008). Relativity: Modern large-scale spacetime structure of the cosmos. World Scientific Publishing Company.
Collas, P. (1977). General relativity in two‐and three‐dimensional space–times. American Journal of Physics, 45(9), 833-837.
Effingham, N. (2020). Time Travel: Probability and Impossibility. Oxford University Press. doi:10.1093/oso/9780198842507.001.0001
Hobson, M. P., Efstathiou, G. P., & Lasenby, A. N. (2006). General relativity: an introduction for physicists. Cambridge University Press.
Kennedy, J. B. (2014). Space, time and Einstein: An introduction. Routledge.
Lewis, D. (1976). The Paradoxes of Time Travel. American Philosophical Quarterly, 13(2), 145–152.
Lovelock, D. (1971). The Einstein tensor and its generalizations. Journal of Mathematical Physics, 12(3), 498-501.
Matthews, M. R. (2013). Introduction: the history, purpose and content of the Springer international handbook of research in history, philosophy and science teaching. In International handbook of
research in history, philosophy and science teaching (pp. 1-15). Dordrecht: Springer Netherlands.
Maudlin, T. (1990). Time-Travel and Topology. PSA: Proceedings Of The Biennial Meeting Of The Philosophy Of Science Association, 1990(1), 303-315. doi: 10.1086/psaprocbienmeetp.1990.1.192712
Norton, J. (2000). ’Nature is the Realisation of the Simplest Conceivable Mathematical Ideas’: Einstein and the Canon of Mathematical Simplicity. Studies in History and Philosophy of Modern Physics,
31B, 135-170.
Norton, J. (2007). General Relativity. Retrieved 15 March 2023, from https://sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/general_relativity/index.html
Norton, J. (2007). Time Travel Universes. Retrieved 14 March 2023, from https://sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/time_travel_universes/index.html
Weingard, R. (1979). General relativity and the conceivability of time travel. Philosophy of Science, 46(2), 328-332. | {"url":"https://www.byarcadia.org/post/einstein-s-philosophy-of-physics-the-possibility-of-time-travel-according-to-general-relativity","timestamp":"2024-11-03T13:56:17Z","content_type":"text/html","content_length":"1051246","record_id":"<urn:uuid:2f2f58c5-02d8-45ae-89f0-c7ddbe362bc4>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00189.warc.gz"} |
%A Schaumburger,Nathan %A Pally,Joel %A Moraru,Ion I. %A Kositsawat,Jatupol %A Kuchel,George A. %A Blinov,Michael L. %D 2023 %J Frontiers in Network Physiology %C %F %G English %K
mathematical,dynamical,Model,prediction,phenotype,Aging,Inflammation,Hormones,IL-6,IGF-1,Frailty,Bistabiity %Q %R 10.3389/fnetp.2023.1079070 %W %L %M %P %7 %8 2023-May-04 %9 Hypothesis and Theory %#
%! Dynamic Model Predicts Mobility Phenotypes %* %< %T Dynamic model assuming mutually inhibitory biomarkers of frailty suggests bistability with contrasting mobility phenotypes %U https://
www.frontiersin.org/journals/network-physiology/articles/10.3389/fnetp.2023.1079070 %V 3 %0 JOURNAL ARTICLE %@ 2674-0109 %X
Bistability is a fundamental biological phenomenon associated with “switch-like” behavior reflecting the capacity of a system to exist in either of two stable states. It plays a role in gene
regulation, cell fate switch, signal transduction and cell oscillation, with relevance for cognition, hearing, vision, sleep, gait and voiding. Here we consider a potential role for bistability in
the existence of specific frailty states or phenotypes as part of disablement pathways. We use mathematical modeling with two frailty biomarkers (insulin growth factor-1, IGF-1 and interleukin-6,
IL-6), which mutually inhibit each other. In our model, we demonstrate that small variations around critical IGF-1 or IL-6 blood levels lead to strikingly different mobility outcomes. We employ
deterministic modeling of mobility outcomes, calculating the average trends in population health. Our model predicts the bistability of clinical outcomes: the deterministically-computed likelihood of
an individual remaining mobile, becoming less mobile, or dying over time either increases to almost 100% or decreases to almost zero. Contrary to statistical models that attempt to estimate the
likelihood of final outcomes based on probabilities and correlations, our model predicts functional outcomes over time based on specific hypothesized molecular mechanisms. Instead of estimating
probabilities based on stochastic distributions and arbitrary priors, we deterministically simulate model outcomes over a wide range of physiological parameter values within experimentally derived
boundaries. Our study is “a proof of principle” as it is based on a major assumption about mutual inhibition of pathways that is oversimplified. However, by making such an assumption, interesting
effects can be described qualitatively. As our understanding of molecular mechanisms involved in aging deepens, we believe that such modeling will not only lead to more accurate predictions, but also
help move the field from using mostly studies of associations to mechanistically guided approaches. | {"url":"https://www.frontiersin.org/journals/network-physiology/articles/10.3389/fnetp.2023.1079070/endnote","timestamp":"2024-11-08T04:47:25Z","content_type":"application/x-endnote-refer","content_length":"3251","record_id":"<urn:uuid:b67e4265-c9e9-4d04-a619-0e9cd608bad0>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00408.warc.gz"} |
Tenths to Feet and Feet to Tenths Converter - GEGCalculators
Tenths to Feet and Feet to Tenths Converter
Tenths to Feet and Feet to Tenths Converter
Tenths to Feet
Feet to Tenths
How to convert feet to 10ths?
To convert feet to tenths, multiply the number of feet by 10.
How many tenths are in a foot?
There are 10 tenths in a foot.
How do you convert feet and inches to feet?
To convert feet and inches to feet, divide the number of inches by 12 and add it to the number of feet.
What is 1 inch in tenths?
1 inch is equal to 0.1 tenths.
Is 10 feet and foot the same?
No, “feet” is the plural form of “foot.” 10 feet is not the same as 1 foot.
What is 8 in in tenths?
8 inches is equal to 0.8 tenths.
How much feet is one foot?
One foot is equal to itself, so it’s 1 foot.
Do you say 10 foot or 10 feet?
You say “10 feet” to refer to a length of 10 feet.
How many feet is 1 tenths of a mile?
There are approximately 528 feet in 1 tenth of a mile.
How can I calculate feet?
You can calculate feet by measuring the length or by using conversion formulas.
How do you convert feet to decimals?
To convert feet to decimals, divide the number of inches by 12 and add it to the number of feet.
How many feet is 40×60?
40 feet multiplied by 60 feet equals 2400 square feet.
How do you read tenths?
Tenths are read as fractions, where one-tenth is equivalent to 0.1.
What is 3 tenths of an inch?
3 tenths of an inch is equal to 0.3 inches.
What decimal is 10 inches of a foot?
10 inches of a foot is equal to 0.8333… in decimal form.
Is 10 inch one feet?
No, 10 inches is not equal to 1 foot. 1 foot equals 12 inches.
How many inches is 10 feet?
10 feet is equal to 120 inches.
Is 1 feet and 1 foot the same?
Yes, “feet” is the plural form of “foot.” So, 1 foot and 1 feet refer to the same length.
What is 10 tenths equal to?
10 tenths is equal to 1 whole unit or 10 feet.
What does 1 tenths mean?
One-tenth represents one part of a whole divided into ten equal parts.
How do I convert inches to tenths?
To convert inches to tenths, divide the number of inches by 10.
Is it 8 foot or 8 feet?
It is “8 feet” to refer to a length of 8 feet.
Is 1 foot 12 feet?
No, 1 foot is not equal to 12 feet. 1 foot is equal to 12 inches.
Is it 5 feet or 5 foot?
It is “5 feet” to refer to a length of 5 feet.
Is 5 foot 10 the same as 6 foot?
No, 5 foot 10 is not the same as 6 feet. 5 foot 10 refers to a height of 5 feet and 10 inches.
Do we say 6 feet or 6 foot?
We say “6 feet” to refer to a length of 6 feet.
What is the difference between foot and feet?
“Foot” is the singular form, while “feet” is the plural form. They both refer to a unit of length, with “feet” being the plural of “foot.”
How many miles are in a tenth?
There are approximately 528 feet in a tenth of a mile.
Is 0.1 a tenth of a mile?
Yes, 0.1 represents one-tenth of a mile.
What is 9 tenths of a mile?
Nine tenths of a mile is equal to approximately 4752 feet.
How many feet is 1 square feet?
1 square foot is equal to 1 foot multiplied by 1 foot, so it’s 1 square foot.
What is 10 ft in square feet?
10 feet in square feet is equal to 100 square feet.
Can I measure feet with my phone?
Yes, you can use apps on your phone or the built-in measuring tools in some smartphones to measure lengths in feet.
Is it 0.5 feet or 0.5 foot?
It is “0.5 feet” to refer to half a foot.
What is the metric conversion for feet?
One foot is approximately 0.3048 meters.
Why is 12 inches a foot?
The concept of 12 inches in a foot is thought to have originated from ancient civilizations, possibly because 12 is divisible by more numbers than 10, making fractional measurements more convenient.
How long is 40 inches in feet?
40 inches is equal to approximately 3.33 feet.
How big is 60 feet by 30 feet?
60 feet by 30 feet is equal to an area of 1800 square feet.
How tall is 48 inches in feet?
48 inches is equal to 4 feet.
How many tenths in a foot?
There are 10 tenths in a foot.
What do tenths look like?
Tenths are represented as fractions or decimals, with each tenth being one-tenth of a whole.
Why are tenths called tenths?
Tenths are called so because they represent one part of a whole divided into ten equal parts.
What does 3 tenths look like?
3 tenths can be represented as 0.3 in decimal form or as the fraction 3/10.
What is 8 inches in tenths?
8 inches is equal to 0.8 tenths.
What does 3 tenths mean?
Three tenths means three parts out of ten, or 0.3 in decimal form.
How do you convert feet and inches into feet?
To convert feet and inches into feet, divide the number of inches by 12 and add it to the number of feet.
How do you convert feet to inches?
To convert feet to inches, multiply the number of feet by 12.
What is the tenths place?
The tenths place is the first digit to the right of the decimal point in a decimal number.
Is one tenth one inch?
No, one-tenth is not equal to one inch. One inch is larger than one-tenth of an inch.
Is 10 feet and foot the same?
No, “feet” is the plural form of “foot.” 10 feet is not the same as 1 foot.
Is 10 inches bigger than 1 foot?
Yes, 10 inches is larger than 1 foot. 1 foot equals 12 inches.
What is the measurement of 10 feet?
The measurement of 10 feet is a length of 10 feet.
Is 1 foot exactly 12 inches?
Yes, 1 foot is exactly equal to 12 inches.
What is 10 feet by 8 feet?
10 feet by 8 feet is equal to an area of 80 square feet.
What is the difference between 10 foot and 10 feet?
“10 foot” is incorrect grammar; it should be “10 feet.” “10 feet” is the correct way to refer to a length of 10 feet.
What is the difference between 2 foot and 2 feet?
“2 foot” is incorrect grammar; it should be “2 feet.” “2 feet” is the correct way to refer to a length of 2 feet.
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Statistics and Commercial Real Estate – October 2013 Newsletter
I was given the book below by my brother-in-law’s wife for my birthday last year. It’s all about statistics. When I read the back cover & realized the subject matter I didn’t think I was going to get
past the first five pages and thought I would put it down after that.
The signal and the noiseWell, turned out I couldn’t put it down… it was one of the most fascinating, well written, well-researched books I had read in a long time. The subject matter was just so
engrossing! And it’s in use daily all around us, most commonly in predicting weather patterns but also in games like baseball and it has many other applications.
The main theme of the book is how to make better predictions by using “Bayes’s Theorem”, Thomas Bayes lived way back… he was born in 1701 & died in 1761 so has been around for a while but I don’t
know how many people know about it.
The basis of his theorem is that one can improve predictions & probabilities based on newer events occurring. Let me give you a quick example from the book:
1. The unexpected underwear in the dresser drawer example.
You are living with a partner & come home from a business trip to discover a strange pair of underwear in your dresser drawer. You will probably ask yourself what is the probability that your partner
is cheating on you? The “condition” is that you have found the underwear, the “hypothesis” you are interested in evaluating is the probability that you are being cheated on. Bayes’s Theorem, believe
it or not, can give you an answer to this sort of question – provided that you know (or are willing to estimate) three quantities:
– First you need to estimate the probability of the underwear’s appearing “as a condition of the hypothesis being true” – that is, you are being cheated upon. Let’s assume for the sake of this
problem that you are a woman and your partner is a man, and the underwear in question is a pair of panties. If he’s cheating on you, it’s certainly easy enough to imagine how the panties got there.
Then again, even (and perhaps especially) if he is cheating on you, you might expect him to be more careful! Let’s say that the probability of the panties’ appearing conditional on his cheating on
you, is 50 percent.
– Second, you need to estimate the probability of the underwear’s appearing “conditional upon the hypothesis being false”. If he isn’t cheating, are there some innocent explanations for how they got
there? Sure, although not all of them are pleasant (they could be his panties). It could be that his luggage got mixed up. It could be that a platonic female friend of his, whom you trust, stayed
over one night. The panties could be a gift to you that he forgot to wrap up. None of these theories is inherently untenable, although some verge on dog-ate-my-homework excuses. Collectively you put
their probability at 5 percent.
– Third and most important, you need what Bayesians call a “prior probability”. What is the probability you would have assigned to him cheating on you before you found the underwear? Of course, it
might be hard to be entirely objective about this now that the panties have made themselves known. (Ideally you establish you priors before you start to examine the evidence.) But sometimes, it is
possible to estimate a number like this empirically. Studies have found, for instance, that about 4 percent of married partners cheat on their spouses in any given year, so we’ll set that as our
If we’ve estimated these values, Bayes’s theorem can then be applied to establish aposterior possibility. This is the number that we’re interested in: how likely is it that we’re being cheated on,
given that we’ve found the underwear? The calculation and the simple algebraic expression that yields it follows:
So, as you can see above, before you ever found the panties, the probability you would assign to your partner cheating on you was 4% but after you found the panties that estimate jumped to 29%. This
might not seem particularly high but it’s largely influenced the low “prior probability”. From the date you found the panties your new “prior” becomes 29% and if you found panties again, the
likelihood that he really is cheating you would keep climbing, in this case the new “posterior possibility” that he is cheating on you would jump to 75%!!! So, as Donald Trump would say “you’re
Well, if you’re still with me… well done! I’m going to start bringing this one in now…. don’t worry, it has to do with commercial real estate!
2. Toll Brothers Buys 600 Acres of land near The Woodlands & TWDC still has +- 1,000 Acres to develop!
Image 2.
So, the big question always for an investor that’s buying land is this “Will it go up in value?”. Let’s see what’s happened historically…
Image 3 – Source: Texas Real Estate Center – A&M University
The red line in the chart at the bottom above is the “nominal” price of land per acre since 1966 and the green line is the inflation-adjusted “real” price per acre. In nominal & real pricing we’ve
had a pretty good run since about 1996 but it’s tapered off a little in the last three or so years – as an average across a very large region. Having looked at the data, I would say that the chances
of prices improving on a “nominal” level are at least 50/50 if you didn’t take any other factor into account.
So… if we used Bayes’s Theorem, we could use 50% as our “prior” estimate.
Now, if we knew that a large company like ExxonMobil was going to bring 10,000 new employees to the area what would that do to land prices within a 10-mile radius of the ExxonMobil campus? I would
estimate that ExxonMobil’s arrival would affect land prices positively about 70% of the time within that zone.
If ExxonMobil was not going to be arrive, our estimate at any “other than usual” effect on pricing would most likely be much less but still positive because the area within the 10-mile radius of the
ExxonMobil location that is West of the I-45 contains The Woodlands and it is growing fast & has been an “engine for growth” for the area. So, probably an estimate of about 25% of the time a tract
would appreciate above the normal rate.
So, we can now run Bayes’ Theorem to see what the chances are of land prices increasing now that ExxonMobil has started building it’s campus…. x = 50%, y = 70%, z = 25% => (xy) divided by xy + z(1-x)
=> 73.7%.
So… once ExxonMobil decided to put in their campus, the chance of the price of a tract of land within 10 miles of that campus increasing in value was 73.7%.
But…. and here’s the interesting thing. The county appraisal values for land around Beltway 8 has increased faster than tracts of land not around Beltway 8 due to the improved access the beltway
provides. Well, if you look at image 2 above you’ll notice that the “Future Grand Parkway” is coming through immediately below ExxonMobil. And, because we know from Beltway 8 history that land within
range of the beltway increases faster than normal we have two factors driving pricing & probability of price increases.
So, we can now update our “prior” from 50% to 73.7% and run the formula again using similar estimates for factors “y” and “z”: x = 73.7%, y = 70%, z = 25% => (xy) divided by xy + z(1-x) => 88.7%
So, the chances of land within 10 miles of the ExxonMobil campus and within reach of the Future Grand Parkway will have an 88.7% chance of increasing in value.
People and companies have been watching this area very closely…. this is why just a few weeks ago Toll Brothers, one of the biggest home builders in the country, just closed a purchase of
approximately 600 Acres of land almost due West of the Creekside subdivision and just East of FM 2978. That 600 Acres is about a mile (as the crow flies) from about a 1,000 Acres that The Woodlands
Development Corporation owns and which has not been developed yet.
We also know that FM 2978 has been scheduled to be widened to handle the high volume of traffic (which is only going to increase when Toll Brothers starts building!).
So…. using Bayes’ Theorem, we can run the expression for each major development or news item to see how subsequent events adjust the probability of price increases (or decreases if there’s bad news).
In the case of the land within 10 Miles of ExxonMobil, to the West of the I-45 and to the North of the Grand Parkway, the prognosis for increases in prices per acre is extremely positive right now.
3. What are the biggest price drivers?
Ok… so we have an exciting way now to evaluate the potential for price appreciation but what types of “events” should we look at? Here are the big ones:
• population growth: this comes with job growth i.e. if big companies move into an are offering many new positions
• new roads or road widenings: land needs to have good access and good visibility to allow development
• big new developments: large land buys, new home subdivisions, large commercial developments
4. Which of these price drivers do we have locally?
So, what do we have in our area around The Woodlands that qualifies? Let’s take a look:
• ExxonMobil: 386 Acres of land purchased bringing 10,000 new employees and expecting to drive an additional 30,000 employees in support industries
• Springwoods: Over 1,000 Acres of land being developed as a “mini-Woodlands”, homes, hospitals, commercial office buildings, retail components, etc.
• Toll Brothers: Purchased approx. 600 Acres for home development between The Woodlands Creekside & FM 2978 (announced Sep. 2013)
• FM 2978: Plans to widen it from 249 all the way North to past FM 1488
• Baker Hughes: $54M USD investment in developing “Western Hemisphere Education Center” in Tomball. Estimate 55,000 employees trained annually at this location.
Any one of these events would usually be cause for celebration in a community… the combined effect of the events listed above is almost unprecedented growth plans happening now and underway.
So, the impact on the “Bayes’s Theorem”, where each subsequent event influences positively (in this case) the probability for future events causing an expected result, becomes self-perpetuating with
all the major events we have happening around us.
4. Which brings me to…..
…. some land we’re selling! Well, I am a broker so you should have expected this to end this way! 🙂
Land Listings
Well….. that’s it for now… hope you have a great week/month until the next time!
If you are interested in buying or selling land or even developing land, please get in touch – we should be able to help.
Blog, CRE: Land, Newsletters
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Clan embeddings into trees, and low treewidth graphs
In low distortion metric embeddings, the goal is to embed a host "hard"metric space into a "simpler"target space while approximately preserving pairwise distances. A highly desirable target space is
that of a tree metric. Unfortunately, such embedding will result in a huge distortion. A celebrated bypass to this problem is stochastic embedding with logarithmic expected distortion. Another bypass
is Ramsey-type embedding, where the distortion guarantee applies only to a subset of the points. However, both these solutions fail to provide an embedding into a single tree with a worst-case
distortion guarantee on all pairs. In this paper, we propose a novel third bypass called clan embedding. Here each point x is mapped to a subset of points f(x), called a clan, with a special chief
point ?(x)e f(x). The clan embedding has multiplicative distortion t if for every pair (x,y) some copy y?e f(y) in the clan of y is close to the chief of x: miny?e f(y)d(y?,?(x))? t· d(x,y). Our
first result is a clan embedding into a tree with multiplicative distortion O(logn/?) such that each point has 1+? copies (in expectation). In addition, we provide a "spanning"version of this theorem
for graphs and use it to devise the first compact routing scheme with constant size routing tables. We then focus on minor-free graphs of diameter prameterized by D, which were known to be
stochastically embeddable into bounded treewidth graphs with expected additive distortion ? D. We devise Ramsey-type embedding and clan embedding analogs of the stochastic embedding. We use these
embeddings to construct the first (bicriteria quasi-polynomial time) approximation scheme for the metric ?-dominating set and metric ?-independent set problems in minor-free graphs.
Original language English
Title of host publication STOC 2021 - Proceedings of the 53rd Annual ACM SIGACT Symposium on Theory of Computing
Editors Samir Khuller, Virginia Vassilevska Williams
Publisher Association for Computing Machinery
Pages 342-355
Number of pages 14
ISBN (Electronic) 9781450380539
State Published - 15 Jun 2021
Externally published Yes
Event 53rd Annual ACM SIGACT Symposium on Theory of Computing, STOC 2021 - Virtual, Online, Italy
Duration: 21 Jun 2021 → 25 Jun 2021
Publication series
Name Proceedings of the Annual ACM Symposium on Theory of Computing
ISSN (Print) 0737-8017
Conference 53rd Annual ACM SIGACT Symposium on Theory of Computing, STOC 2021
Country/Territory Italy
City Virtual, Online
Period 21/06/21 → 25/06/21
• Clan Embedding
• Compact Routhing Scheme
• Metric $\rho$-dominating set
• Metric $\rho$-isolated set
• Metric embeddings
• Minor-free Graphs
• Ramsey Type Embedding
• Treewidth
ASJC Scopus subject areas
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VBA Row Count
What is Excel VBA Row Count?
In Excel VBA, the Row count refers to the number of rows within a worksheet or a specific range of cells. It is commonly used to determine the size or extent of a data set. To obtain the Excel
VBA Row Count, you can use the Rows.Count property for VBA row count visible cells.
An example showcasing the row count of a sheet is shown below:
It counts all the rows available in “Sheet1” and prints the value in the Immediate tab, resulting in the output below:
You can modify this VBA code to suit your needs, such as obtaining the row count within a particular range or performing further operations based on the row count.
Key Takeaways
• In Excel VBA Row Count, the Rows.Count returns the total number of rows in a worksheet. It represents the maximum row index that can be used in a worksheet.
• The value returned by Rows.Count is based on the maximum number of rows allowed in the Excel version being used. In the most recent versions of Excel, this value is 1,048,576 rows.
• The actual last used row may be less than the Rows.Count if there are empty rows below the last used row.
• When using Rows.Count, consider the context in which it is used. It is commonly used in combination with other methods or properties to perform operations on specific ranges, such as finding the
last used row in a column or looping through rows of data.
How to count rows in VBA?
Consider an example table consisting of multiple rows and columns where you need to find the number of rows in a specific range defined by the user using the VBA Row Count array.
Step 1: Go to the “Developer tab” section in the toolbar and click the “Visual Basic” option. Now, the VBA Editor opens to add functions and Sub procedures. Then, click the “Insert” and “Module”
buttons to create a new module or blank page.
Step 2: Define a sub-procedure to find the row count of a given range from cells A1-F10.
Step 3: Initialize three variables, RowCount to store the count of rows, ‘w’ to specify which sheet this sub-procedure will work on, and rng to define the range of cells to be worked on in this
Step 4: Set ‘w’ as the current worksheet to be working in, rng as the range of cells we’ll work on, and RowCount as the value returned by the Rows.Count function.
Step 5: Print the RowCount value using the Debug.Print function. It will print the output in the Immediate tab.
Sub RowCountforSpecificRange()
Dim RowCount As Long
Dim w As Worksheet
Dim rng As Range
Set w = ThisWorkbook.Worksheets(“Sheet1”)
Set rng = Range(“A1:F10”)
RowCount = rng.Rows.Count
Debug.Print “Row Count: ” & RowCount
End Sub
Step 6: Press the run arrow button or F5 to run the program. The output is shown below:
Here, we can see, irrespective of the number of columns in the table, Rows.Count returns only the number of rows.
Excel VBA – All in One Courses Bundle (35+ Hours of Video Tutorials)
If you want to learn Excel and VBA professionally, then Excel VBA All in One Courses Bundle (35+ hours) is the perfect solution. Whether you’re a beginner or an experienced user, this bundle covers
it all – from Basic Excel to Advanced Excel, Macros, Power Query, and VBA.
Let us look at some examples on how to use the Excel VBA Row count function in our examples.
Example #1
Assume an example where you want to implement a Boolean or gate for two inputs in an Excel Table. To do so, we need to perform a FOR loop that runs through the entire table, checking inputs in both
columns “A” and “B” and printing the output of the Boolean ‘OR’ function in column “C.”
Step 1: Create a sub-procedure that implements the Boolean OR gate function.
Step 2: Then, define two variables lastRow and ‘,’’ where lastRow returns the row count of the table, and i is the iterative variable to be used in the For loop.
Step 3: Set the row count variable as lastRow. For the table to be dynamic whenever there’s a change in row count, we use the Excel VBA row count xlUp function. The xlUp function returns the last
active row in a select column with which we can find the count and the number of iterations needed to get the desired results.
Step 4: Initialize a FOR loop from 0 to the total row count available in the table, then check if either value in columns A and B has the value 1, which will print the value 1 in column C, or else,
it prints 0.
Sub OrGate()
Dim lastRow As Long
Dim i As Long
lastRow = Range(“A” & Rows.Count).End(xlUp).Row
For i = 1 To lastRow
If Range(“A” & i).Value > 0 Or Range(“B” & i).Value > 0 Then
Range(“C” & i).Value = 1
Else: Range(“C” & i).Value = 0
End If
Next i
End Sub
Step 5: Run the module using F5. The output is shown below.
Example #2
Consider an example where you need to copy a table from one place to another in the same sheet.
To do this, we need to find the size of the table by returning the VBA row count array, using which we’ll copy the table from column A to column K in sheet 1.
Step 1: Create a sub-procedure that performs VBA copy-paste.
Step 2: Define ‘ro,’ which will return the VBA row count visible cells. Then, define ‘rng’ as the range of the table, source as the table which will be copied, and destination as the part in the
sheet where the table will be posted.
Step 3: Assign ‘rng’ as the range of the given table and ‘ro,’ which returns the count of rows.
Step 4: Set the source and destination cells. For the source, we assign the range of the table from columns A to F for the row count.
Step 5: Perform the Excel VBA Copy-paste task using the source and destination.
Sub CopyTables()
Dim ro As Long
Dim rng As Range
Dim source As Range
Dim destination As Range
Set rng = Range(“A1:F10”)
ro = rng.Rows.Count
Set source = Range(“A1:F” & ro)
Set destination = Range(“K1”)
source.Copy destination
End Sub
Step 6: Run the above code using F5 or by pressing the green triangle on the VBA toolbar. We will get the output shown below. The copied table will be pasted from cell K1 to cell P10:
Example #3
In this example, we need to simulate a Boolean AND gate for three inputs and print ‘TRUE’ or ‘FALSE’ for the Excel VBA Row Count.
For this, we run a FOR loop through the table to check each value and apply the function depending on whether the conditions are true or false. To execute the FOR loop, we need the number of rows,
which can be found by the Excel VBA Row Count xlDown function.
Step 1: Initialize a sub-procedure which will implement the Boolean AND gate function.
Step 2: Initialize two variables, ‘l,’ which will return the number of rows, and ‘c,’ an iterative variable for the FOR loop.
Step 3: Find the number of rows using the Excel VBA Row Count xlDown function.
Starting from cell A1 (Range(“A1”)), the End(xlDown) method is applied, which simulates pressing the ‘Ctrl + Down Arrow’ key combination in Excel. It causes Excel to move down from the starting cell
until it reaches the last used cell in the column. Finally, the .Row property retrieves the row number of the last used cell.
Step 4: Initialize a for loop from cell 1 to the last used row in the table and check if the values are more significant than 0 in all columns using the Excel VBA AND function.
If all values are greater than 0, it prints as “TRUE”; else, it prints as “FALSE.”
Sub AndGate()
Dim l As Long
Dim c As Long
l = Range(“A1”).End(xlDown).Row
For c = 1 To l
If Range(“A” & c).Value > 0 And Range(“B” & c).Value > 0 And Range(“C” & c).Value > 0 Then
Range(“D” & c).Value = True
Else: Range(“D” & c).Value = False
End If
Next c
End Sub
Step 5: Run the module using F5 or click the green triangle on the VBA toolbar. The output is as shown:
Important Things To Note
• Use the Rows.Count property to get the total number of rows in a worksheet or range. It provides a dynamic way to determine the row count, adapting to changes in the data.
• Assign the row count to a variable for further use or store it in a specific location.
• Verify if the row count is greater than zero before performing operations that depend on the existence of data.
• Avoid hard-coding row counts unless you have a specific reason. Instead, use dynamic methods to determine the row count, allowing your code to adapt to changes in the data.
• It’s important to handle cases where the worksheet has no data. In such cases, the Rows.Count will still return the maximum number of rows, but you may need to check if there is any actual data
in the worksheet using methods like UsedRange or specific column checks.
• Don’t assume that the last row with data is always the same as the row count. For example, it’s possible to have blank rows in between or data that extends beyond the last used row.
• Avoid relying solely on the row count for data validation. Instead, always validate the data range using appropriate techniques, such as checking for empty cells or defined ranges.
Frequently Asked Questions (FAQs)
1. How do I count non-blank rows in Excel VBA?
To count non-blank rows in Excel VBA, you can use the SpecialCells method with the xlCellTypeConstants argument. This method allows you to select cells with constant values, excluding empty cells.
2. How to count Filter Rows in VBA?
To count filtered rows in Excel VBA, you can use the SUBTOTAL function in combination with the VisibleCells property. For example:
• Set a Range object to the desired range of cells.
• Apply a filter to the range using the AutoFilter method.
• Use the SpecialCells(xlCellTypeVisible) property to select only the visible (filtered) cells.
• Use the SUBTOTAL function with the COUNTA function as the first argument and the visible cells range as the second argument.
• Retrieve the count using the VBA Value property.
3. What does cell row count 1 mean in VBA?
In VBA, Cells(row count, 1) refer to a specific cell within a worksheet.
An explanation of each component is written below:
• Cells is a method used to reference cells within a worksheet.
• Row count represents the row number of the cell you want to refer to.
• 1 indicates the column number of the cell you want to refer to. Here, one refers to the first column (column A).
Download Template
This article must be helpful to understand the VBA Row Count, with its formula and examples. You can download the template here to use it instantly.
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This has been a guide to VBA Row Count. Here we learn how to count rows using Rows.count property in Excel VBA with examples and downloadable template. You may learn more from the following articles | {"url":"https://www.excelmojo.com/vba-row-count/","timestamp":"2024-11-10T10:56:04Z","content_type":"text/html","content_length":"216763","record_id":"<urn:uuid:9bed1f94-3468-4c09-aa5d-affaf23d0347>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00700.warc.gz"} |
What's the best way to interleave two Python lists?
[NOTE: I wrote this in January 2009, but didn't publish it. Originally, I planned to provide a short discussion about each of the potential solutions listed below, which I never got around to
doing. Anyway I just noticed my draft and decided to go ahead and publish it without adding any more discussion. The code snippets seem fairly self explanatory. If anyone has any comments on the
various solutions, I would be very interested in hearing them.]
Until early 2009, I had to add the following
file to build numpy or scipy on my 64-bit Fedora Linux box:
library_dirs = /usr/lib64
To make numpy aware of the default location required me to add
(which I will refer to as
for brevity) in
Where do 64-bit libraries belong?
directory is the default location for 64-bit libraries on Red Hat-based system. Unfortunately, not all Linux distributions conform to this
; but, fortunately, most distributions that don't use
as the default location for 64-bit libraries at least create a
symlink pointing to whatever their default location happens to be. So it appears I can assume that if I am on a 64-bit machine, looking in
should work in most cases.
Since I only wanted to add the
path on 64-bit machines, I changed the assignment to:
lib_dirs = libpaths(['/usr/lib'], platform_bits)
is 32 and
['/usr/lib64', '/usr/lib']
when it is 64. I used the
module to set
# Determine number of bits
import platform
_bits = {'32bit':32,'64bit':64}
platform_bits = _bits[platform.architecture()[0]]
An outline of the solution
So far everything has been pretty straight-forward. Now all that is left is to write libpaths.
def libpaths(paths, bits):
"""Return a list of library paths valid on 32 or 64 bit systems.
paths : sequence
A sequence of strings (typically paths)
bits : int
An integer, the only valid values are 32 or 64.
>>> paths = ['/usr/lib']
>>> libpaths(paths,32)
>>> libpaths(paths,64)
['/usr/lib64', '/usr/lib']
if bits not in (32, 64): raise ValueError
# Handle 32bit case
if bits==32: return paths
# Handle 64bit case
return ????
How to skin the cat?
So we finally arrive at the motivation for this post. At this point, I started thinking that if I had two equal-sized lists that there should be a simple function for interleaving the elements of the
two lists to make a new list. Something like
. But
returns a list of tuples. After discussing this with several people (Fernando Pérez, Brian Hawthorne, and Stéfan van der Walt), we came up with several different solutions.
from itertools import cycle
paths64 = (p+'64' for p in paths)
return list((x.next() for x in cycle([iter(paths),paths64])))
def _():
for path in paths:
yield path
yield path+'64'
return list(_())
out = [None]*(2*len(paths))
out[::2] = paths
out[1::2] = (p+'64' for p in paths)
return out
out = []
for p in paths:
return out
out = []
for p in paths:
out.extend([p, p+'64'])
return out
return [item for items in zip(paths, (p+'64' for p in paths)) for item in items]
from operator import concat
return reduce(concat, ([p, p+'64'] for p in paths))
I liked Solution 5 the best and it is what I
An itertools recipe
While we were looking for a solution, Fernando and I came up with the following recipe:
from itertools import cycle,imap
def fromeach(*iters):
"""Take elements one at a time from each iterable, cycling them all.
It returns a single iterable that stops whenever any of its arguments is exhausted.
Note: it differs from roundrobin in the itertools recipes, in that roundrobin continues until all of its arguments are exhausted (for this reason roundrobin also needs more complex logic and thus has more overhead).
>>> list(fromeach([1,2],[3,4]))
[1, 3, 2, 4]
>>> list(fromeach('ABC', 'D', 'EF'))
['A', 'D', 'E', 'B']
return (x.next() for x in cycle(imap(iter,iters)))
8 comments:
I also like solutions 5 as a general solution, though stylistically, I prefer "map(out.extend, ([p, p + '64'] for p in paths))" to the for loop. I think yours is the more pythonic solution (and
might even be more efficient). I like to use map if all a for loop does is call a function on each element of an iterable, since I think it's a bit clearer and it saves a line of code.
Another solution (a modification of solution 7) is "return sum(([p, p + '64'] for p in paths), [])", which saves an import. I'd probably go with that since it's shorter, and efficiency only
matters if the lists are extremely long or this function gets called a lot, neither of which is true in your application.
A variation of solution 6:
[item for items in ([p, p + '64'] for p in paths) for item in items]. No need for the zip just use a generator expression.
Good thing there's only one way to do it with Python... :-)
Here's another simple one (note that you are throwing away the result of the map statement).
1> A = [1,2,3,4,5]
2> B = [6,7,8,9,10]
3> out = []
4> map(out.extend,zip(A,B))
<4> [None, None, None, None, None]
5> out
<5> [1, 6, 2, 7, 3, 8, 4, 9, 5, 10]
Or using a bit from the previous comment (and relating it to your example):
out = []
map(out.extend,([p, p + '64'] for p in paths))
another variation of solution 7:
reduce(lambda out, p: out + [p, p+'64'], paths, list())
I would use itertools.chain:
from itertools import chain
a = list('abcde')
b = list('12345')
['a', '1', 'b', '2', 'c', '3', 'd', '4', 'e', '5']
It can be extended to merge more than two lists:
c = list('XYZZY')
If the lists are long, or are iterables, or you want to merge the lists lazily, use iterttools.izip:
from itertools import chain, count, izip
for value in chain(*izip(count(),a+b+c)):
print value
Since you liked zip, why not to like reduce (although iirc Guido doesn't like it ;-)), so for the actual interleave just use both functions you would like:
*In [1]: reduce(tuple.__add__, zip([1,2],[11,21]))
Out[2]: (1, 11, 2, 21) | {"url":"http://blog.jarrodmillman.com/2010/10/whats-best-way-to-interleave-two-python.html?showComment=1287017052117","timestamp":"2024-11-06T17:10:01Z","content_type":"text/html","content_length":"61815","record_id":"<urn:uuid:2591fd87-10d4-4b96-af73-d617851f00f0>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00016.warc.gz"} |
8th Grade Math Early Finishers
This resource is my 8th Grade Math Early Finishers (Paper Version). It includes NO PREP! Just print and go! CLICK THE PREVIEW!!!
There are 42 pages. Each page has a word search, unscramble the words, and an activity on the bottom (color by code, practice problems, or etc). These early finishers can be used when students finish
a test early, on sub days, as homework, as warmups, and so much more!
The concepts included are:
(1) The Real Number System
(2) Rationals vs Irrationals
(3) Fractions into Repeating Decimals
(4) Repeating Decimals into Fractions
(5) Square Roots & Cube Roots
(6) Approximating Non-Perfect Squares
(7) Exponent Properties
(8) Adding and Subtracting in Scientific Notation
(9) Multiplying and Dividing in Scientific Notation
(10) Equations with Variables on Both Sides
(11) Equations with Parentheses
(12) Number of Solutions to Linear Equations
(13) x- and y-Intercepts Given a Table
(14) x- and y-Intercepts Given a Graph
(15) x- and y-Intercepts Given an Equation
(16) Graphing using Intercepts
(17) Types of Slope
(18) Slope Given a Graph
(19) Slope Given a Table
(20) Slope Given Two Points
(21) Slope-Intercept Form
(22) Writing Linear Equations from a Graph
(23) Writing Linear Equations Given 2 Points
(24) Systems of Equations (Graphing)
(25) Systems of Equations (Substitution)
(26) Systems of Equations (Elimination)
(27) Function Notation & Evaluating a Function
(28) Linear vs Non-Linear
(29) Rigid Transformations
(30) Dilations
(31) Similarity
(32) Congruence
(33) Angle Relationships
(34) Triangle Angle Sum Theorem
(35) Triangle Exterior Angle Theorem
(36) The Pythagorean Theorem
(37) Pythagorean Theorem in the Plane
(38) Volume: Cone, Cylinder, & Sphere
(39) Intro to Scatter Plots
(40) Constructing Scatter Plots
(41) Line of Best Fit
(42) Two-Way Tables
You can use the assignment as a review at the end of the year OR you can use it throughout the entire school year!
Questions? Please contact me at mathindemand@hotmail.com. Thanks so much!
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EViews Help: mtos
Convert matrix object to a series or group.
atrix-TO-eries object: convert the data in a matrix object to a series, alpha, or group.
mtos(vector, series[, sample])
mtos(svector, alpha[, sample])
mtos(matrix, group[, sample, prefix])
mtos(matrix, group[, prefix])
The number of observations in the sample must match the row size of the matrix to be converted. If no sample is provided, the matrix is written into the series using the current workfile sample.
For the matrix forms of the command, the prefix parameter is a string. If the target group object does not exist, the group is created and populated with series named <prefix>1, <prefix>2, etc. If
the prefix is omitted, the default prefix is “SER”.
converts the first column of the matrix MOM to the first series in the group GR1, the second column of MOM to the second series in GR1, and so on. The current workfile sample length must match the
row length of the matrix MOM. If GR1 is an existing group object, the number of series in GR1 must match the number of columns of MOM. If a group object named GR1 does not exist, EViews creates GR1
with the first series named SER1, the second series named SER2, and so on.
series col1
series col2
group g1 col1 col2
sample s1 1951 1990
The first two lines declare series objects, the third line declares a group object, the fourth line declares a sample object, and the fifth line converts the columns of the matrix M1 to series in
group G1 using sample S1. This command will generate an error if M1 is not a
“Matrix Language”
for further discussion and examples of the use of matrices.
See also
, and | {"url":"https://help.eviews.com/content/commandcmd-mtos.html","timestamp":"2024-11-11T21:46:49Z","content_type":"application/xhtml+xml","content_length":"11097","record_id":"<urn:uuid:cc177303-256f-49de-b858-7a0fad1b246b>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00444.warc.gz"} |
Radiometric Adjustment of Aerial Imagery
Miriam MN Nærum
Kongens Lyngby 2014 DTU Compute-M.Sc.-2014
Phone +45 45253031, Fax +45 45881399 reception@compute.dtu.dk
www.compute.dtu.dk DTU Compute-M.Sc.
The goal of this thesis is to develop a colour correction to a number of overlap- ping aerial photographs.
Maps created from aerial photographs are used for many practical purposes, for instance environmental investigations and city planning. In order to make maps of large areas, a mosaic from several
photographs is created, and therefore the colours in the images should match to avoid visible seamlines between them.
COWI A/S has provided 22 overlapping orthophotos from aerial photos, which are used to investigate dierent methods of radiometric colour correction.
Three dierent methods are investigated: Global histogram matching, global pixelwise matching, and global gradual matching.
In histogram matching the histograms in two neighbouring orthophotos are matched, and a linear transformation is estimated for each of the 22 orthophotos simultaneously in global histogram matching.
Then global pixelwise matching, where linear transformations are estimated by simple pixelwise correspondence, is investigated.
The third method described is global gradual matching, where the colour correc- tion is performed under the assumption that there is a gradual change in colours over each orthophoto. In global
gradual matching the results are improved by using boundary conditions and change detection.
Change detection is used to remove pixels that contain e.g. moving objects, or tall objects photographed from dierent angles, from the model estimation. In all three models a regularization term is
added, such that a colour transforma- tion, which is too large, does not occur.
In order to evaluate the quality of the results three measures are dened: The seamline measure, the saturation, and the contrast.
Experiments are performed to determine the optimal regularization, which show that it should be chosen as a trade-o, between making the seamlines less dis- tinct, and obtaining a too low contrast for
global histogram matching and global pixelwise matching. For global gradual matching the connection between the regularization of the model coecients, the regularization used on the colour change in
the boundary, and change detection is investigated.
The experiments show that the best results are obtained, when global gradual matching is used with boundary conditions and change detection.
Målet med dette speciale er at udvikle metoder til farvekorrektion til ere over- lappende luftfotos.
Kort dannet ved hjælp af luftfotos bliver brugt til mange praktiske formål, for eksempel til miljøundersøgelser og byplanlægning. For at lave kort over større områder, dannes en mosaik ud fra ere
fotograer, og derfor bør farverne i billederne matche for at undgå synlige seamlines mellem dem.
COWI A/S har stillet 22 overlappende ortofotos fra luftfotos til rådighed til brug for undersøgelserne i projektet.
Tre forskellige metoder undersøges: Global histogrammatching, global pixelvis matching og global gradvis matching.
I histogrammatching matches histogrammerne i to nabobilleder, og en lineær transformation bliver estimeret for hvert af de 22 ortofotos samtidigt, og dette er grundlaget for den første metode, global
Den næste metode, der undersøges, er global pixelvis matching, hvor lineære transformationer estimeres ud fra en simpel pixel-til-pixel matching.
Den tredje metode, der er beskrevet, er global gradvis matching, hvor farve- korrektionen udføres under antagelse af, at der er en gradvis farveovergang i hvert ortofoto. I global gradvis matching
bliver resultaterne forbedret ved brug af randbetingelser og change detection.
Change detection bruges til at fjerne pixels, som indeholder for eksempel ob- jekter, der ytter sig, eller høje objekter der fotograferes fra forskellige vinkler, så der ses bort fra disse pixels i
estimeringen af modellen. I alle tre modeller tilføjes regulering, således at der ikke opstår for stor farvetransformation.
For at bestemme kvaliteten af resultaterne deneres tre mål: Seamline measure, saturation, og kontrast.
Den optimale regulering bestemmes ved hjælp af en række eksperimenter, som viser, at reguleringen bør vælges som et kompromis mellem at gøre seamlines mindre synlige, mod at kontrasten bliver for lav
ved global histogrammatching og ved global pixelvis matching. For global gradvis matching bliver sammen- hængen mellem reguleringen af modellens koecienter, reguleringen fra rand- betingelserne og
change detection undersøgt.
Eksperimenterne viser, at de bedste resultater ndes, når global gradvis mat- ching benyttes med randbetingelser og change detection.
This thesis was prepared at Department of Applied Mathematics and Computer Science at the Technical University of Denmark in fullment of the requirements for acquiring an M.Sc. in Mathematical
Modelling and Computing. The thesis was produced in collaboration with COWI A/S, Parallelvej 2, DK-2800 Kongens Lyngby.
The thesis deals with radiometric colour adjustment in orthophotos developed from aerial photos.
The thesis is structured as follows: There is an introduction in Chapter 1, Chapter 2 contains a description of previous work, and in Chapter 3 there is a description of the used data. Then follows
Chapter 4 with the theory of the methods for colour correction. In Chapter 5 these methods are investigated by performing several experiments, and the results are then discussed in Chapter 7. In
Chapter 6 some suggestions for future development are presented. The conclusions are listed in Chapter 8. Finally at the end of the thesis is the appendix and the bibliography.
Lyngby, 02-January-2014
Miriam MN Nærum
I would like to thank my supervisor associate professor Henrik Aanæs and as- sociate professor Anders Bjorholm Dahl from Department of Applied Mathe- matics and Computer Science. I would also like to
thank chief specialist Søren Andersen, engineer, consultant, coordinator Regin Møller Sørensen, and project director Søren Vosgerau Jespersen from COWI A/S, technical director David Child from COWI
Mapping UK, and geospatial software developer Lars Hansen from CDT3 Ltd.
Abstract i
Resumé iii
Preface v
Acknowledgements vii
1 Introduction 1
2 Previous Work 7
3 Data 9
3.1 Data Overview . . . 11
3.2 Aerial Photos . . . 13
4 Method 17 4.1 Mosaicking . . . 17
4.1.1 Downsampling . . . 19
4.2 Neighbourhood . . . 20
4.3 Histogram Matching . . . 21
4.4 Global Histogram Matching . . . 25
4.4.1 Reference Image . . . 27
4.4.2 Regularization . . . 29
4.5 Global Pixelwise Matching . . . 30
4.6 Global Gradual Matching . . . 31
4.6.1 Interpolation Method . . . 31
4.6.2 Multiplication Method . . . 33
4.6.3 Addition Method . . . 40
4.6.4 Logarithm Method . . . 42
4.6.5 Division Method . . . 44
4.7 Boundary . . . 48
4.8 Change Detection . . . 50
4.9 Quantication . . . 53
4.9.1 Gradient Based Quantication of Seamlines . . . 53
4.9.2 Saturation . . . 61
4.9.3 Contrast . . . 62
4.9.4 Trade-o . . . 63
4.10 Residuals . . . 63
4.11 Computational Optimization . . . 64
5 Results 67 5.1 Residuals . . . 67
5.2 Neighbourhood . . . 69
5.3 Change Detection . . . 72
5.3.1 Threshold, High Damping Parameter . . . 72
5.3.2 Pixel Ratio and Threshold . . . 74
5.3.3 Convergence Limit . . . 74
5.3.4 Change Detection Weights . . . 75
5.3.5 Examples with and without Change Detection . . . 76
5.3.6 Threshold, Low Damping Parameter . . . 78
5.3.7 Quantication of Change Detection . . . 80
5.4 Quantication . . . 81
5.5 Histogram Matching . . . 82
5.6 Global Histogram Matching . . . 83
5.6.1 Regularization . . . 83
5.6.2 Quantication . . . 86
5.6.3 Residuals . . . 91
5.7 Global Pixelwise Matching . . . 93
5.8 Global Gradual Matching . . . 97
5.8.1 Multiplication Method . . . 97
5.8.2 Division Method . . . 99
6 Future Work 117 7 Discussion 121 8 Conclusion 127 A Appendix 129 A.1 Correction to Rasmussen 2010 . . . 129
A.2 Boundary Conditions . . . 130
Bibliography 135
In many situations it is relevant to use a graphical map showing details of an area. Graphical maps are used for instance by local authorities to nd unauthorized buildings as house extensions, sheds
etc., for construction of roads and railways, by farmers to document eld boundaries, to measure vegetation e.g. rosehip or hogweed for removing weed, and investigations to protect the environment.
Furthermore the graphical maps provides basis for construction of technical maps and aerial maps. [16] [8]
A graphical map is created from a number of aerial photographs. These pho- tographs are taken from a plane, ying over the desired area a number of times, taking a number of overlapping photos. The
dierent light and weather con- ditions during the ight cause large changes in the colours in the aerial pho- tographs, and when they are combined, some will have dierent colours than others, although
they may cover some of the same area as illustrated in Fig- ure 1.1. The dierences will be seen as distinct lines, called seamlines, between neighbouring photographs. Furthermore, the dierences may
be seen as dierent shadows in images showing the same area, see Figure 1.3. Another problem is that moving objects may not have the same position in two overlapping images, see Figure 1.2.
(a) Orthophoto 3 (b) Orthophoto 11 (c) Orthophoto 3 and 11
Figure 1.1: The gure shows the seamline between two orthophotos in a small area of the graphical map. The black area in (a) is outside orthophoto 3.
(a) Orthophoto 5 (b) Orthophoto 6
Figure 1.2: Small area of two images showing a moving object.
(a) Orthophoto 18 (b) Orthophoto 20
Figure 1.3: Small area of two images showing trees photographed from dierent an- gles.
In this thesis, a number of methods for correcting the colours, in order to remove the dierences between the photos, will be investigated. Only radiometric colour correction is investigated, i.e. the
colour correction is only estimated from the pixels in the images and no information about e.g. the time of day, the position of the air plane, weather, etc. is taken into account. Some colour
correction have been performed on the data initially by COWI A/S, based on non-radiometric information.
Before correcting the colours in the images, an orthorectication is performed.
This is a process, in which the aerial photos are transformed into orthophotos, which have the property that in every pixel the photo appears to be taken from directly above [5]. Each orthophoto is
also georeferenced, which means that the geographical positions of the orthophotos are found [20]. The overlapping orthophotos are then combined into a single image of the entire area by using a
mosaicking method.
In this thesis the used colour correction methods are based on evaluation of the overlap between neighbouring orthophotos. An algorithm is used to exclude some of the overlaps to reduce the necessary
amount of data, using either 4 or 8 neighbouring orthophotos for each orthophoto. A method is presented, which uses the colour histogram in the overlaps to match neighbouring orthophotos.
The global histogram matching algorithm is presented, as an algorithm which matches all the used overlaps simultaneously. This algorithm is compared to an algorithm called global pixelwise matching.
Global pixelwise matching matches each pair of pixels in the overlaps instead. The third algorithm presented is based on the assumption that there is a linear change in the light from one side
of an orthophoto to the other caused by the position of the sun. This means that the correction is performed by transformation of the colour values, using a bilinear function for each orthophoto.
In order to be able to compare the results a quantication method is devel- oped, by using three measures consisting of a measure to quantify visibility of seamlines, a measure of the saturation, and
a measure of the contrast. Another measure, residuals, is provided in order to quantify the colour dierences lo- cally. This is a spatial measure of the colour dierences, based on the standard
In the correction methods change detection is used to remove objects that have moved, and therefore have dierent positions in dierent orthophotos. Further- more boundary conditions can be added to
global gradual matching.
The results are computed using aerial photographs, taken over a small area of Bornholm, provided by COWI A/S.
Figure 1.4: The original test area
Previous Work
Colour correction of aerial photographs has been investigated in earlier projects and scientic papers.
A number of requirements to ensure acceptable orthophotos is stated by a task group under Geoforum Denmark in [2] and by Ordnance Survey in [10]. There are demands for e.g. the size of the overlap
between the images, geometric qual- ity, the contrast, and the resolution. In [2] there is also a description of the basic methods used for mosaicking and feathering.
A colour transformation from RGB (Red, Green, and Blue) to HSV (Hue, Sat- uration, and Value) is presented by Naoum et al. in [9] and Tsai et al. in [18]
as a tool used for dierent methods of image colour enhancement.
The change detection method, Iterative Reweighted Multivariate Alteration De- tection (IR-MAD), which uses canonical correlation analysis to maximize the dierence between overlapping images, has been
described by Aasbjerg Nielsen in [11]. A linear multivariate alteration detection transformation is created, which highlights the areas with low correlation. For each pixel, the probabil- ity that a
change has occurred, is calculated and used as a weight for each pixel. The weights are updated in each iteration of the IR-MAD algorithm.
The method is invariant to linear transformations between the colours of the overlapping images caused by e.g. weather, sunlight, or other changes.
Mosaicking is used to merge orthophotos to form a mosaic, such that the seam- lines between overlapping orthophotos are as indistinct as possible. Dier- ent mosaicking methods are discussed by
Ødegaard Nielsen in [12]. Ødegaard Nielsen also describes a method of feathering to decrease the visibility of seam-
Colour adjustment of overlapping orthophotos has previously been investigated by Ødegaard Nielsen in [12] by matching the histograms of a pair of overlap- ping orthophotos. The overlapping areas are
transformed by matching their histograms to obtain a linear transformation. Rasmussen has extended the method in [15] to match a grid of overlapping orthophotos iteratively. A further extension is
presented namely global histogram matching by matching a number of overlapping images simultaneously. Regularization is used to prevent the im- ages from changing too much compared to the original
colours. An example is presented, which proposes to use image segmentation, and make dierent colour transformations for the dierent classes to obtain a higher saturation. It is also suggested to let
the transformation vary over each orthophoto.
In order to determine the quality of the results from colour correction a panel test is used to evaluate the results manually in [15].
A method for removing scan-angle distortion caused by dierent illumination, atmospheric conditions and reective properties of objects is proposed by Palu- binskas et al. in [13]. The method is a
correction algorithm, which is used for line scanner images made with a wide eld of view. A model is presented that calculates the radiance of an object as a function of the scan angle. The im- age
is initially classied using an unsupervised classication method, and linear regression is performed from the resulting clustering in order to estimate the needed parameters.
An approach to nd the optimal seamline between two images has been inves- tigated by Sadeghi et al. [17]. Dierent weight functions have been tested and the best candidate is for each pixelpgiven by
W(p) =k∇X[ij](p)− ∇X[ji](p)k^1[1] , (2.1) where Xij is the overlap between orthophoto i and j in orthophoto iand Xji
is the corresponding overlap in orthophotoj. This weight function is preferred since it is robust even if orthophotoiandj have dierent light conditions.
A method for making a colour correction in the overlap is presented. A correc- tion function on the overlapping area is estimated by solving Poisson's Equation with dierent boundary conditions.
This section contains a description of the data provided by COWI A/S. There is a denition of the used notation and an overview of the provided orthophotos.
Furthermore, some challenges with aerial photography are presented, due to e.g.
ight position and shadows.
Throughout this thesis a number of images are used to test the dierent ap- proaches in order to adjust the colours to form a geographical map. These im- ages are provided by COWI A/S and consist of
22 orthophotos. The orthophotos are constructed by performing orthorectication, where geometric distortion is removed and a topographic relief is used to make a planimetric geometry [5].
This means that any area of an orthophoto appears to be photographed from orthogonally above.
The data is provided in 4361 RGB images of 625 by 625 pixels called tiles. The tiles are placed in a chess board pattern with no overlaps. Each orthophoto consists of a number of tiles as shown in
Figure 3.1.
Figure 3.1: The gure shows orthophoto no. 2 with the tiles marked by a dark red grid. The tiles outside the orthophoto are marked in black.
In this project the following denitions will be used:
• Orthophoto no. iis dened asX[i]
• The overlap between orthophotoiandj in orthophotoiis dened asX[ij]
• The overlap between orthophotoiandj in orthophotojis dened asX[ji] as illustrated in Figure 3.2. Orthophotoiis represented by a3×ni matrixXi, where each row contains the pixel values of one of the
three colour bands, and ni is the number of pixels in orthophotoi. Xij is a3×nij matrix, wherenij is the number of pixels in the overlap.
Figure 3.2: The gure illustrates the denitions of variables that dene orthophotos and the corresponding overlaps.
3.1 Data Overview
In order to get an overview of the provided data it is investigated which or- thophotos the data set consists of. For this purpose a 3D-matrixD∈N^n^r^×n^c^×n^v is constructed, wheren[r]×n[c] denotes
the size of the area, denoted in tiles, and n[v] is the number of orthophotos. From this matrix it is observed that there are 22 orthophotos, and they are situated as shown in Figure 3.3. In the gure
each orthophoto is represented by a colour, and consequently the overlaps are shown with colours, which are combinations of these colours. In this example parts of the tiles are not covered by the
orthophoto, and they are therefore black or white. This will be taken into consideration in the calculations by using masks to exclude these areas.
It is deduced from the gure that the data is a segment of the route of an air plane, and that the data consists of 3 parallel lanes, as shown by the arrows in Figure 3.4.
Figure 3.3: The gure shows how the32×32tiles are situated in dierent orthophotos.
Each colour marks a dierent orthophoto.
Figure 3.4: The gure shows how the tiles are situated in dierent orthophotos. Each colour marks a dierent orthophoto and the direction of the 3 lanes in the route of the air plane is shown in red
3.2 Aerial Photos
In order to construct graphical maps, a number of aerial photos are taken by ying over the desired area a number of times, taking a number of overlapping photos. These overlapping photos are then
transformed into orthophotos, which have the property that in every pixel the photo appears to be taken from di- rectly above, as described at the beginning of this chapter. The overlapping
orthophotos are then combined into a single image of the entire area by using a mosaicking method, as described in Section 4.1.
Orthophotos are taken at dierent times of the day and dierent times of the
year, and under dierent weather conditions. This means that some orthophotos are brighter than others that cover the same area. For this reason the combined graphical map will clearly show the
borders between the dierent orthophotos.
The photographs are taken, such that there is a large overlap between images within the same lane, and a smaller overlap between images in dierent lanes.
Dierent demands can be set for the quality. For instance it is specied by Ordnance Survey [10] that they require a minimum of 55% within ight lanes, and a minimum of 20% between ight lanes.
The time of day can have much eect on the colours in orthophotos. The only light source used for the imaging is the sun, and therefore the colours in the image are dependent on the relative angle to
the sun. Dierent angles relative to the sun will change the overall brightness of the image.
An object with a reective surface will appear brighter, when the air plane is positioned, such that the angle of incidence, i.e. the angle between the line from the sun to a reective object and the
normal of the object surface, is equal to the angle of reection, i.e. the angle between the line from the plane to the reective object and the normal of the object surface. Therefore such an object
will appear dierently, depending on the position of the plane. An example of such an object could e.g. be a tin roof, which is highly reective and not parallel to the ground. [12]
A similar problem can arise, if a tall object, e.g. a building or a tree, is pho- tographed from dierent angles. From one angle the object will cover more of its shadow and the ground than from
another angle. This is illustrated in Figure 3.5.
(a) Orthophoto 19 (b) Orthophoto 9
Figure 3.5: A single tile from two dierent orthophotos taken at dierent angles.
Another part of the colour adjustment is that it is important to avoid over- and underexposed areas. If a reective surface becomes so bright that the structure is indistinguishable or if a shadow
becomes so dark that it is impossible to observe any details on the ground, the image is not acceptable [8]. An example of too dark a shadow is shown in Figure 3.6 where it is very dicult to
distinguish the ground from the part of the roof that lies in shadow.
Figure 3.6: A small part of orthophoto 11 where the shadow on the roof and on the rightmost part of the courtyard are indistinguishable
In this chapter the theory for some methods for radiometric colour correscion are described.
Initially the theory of mosaicking and histogram matching is presented as a tool for the three developed colour correction methods: Global histogram matching, global pixelwise matching, and global
gradual matching. The theory behind each of these methods is described, and two methods, used to improve the results, are specied. Finally some quality measures are presented.
4.1 Mosaicking
Mosaicking has a large inuence on the quality of the resulting graphical map, since it determines the position of the seamlines. For practical use minimum cost methods are used to place the
seamlines, and feathering is used to disguise seamlines, but in this thesis a crude mosaicking algorithm is used without feath- ering. The mosaicking in this thesis is performed using masks that
determines the position of the data in the tiles. The computational time is reduced by downsampling the tiles.
In order to make a map of a large area, a number of orthophotos are combined.
This process is called mosaicking, and it has great inuence on the result. If there is too large a dierence between two orthophotos, where the seamline is placed, it will become very distinct. This
can be limited by placing the seamlines
where the dierences are small. At COWI A/S a minimum cost algorithm is used to nd the optimal positions of the seamlines. However, sometimes it is not possible for the algorithm to nd a suitable
position, and some of the seamlines are therefore placed manually. This is done by placing seamlines at roads, streams, and along eld boundaries. It is important that seamlines are not placed too
close to buildings and other tall objects, since a tall object seen from dierent angles, may be covered by dierent pixels, and may therefore be shown twice. [8], [2], [6]
When the seamlines have been placed, the visibility of the seamlines is reduced by using feathering. This is a process, where the colour dierence between each side of the seamline is reduced by
smoothing a small surrounding area at both sides of the seamline. In city areas the feathering is performed on a narrow area, contrary to a wider area used in elds and forests. [8], [2]
In this thesis feathering is not performed, and a simple mosaicking algorithm is used. Each orthophoto is simply added to the graphical map by a user specied order, such that the rst orthophoto has
the highest priority, and the second is added only in the area that is not covered by the rst orthophoto. The third orthophoto is placed, where the area is not covered by either the rst or the second
orthophotos, etc.
As each orthophoto is placed in the graphical map, the area it covers is recorded in a reference map. The reference map can therefore be used to specify where each of the orthophotos in the graphical
map is visible. The algorithm is sum- marised in Algorithm 1 and an example of the mosaicking of the test area and the corresponding reference map are shown in Figure 4.1. In this case the or-
thophoto priority list is given by
P = (1,2,3,4,5,6,7,15,16,17,18,19,20,21,22,8,9,10,11,12,13,14) . (4.1) The orthophotos have been given this order to ensure that a large number of orthophotos are visible, since this provides more
visible seamlines. Figure 4.1b shows that 17 of the available 22 orthophotos are visible.
The chosen order in the sequence can have large inuence on the result, since it determines how much of an orthophoto is shown in the resulting graphical map.
Algorithm 1 Simple mosaicking algorithm
Require: Orthophoto priority listP and Transformation booleanT
1: DeneG: Graphical map, size of photographed area
2: DeneR: Reference map, size of photographed area
3: for all Orthophotos∈P do
4: if T then
5: Insert orthophotoA[i]·P[i] whereGis empty
6: else
7: Insert orthophotoP[i] whereGis empty
8: end if
9: Insert the position of orthophotoPi intoR
10: end for
11: return Gand R
(a) (b)
Figure 4.1: The gure shows (a) the original test area and (b) the corresponding reference map.
4.1.1 Downsampling
In order to decrease the necessary computational time all the tiles are downsam- pled before the computations are performed. The images are downsampled to a smaller number of pixels by using bicubic
interpolation, i.e. the pixels in the downsampled image are computed as a weighted average of the neighbouring pixels in the original image.
If a tile is only partly covered by an orthophoto, the rest of the image is black or white. Therefore a mask is created that removes these parts of the image.
However, due to rounding errors and the interpolation used in the downsampling process, a part of the resulting graphical map will contain undened small white or black parts near the border of the
orthophoto. This is removed by performing dierent morphological operations.
Downsampling is discussed further in Section 4.11.
4.2 Neighbourhood
In order to reduce the computational time, model estimation is performed only on a selected set of overlaps. Since the images are approximately placed in a grid, 4-neighbourhood and 8-neighbourhood
are used to exclude some of the overlaps.
In some cases there are a lot of overlaps between the orthophotos dependent on how they are produced. In this case the photographs are taken with relatively small intervals as described in section 3.
Therefore there are 109 overlaps be- tween the 22 orthophotos. A high number of overlaps can be an advantage since each overlap adds some information to the algorithm, which makes the model more
precise. However, it is not practical to use so many overlaps, because the great number of matchings is very time consuming. The linear system of equa- tions would also be less sparse, and therefore
more computational power will be necessary.
Therefore only a selected number of overlaps are used in the implementation.
Since the orthophotos have been made from an air plane, they are aligned in three parallel lanes, approximately equidistantly. This means that the centers of the orthophotos are approximately
situated in a grid as illustrated in Figure 4.2. Due to the fact that the orthophotos are situated in lanes, each orthophoto should be dependent on information in its own lane and in neighbouring
Figure 4.2a shows how neighbouring pairs of orthophotos are selected by using 4-neighbourhood (also called city-block/Manhatten neighbourhood). Each or- thophoto that does not lie on the boarder has
4 neighbours, and this method only uses 34 of the possible 109 overlaps. In Figure 4.2b neighbouring orthophotos are selected using 8-neighbourhood (also called checkerboard neighbourhood).
In this method each orthophoto that is not on the boarder has 8 neighbours, and there are therefore more overlaps between the three lanes. This method uses 58 overlaps of the possible 109.
(a) (b)
Figure 4.2: The gure shows (a) 4-neighbourhood and (b) 8-neighbourhood of the orthophoto centres.
In Section 5.2 some experiments have been made to show the eect of using 4-neighbourhood and 8-neighbourhood using global histogram matching.
If 8-neighbourhood is chosen the model estimation is based on more overlaps, but the computational time will also be higher than if 4-neighbourhood is used.
4.3 Histogram Matching
Histogram matching is a method that uses the overlap between two images to give neighbouring orthophotos matching colours. This is done by matching the two colour histograms in the overlap, and use
the result to estimate a lin- ear transformation. Histogram matching is basis for the rst colour correction method in this thesis, global histogram matching.
Histogram matching is uset for colour correction, since it is invariant to possible errors in orthorectication and georeferencing [15].
Initially histogram matching has been performed as described in [15] and [3].
In histogram matching a model is estimated to ensure that the histograms of two images match in the overlap. This means that the two images will have
approximately the same amount of each colour. The model is dened by [3] as
cm(vout) =c(vin) , (4.2)
wherec(vin)is the cumulative histogram of the input imagevinandcm(vout)is the cumulative model histogram of the output imagevout. In order to nd the output image the combined process is performed
v[out] =c^−1[m] (c(v[in])) . (4.3) Since the images consist of a discrete number of colours the inverse of the cu- mulative model histogram is found by making a lookup table, where each entry tis
dened by
vout = min
t |cm(t)−c(vin)| (4.4)
This process is illustrated by the sketch in Figure 4.3. The gure illustrates that for each colour valueciin the cumulative histogram of the input image the corresponding colourcjin the cumulative
histogram of the output image, which is approximately at the same number of pixels in the cumulative histogram is found. This is done for all 256 colour values for each band red, green, and blue in
the input image, and the results are inserted into the lookup table.[12]
An example of the histograms in the overlap between orthophoto 1 and 2 is shown in Figure 4.4a, and the histograms after the matching process is shown in Figure 4.4b. It can be observed from the gure
that in the resulting histograms some peaks have been added where the original histograms should be higher, and some intensities have been removed to make the original histograms lower.
Figure 4.3: The gure shows a sketch of the process used to create the lookup table.
For each colour valueciin the cumulative histogram of the input image (marked in red) the corresponding colourcjin the cumulative histogram of the output image (marked in blue), which is
approximately at the same number of pixels in the cumulative histogram is found.
Figure 4.4: Histograms of the overlap between orthophoto 1 and 2 (a) before and (b) after the histogram matching
Once the histogram matching has been performed, a model which describes the colour transformation is estimated. This is done such that the model can be applied for the entire overlapping orthophoto.
For this a linear model is used, based on a least squares estimate.
Rasmussen has suggested three dierent linear colour correction models in [15].
All three have the general form
AI1=I2 , (4.5)
where the 3×nmatrixI1 is the image to be transformed, the3×nmatrixI2
is the reference image it is matched to, and nis the total number of pixels. In one of the modelsA is simply given by a full matrix
a b c d e f g h i
. (4.6)
The least squares estimate is then dened by
A=I2I[1]^T(I1I[1]^T)^−1 . (4.7) The second model presented is similar, but with an oset inserted such that
a b c α d e f β g h i γ
. (4.8)
The third model is a diagonal model given by
α 0 0 0 β 0 0 0 γ
. (4.9)
The rst diagonal element in the transformation matrix is estimated by α =
mean(R[1]), where R[1] andR[2] are the pixel values in the red band in I[1] and I[2], respectively, and similar estimations are made for the green and the blue band, respectively.
The two models in Equation (4.6) and (4.8) respectively, suggested by Ras- mussen in [15] are models that include o-diagonal elements. However, since the histogram matching has been performed
bandwise, any possible dependency between the three bands has been removed. Therefore only the diagonal model should be used. However, since this was discovered late in the project, the following
derivations have been performed using the model shown in Equation (4.6). The results show that the estimated o-diagonal elements are close to zero, and they have therefore little importance.
The histogram matching is extended for later use in the method global his- togram matching, where it is used globally, in order to histogram match all the orthophotos to their neighbouring images
4.4 Global Histogram Matching
In histogram matching one image is transformed to match another image by using the histograms of the overlap between them. A global histogram match- ing algorithm is used to nd the colour
transformation of all the orthophotos simultaneously. Initially a reference image is used, but later in this secdtion regularization is used to penalize the transformation to ensure that the colour
transformation is not too large.
The colour correction of all orthophotos simultaneously can be done by matching overlapping images in pairs succeedingly, but this is dependent on the sequence of images in the computation, and can
lead to inconsistencies. Therefore a global histogram matching algorithm, as described in [15], is used. This algo- rithm computes the transformations between all image pairs simultaneously by
computing the least squares solution to a linear system of equations.
The global histogram matching algorithm described in [15] is used to nd a transformation matrix Ai for each orthophotoi, which is represented by a 3×ni matrixXi, where each row is the pixel values of
the three colour bands, and ni is the number of pixels in the orthophoto. The pixels in the overlap between orthophoto i and orthophoto j is denoted Xij, which is a 3×nij
matrix, wherenij is the number of pixels in the overlap. In order to compute all the transformation matrices simultaneously, the algorithm attempts to solve the optimization problem, which minimizes
the expression given by
F = X
Gij (4.10)
= X
kAiXij−AjYijk^2[F] , (4.11)
whereN(i)is the set of neighbours to orthophotoiandk·k^2[F] is the Frobenious norm, which is dened by kMk^2[F] =P
jm^2[ij] =tr(M M^T), and Xij and Yij
is the overlap between orthophotoi andj as shown in Figure 3.2 in Chapter 3 before and after the histogram matching, respectively.
As the expression states, the algorithm attempts to minimize the sum of all G[ij] = kA[i]X[ij]−A[j]Y[ij]k^2[F]. In other words, the goal is to minimize the dif- ference between the transformed
original orthophoto i in the overlap and the transformed histogram matching of the other orthophotoj in the overlap. The expression in (4.11) is dierentiated and set equal to zero, which yields a
linear system of equations.
If all orthophotos overlap, the linear system of equations is given by [15]^1
K12+K13+. . . −L12 −L13 · · ·
−L21 K21+K23+. . . −L23 · · ·
−L31 −L32 K31+K32+. . . · · ·
... ... ... ...
A^T[1] A^T[2] A^T[3] ...
= 0(4.12)
where K[ij] = 2X[ij]X[ij]^T + 2Y[ji]Y[ji]^T and L[ij] = 2X[ij]Y[ij]^T + 2Y[ji]X[ji]^T. If a pair of orthophotos do not overlap, the correspondingK[ij] andL[ij] are simply 0.
This system of linear equations can simply be solved by setting all transforma- tion matricesAi= 0. This is of course not a viable option and it is avoided by letting one or more orthophotos be
reference images. This means that they are not transformed and the corresponding transformation matrix should therefore be the identity matrix. This is ensured by simply letting the left and the
right side be identity matrices. If e.g. orthophoto 1 is a reference image, the linear system of equations is modied to [15]
I 0 0 · · ·
0 K[21]+K[23]+. . . −L23 · · · 0 −L[32] K[31]+K[32]+. . . · · ·
... ... ... ...
A^T[1] A^T[2] A^T[3] ...
I L[21] L[31] ...
. (4.13)
The global histogram matching algorithm is implemented as described in Algo- rithm 2.
1There is a correction to the system of linear equations in [15], such that the transformation matrices are written asA^T[i] instead ofAi. This is explained further in appendix A.1
Algorithm 2 Global histogram matching [15]
1: For each pair of overlapping imagesi and j extract the overlapping pixels X[ij] andX[ji]and nd the corresponding histogram matchingsY[ij] andY[ji]
2: Set the right hand side valuesRHSto 0 and construct left hand side matrix LHSas follows:
3: for all imagesido
4: Add 3 rowsLHSi toLHS, constructed as follows:
5: for all neighboursj do
6: AddKij to the columns corresponding toAi 7: Add−Lij to the columns corresponding toAj 8: end for
9: end for
10: for all reference imagesido
11: Replace the image's 3 rows withIA^T[i] = I and move all other values in the columns forAi to the right hand side while negating them.
12: end for
13: Find transformationsA=
A^T[1] A^T[2] A^T[3] ...
as solution to
LHS·A=RHS⇒A= (LHS^TLHS)^−1LHS^TRHS (4.14)
14: Apply transformations to images
In global histogram matching all the transformation matrices are estimated for all the orthophotos simultaneously by nding the least squares solution of a sys- tem of linear equations. The algorithm
for doing this minimizes the dierences between the histograms in each overlap.
4.4.1 Reference Image
The global histogram matching algorithm is greatly inuenced by the choice of reference image. This is due to the fact that the histograms of all orthophotos are matched to each other, and as the
histogram of the reference image does not change, then all other histograms must change.
The inuence of the choice of reference image is been illustrated in Figure 4.5.
The result of global histogram matching using orthophoto 10 as reference image
is shown in Figure 4.5b. It can be observed that the image contains more yellow colours, and that some of the seamlines are more distinct, compared to the result using orthophoto 1 as reference image
in Figure 4.5a.
(a) (b)
Figure 4.5: The gure shows (a) the result of a global histogram matching using orthophoto 1 as reference image and (b) the result of a global histogram matching using orthophoto 10 as reference
image. Both computations are made using 4-neighbourhood.
The observations extracted from Figure 4.5 can be conrmed by calculating the three quantication measures. The measures for the two graphical maps are computed and shown in Table 4.1. The seamline
measures in the table state that the seamlines are more distinct in the result shown in Figure 4.5b. This conrms the previous statement from the observations of the gure.
The measures in the table also state that the saturation is higher when reference image 1 is used. This is conrmed by the gure since it can be observed that Figure 4.5b has more grey colours than
Figure 4.5a. It can also be observed that the contrast in Figure 4.5b is higher than in Figure 4.5a. This is in accordance with the contrast measures in the table.
Reference image 1 Reference image 10
Seamline measure 11.24 14.16
Saturation 0.353 0.268
Contrast 0.0885 0.127
Table 4.1: The table shows the three measures for two results of the global his- togram matching with dierent reference images. Both examples are computed using 4-neighbourhood.
4.4.2 Regularization
The example in Section 4.4.1 shows that when a reference image is used it has a large inuence on the result. The colours of the additional orthophotos are forced to match the colours of the reference
image, but it is only necessary that the orthophotos match each other.
Therefore a regularization term is introduced to replace the use of a reference image [15]. A damping parameter λ is used to penalize the sum of squared dierences between the transformation matrix
and the identity matrix, thus penalizing too large changes in the colours. This means that the regularization term is given by
λkI−A[i]k^2[2] , (4.15)
In order to prevent the dependency on choice of reference image the model is altered such that all images have equal importance. Then the linear system of equations is given by [15]
2λI+K12+K13. . . −L12 −L13 · · ·
−L21 2λI+K[21]+K[23]. . . −L23 · · ·
−L31 −L32 2λI+K[31]+K[32]. . . · · ·
... ... ... ...
A^T[1] A^T[2] A^T[3] ...
2λI 2λI 2λI...
where K[ij] = 2X[ij]X[ij]^T + 2Y[ji]Y[ji]^T and L[ij] = 2X[ij]Y[ij]^T + 2Y[ji]X[ji]^T. If a pair of orthophotos do not overlap, the corresponding K[ij] andL[ij] are simply 0.
In this altered model no image is chosen as reference image, but the colours of all images are moving as close to each other as possible, minimizing the dierence in the colour histograms between the
overlapping orthophotos.
The results obtained using regularization vary depending on the choice of damp- ing parameter. A number of examples have been made to investigate this eect.
The examples are described in Section 5.6.1. The examples conrm that with a large value of the damping parameter only a small change is permitted, and with a small value of the damping parameter
there are large changes in the colours.
It should be noted that too small values of the damping parameter may result in negative pixel values. An example is described in Section 5.6.2.2.
It is seen that global histogram matching is very dependent on the choice of ref- erence image. Therefore regularization is used to penalize the amount of colour transformation. The regularization is
dependent on the choice of a damping parameter λ.
4.5 Global Pixelwise Matching
Pixelwise matching is developed as a method, where each pair of pixels are matched directly.
In pixelwise matching each pixel in an overlap from orthophoto i is matched directly to the pixel in the corresponding position in orthophoto j. In order to avoid some of the possible errors of
orthorectication and georeferencing, a blurring of the image should be performed initially. Blurring is a process where the colour of a pixel is changed, such that it is inuenced by the neighbouring
pixels [7]. However, since the implementation in this thesis uses downsampling to reduce the computational time, the blurring is already performed by the downsampling process.
In this case the pixelwise matching is done by selecting all pixels of the same colour in orthophoto i, and calculating the mean colour of the corresponding pixels in orthophotoj for each band red,
green, and blue.
The reason for choosing this pixelwise matching method is that there are on average2.43·10^7 pixels in an overlap, and only 256 levels of each colour band.
Because of this the probability that each colour is fairly represented in each overlap is relatively high. This means that the probability that there are only a few outliers of the corresponding
pixels in orthophoto j is relatively high. The pixelwise matching is implemented in a global pixelwise matching algorithm corresponding to the global histogram matching algorithm.
A regularization term is added to the optimization problem, which penalizes the distance between the transformation matrices and the identity matrix. This is done similarly to the regularization for
global histogram matching in Section 4.4.2.
Some experiments have been performed, and the result of the global pixelwise matching algorithm is compared to the result of the global histogram matching
algorithm in Section 5.7.
Global pixelwise matching minimizes the sum of squared dierences between pixels in an overlap, unlike global histogram matching. As for global histogram matching, global pixelwise matching is very
dependent on the choice of the damping parameterλ. The experiments show that there is not much dierence between results of the two methods for this test area. However, the computa- tional time is
smaller for pixelwise matching, and this method allows the colour transformation to take dependencies between the colour bands into account.
4.6 Global Gradual Matching
In Section 4.4 global histogram matching is used to compute a transformation matrix for each of the 22 orthophotos. However, since each of the orthophotos cover a large area, there may be other light
conditions in one end of the photo than in the other. Therefore it would be prudent to change the model, such that it is able to take the local variation into account.
Five dierent models are investigated, which lead to: The interpolation method, the multiplication method, the addition method, the logarithm method, and the division method. Only the multiplication
method and the division method are used for experiments in this thesis.
4.6.1 Interpolation Method
In this method the linear colour transformations, estimated by global histogram matching or global pixelwise matching, are used to make an interpolation be- tween the transformation of the
neighbouring orthophotos.
With histogram and pixelwise matching, a transformation matrix is obtained for each orthophoto. Therefore, it may be prudent to calculate 4 or 8 (depend- ing on the neighbourhood) dierent
transformations and make an interpolation between them. This is done by computing a distance transformation and using it to weigh the dierent transformations in the interpolation. This ensures that
the pixels in an overlap only use the transformation found at the given overlap, and that the weight of this transformation is decreased the further away it is from the overlap.[15]
The purpose of using gradual transformation is to allow local variation within
each orthophoto. In order to model the local variation it is no longer sucient to compute a single linear transformation for an entire orthophoto. The local variation can be modelled by letting each
of the neighbouring orthophotos have dierent linear models obtained from the their respective overlaps.
Initially histogram matching is performed for each overlap as described in section 4.3. Once the linear models for the respective overlaps have been obtained, models should be found for the rest of
the pixels in the orthophoto. It seams plausible to make an interpolation between the obtained linear models. The interpolation is performed such that if a pixel is close to overlapk, the model of
the pixel has a large component from the model of the closest overlapAk. The interpolation in pixelj is given by [4]
yj = α1A1xj+α2A2xj+. . .+αPAPxj , (4.17) where P is the number of overlaps, α[r], where r ∈ {1,2, . . . , P} denotes the interpolation coecients, A[r] is the transformation matrix computed as shown
in Equation (4.16) for the respective overlaps, andx[j] is the pixel colour values in the orthophoto in question.
The interpolation coecients are found by making a distance transformation of each overlap. This method is performed by using a binary image. In this case the overlap will have value 1, and the rest
of the orthophoto will have value 0. A distance transformation is performed such that for each pixel in the orthophoto the Euclidian distance to the pixels with the value 1 is calculated. In this
case the distance transformation is used to assign coecients in the interpolation, and therefore the results are reversed, such that the values decrease, the further the pixels are from the overlap.
Afterwards the results are normalized between the orthophotos.
The algorithm is summarized in Algorithm 3
Algorithm 3 Gradual Transformation
1: for all imagesido
2: Find the neighboursN(i)using e.g. 4- or 8-neighbourhood
3: Locate the overlap between each pairiandj∈ N(i)
4: for all overlapsXij do
5: Perform a histogram matching between X[ij] and X[ji] and estimate a linear model
6: Make a distance transformationD[ij] fromX[ij]
7: NormalizeD[ij]
8: end for
9: Make an interpolation between the transformation matrices obtained with eachk∈ N(i)
10: Apply the obtained transformation toi
11: end for
The downside of using interpolation between previously created transformation matrices is that during the estimation of the matrices, the spatial nature of the algorithm is not taken into account.
This means that the coecients in each transformation matrix are already determined, and as the distance from each pixel to the neighbouring orthophotos is initially determined, the model no longer
gives a gradual estimation. Therefore this model was not investigated any further.
4.6.2 Multiplication Method
In the multiplication method a bilinear function is estimated and multiplied on each colour band. The function is estimated, such that the sum of squared dierences in the overlap is minimized. A
regularization term is added to prevent too large dierences.
The bandwise linear model, described in Section 4.3, given by
α 0 0 0 β 0 0 0 γ
, (4.18)
uses the same colour model, with constantα, β,and γ, for all the pixels in an orthophoto. The coecients can be computed by utilising that α= ^mean(R[mean(R]^y^)
for the red channelRin the original imageR[x]and the histogram matched image Ry, respectively. Similar computations can be performed to obtainβandγ[15].
In order to take the position of each pixel into account, the colour model is changed, such that the three componentsα, β,andγdepend on the coordinates (x, y)of each pixel. The three components in
the colour model are computed using a bilinear model, depending on the position, which for the red colour band, is given by
α(x, y) =ax+by+c , (4.19) wherea, b,andcare constant for each orthophoto, and(x, y)is the pixel position.
Other models could be used, but this model is the simplest method that is dependent on the pixel position, and it will make the computations relatively simple. This expression is used to compute the
component for the red band, and similar computations are performed for the green and the blue band.
In order to perform a gradual transformation the three coecients a, b, and c are estimated, such that the sum of squared dierences between orthophoto i and orthophoto j after the transformation is
minimized for the pixels in each overlap. This means that initially the overlap in orthophotoiis matched to the overlap in orthophotoj, which is held constant. The minimization problem is then given
minβ f =
r^(k)[i] (ax^(k)+by^(k)+c)−r^(k)[j] ^2
r^(k)[i] ax^(k)+r[i]^(k)by^(k)+r[i]^(k)c−r^(k)[j] ^2
= kRiβ−r[j]k^2[2] , (4.22)
Ri =
r[i]^(1)x^(1) r^(1)[i] y^(1) r[i]^(1)
... ... ...
r^(k)[i] x^(k) r[i]^(k)y^(k) r^(k)[i]
... ... ...
r^(n)[i] x^(n) r[i]^(n)y^(n) r[i]^(n)
β =
a b c
, (4.24)
andr[i] andr[j] are vectors containing the values in the red band in the pixels of the overlap in orthophotoiandj, respectively, andnis the number of pixels in the overlap. β is in this context the
three coecients inα(x, y)and should not be confused withβ in Equation (4.18).
4.6.2.1 Global Gradual Matching for Multiplication Method
With the method described in the previous section it is possible to compute a linear model depending on the pixel position for an entire orthophoto. However, since each overlap is used in the
calculations, the orthophoto is close to matching each of the overlaps, but not necessarily the neighbouring orthophotos. This is due to the fact that other linear models are applied to the
neighbouring orthophotos. Therefore, a global method should be developed to estimate all the coecients at once, as was done with the global histogram matching described in Section 4.4 [15].
Since the transformation from global gradual matching is estimated by using several orthophotos, a new variable, similar toR[i]in Equation (4.23), for overlap X[ij] is denoted
R[ij] =
r^(1)[ij] x^(1)[ij] r[ij]^(1)y[ij]^(1) r^(1)[ij]
... ... ...
r[ij]^(k)x^(k)[ij] r^(k)[ij] y[ij]^(k) r[ij]^(k)
... ... ...
r[ij]^(n)x^(n)[ij] r^(n)[ij] y[ij]^(n) r^(n)[ij]
. (4.25)
The subscript on the position coordinates (x, y) are due to the fact that the position in each pixel is relative to its orthophoto. This means that origo is at the bottom left corner of the
orthophoto. This will have the eect that the coecients of the linear model in each orthophoto are of the same order of mag- nitude. The linear function is created for each orthophoto with the
coecient vector, similar to β in Equation (4.24), dened by
β[i] =
. (4.26)
The optimal solution will make a gradual transformation for each orthophoto which minimizes the dierence between the orthophotos in their respective over- laps. Since both orthophotosiandj are
transformed, the squared dierence in the overlap between them must be given by
Gij = kRijβi−Rjiβjk^2[2] . (4.27)
The total sum of the dierences in the overlaps is given by
F = X
Gij (4.28)
Gij = kRijβi−Rjiβjk^2[2] (4.29)
= (Rijβi−Rjiβj)^T(Rijβi−Rjiβj) (4.30)
= β[i]^TR^T[ij]R[ij]β[i]−β[i]^TR^T[ij]R[ji]β[j]−β^T[j]R^T[ji]R[ij]β[i]+β^T[j]R^T[ji]R[ji]β[j] (4.31)
= β[i]^TR^T[ij]R[ij]β[i]+β[j]^TR^T[ji]R[ji]β[j]−2β[i]^TR^T[ij]R[ji]β[j] (4.32)
= β[i]^TK[ij]β[i]+β^T[j]K[ji]β[j]−2β[i]^TL[ij]β[j] , (4.33) where
K[ij] = R^T[ij]R[ij] (4.34)
L[ij] = R^T[ij]R[ji] . (4.35)
It should be noticed thatK[ij] is symmetric, which means thatK[ij] =K[ij]^T. In order to nd the optimal value of the expression in Equation (4.28), the derivative of the quadratic program is
calculated, such that [14]
∂β[i] = (Kij+K[ij]^T)βi−2Lijβj (4.36)
= 2K[ij]β[i]−2L[ij]β[j] (4.37)
∂β[i] = 2Kijβi−2Lijβj (4.38)
= X
2Kijβi−2Lijβj−2Lijβj+ 2Kijβi (4.39)
= X
4Kijβi−4Lijβj . (4.40) In order to minimizeF the derivative is set equal to zero, such that
= 0⇒ (4.41)
4Kijβi−4Lijβj = 0 . (4.42)
The second order derivative is given by
∂β^2[i] = X
4Kij>0 (4.43) | {"url":"https://9pdf.org/document/yng6pwx0-radiometric-adjustment-of-aerial-imagery.html","timestamp":"2024-11-13T18:17:54Z","content_type":"text/html","content_length":"214068","record_id":"<urn:uuid:de86c5ac-b994-4007-b9d5-fa8414bb8da7>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00644.warc.gz"} |
How many cubic meters are in a cubic meter?
How many cubic meters are in a cubic meter?
Cubic Meter to Meter
Cubic Meter Meter
2 1.2599
3 1.4422
4 1.5874
How many cubic centimeters are there in a cubic meter * 1 point?
Cubic Meters to Cubic Centimeters table
Cubic Meters Cubic Centimeters
1 m³ 1000000.00 cm³
2 m³ 2000000.00 cm³
3 m³ 3000000.00 cm³
4 m³ 4000000.00 cm³
How many km are in km3?
1 km3 = 1000 meter 1e-9 km3
7 km3 = 1912.9312 meter 3.4300000000000004e-7 km3
8 km3 = 2000 meter 5.12e-7 km3
9 km3 = 2080.0838 meter 7.289999999999998e-7 km3
10 km3 = 2154.4347 meter 0.0000010000000000000002 km3
How do you calculate km3?
Formulas in words
1. By multiplication. Number of cubic kilometre multiply(x) by 1000000000, equal(=): Number of cubic metre.
2. By division. Number of cubic kilometre divided(/) by 1.0E-9, equal(=): Number of cubic metre.
3. By multiplication. 143 km3(s) * 1000000000 = 143000000000 m3(s)
4. By division.
How many kilometers are in a cubic kilometer?
A cubic kilometer (km3) was a unit of volume defined as a cube measuring one kilometer on each side.
Is cubic meter volume?
A cubic metre (often abbreviated m3 or metre3) is the metric system’s measurement of volume, whether of solid, liquid or gas. It is also equivalent to 35.3 cubic feet or 1.3 cubic yards in the
Imperial system. A cubic metre of water has a mass of 1000 kg, or one tonne.
Which is bigger cubic centimeters or cubic meters?
Enter the volume in cubic meters below to get the value converted to cubic centimeters. Do you want to convert cubic centimeters to cubic meters?…Cubic Meter to Cubic Centimeter Conversion Table.
Cubic Meters Cubic Centimeters
0.1 m³ 100,000 cm³
1 m³ 1,000,000 cm³
What is the formula to convert meters to kilometers?
METER TO KILOMETER (m TO km) FORMULA. To convert between Meter and Kilometer you have to do the following: First divide 1 / 1000 = 0.001. Then multiply the amount of Meter you want to convert to
Kilometer, use the chart below to guide you.
How do you convert meters cubed to kilograms?
Write out the weight/volume conversion ratio as a fraction, with cubic meters on top and kilograms on the bottom. Then multiply this by the number of cubic meters you’re converting into kilograms.
How many cubic meters are in a kg?
Amount: 1 cubic meter (m3) of volume. Equals: 2,406.53 kilograms (kg – kilo) in mass. Converting cubic meter to kilograms value in the concrete units scale. | {"url":"https://www.shakerdesignproject.com/mixed/how-many-cubic-meters-are-in-a-cubic-meter/","timestamp":"2024-11-14T21:02:37Z","content_type":"text/html","content_length":"59342","record_id":"<urn:uuid:380560e4-0e21-4410-97ef-cd0c2bdd4c2c>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00325.warc.gz"} |
Stephanie van Willigenburg (Univ. of British Columbia)
In the intersection of algebra, combinatorics, algebraic geometry and more, are functions called skew Schur functions. These functions are invaluable in translating problems from one area to another
in order to make them more accessible. For example, going from algebra to combinatorics, “When are two skew Schur or Weyl modules equivalent over C?” becomes “When are two skew Schur functions equal?
”, which becomes “When are two pictures of boxes the same?”
In this talk we'll attack the former by studying the latter. More precisely, we'll introduce skew Schur functions pictorially before determining conditions under which these pictures are “the same”,
and then see where else in combinatorics these conditions arise.
No prior knowledge of any of the above is required.
This is joint work with Vic Reiner and Kris Shaw. | {"url":"https://www2.math.binghamton.edu/p/seminars/comb/abstract.200604wil","timestamp":"2024-11-11T23:45:21Z","content_type":"text/html","content_length":"17808","record_id":"<urn:uuid:91d8009d-f388-4a17-a2eb-aef2a553f236>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00723.warc.gz"} |
A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides...
Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides...
A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 18.5 minutes at an average speed of 5.71 m/s.
During the second part, she rides for 35.3 minutes at an average speed of 3.38 m/s. Finally, during the third part, she rides for 7.88 minutes at an average speed of 14.5 m/s.
Show more
Homework Categories
Ask a Question | {"url":"https://studydaddy.com/question/a-bicyclist-makes-a-trip-that-consists-of-three-parts-each-in-the-same-direction","timestamp":"2024-11-02T06:04:58Z","content_type":"text/html","content_length":"26275","record_id":"<urn:uuid:b04c65f3-f24a-4d4c-be32-1b193836eb37>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00458.warc.gz"} |
Albert Einstein - The Gardian
Early Life:
Albert Einstein, born on March 14, 1879, in Ulm, Germany, was a theoretical physicist whose contributions revolutionized the understanding of space, time, and energy. Growing up in a middle-class
Jewish family, Einstein showed an early interest in science and mathematics.
Educational Journey:
Einstein's academic prowess became evident during his years at the Swiss Federal Institute of Technology in Zurich, where he studied physics and mathematics. In 1905, often referred to as his
"miracle year," he published four groundbreaking papers that laid the foundation for modern physics.
The Special Theory of Relativity:
Einstein's most famous equation, $E=mc^2$, emerged from his Special Theory of Relativity, published in 1905. This theory redefined concepts of space and time, demonstrating that they are not absolute
but are intertwined in a four-dimensional continuum known as spacetime.
The General Theory of Relativity:
In 1915, Einstein presented the General Theory of Relativity, providing a new understanding of gravity as the curvature of spacetime caused by mass and energy. This theory has been crucial for
advancements in astrophysics and cosmology, including our understanding of black holes and the expansion of the universe.
Nobel Prize in Physics (1921):
Einstein was awarded the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect, which laid the groundwork for the development of quantum theory. However, his work in quantum
mechanics put him at odds with some of its principles, leading to famous debates with physicist Niels Bohr.
Humanitarian and Political Activism:
Beyond his scientific pursuits, Einstein was a vocal advocate for civil rights, pacifism, and humanitarian causes. Fleeing Nazi Germany in 1933, he settled in the United States, where he continued
his scientific research and became an influential figure in American academia.
Later Years and Legacy:
Einstein spent his later years searching for a unified field theory, attempting to combine electromagnetism and gravity into a single framework. Although he did not achieve this goal, his work paved
the way for future developments in theoretical physics.
Albert Einstein passed away on April 18, 1955, leaving an indelible mark on the scientific landscape. His theories and contributions continue to shape our understanding of the universe, and his name
is synonymous with genius, innovation, and the boundless possibilities of the human mind.
Conclusions on Albert Einstein: A Singular Force in Science and Humanity
Albert Einstein's legacy stands as an unparalleled force in the realms of both science and humanity. His groundbreaking contributions to theoretical physics, particularly the Special and General
Theory of Relativity, revolutionized our understanding of the fundamental fabric of the universe.
Einstein's intellectual prowess, evident from his early years, catapulted him into the scientific forefront during what is famously termed his "miracle year" of 1905. The iconic equation $E=mc^2$
encapsulates the essence of his Special Theory of Relativity, forever altering the way we perceive energy, mass, and the interwoven dimensions of spacetime.
Beyond his scientific achievements, Einstein emerged as a vocal proponent of humanitarian causes and civil rights. His principled stance against oppression and his unyielding commitment to pacifism
reflected a profound concern for the well-being of humanity. Fleeing the Nazi regime and finding refuge in the United States, Einstein continued to be a guiding voice against intolerance and
While awarded the Nobel Prize in Physics for his explanation of the photoelectric effect, Einstein's intellectual journey was not without challenges. Engaging in spirited debates with fellow
physicists, such as Niels Bohr, he became an emblematic figure in the evolving landscape of quantum mechanics.
Einstein's later years, marked by a quest for a unified field theory, demonstrated his unrelenting pursuit of understanding the intricacies of the universe. Although he did not achieve this elusive
goal, his contributions paved the way for subsequent advancements in theoretical physics.
The name Albert Einstein has become synonymous with genius, innovation, and the capacity of the human mind to unravel the mysteries of the cosmos. His impact on science and society is immeasurable,
and his legacy endures as an inspiration for generations of scientists, thinkers, and advocates for a better world.
Albert Einstein has been referenced in various books, films, TV series, and websites. Here are a few notable examples:
• "Einstein: His Life and Universe" by Walter Isaacson
• "The Universe and Dr. Einstein" by Lincoln Barnett
• "Einstein: A Biography" by Jürgen Neffe
• "A Beautiful Mind" (2001) — While primarily focused on John Nash, the film briefly mentions Einstein's theories as influential in the field of mathematics.
• "The Imitation Game" (2014) — Although centered around Alan Turing, Einstein's work is acknowledged as a pivotal element in the development of modern science.
TV Series:
• "Genius" (2017–2018) — Season 1 of this anthology series delves into the life of Albert Einstein, played by Geoffrey Rush. Season 2 focuses on Pablo Picasso.
• "Cosmos: A Spacetime Odyssey" (2014) — In this documentary series, hosted by Neil deGrasse Tyson, Einstein's theories are explored in the context of the evolution of our understanding of the
• Encyclopaedia Britannica — Articles on physics and relativity often reference Albert Einstein's groundbreaking contributions.
• NASA — Einstein's theories are frequently mentioned in discussions about space, time, and gravity on the NASA website.
These references only scratch the surface, as Albert Einstein's influence is pervasive in literature, media, and educational content, reflecting the profound impact of his work on the scientific
community and popular culture. | {"url":"https://the-gardian.com/news/item/169622-albert-einstein","timestamp":"2024-11-06T21:00:10Z","content_type":"text/html","content_length":"53915","record_id":"<urn:uuid:eb72f9ef-1f6f-4e73-8b02-265187a50270>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00123.warc.gz"} |
The waves and the wind. Calculation of wave characteristics
Calculation of wave characteristics
We were asked to create a calculator for the "calculation of the wave height and intervals between waves (frequency)?".
Intuition suggests that there is some relationship between the force of the wind and waves. Since I don't know much about wave theory, I've got to study it.
The result of my study is a calculator below, along with my thoughts on the subject. The calculator does not calculate, or more precisely, does not predict the height of the waves - is a separate
issue covered here - The waves and the wind. Wave height statistical forecasting.
Obviously, the waves on the sea can not be described by a single sine wave, as they are formed as a result of imposing a plurality of waves with different periods and phases. For example, look at the
picture below, which shows the wave resulting from the imposition of three different sine waves.
"Wave disp" by Kraaiennest - Own work. Licensed under GFDL via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:Wave_disp.gif#mediaviewer/File:Wave_disp.gif
Source: "Wave disp" by Kraaiennest - Own work. Licensed under GFDL via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:Wave_disp.gif#mediaviewer/File:Wave_disp.gif
Therefore, for analysis of the sea state, energy spectrum is usually built, energy units lotted on the Y-axis, and frequency on the X-axis, thus obtaining energy density - the amount of energy
carried by waves with a corresponding frequency range. As it turned out, under the wind's influence, the shape of the energy spectrum is changed. The stronger the wind, the more strongly peak in the
spectrum expressed - waves of certain frequencies carrying the most energy. In the picture below have drawn its approximate look as best I could.
The energy distribution of the frequency spectrum, depending on the wind strength
The frequencies where a peak is observed are called dominant. Accordingly, you can make your life easier and calculate the waves' characteristics only for the dominant frequency. The practice has
shown that it will be enough to give a good approximation to reality.
What concerns waves' characteristics, the linear wave theory comes to help, namely, the calculation of gravitational waves in the linearized approximation. To make it clearer what I mean, let's give
some definitions further from
Capillary waves —the name of various waves generated at the interface between a liquid and a gas or liquid and liquid. The lower part of waves is called trough, higher— crest.
Gravity waves on the water — kind of waves on the surface of the liquid in which the force returning the deformed surface of the liquid to a state of balance are simply the force of gravity related
to the height difference of crests and troughs in the gravitational field.
Wave dispersion — in the wave theory, the difference in phase velocities of linear waves according to their frequency. That is, different wavelengths (respectively of different frequency) have
different speeds in an environment that has clearly demonstrated experience with the refraction of light in a prism. This is important for further discussion
Wavenumber — is the 2π radians to the wavelength ratio: $k \equiv \frac{2\pi}{\lambda}$.Wavenumber can be represented as the difference in wave phase (in radians) at the same time in the spatial
points at a distance per unit length (one meter) or a number of spatial periods (crest) of waves arriving at 2π meters.
Using the definition of the wavenumber, we can write the following formula:
Phase velocity (crest velocity )
Wave period(expressed in terms of angular frequency)
The picture to draw attention - кred dot shows the phase velocity, green - the group velocity (the velocity of the wave packet)
"Wave group" by Kraaiennest - Own work. Licensed under GFDL via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:Wave_group.gif#mediaviewer/File:Wave_group.gif
Source: "Wave group" by Kraaiennest - Own work. Licensed under GFDL via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:Wave_group.gif#mediaviewer/File:Wave_group.gif
The dispersion law
The key element in the calculation of the characteristics of the wave is the concept of the dispersion law or dispersion relation (ratio)
The dispersion law or dispersion equation(ratio) in the wave theory - is the relationship of frequency and wave vector (wave number).
In general terms, this relationship is written as
This ratio of water derived in the linear wave theory for the so-called free surface, i.e., the surface of the liquid, not limited to the walls of the vessel or bed, and looks as follows:
$\omega^2=gk \tanh(kh)$,
g - acceleration of free fall,
k - wave number,
tanh - hyperbolic tangent,
h - distance from the liquid surface to the bottom.
It is possible to simplify further the formula based on the graph of the hyperbolic tangent. Note that for kh, tends to zero, the hyperbolic tangent can be approximated by its argument, i.e., kh
value, and if kh, tends to infinity, the hyperbolic tangent kh tends to one. The latter case, obviously, relates to a very great depth. Is it possible to evaluate how they need to be big? If you take
the hyperbolic tangent of pi, its value is approximately equal to 0.9964, which is already quite close to one (the number Pi is taken for the formula's convenience). Then
$kh\geq\pi \Rightarrow \frac{2 \pi h}{\lambda}\geq\pi \Rightarrow h\geq\frac{\lambda}{2}$.
To calculate the characteristics of the wave, water can be considered as deep if the depth is more than at least half the wavelength, and in most places of the world ocean, this condition is met.
In general, based on the hyperbolic tangent graph, the following classification of waves for the relative depth is used (ratio of depth to wavelength).
1. Waves in deep water
The depth is more than half of wavelength, the hyperbolic tangent approximates by one
$h\geq\frac{\lambda}{2}, \tanh(kd)\approx1$
2. Waves on the transitional depths
The depth from one-twentieth to one-half wavelength, the hyperbolic tangent can not be approximated
$\frac{\lambda}{20} \leq h \leq \frac{\lambda}{2}, \tanh(kd)=\tanh(kd)$
3. Waves in shallow water
The depth less than one-twentieth of the wavelength, the hyperbolic tangent approximated by its argument
$h\leq\frac{\lambda}{20}, \tanh(kd)=kd$
Consider the ratio for these cases:
The case of shallow water
The equation takes the form
The group velocity for the case of shallow water
According to the theory, in shallow water, the wave does not have to have dispersion because the phase velocity is independent of frequency. However, we must remember that nonlinear effects in
shallow water are beginning to work associated with the increase in wave amplitude. It starts happening when the amplitude of the wave is comparable to its length. One of the characteristic effects
of this mode is the appearance of fractures on the wave tops. Also, there is a possibility of wave breaking - a well-known surf. These effects are not yet amenable to precise analytic calculation.
Transitional depths case
The equation is not simplified, and then
$c=\frac{gT}{2\pi}\tanh(\frac{2\pi d}{\lambda})$
$\lambda=\frac{gT^2}{2\pi}\tanh(\frac{2\pi d}{\lambda})$
The group velocity for transitional depths
$c_g=\frac{1}{2}(1+\frac{4 \pi \frac{d}{\lambda}}{\sinh(4 \pi \frac{d}{\lambda})})c$
Note that the equation of the wavelength is transcendental, and find its solution should be found numerically. For example, usingFixed-point iteration method
Deep water case
The equation takes the form
The group velocity in the deep water case
So, by measuring the period of a wave with sufficient accuracy, we can calculate the phase velocity, group velocity, and wavelength. And the measurement of the wave period can be made, for example,
by timing the passage time of wave crests with a stopwatch, i.e., the period - this is the most reasonable thing that can be measured without special instruments. If you are somewhere near the coast
- it is necessary to know the depth. If the depth is obviously big, you can use formulas for the deep water, in which the depth, as a parameter, is not included. Since we have a computer processing
power at hand, the calculator does not use the simplified formula, finding wavelength by iteration method (method will converge, as the function's derivative is less than one).
Now, back to the wind. The wind, constantly blowing in one direction, generates the waves by transmitting its energy to them.
And quite obvious, to transmit energy to waves, the wind should blow faster, or at least at a rate equal to the wave's phase velocity.
Here comes a fully developed sea formulation. Fully developed sea - wave, which achieved maximal values at a given wind. The wave is in equilibrium concerning energy - as much energy is transmitted,
as much goes into the motion. Not every wave reaches such a state, as it's required for the wind to constantly blow over the entire surface, which the wave passes for some time. And the stronger the
wind, the more time and distance required for the formation of such a wave. When it's formed, it will certainly catch up with the phase velocity of the wind speed.
Similar calculators
PLANETCALC, The waves and the wind. Calculation of wave characteristics | {"url":"https://planetcalc.com/4406/","timestamp":"2024-11-12T20:22:09Z","content_type":"text/html","content_length":"50758","record_id":"<urn:uuid:5def90d5-5f11-4e2f-a9c8-ca1baeb79465>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00470.warc.gz"} |
Random geometric graphs and their applications in neuronal modelling
Random graph theory is an important tool to study different problems arising from real world.
In this thesis we study how to model connections between neurons (nodes) and synaptic connections (edges) in the brain using inhomogeneous random distance graph models. We present
four models which have in common the characteristic of having a probability of connections
between the nodes dependent on the distance between the nodes. In Paper I it is described a
one-dimensional inhomogeneous random graph which introduce this connectivity dependence
on the distance, then the degree distribution and some clustering properties are studied. Paper
II extend the model in the two-dimensional case scaling the probability of the connection both
with the distance and the dimension of the network. The threshold of the giant component
is analysed. In Paper III and Paper IV the model describes in simplied way the growth of
potential synapses between the nodes and describe the probability of connection with respect
to distance and time of growth. Many observations on the behaviour of the brain connectivity
and functionality indicate that the brain network has the capacity of being both functional
segregated and functional integrated. This means that the structure has both densely inter-
connected clusters of neurons and robust number of intermediate links which connect those
clusters. The models presented in the thesis are meant to be a tool where the parameters
involved can be chosen in order to mimic biological characteristics.
Original language English
Qualification Doctor
• Mathematical Statistics
Awarding Institution • University of Lausanne
• Turova, Tatyana, Supervisor
Supervisors/Advisors • Chavez, Valerie, Supervisor, External person
Award date 2018 Sept 27
Place of Publication Lund
ISBN (Print) 9789177537984
ISBN (electronic) 9789177537991
Publication status Published - 2018 Sept
Bibliographical note
Defence details
Date: 2018-09-27
Time: 09:00
Place: Lecture Hall MH:R, Matematikcentrum, Sölvegatan 18, Lund
External reviewer(s)
Name: Britton, Tom
Title: Professor
Affiliation: Stockholm University, Sweden
Subject classification (UKÄ)
• Mathematics
• Probability Theory and Statistics
Free keywords
• random graph
• Neural Network
• Probability
• Inhomogeneous random graph
• random distance graph
• random grown networks
Dive into the research topics of 'Random geometric graphs and their applications in neuronal modelling'. Together they form a unique fingerprint.
• Ajazi, F., Napolitano, G. M. &
Turova, T.
2017 Dec 1
In: Journal of Applied Probability. 54
p. 1278-1294 17 p.
Research output: Contribution to journal › Article › peer-review
• Ajazi, F., Napolitano, G.,
Turova, T.
& Zaurbek, I.,
In: BioSystems. 136
Online 14 September 2015
p. 105-112
Research output: Contribution to journal › Article › peer-review
• Podgórski, K. (Examiner)
2018 Sept → …
Activity: Examination and supervision › Examination | {"url":"https://portal.research.lu.se/en/publications/random-geometric-graphs-and-their-applications-in-neuronal-modell","timestamp":"2024-11-10T18:15:31Z","content_type":"text/html","content_length":"82044","record_id":"<urn:uuid:8bac7601-9f59-4c89-8e06-9fcd29f01d80>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00010.warc.gz"} |
Infinity ain't everything
Steve Esser
drops an interesting comment at
If the actual universe is infinite, wouldn’t it contain all possibilities?
Not necessarily. A collection may be infinite whilst making systematic exclusions. Consider the set of even numbers. It's infinite in number, but it doesn't contain all the numbers. Granted, the odd
numbers never really had a chance, in this artificial case. So perhaps the thought is instead that an infinite expanse will contain every finite pattern that it possibly could. This seems more
intuitive. But in fact this too turns out to be false.
Consider an infinite sequence of coin-flips. Suppose it is a fair coin, so for each individual flipping event, there is a 50/50 chance of it landing heads or tails. So one obvious possibility is for
a "tails" event to occur. But is it guaranteed to occur, at some point in the infinite sequence? Well, no. There is, after all, some (infinitesimal) chance that the coin will land heads on every
single flip whatsoever! In this case, all the other possibilities that might have obtained, won't. So, infinity is no guarantee of actualizing every possibility.
7 comments:
1. Are you sure you can generalize the coin-flip reasoning to an actual infinity? I don't have a full grasp of the mathematics, but the impression I have is that infinitesimals are only good when
you're approaching infinity, not at actual infinity.
So in this case, the probability of at least one coin being tails approaches one as the number of coin flips approaches infinity, retaining the infinitesimal possibility of all the flips being
heads. However, for an infinite number of flips, the probability of at least one coin being tails is one, and there is a zero probability of all the flips being heads.
Intuition says that there is always the chance of getting yet another heads, but intuition tends to break down with true infinity.
2. I'm not sure. But note that the "all heads" outcome is not any less likely than any other particular (fully specified) outcome. And some or other particular outcome must result. So it can't be
impossible. (It may be "probability zero" in some weird mathematical sense that doesn't entail actual impossibility.)
Here's another example: a random sequence of natural numbers isn't guaranteed to "eventually" contain the number 7. For there's some (again, infinitesimal) chance that the sequence will just
happen to mimic the sequence of even numbers. And in that case, wait as long as you like, the number 7 will never, ever, show up.
3. > But is it guaranteed to occur, at some point in the infinite sequence? Well, no.
In your example is it not as or more guaranteed than anything else you have ever guaranteed in your life? I also presume the x/infinity probably isn't large enough to make Steve feel any better
about it.
4. Ha, yeah, fair point!
5. Hi Richard. I think your first point is right that one can restrict the class of possibilities in the infinite set. In the context of a world, there still may be limits of this sort, I guess. For
instance if laws of nature are fixed and not variable. This is certainly a difficult topic.
- Steve Esser
6. One has to ask what one means by an actual universe being infinite. Are we speaking temporally, spatially or something like the many-worlds interpretation of quantum mechanics?
After all if we're just talking about temporal then we'd have very limited possibilities if the universe just reaches heat death. That is still infinite but hardly encompasses all possibilities.
The spatially infinite universe also would seem to limit possibilities. Certainly not everything would happen.
So it seems by "infinite" one is talking in a more Lewis sense. In a Lewis sense what the person says is true but almost no one is talking about Lewis' sort of realism towards possibilities when
they talk about infinite. Certainly the MWI is more narrow.
I think Tipler goes in the direction of dealing with finite patterns. But as you note that is false. (I'm not sure what exactly Tipler's position is, I should note)
7. There is a very precise way in which infinite sets can be of “different sizes”. Notably, the set of real numbers is bigger than the set of counting numbers. What this means is that if you try and
create a one to one correspondence between counting numbers and real numbers, it is always possible to show that you have missed some real numbers even though your list of counting numbers is
infinitely long. Look at Cantor’s diagonal argument” if you want to see how this is done—it is a clever proof.
So yes, it is possible to have infinite sets that do not contain all the elements of larger infinite sets. I guess the question then is would infinite space/time be such a set? I know physicists
have considered the question, but I don’t know if current theory points to much of an answer.
Visitors: check my comments policy first.
Non-Blogger users: If the comment form isn't working for you, email me your comment and I can post it on your behalf. (If your comment is too long, first try breaking it into two parts.)
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Volume 24: The 2018 Election, Who Projected It Best?
A log-loss comparative analysis of quantitative and qualitative 2018 U.S. House of Representatives election projections
“Well, how did your projections do?” – Dale Cohodes.
It will come as a shock to nobody that I maintained a personal set of projections for the recently completed elections to the House of Representatives. It may surprise you more to know that reviewing
my projections alongside the so-called “professionals” gives us an excellent opportunity to think through one of our favorite topics—probability. Elections are an interesting class of random event:
probabilistic with a single trial and a discrete outcome. The tools we have to predict their outcome—polls, demographics, past voting patterns—result in distributions that include deviations from a
mean. But no matter how much we’d like to, we cannot re-run the recent election in Georgia’s 7th or North Carolina’s 9th Congressional district, even though each was decided by fewer than 1,000
votes. And no matter how small the margin, the candidate with a plurality of the votes wins; a margin of 10,000 votes or 1,000,000 votes results in the same practical outcome. Elections are
fundamentally different from random processes like flipping a coin or tomorrow’s high temperature.
Because of this, a simple question about forecast quality can be extended to provide insight into the general nature of probabilistic forecasts.
• What’s a good probabilistic forecast? • Whose House projections were the best?
What’s a good probabilistic forecast?
Let’s start with the basics. We define a probabilistic forecast as a statement of the likelihood of the occurrence of a discrete event, made by a person (the forecaster) before the event is decided.
(1) When a sports handicapper says the Wolverines have a 75% chance of winning their next game, that is a probabilistic forecast. When your local weatherman says there is a 50% chance of rain
tomorrow, that is also a probabilistic forecast.(2)
We already know that probabilistic thinking is a skill the human mind does not necessarily possess. We are not good at translating concepts like “possible,” “likely,” and “almost certain” into
quantitative likelihoods of occurrence. If we are told that the probability of something happening is 80%, and it doesn’t occur, we are frequently quite distraught. And maybe we should be. But a
forecaster who places an 80% probability on an event that always happens is also not doing a very good job. Saying that there is an 80% chance of the sun rising tomorrow is not a show of forecasting
skill, but rather a lack thereof.
So how do we know a good probabilistic forecast when we see one? Consider a weatherman(3) who says that there is a 50% chance of rain on Tuesday. If it rains, then the weatherman wasn’t wrong; it was
clearly something in the realm of possibilities. But the rub is that, if the same prediction is made the following day, and the sun in fact comes out, the forecast is equally good—and equally bad.
Over the two-day span, the forecasts did not add any informational value. A weather forecast that says day after day after day that the chance of rain is 50% is useless. Such a weatherman would soon
be exited from your local television station, and they should be.
But let’s move to Phoenix, where it rains only 10% of the time on average.(4) Now a forecast showing a 50% chance of rain that is borne out is a fantastic one. On the other hand, if it doesn’t rain,
then it isn’t such a bad prediction, as it almost never rains in Phoenix. A 2-day forecast showing a 50% chance of rain each day, one day of which is borne out, has a lot of value in the desert.
Which brings us to a principle: the quality of a forecast depends on how different it is from the probability that would be assigned in its absence.(5) Showing a few different sets of Phoenix
predictions gives us more information on which weathermen should keep their jobs.
First, let’s check against our prior. It rains in Phoenix 10% of the time, and we had one shower in ten days. Check; our expectations for long-run rain held out.(6)
Let’s start with Weatherman Ugly. These were some bad predictions. Not only did he think rain was likely on five dry days, but he also put a probability of 0% on the one day where it did rain.(7)
This man is bad at his job; listening to him is literally worse than just going with the long-term average of 10% chance of rain every day.
Which is precisely what our Bad Weatherman did. These predictions were not so bad as his Uglier brother-in-forecasting, but they are also essentially useless. You don’t need a degree in meteorology
or fancy weather radar to make these predictions. He should still be fired.
On the other hand, our Good Weatherman in fact did some strong work. It rained on one of the three days on which he thought it might rain; 33% realization on a 40% prediction isn’t bad. He also
confidently predicted no rain on seven days and was correct on each. Using these predictions is far superior to simply relying on the long-run average.
Before we finally describe our metric for the quality of a probabilistic forecast,(8) let’s run through one more set of forecasters. For this, we go back to a wet sub-tropical climate where we can
expect rain 50% of the time.
Our Bad Forecaster…well he’s still doing his thing, going with the historical average. I’m getting tired of this guy. Our Good Weatherman puts up a reasonable showing. When he predicts a 60% chance
of rain, it rains; when he predicts a 40% chance of rain, it doesn’t. It almost seems that he is better at this job than he thinks he is. The days are segregated properly, but he lacks confidence.
And this is made our Better forecaster better. Even though each individual day is still far from certain, these predictions are clearly better than the previous set. Perhaps these predictions should
also be more confident; after all, on days where rain was likely, she was right 100% of the time, not 80%. But we’ve come far enough to state two principles of probabilistic forecasting:
1. A probabilistic forecast is “good” if it is better than a relevant, uninformed estimate.
2. A more certain forecast is better than a less certain forecastif it is correct.
Now for our very simple metric of the quality of a probabilistic forecast: log-loss.(9) For a probabilistic forecast with probability p,
Log-loss = -1 * log ( p )
If the event does not occur,
Log-loss = -1 * log ( 1 – p )
That’s it. Just be sure that for your “log,” you use the natural log of base e. Also, don’t try to use a probability of 0% of 100%; use a very small number of your choice.(10) Rather than describe
what this looks like, let’s visualize it:
The first thing that we see is that log-loss is a penalty. See those big numbers for low probability events that occur (and high probability events that don’t)? You don’t want to be out there. Don’t
make especially confident predictions that don’t come true. The two lines intersect at a probability of exactly 50%. Estimating a 50% probability on a coin flip is equally good or bad no matter what
you forecast, or what happens.
Log-loss is especially useful when you sum it over several probabilistic forecasts.(11) We can do so for the first set of weather forecasters we considered earlier.
As we expected, our Good Forecaster did the best. One drawback of log-loss is that the number has little objective meaning. Our Good Weatherman had a total log-loss of 1.945, but that means nothing
on its own. However, when we compare multiple sets of forecasts,(12) we can start to have some qualitative and quantitative opinions. Our Good Forecaster is much better than our Bad Forecaster, and
Ugly is way off base.
Whose House projections were the best?
And now we apply what we’ve learned.
Elections to the United States House of Representatives are a dream for forecasters and statisticians alike. A large natural experiment, with 435 simultaneous trials, quantitative results, a large
data set, local specifics to learn, and many other things that just can’t be quantified; each of these was present on November 6, 2018.
Even forgetting those running for office, managing the two political parties, and spending the money, there was a lot of interest in the results of these elections. Therefore, many people (and
groups) attempted to forecast the results. Not only does forecasting provide a public service of some value, it also provides a lot of hits on one’s website.
Broadly speaking, there are two types of election forecasters. Quantitative forecasters look for publicly available information like polls and fundraising, as well as endemic variables like economic
conditions. Using some type of fitting, they decide which of these variables are predictive of upcoming House elections results. They also attempt to determine the best way to “mix” these variables
together to predict results. Because of the nature of their forecasting, they typically offer numerical, probabilistic predictions: Candidate A has a 75% chance of winning. They might also predict
vote shares: Candidate B is projected to win 45% of the total 2-party vote.
Qualitative forecasters use the same variables but add in other factors, such as knowledge of the candidate or the district that isn’t quantifiable. Typically, they offer qualitative forecasts; for
example, “Candidate C is likely to win the election”. I created my own set of qualitative forecasts.
If you didn’t skip the first section, you are probably getting excited, because this is the perfect setup for a log-loss analysis. The only slight hitch is that we are forced to change the
qualitative rankings into probabilities. I used the following mapping, expressed as the probability of the Democrat winning the given seat:(13)
Figure 1 - Race Rating to Probabilities
Recall that the log-loss analysis runs into trouble with probabilities of 0% and 100%, hence the small deviations for Safe predictions.
As a first cut of the data, we can look at each forecaster to see their distribution of projections. (See the appendix for descriptions of the professional forecasters shown in Figure 2.) Note that
only Inside Elections and I used the “Lean” designation. Sabato’s Crystal Ball took a stand just before the election, forbidding the use of the “Toss Up” designation.(14)
Figure 2 - Seat Projection Distributions
We can already glean a few interesting pieces of data. IE and CNN (and RCP to a lesser extent) had many more seats viewed as Safe R; other forecasters took seriously the polls showing at least the
possibility of a truly massive Democratic wave. Similarly, RCP had very few Safe D seats, while Crosstab, 538, and I had many. I’m not quite sure why RCP considered the result in New York’s 18th
district or Iowa’s 2nd to be in doubt; this hurt their performance. Like the quantitative forecasters, I was not afraid to call some seats previously held by Republicans as “Safe D.” For example,
there was no doubt in my mind that Democrats would win two seats in Pennsylvania that had changed due to redistricting. Also, Sabato’s proscription of the Toss Up rating greatly increased the number
of Lean D seats in his projection.
Using this raw data, as well as the probabilities in Figure 1, we can translate into a projected number of Democratic seats won for each forecaster.(15) We can also see the number of seats in which
each party was favored by a forecaster. I call the former metric the Mean and the latter the Median.(16)
Figure 3 - Projected Seats Won
The totals won fell within a reasonably tight range: 7 on median and 9 on mean. It bears noting that each forecaster predicted a skewed distribution, with the mean Democratic seats won greater than
the median. This is again due to the potential wipeout of the GOP House caucus; evidence of this possibility was seen by both the qualitative and quantitative forecasters. A GOP enthusiasm plunge
could have put normally safe seats at risk. Quantitatively, there is also a tradeoff of the aggressive gerrymanders created by the GOP state governments in 2010. A greater likelihood of gaining a
majority in the House is balanced somewhat by the potential of losing more seats than expected in extreme scenarios.(17)
With the description out of the way, we can look and see how everybody did. Remember that log-loss analysis is like golf: a lower score is better.
Figure 4 - Log-Loss Results
And there it is: Larry Sabato’s team at the University of Virginia emerges victorious, with some room to spare. 538 is the runner-up, and my projections pull in a solid bronze. Referencing Figure 1
again, it is interesting that our winner was the only contestant that made a call on every race; fortune truly favors the bold. I was happy with my result against the professionals, but it is
important to remember that I had a key advantage. Not only could I use all the data that they used, but I could—and did—use their ratings and commentary. Without this, I could not have been
That said, it is interesting to consider a few contests specifically. For each of the 435 seats I calculated the average log-loss for all forecasters. Comparing each forecasters individual log-loss
to this average is a measure of the relative quality of their forecast for a given seat. A “Safe” prediction is easy to make when it agrees with everybody else. A correct “Likely” call when everybody
else was only Leaning adds value.
My best call was in Oklahoma’s 5th district, where incumbent Republican Steve Russell was considered Safe by three forecasters. Fortunately for Democrat Kendra Horn, he was not safe, in what was
probably the biggest upset of the night. My “Likely R” rating was a good one. I’d seen a lot of Democratic strength in local Oklahoma elections. I also thought that Democratic strength in upscale and
educated suburbs might extend to Oklahoma City. Because of the large penalty for incorrectly calling races Safe, this seat was the most critical of the night forecasters to get right.
I also made a good call in New York’s 11th district. I don’t live in this district, but its border is only a few miles from my home. I’ve also met with Congressman-elect Max Rose, know people who
worked for his campaign, and saw him break through the crowded NYC airwaves with a ton of positive coverage. I rated this one Tilt R, while every other forecaster said Lean R. Two weeks before the
election, most predicted Likely R. Local knowledge seemed to help me in other places as well; two competitive Michigan districts, both located near where I grew up, were also among my best calls
(Tilt D in Michigan 8th; Likely D in Michigan 11th).
Of course, I was far from perfect. I didn’t think Lucy McBath was likely to win the Georgia 6th, site of a close Democratic loss in a special election in 2017. I had this one Lean R. To make matters
worse, I thought Republican Rob Woodall in the neighboring Georgia 7th was at real risk. Despite my Toss Up rating, Congressman Woodall snuck back into Congress in one of the night’s closest races.
Outside of Metro Atlanta, I also fared poorly in upstate New York. I didn’t expect this to be an area of significant Democratic strength, keeping the 19th and 22nd districts as pure Toss Ups. To make
matters worse, I thought Republican incumbents were at risk in the 21st (Stefanik, Likely R), 23rd (Reed, Likely R), 24th (Katko, Lean R) and 27th (Collins, Lean R). Even with my call in the 11th, I
barely broke even in the Empire State despite living here.
South Carolina’s 1st district, another surprise on election night, bears mentioning. This district, a conservative one represented by former Governor Mark Sanford of Appalachian Trail fame,(18) was
not ready to accept unapologetic Trumpist Katie Arrington. Even Crosstab, the qualitative forecaster at the very low end of Safe R seats, did not see this one coming. California’s 21st, which
Democrat TJ Cox stormed back to win as California conducted its usual deliberate count, was also home of a wide split in opinion. I fell on the wrong side, thinking David Valadao would hang on
(Likely R). Iowa’s 3rd district, previously held by noted non-entity David Young,(19) seemed like an easier pickup to me (Lean D) than New Jersey’s 7th, featuring the popular incumbent Leonard Lance
(Tilt D). I also didn’t get ahead of myself in Illinois, keeping the 13th and 14th districts at Lean R. Democrat Lauren Underwood won the latter.
While I could go on for this whole article listing districts,(20) this is as good a stopping point as I’m likely to find. Should you be interested in knowing more, all the picks are available in a
spreadsheet, which is linked in the appendix.
Elections are important for more than simply log-loss analyses of their forecasters. You may be interested in my opinions about this one’s winners, losers and what it all means.
Our views on elections are shaped, partially, by the coverage on election night. This coverage, in turn, is improperly shaped by the order in which states report election results. On election night
2018, an early cable news message set in that the night was not going to be a good one for Democrats based on early returns in Kentucky, Indiana, and Florida. With this message set in, many think
that the election was a close one; after all, Democrats won the House and Republicans the Senate, an effective tie?
Like many things you see on television, this is not really true. Democrats won the national vote in the House by almost 9%. The Senate, where the GOP claimed a great victory? Democrats won the vote
there by about 20%.(21) This is the largest margin of victory in any midterm, by either party, since the Democratic victory in 1974 during the Watergate aftermath.
Also pierced forever is the idea of Donald Trump as a person with some magical political skill. He spent the month before the election traveling the country, holding rallies, attempting to make the
election about him. Unfortunately, the candidates he stumped for did not perform especially well. All those trips to West Virginia did nothing to help Patrick Morrisey. Big attendance in Big Sky
Montana didn’t prevent Democrat Jon Tester from gaining a comfortable victory. Thousands of Republicans showed up for a Trump rally in Elko, Nevada, with Senator Dean Heller. Democrats skipped the
show and swamped the polls. Trump stumped for winning candidates as well, but there is no evidence that candidates that he stumped for outperformed.
Democrats had disappointments as well. Rick Scott and Ron Desantis will represent Florida in the Senate and the state house. Beto O’Rourke’s tens of millions left him 3% short of defeating Ted Cruz.
Claire McCaskill in Missouri and Joe Donnally in Indiana couldn’t overcome the increasing red hue of their respective states. But America spoke in a voice that was as clear as she ever uses in an
election. That voice was, at all levels, a broad and deep rejection of Donald Trump and Trumpism.
Congratulations on making it to the very first LobbySeven Commentary appendix. It amazes me that this is the first time I’ve broached such sacred ground.
As discussed, I used the projections of seven “professional” forecasters and my own. The pros are:
I have a few others that I would have included but they didn’t provide the data in any reasonable format. But they still did good work! And if you will type it all out in Excel format, I’ll add them
How do you know that my projections were done before the election, rather than backfilled? Well, I now wish I’d posted them before, as a record, but you’ll just have to trust that this would be a lot
of work to pretend that I’d falsified a set of projections that, frankly, were barely middle of the road.
As mentioned, one of the choices I was forced to make was a translation between qualitative race ratings (i.e., Lean, Likely) and probabilities. Some of the forecasters listed their probabilities; I
didn’t use them, putting everybody on the same scale. Maybe this was a mistake, but it was easier to code and I liked the consistency. For the two quantitative forecasters (538 and Crosstab) I ran
two sets of projections. In the first, I “bucketed” their probabilities into the relevant qualitative rating, and then used the “consistent” scale. I did this because they have a lot of races at
probabilities of 98% or 99% toward the favorite, and I didn’t want them to be negatively impacted by the small divergence from safe. In fact it had the opposite effect; both of the quantitative
forecasters did better if you use their straight probabilities. I’ve used only these in this paper, but both sets are in the spreadsheet.
At the time of this writing, there are only one House seat has not been decided: the North Carolina 9th. It appears that there was some type of massive election fraud in the absentee ballots in this
district. It was allegedly conducted by an operative aligned with the Republican candidate, and that something similar happened when the GOP incumbent was defeated in the primary. The state board of
elections made the bipartisan, unanimous decision to not certify the election results. I have no idea what will happen here; there could be a new election ordered, or the House might refuse to seat
Republican Mark Harris (who benefitted from the skullduggery). The race is an exceptionally close one no matter what; the uncertified vote differential is about 900. I am not including it in the
log-loss analysis; forecasters were not really thinking about this type of situation when placing their bets. That being said, if it fell D, Sabato would come out even further ahead.
My sheet therefore has a Congress of 235 Democratic seats, 199 GOP, and 1 ¯\_(ツ)_/¯.
I think from here you can follow the remainder of my work, and I hope you do.
(1) We are going to stay mostly in the realm of events that have only two possible outcomes, such as elections in the United States or getting in a car accident on your way home. You can have
probabilistic forecasts that are not binary, like a soccer game, which frequently ends in a tie. There are also predictions that are not probabilistic. For example, a single-value forecast, like
tomorrow’s high temperature, is not probabilistic.
(2) Sports gambling and weather forecasting are by far the two most commonly used examples of probabilistic forecasting.
(3) I try to write gender-neutrally, but “weatherperson” just doesn’t work. I’ll make it up to you by describing later how many women were newly elected to Congress.
(4) All weather statistics are fictitious, made up by me.
(5) I’m avoiding Bayesian language to the extent possible, but this all translates. The informational value of a probabilistic forecast is proportional to the extent that it differs from your
Bayesian prior.
(6) Of course, ten days is too small a sample size to test a prior as strong as long-run average weather patterns. It is eminently possible to flip a coin ten times and get ten heads. You can also
roll a fair die ten times and get five 3s.
(7) We all know that we shouldn’t ever be putting probabilities of literally 0% or 100% on any prediction. I did so for simplicity here; feel free to assume your own arbitrarily small number.
(8) I can sense your excitement from here.
(9) The version I’m showing is for a binary prediction. There is a simple extension for predictions with multiple possible states, but this is enough math. Get to the election forecasts already!
(10) I’m using 0.001 or 0.1% as my arbitrarily small number for log-loss calculations. We math people call this an epsilon.
(11) Like 435 elections for the U.S. House of Representatives.
(12) Such as several different forecasters predicting the results of said elections.
(13) The analysis is robust for small changes in these probabilities, as well as small changes in epsilon.
(14) For the quantitative forecasters, I translated them back to ratings based on the same key used in Figure 1.
(15) As a reminder, a party needs 218 House seats to have a majority.
(16) Strictly speaking, this is not a median, but it lent a nice symmetry to the verbiage. Toss Ups counted as 0.5 for each side, in case you were wondering.
(17) It depends on assumptions, but in the current maps the Democrats would be expected to win excess seats if they won the national House vote by around 10%.
(18) Look him up if you don’t know about this.
(19) Not to be confused with Don Young of Alaska, who faced down a strong challenge to win his seat for the 24th time.
(21) Yes, this is skewed by California running only two Democrats in the election. But even correcting for this, the national Democratic victory in the Senate would still be around 12%. And if the
GOP’s strongest case for their electoral success is that the numbers looked bad because they couldn’t get enough votes to get a candidate into the Senate race in our nation’s most populous state,
then they have officially lost the argument. | {"url":"https://www.lobbyseven.com/single-post/2018/12/17/Volume-24-The-2018-Election-Who-Projected-It-Best","timestamp":"2024-11-12T09:49:59Z","content_type":"text/html","content_length":"1050371","record_id":"<urn:uuid:f68aeb3b-616d-4950-b964-242be4d1062b>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00049.warc.gz"} |
MAT 271 - Probability and Statistics
Course Objectives
Inform and teach to engineering students the main probability and statistical methods with techniques for gaining interpretation of their interesting area data.
Course Description
Product rule, permutation, combination, concept of Probability (Kolmogorov axioms), conditional probability and independency, random variables, Probability density function, distribution function,
discrete disributions: Bernoulli, Binomial, Poisson, continuous distributions: Normal, Gamma, Exponential, Expectation, Moment generating function, mean, variance, standart deviation, covariance,
correlation, Chebchev’s inequality, Estimator and its properties, maximum likelihood estimators, Confidence intervals, Hypothesis testing, One and two sample test for means, Regression.
Course Coordinator
Ahmet Ergin
Course Language | {"url":"https://ninova.itu.edu.tr/en/courses/faculty-of-science-and-letters/29656/mat-271/","timestamp":"2024-11-06T07:48:34Z","content_type":"application/xhtml+xml","content_length":"7824","record_id":"<urn:uuid:b7f92dc9-854c-4156-aff2-aea05be68a2b>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00414.warc.gz"} |
Classical Aspects of General Relativity
The following research areas are of interest for this research line:
• analysis and interpretation of exact solutions of Einstein’s field equations;
• spacetime splitting techniques, measurement process and the role of the observer in General Relativity;
• particle dynamics in certain gravitational backgrounds (either test particles with scalar structure: the mass, or particles with internal structure: spinning test particles and particles with
multipolar structure, quadrupolar and beyond);
• gravitational perturbations and gravitational self-force;
• gravitational waves, with applications to binary systems and effective one-body model.
Group leader: Donato Bini (Istituto per le Applicazioni del Calcolo “M. Picone,” CNR, Rome
Group members: Robert Jantzen (Villanova University, PA, USA)
Previous members: C. Cherubini, F. de Felice, A Geralico, R. Ruffini
Brief description
Figure taken from a recently submitted paper on Godel spacetime, concerning special type of planar orbits.
Figure taken from a recently submitted paper on two-body scattering at the first post-Minkowskian level.
Recent publications
1) Bini D., Geralico A., Jantzen R.T. "Black hole geodesic parallel transport and the Marck reduction procedure", Phys. Rev. D, 99, 064041 (2019). e-print arXiv:1807.10085, DOI:10.1103/
2) Bini D., Geralico A., Plastino W. "Cylindrical gravitational waves: C-energy, super-energy and associated dynamical effects", Class. Quantum Grav., 36, 095012 (2019). e-print arXiv:1812.07938
[gr-qc], DOI: 10.1088/1361-6382/ab10ec
3) Nagar A., Messina F., Rettegno P., Bini D., Damour T., Geralico A., Akcay S., Bernuzzi S. "Nonlinear-in-spin effects in effective-one-body waveform models of spin-aligned, inspiralling, neutron
star binaries", Phys. Rev. D 99, 044007 (2019). DOI:10.1103/PhysRevD.99.044007, [arXiv:1812.07923 [gr-qc]]
4) Bini D., Geralico A., Jantzen R.T., Plastino W. "Godel spacetime: elliptic-like geodesics and gyroscope precession", Phys. Rev. D, 100, 084051, (2019). DOI:10.1103/PhysRevD.100.084051, e-print
arXiv:1905.04917 [gr-qc]
5) Bini D., Geralico A., Gionti G., Plastino W., Velandia N. "Scattering of uncharged particles in the field of two extremely charged black holes", Gen. Rel. Gravitation, 51, 153, (2019). e-print
arXiv:1906.01991 [gr-qc], DOI:doi.org/10.1007/s10714-019-2642-y
6) Bini D. and Geralico A. "New gravitational self-force analytical results for eccentric equatorial orbits around a Kerr black hole: redshift invariant", Phys. Rev. D, 100, 104002, (2019).
DOI:10.1103/PhysRevD.100.104002, e-print arXiv:1907.11080 [gr-qc]
7) Bini D. and Geralico A. "New gravitational self-force analytical results for eccentric equatorial orbits around a Kerr black hole: gyroscope precession", Phys. Rev. D, 100, 104003, (2019).
DOI:10.1103/PhysRevD.100.104003, e-print arXiv:1907.11082 [gr-qc]
8) Bini D. and Geralico A. "Analytical determination of the periastron advance in spinning binaries from self-force computations", Phys. Rev. D, 100, 121502, (2019). e-print arXiv:1907.11083 [gr-qc]
9) Bini D., Damour T. and Geralico A. "Novel approach to binary dynamics: application to the fifth post-Newtonian level", Phys. Rev. Lett., 123, 231104, (2019). DOI:10.1103/PhysRevLett.123.231104,
e-print arXiv:1909.02375 [gr-qc]
10) Bini D., Damour T. and Geralico A. "Scattering of tidally interacting bodies in post-Minkowskian gravity", Phys. Rev. D., 101, 044039, (2020) | {"url":"https://www.icranet.org/index2.php?option=com_content&task=view&id=1292&pop=1&page=2","timestamp":"2024-11-06T09:24:19Z","content_type":"application/xhtml+xml","content_length":"7997","record_id":"<urn:uuid:905706f7-be2f-4bb3-89ac-d51601f86a8c>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00510.warc.gz"} |
The Stacks project
Example 52.16.4. Let $k$ be a field. Let $A = k[x, y][[t]]$ with $I = (t)$ and $\mathfrak a = (x, y, t)$. Let us use notation as in Situation 52.16.1. Observe that $U \cap Y = (D(x) \cap Y) \cup (D
(y) \cap Y)$ is an affine open covering. For $n \geq 1$ consider the invertible module $\mathcal{L}_ n$ of $\mathcal{O}_ U/t^ n\mathcal{O}_ U$ given by glueing $A_ x/t^ nA_ x$ and $A_ y/t^ nA_ y$ via
the invertible element of $A_{xy}/t^ nA_{xy}$ which is the image of any power series of the form
\[ u = 1 + \frac{t}{xy} + \sum _{n \geq 2} a_ n \frac{t^ n}{(xy)^{\varphi (n)}} \]
with $a_ n \in k[x, y]$ and $\varphi (n) \in \mathbf{N}$. Then $(\mathcal{L}_ n)$ is an invertible object of $\textit{Coh}(U, I\mathcal{O}_ U)$ which is not the completion of a coherent $\mathcal{O}_
U$-module $\mathcal{L}$. We only sketch the argument and we omit most of the details. Let $y \in U \cap Y$. Then the completion of the stalk $\mathcal{L}_ y$ would be an invertible module hence $\
mathcal{L}_ y$ is invertible. Thus there would exist an open $V \subset U$ containing $U \cap Y$ such that $\mathcal{L}|_ V$ is invertible. By Divisors, Lemma 31.28.3 we find an invertible $A$-module
$M$ with $\widetilde{M}|_ V \cong \mathcal{L}|_ V$. However the ring $A$ is a UFD hence we see $M \cong A$ which would imply $\mathcal{L}_ n \cong \mathcal{O}_ U/I^ n\mathcal{O}_ U$. Since $\mathcal
{L}_2 \not\cong \mathcal{O}_ U/I^2\mathcal{O}_ U$ by construction we get a contradiction as desired.
Note that if we take $a_ n = 0$ for $n \geq 2$, then we see that $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{L}_ n)$ is nonzero: in this case we the function $x$ on $D(x)$ and the function $x + t
/y$ on $D(y)$ glue. On the other hand, if we take $a_ n = 1$ and $\varphi (n) = 2^ n$ or even $\varphi (n) = n^2$ then the reader can show that $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{L}_ n)$
is zero; this gives another proof that $(\mathcal{L}_ n)$ is not algebraizable in this case.
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Signs for Sums
We’re so familiar with the standard set of symbols for arithmetic operations that appears on every calculator keyboard that we hardly ever stop to think who created them, or when.
Considering that people have been keeping records on everything from wax and clay tablets to animal skins and tree bark for at least 4000 years, it’s a bit of a shock to discover that our symbols for
operations like addition and subtraction are less than 500 years old. But it’s only in such comparatively recent times that most calculations have been done by making marks on paper.
An older method was to use a counting frame such as the abacus. There was a long-running controversy in medieval times about which was faster, the counting frame or pencil and paper, and competitions
were held between the two systems to try to decide the matter. There were even names for the disputing groups, abacists and algorists. The second word is closely related to our modern algorithm for a
step-by-step recipe for carrying out some calculation. Even now, the abacus is far from defunct in many societies, because in practised hands it’s very quick for simple calculations — and unlike the
calculator, it doesn’t need batteries.
For those who did calculations using symbols, it was common in medieval times to indicate plus and minus by the letters p and m, each with a bar or a wavy line over the top, a system that grew up in
Italy. Our modern symbols for these operations didn’t appear until the late fifteenth century. They first turn up in a textbook on commercial arithmetic which Johann Widman published at Leipzig in
1489 under the title Rechnung uff allen Kauffmanschafften. But he didn’t use them as we do now, but as symbols for surpluses or deficits in business problems (though some historians still argue about
what he did mean by them). There’s some evidence they were around in the commercial world before Widman adopted them, for example as a quick way for merchants to mark barrels to indicate whether they
were full or not. It seems that the plus sign started out as an abbreviated scribes’ way of writing the Latin et, “and”, but nobody seems to know for sure where the minus sign came from. They were
gradually adopted as standard symbols throughout Europe in the century after Widman’s book came out.
The person most responsible for introducing them to England, and so eventually to the English-speaking world, was Robert Recorde, a mathematician of the sixteenth century, perhaps the only one of any
stature in a century that saw very few English workers of note in the field. His mathematical works were written in English. At the time this was still extremely uncommon (more than a hundred years
later, Newton automatically wrote his books in Latin, as did many scholars even after him). As a result, Recorde’s books stayed in print as standard texts for at least a century after his death, and
they were correspondingly influential. His most famous work was the Whetstone of Witte of 1557. In it he introduced these newfangled signs: “There be other 2 signes in often use of which the first is
made thus + and betokeneth more: the other is thus made - and betokeneth lesse”.
In the same book he introduced our modern equals sign: “I will sette as I doe often in woorke use, a paire of paralleles, or Gemowe [that is, twin] lines of one length, thus: ======, bicause noe 2.
thynges, can be moare equalle”. His equal signs were actually quite big, about five or six times the length of ours today. They varied in length, seemingly being sized to fit the space available, and
were made up of shorter type characters, which look very like our modern equals sign (it seems the printer had this character in his case, perhaps as decorative type, or as a variant on a hyphen). It
took more than a century for Recorde’s sign to oust rival schemes, such as the curly symbol of Descartes (which was probably the astrological sign for Taurus turned on its side), and for it to be
shortened to match the lengths of the other symbols.
The word whetstone in the title of Recorde’s book, by the way, was a pun on the word coss, then used in English for the unknown thing in algebra (and hence the cossic art or the rule of coss for
algebra). This word had come through French from the Italian cosa as a translation of the Arabic shai, “a thing”, but Recorde probably got it from German, where it was also used. The pun arises
because in Latin cos means a whetstone. Recorde may have written in English, but he still expected his audience to appreciate a trilingual pun!
Our modern multiplication sign (×) seems to have been invented by the British mathematician William Oughtred. He used it in his Clavis Mathematicae (Key to Mathematics), which was written about 1628
and published in London in 1631. It had turned up in an appendix to a posthumous work by the Scottish mathematician John Napier a decade before, but the suggestion is that Oughtred wrote that, too.
The symbol for division, ÷, was first used by Johann Rahn in Teutsche Algebra in 1659, though it may be that John Pell, who edited Rahn’s book, could really be responsible for introducing it. The
symbol wasn’t new. It had been used to mark passages in writings that were considered dubious, corrupt or spurious. Sometimes a dash had been used instead, so in historical origin the division sign
could be a dash with a dot above and below it. It has also been suggested that it’s a minus sign with added dots; this is supported by its surviving in Denmark until recently, very confusingly, as a
symbol for subtraction, not division, though it has fallen out of use as Danes adopt our symbols under pressure to conform with international usage.
The division sign was in Rahn’s time known either as the obelus or sometimes the obelisk, from a Greek word meaning a roasting spit. The idea seems to have been that such dubious matter was thrust
through, as with a spit; the word is the same as that for a tapering pillar, another object with a pointed end. Confusingly, the word obelus was later used for the printer’s character we often call a
dagger, another symbol with a point.
As an aside, when people began to write computer languages from the 1950s on, they were hampered by the absence of mathematical symbols in the character sets of the time: they had the plus, minus and
equals signs, but not those for multiplication or division. So computer scientists had to improvise by borrowing the asterisk for multiplication and the forward slash for division. This latter mark
has a number of aliases, being known also as the solidus, oblique or virgule, among other names. So now we have two sets of symbols for these operations, though the computer characters hardly impinge
on the lives of most of us.
Like the words in our language, the signs for arithmetic have been invented and have become standard through historical accident, influenced by a very few pioneers. They could so easily have been
Mark Brader’s help is gratefully acknowledged. He supplied extra information from his own research, and commented on a draft of this piece. Thanks also to Dermod Quirke and Brian Holser for
additional comments. Any mistakes are mine, of course. | {"url":"http://www.worldwidewords.org/articles/signs.htm","timestamp":"2024-11-10T21:30:12Z","content_type":"text/html","content_length":"11946","record_id":"<urn:uuid:87595df0-7a16-48dd-b81b-492e23c545ed>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00812.warc.gz"} |
sum of hermitian squares
Abstract. NCSOStools is a Matlab toolbox for – symbolic computation with polynomials in noncommuting variables; – constructing and solving sum of hermitian squares (with commutators) programs for
polynomials in noncommuting variables. It can be used in combination with semidefinite programming software, such as SeDuMi, SDPA or SDPT3 to solve these constructed programs. This paper provides …
Read more
On the nonexistence of sum of squares certificates for the BMV conjecture
The algebraic reformulation of the BMV conjecture is equivalent to a family of dimensionfree tracial inequalities involving positive semidefinite matrices. Sufficient conditions for these to hold in
the form of algebraic identities involving polynomials in noncommuting variables have been given by Markus Schweighofer and the second author. Later the existence of these certificates has been …
Read more | {"url":"https://optimization-online.org/tag/sum-of-hermitian-squares/","timestamp":"2024-11-13T15:45:25Z","content_type":"text/html","content_length":"86266","record_id":"<urn:uuid:41cdcefb-d23f-4522-9849-54832ce4173b>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00084.warc.gz"} |
Take Your Dog to Work Day
Every day grown-ups go to work — maybe at an office, restaurant, school, or construction site. Well, the Friday following Father’s Day – today! – is Take Your Dog to Work Day, and we’re wondering how
well that’s going to go. Sure, dogs are friendly and helpful, and if they could type emails or drive a truck, they’d happily do it for us. But most visiting dogs are probably better at sniffing
people’s feet and chewing the furniture. It’s just as well they go to work only one day of the year.
Wee ones: Dogs have 4 legs. Get down on your hands and knees like a dog, and bark 6 times!
Little kids: If today is Friday and your dog also went to work yesterday, what day was that? Bonus: If you and your dog take turns typing, and you type the 1st sentence, then your dog types the 2nd,
then you type the 3rd and so on…who types the 12th sentence?
Big kids: If you work as a chef and your dog eats every 4th burger you flip, how many does she eat out of 12 burgers you cook? Bonus: If a dog has 2 people working for him, and they each have 3
people working for them, and those people each bring 4 dogs, how many dogs work with that top dog?
Wee ones: Count your barks: 1, 2, 3, 4, 5, 6!
Little kids: Thursday. Bonus: Your dog.
Big kids: 3 burgers, whether she starts on the 1st or some other burger. Bonus: 24 dogs, brought by 6 people (3+3). | {"url":"https://bedtimemath.org/take-your-dog-to-work-day/","timestamp":"2024-11-14T11:55:49Z","content_type":"text/html","content_length":"87273","record_id":"<urn:uuid:3a973c69-913d-49f9-b915-63fe3d417df4>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00673.warc.gz"} |
Problem with wavelet code
Answers (1)
Answered: Suraj Kumar on 28 Aug 2024
Dear MATLAB users
I have the following code which is used to select limited number of point from a vector and calclate the wavelet of the original vector and the new-sampled vector from the points we selected then
calculate the wavelet energy ratio between the two vectors:
% Adjustable parameters
frequency = 58; % Frequency of the sine function
speed_rpm = 6000; % Rotation speed in rpm
duration = 60 / speed_rpm; % Duration of the signal for both high and low sampling rates
sampling_rate_high = 1668; % High sampling rate
num_samples_low = 16; % Number of samples for signal_low
% Time vector for signal_high
t_high = linspace(0, duration, duration * sampling_rate_high);
% Generate sine function for signal_high
signal_high = sin(2 * pi * frequency * t_high);
% Calculate the average of signal_high
average_signal_high = mean(signal_high);
% Select a subset of values from signal_high
step = floor(length(signal_high) / num_samples_low);
selected_indices = 1:step:length(signal_high);
selected_values = signal_high(selected_indices);
% Normalize selected subset to match the average of signal_high
average_selected_values = mean(selected_values);
normalized_selected_values = selected_values - (average_selected_values - average_signal_high);
% Linear interpolation to match the length of t_low
t_low = linspace(0, duration, num_samples_low);
normalized_selected_values_interp = interp1(linspace(0, duration, length(normalized_selected_values)), normalized_selected_values, t_low, 'linear', 'extrap');
% Construct signal_low
signal_low = normalized_selected_values_interp;
% Compute wavelet transform for both high and low sampling rates
scales = 1:64; % Choose appropriate scales for wavelet analysis
coefficients_high = cwt(signal_high, scales, 'db4');
coefficients_low = cwt(signal_low, scales, 'db4');
% Calculate the wavelet energy ratio
energy_ratio = zeros(1, length(scales));
for i = 1:length(scales)
energy_ratio(i) = sum(abs(coefficients_low(i, :)).^2) / sum(abs(coefficients_high(i, :)).^2);
% Plotting
% Original signal (signal_high)
subplot(3, 2, 1);
plot(t_high, signal_high, 'b');
title('Original Signal (High Sampling Rate)');
% Signal_low
subplot(3, 2, 2);
stem(selected_indices, selected_values, 'r', 'Marker', 'o');
hold on;
plot(t_low, signal_low, 'b');
title('Signal Low Sampling Rate');
legend('Selected Values', 'Interpolated Signal');
% Wavelet transform for signal_high
subplot(3, 2, 3);
imagesc(t_high, scales, abs(coefficients_high));
title('Wavelet Transform (High Sampling Rate)');
% Wavelet transform for signal_low
subplot(3, 2, 4);
imagesc(t_low, scales, abs(coefficients_low));
title('Wavelet Transform (Low Sampling Rate)');
% Energy ratio as a function of scale
subplot(3, 2, [5 6]);
plot(scales, energy_ratio, 'LineWidth', 1.5);
title('Wavelet Energy Ratio');
ylabel('Energy Ratio');
% Display the average energy ratio
disp(['Average Wavelet Energy Ratio: ', num2str(mean(energy_ratio))]);
I am facing a problem, I am playing with the adjustable paramters but when I put the sampling rate lower than 1668 the code stop to work:
also when I change the speed this affect the number of samples, I know I missing something but I have no idea about it.
speed_rpm = 6000; % Rotation speed in rpm
duration = 60 / speed_rpm; % Duration of the signal for both high and low sampling rates
sampling_rate_high = 1668; % High sampling rate
num_samples_low = 16; % Number of samples for signal_low
the second part of my question:
I would like to implement a code that reduce the number of samples taken into consideration and define the appropriate positions (unifrom or non uniform arraangement) without losing the information
of the original signal. for this I am using the wavelet energy ratio as an indicator but to compute all the possible arrangement it would be computationally cost so I am thinking about integrating
Genetic Algorithms GA. to make it but I have no idea about it, I know just the theory of GA so if someone can set me on the path I will be very greatful for it.
Thanks a lot for reading and answering :) .
2 views (last 30 days)
Problem with wavelet code
Hi Mohammed,
From what I gather you want to ensure the working of signal processing correctly across different sampling speeds and aim to utilize Genetic Algorithms (GA) to optimize the selection of sample
To accomplish this, you can consider following these steps and refer to the attached code snippets:
1. Create a signal based on given frequency and sampling rate and compute its continuous wavelet transform (CWT) to analyse its time-scale characteristics.
% Generate sine function for signal_high
signal_high = sin(2 * pi * frequency * t_high);
average_signal_high = mean(signal_high);
% Compute wavelet transform for the high sampling rate
coefficients_high = cwt(signal_high, scales, 'db4');
2. Configure a Genetic Algorithm (GA) to find the optimal indices for down sampling. The GA aims to maximize the preservation of wavelet energy by selecting the best indices from ‘signal_high’.
% Define the fitness function
function energy_ratio = wavelet_energy_fitness(selected_indices, signal_high, scales, average_signal_high, duration, num_samples_low)
selected_indices = round(selected_indices);
selected_values = signal_high(selected_indices);
average_selected_values = mean(selected_values);
normalized_selected_values = selected_values - (average_selected_values - average_signal_high);
% Interpolate to match the length of t_low
t_low = linspace(0, duration, num_samples_low);
normalized_selected_values_interp = interp1(linspace(0, duration, length(normalized_selected_values)), normalized_selected_values, t_low, 'linear', 'extrap');
% Compute wavelet transform for low sampling rate
coefficients_low = cwt(normalized_selected_values_interp, scales, 'db4');
coefficients_high = cwt(signal_high, scales, 'db4');
energy_ratio = zeros(1, length(scales));
energy_ratio(i) = sum(abs(coefficients_low(i, :)).^2) / sum(abs(coefficients_high(i, :)).^2);
energy_ratio = -mean(energy_ratio);
3. Use the GA to determine the optimal indices ‘selected_indices_opt’. Select and normalize these indices from ‘signal_high’ to create a down sampled signal ‘signal_low_opt’.
% Use the GA to find optimal indices
num_samples_high = length(signal_high);
[selected_indices_opt, fval] = ga(@(indices) wavelet_energy_fitness(indices, signal_high, scales, average_signal_high, duration, num_samples_low), ...
num_samples_low, [], [], [], [], ...
ones(1, num_samples_low), num_samples_high * ones(1, num_samples_low), ...
% Construct signal_low using optimal indices
selected_values_opt = signal_high(round(selected_indices_opt));
average_selected_values_opt = mean(selected_values_opt);
normalized_selected_values_opt = selected_values_opt - (average_selected_values_opt - average_signal_high);
For better understanding, you can refer to the output below:
For more insights, kindly refer to the documentation links below:
Hope this helps! | {"url":"https://www.mathworks.com/matlabcentral/answers/2096751-problem-with-wavelet-code","timestamp":"2024-11-03T19:09:35Z","content_type":"text/html","content_length":"150582","record_id":"<urn:uuid:ad48f83c-3e6b-4c92-82ef-6b3ce500fb98>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00858.warc.gz"} |
This article is not detailed enough and needs to be expanded. Please help us by adding some more information.
Laconic is a programming language that compiles to a one-tape two-symbol Turing machine. The goal for its creation was to create two-symbol Turing machines with very few states (a golfed Turing
machine) that does something interesting when started on a blank tape. Laconic is a strongly-typed language that supports recursive functions.
See also
External resources | {"url":"https://esolangs.org/wiki/Laconic","timestamp":"2024-11-13T19:08:10Z","content_type":"text/html","content_length":"19340","record_id":"<urn:uuid:1e2c0b19-2c76-4cd6-9031-e266dc6b7ae8>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00825.warc.gz"} |
Anderson, D. R. (2008) Model-based Inference in the Life Sciences: a primer on evidence. Springer: New York.
Burnham, K. P., Anderson, D. R. (2002) Model Selection and Multimodel Inference: a practical information-theoretic approach. Second edition. Springer: New York.
Dail, D., Madsen, L. (2011) Models for estimating abundance from repeated counts of an open population. Biometrics 67, 577--587.
Lebreton, J.-D., Burnham, K. P., Clobert, J., Anderson, D. R. (1992) Modeling survival and testing biological hypotheses using marked animals: a unified approach with case-studies. Ecological
Monographs 62, 67--118.
MacKenzie, D. I., Nichols, J. D., Lachman, G. B., Droege, S., Royle, J. A., Langtimm, C. A. (2002) Estimating site occupancy rates when detection probabilities are less than one. Ecology 83,
MacKenzie, D. I., Nichols, J. D., Hines, J. E., Knutson, M. G., Franklin, A. B. (2003) Estimating site occupancy, colonization, and local extinction when a species is detected imperfectly. Ecology 84
, 2200--2207.
Mazerolle, M. J. (2006) Improving data analysis in herpetology: using Akaike's Information Criterion (AIC) to assess the strength of biological hypotheses. Amphibia-Reptilia 27, 169--180.
Royle, J. A. (2004) N-mixture models for estimating population size from spatially replicated counts. Biometrics 60, 108--115.
Schwarz, G. (1978) Estimating the dimension of a model. Annals of Statistics 6, 461--464. | {"url":"https://www.rdocumentation.org/packages/AICcmodavg/versions/2.3-1/topics/bictabCustom","timestamp":"2024-11-08T13:42:29Z","content_type":"text/html","content_length":"81898","record_id":"<urn:uuid:fe834122-d4b4-4a94-af75-55121eb0b8dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00240.warc.gz"} |
Astronomical Units Calculator - Online Calculators
To calculate the distance using astronomical units, multiply the AU value by 149,597,870.7 km, which is the average distance from the Earth to the Sun.
Astronomical units (AU) are a unit of distance commonly used to measure the space between celestial bodies, such as planets and the Sun. One astronomical unit represents the average distance between
the Earth and the Sun, approximately 149,597,870.7 kilometers. This unit is crucial for simplifying measurements in space and astronomy.
$D = AU \times 149,597,870.7$
• $D$ = Distance in kilometers
• $AU$ = Astronomical Unit (value of AU)
• 149,597,870.7 = Number of kilometers in one astronomical unit
How to Calculate ?
1. Determine the AU value you want to convert into kilometers.
2. Multiply the given AU value by 149,597,870.7 (which is 1 AU in kilometers).
3. The result will give the distance in kilometers.
Solved Calculations
Example 1:
• Astronomical Units (AU): 2 AU
Parameter Value
Astronomical Unit (AU) 2 AU
Distance (D) 299,195,741.4 km
Answer: The distance is 299,195,741.4 kilometers.
Example 2:
• Astronomical Units (AU): 5.2 AU
Parameter Value
Astronomical Unit (AU) 5.2 AU
Distance (D) 777,909,128.64 km
Answer: The distance is 777,909,128.64 kilometers.
What is Astronomical Units Calculator ?
The Astronomical Units Calculator is a valuable tool for anyone working in astronomy or related fields. It helps convert distances in space, specifically measuring astronomical units (AU), into more
familiar terms like kilometers or light-years. An astronomical unit is defined as the average distance between the Earth and the Sun, approximately 149.6 million kilometers.
When using this calculator, you can easily convert distances expressed in AU to kilometers or other units. For example, if you input 1 AU, the calculator will show that this is about 149.6 million
kilometers. If you want to convert 5.2 AU to kilometers, you simply enter the value, and it will provide the equivalent distance.
Astronomers often use this measurement when discussing distances between celestial objects. It simplifies complex calculations and provides a clearer understanding of the vastness of space. With this
tool, you can also explore the relationship between astronomical units and light-years, enhancing your comprehension of cosmic distances.
Final Words:
Whether you’re calculating the distance to another planet or understanding how far stars are from Earth, the Astronomical Units Calculator makes these conversions straightforward and accessible. | {"url":"https://areacalculators.com/astronomical-units-calculator/","timestamp":"2024-11-03T02:57:32Z","content_type":"text/html","content_length":"107591","record_id":"<urn:uuid:8d1cc1e1-3de1-456b-900c-f9cc0cb69a57>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00712.warc.gz"} |
5 Best Ways to Find the Second Largest Number in a Python List
π ‘ Problem Formulation: We intend to solve a common programming challenge: finding the second largest number in a list. Given an input list, for instance, [3, 1, 4, 1, 5, 9, 2], the desired output
is the second largest unique value, which in this case should be 4.
Method 1: Sort the List and Select the Second Last Element
One straightforward method involves sorting the given list and then picking the second last element. This ensures that the largest number is at the end, and the second largest is just before it. This
method is simple, but not the most efficient for large lists, since sorting can be computationally intensive.
Here’s an example:
numbers = [7, 5, 6, 3, 8, 9]
second_largest = numbers[-2]
Output: 8
This code snippet sorts the list of numbers and extracts the second last element, which is the second largest in the sorted list.
Method 2: Use the ‘heapq’ Module
Utilizing Python’s heapq module to find the second largest number is efficient, as it converts the list to a heap in linear time and then retrieves the largest values. While this method is quick,
it’s less intuitive and requires knowledge of the heapq module.
Here’s an example:
import heapq
numbers = [7, 5, 6, 3, 8, 9]
largest_nums = heapq.nlargest(2, numbers)
second_largest = largest_nums[1]
Output: 8
This code uses heapq.nlargest() to find the two largest numbers and then selects the second one from that list.
Method 3: Use a Set to Remove Duplicates and Then Sort
This method involves converting the list to a set to remove any duplicates and then sorting the remaining elements to find the second largest. This is useful when the list has duplicates, but it has
some performance overhead due to sorting.
Here’s an example:
numbers = [7, 5, 6, 3, 8, 9, 8]
unique_numbers = sorted(set(numbers))
second_largest = unique_numbers[-2]
Output: 8
This snippet first removes duplicates by converting the list to a set, sorts the unique numbers, then selects the second largest.
Method 4: Iterate Through the List
Iterating through the list to keep track of the largest and second largest numbers can be a more efficient way for finding the second largest number without sorting. Ideal for unsorted lists with a
large number of elements where sorting is not desired.
Here’s an example:
numbers = [7, 5, 6, 3, 8, 9]
first, second = float('-inf'), float('-inf')
for n in numbers:
if n > first:
second = first
first = n
elif first > n > second:
second = n
Output: 8
The code iterates through the list, updating the first and second largest numbers accordingly without sorting the entire list.
Bonus One-Liner Method 5: Using Max with a Filter
This elegant one-liner uses Python’s built-in max() function with a filter to directly find the second largest number. It’s sleek and readable but not the most efficient due to double traversal over
the list.
Here’s an example:
numbers = [7, 5, 6, 3, 8, 9]
second_largest = max(filter(lambda x: x != max(numbers), numbers))
Output: 8
The snippet finds the maximum number to exclude it and then finds the maximum again among the remaining numbers.
• Method 1: Sorting and Selecting. Simple and direct. Not efficient for large lists.
• Method 2: Heapq Module. Efficient and works well for large lists. Requires additional module knowledge.
• Method 3: Set and Sort. Handles duplicates well. Sorting may add overhead.
• Method 4: List Iteration. Efficient for unsorted lists. In-place without extra memory.
• Method 5: Max with Filter. A concise one-liner. Not the most efficient, double list traversal. | {"url":"https://blog.finxter.com/5-best-ways-to-find-the-second-largest-number-in-a-python-list/","timestamp":"2024-11-06T11:26:53Z","content_type":"text/html","content_length":"69522","record_id":"<urn:uuid:23790cae-e662-46ed-abb0-b2c189540164>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00440.warc.gz"} |
59 km to miles
1: The Basics of Kilometers and Miles
Kilometers and miles are two common units of measurement used to quantify distances. While both are used to measure distance, they are used in different parts of the world. Kilometers are
predominantly used in countries that have adopted the metric system, while miles are mainly used in countries that follow the imperial system of measurement.
One kilometer is equal to 0.621371 miles. Conversely, one mile is equivalent to approximately 1.60934 kilometers. This conversion factor is essential in order to convert distances from one unit to
another. Understanding the basics of kilometers and miles is important in various fields, such as travel, sports, and engineering, where distance measurements play a significant role.
2: Why Convert Kilometers to Miles?
When it comes to measuring distance, different countries use different units of measurement. While many countries, including the United States, use miles to measure distance, other countries, such as
the majority of countries in Europe, use kilometers. This can often lead to confusion and inconvenience when traveling or interacting with individuals from different regions. That is why it is
important to understand how to convert kilometers to miles and vice versa.
One of the main reasons why it is necessary to convert kilometers to miles is for seamless communication and understanding. Imagine you are planning a road trip with your friends and you are
discussing the distance between two cities. If one person mentions that a particular city is 100 kilometers away, while another person is more familiar with miles and prefers to work in that unit,
the confusion can easily arise. By knowing how to convert kilometers to miles, you can avoid misunderstandings and ensure everyone is on the same page. Additionally, for travelers who need to
navigate their way through unfamiliar cities, understanding mileage in miles can help in estimating the time and effort required for traveling between destinations.
3: The Conversion Formula Explained
When it comes to converting kilometers to miles, the conversion formula is quite simple. The key is understanding the relationship between these two units of measurement.
To convert kilometers to miles, you can use the following formula: miles = kilometers * 0.621371. This formula is derived from the fact that one mile is equal to approximately 0.621371 kilometers.
So, by multiplying the number of kilometers by this conversion factor, you can quickly obtain the equivalent distance in miles. For example, if you have 10 kilometers, you can multiply it by 0.621371
to get approximately 6.21371 miles. Easy, right?
4: Common Examples of Kilometers to Miles Conversion
When it comes to converting kilometers to miles, there are several common examples that you may encounter. For instance, if you’re planning a road trip and need to know the distance in miles, but
your GPS is set to kilometers, the conversion is crucial. Let’s say your GPS tells you that the next town is 100 kilometers away. To convert this to miles, you can use the conversion factor of
0.6214, which means that one kilometer is equivalent to 0.6214 miles. By multiplying 100 kilometers by this conversion factor, you’ll find that the town is approximately 62.14 miles away.
Similarly, if you’re a fitness enthusiast who tracks your jogging distance in kilometers but prefer to calculate your progress in miles, knowing how to convert is essential. Let’s say you’ve just
completed a run of 5 kilometers, and you want to determine the equivalent distance in miles. By multiplying 5 kilometers by the conversion factor of 0.6214, you’ll find that you’ve jogged roughly
3.11 miles. This conversion allows you to compare your running achievements with others who track their distances in miles, making it easier to participate in virtual races or fitness challenges.
5: Quick Tricks for Approximate Conversion
Approximate conversions between kilometers and miles can come in handy when you need a quick estimate. While they may not be completely accurate, they can give you a rough idea of the distance
involved. One handy trick is to multiply the number of kilometers by 6 and then divide by 10 to get the approximate equivalent in miles. For example, if you have 30 kilometers, multiplying it by 6
gives you 180, which you can then divide by 10 to get 18 miles.
Another simple trick is to use a 1:0.6 ratio for a quick conversion. This means that for every kilometer, you can simply multiply it by 0.6 to get the rough equivalent in miles. For instance, if you
want to know how many miles are in 15 kilometers, you can multiply 15 by 0.6 to get 9 miles. While these tricks may not give you the exact values, they can be quite helpful when you need a quick
estimate in day-to-day situations. | {"url":"https://convertertoolz.com/km-to-miles/59-km-to-miles/","timestamp":"2024-11-09T13:29:18Z","content_type":"text/html","content_length":"41478","record_id":"<urn:uuid:cc691aab-4c75-4d18-ad73-56a319a05beb>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00635.warc.gz"} |
Competition filters
Competition filters¶
To submit a strategy for the contest, click on the Submit button in your Development area:
You also have the option to directly submit code from your development environment in Jupyter Notebook or JupyterLab.
Upon submission, our servers will check your code. The status of this check will be displayed in the Competition section of your account, under the Checking tab:
If your algorithm passes these checks (filters), it will be admitted to the Contest and you can find it under the Candidates tab. If it fails these checks, it will be listed under the Filtered tab,
where you can inspect the logs and understand the reason for the error.
Technical filters¶
Source file must exist¶
An error message stating that the strategy.ipynb file was not found is connected to a non-standard name for the file containing your strategy. This file must be named strategy.ipynb.
Execution failed¶
If you see an error message stating that the execution of strategy.ipynb failed, then you should check the logs as they will contain the necessary information.
You should check the logs (server logs and html columns) as they will contain the necessary information.
Pay special attention to the dates in the logs: you can use this information to reproduce the problem in the precheck.ipynb file you find in your root directory. Substitute these dates when calling
Weights must be written¶
If you see an error message stating that the call to the write_output function was skipped (example: Missed call to write_output), then your strategy does not save the final weights. Your last call
in the strategy.ipynb file should be qnt.output.write(weights) (or qnt.backtest(…) if you use Multi-Pass Backtesting), assuming that you used weights for the final allocation weights.
All data must be loaded¶
An error message stating that data are loaded only until a certain day is due to the fact that you are loading the data cropping the number of days. Do not crop data when you submit, as your system
needs to run on a daily basis on new data.
qndata.futures.load_data(min_date="2006-01-01", max_date="2008-01-01")
Weights must be generated for all trading days¶
An error message stating that the strategy does not display weights for all trading days means that weights for some days are not generated, for example because of a drop operation. This problem can
be avoided using the function qnt.output.check(weights, data, “futures”), assuming that you are working with futures and you are generating weights on data.
Weights are not generated at the beginning of the time series¶
Your strategy must produce non-zero weights starting from the date defined for each corresponding contest:
• NASDAQ-100 Contest - Trading should begin from January 1, 2006.
• Futures Contest - Trading should begin from January 1, 2006.
• Bitcoin Futures - Trading should start from January 1, 2014.
• Crypto Top-10 Long - Trading should commence from January 1, 2014.
To ensure compliance, review your strategy code:
1. Verify the data range being loaded. Define the appropriate time frame as follows:
futures = qndata.futures.load_data(min_date="2006-01-01")
2. Ensure the data range being saved:
The Sharpe ratio’s computation period commences from the date when the first non-zero weights are identified. If, for instance, your algorithm begins generating weights on Bitcoin Futures from
January 1, 2017, it will not be accepted because the In-Sample period is effectively too brief.
This issue often arises when utilizing technical analysis indicators which necessitate a warm-up period. You can check the date using:
min_time = weights.time[abs(weights).fillna(0).sum('asset')> 0].min()
The value of min_time should be equal to or later than the starting date specified in the rules for the respective competition.
If min_time is later than the starting date, it’s recommended to fill the starting values of the time series with non-zero values. For instance, you could use a simple buy-and-hold strategy.
def get_enough_bid_for(data_, weights_):
time_traded = weights_.time[abs(weights_).fillna(0).sum('asset') > 0]
is_strategy_traded = len(time_traded)
if is_strategy_traded:
return xr.where(weights_.time < time_traded.min(), data_.sel(field="is_liquid"), weights_)
return weights_
weights_new = get_enough_bid_for(data, weights)
weights_new = weights_new.sel(time=slice("2006-01-01",None))
For additional information regarding the calculation method, please refer to the source code of the library, specifically the qnt.output.calc_sharpe_ratio_for_check method.
An error message stating that the strategy calculation exceeds a given time implies that you need to optimize the code and reduce the execution time. Futures systems should be evaluated in 10 minutes
and Bitcoin futures/Crypto long systems in 5 minutes of time.
Number of strategies¶
An error message stating that the limit for strategies has been exceeded is connected to the number of running strategies in your area. You can have at most 50 of them and you should select 15 for
the contest.
A copy of a template will NOT be eligible for a prize. | {"url":"https://quantiacs.com/documentation/en/user_guide/passFilters.html","timestamp":"2024-11-02T16:58:32Z","content_type":"text/html","content_length":"44806","record_id":"<urn:uuid:93a90729-9714-47ad-8dc0-c4c8c41459c8>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00570.warc.gz"} |
The importance of scientifically literate citizens
At the Concord Consortium our goal is to prepare students to ask questions and use mental models to answer them. Students who develop this habit of mind early on will, we hope, become engaged and
scientifically literate adults. And surely they will not lack for important questions to ask!
Here’s an example: According to a recent study published by the National Academy of Sciences, global sea level has increased by more than two inches in this century alone! Why is that happening?
People who live on low-lying islands or in coastal cities around the world would really like to know.
Representative Mo Brooks (R-AL), a member of the House Science and Technology Committee and Vice Chair of its Subcommittee on Space, recently proposed a model for this phenomenon. He offered the
opinion that a significant cause of the rise in sea level is falling rocks and other erosion, pointing specifically to the California coastline and the White Cliffs of Dover. This debris “forces the
sea levels to rise because now you have less space in those oceans because the bottom is moving up,” he explained.
Is he right? Can erosion be causing the rise in sea level? Most important: do you have to be a scientist to address that question?
Actually, anyone can do it. All it takes is a little physics, a little math, and Google.
First, the physics:
1. When you throw a rock in the ocean, the volume of the ocean goes up by exactly the volume of the rock because…
2. the water displaced by the rock pushes on the surrounding water and ends up being spread evenly across the surface of the ocean (remember, water is a liquid!).
3. So the increase in sea level ends up as a thin layer of water—a layer whose volume is equal to that of the rock itself. And the volume of that little layer of water is the vertical rise in sea
level times the surface area of the ocean, and that equals the volume of the displaced water, which is just the volume of the rock itself.
Let’s write that up as an equation:
(volume-of-rock) = (surface area of ocean) x (increase in sea level)
Now we need to know the surface area of the ocean. We could estimate it (4/5 of the Earth’s area is ocean, the radius of the Earth is 4000 miles…), or we could Google it.
From Google: The surface area of the Earth’s oceans is 510 million square kilometers.
So to make the sea level rise by one inch we would need to throw in a lot of rocks—or one REALLY BIG rock—whose volume is 1 inch times that surface area. How big is that? Here comes the math!
Let’s put everything in feet, so we can compare. A kilometer is 1000 meters and a meter is about 3.28 feet, so a kilometer is 3280 feet, which makes a square kilometer roughly 10.8 million square
feet, which makes 510 million square kilometers, which works out to 5500 million million or 5.5 X 10^15 square feet.
So the volume of rock required to raise the ocean level by one inch (1/12 of a foot) is
(5.5/12) X 10^15 cubic feet or 0.46 X 10^15 cubic feet or 460 trillion cubic feet
How big is 460 trillion? Is it a mountain or a molehill? Turns out, more like a mountain.
Back to Google: The volume of Mount Everest (starting from its base, not from sea level) is 2.1 trillion (2.1 X 10^12) cubic feet. So to cause a one-inch rise in sea level you would need to push into
the sea
460 / 2.1 = 220 Mount Everests
That’s a lot of rock! Can erosion possibly account for the equivalent of 220 Mount Everests in just a few years? Back to Google…
There’s not a lot of information concerning falling rocks, it turns out, but topsoil erosion is a major concern to a lot of people, so we do know something about that. The European Commission’s Joint
Research Centre on Sustainable Resources estimates that 36 billion tons of soil are washed away, worldwide, every year. A cubic foot of rock weighs 150 pounds so all those Mount Everests (2.1
trillion cubic feet’s worth) weigh over 300 trillion pounds. At that rate, it would take almost 9000 years for soil erosion to raise the ocean level by an inch. If rocks and other non-soil debris
contributed a similar amount it would still take thousands of years.
The Next Generation Science Standards call for students—and that really applies to all of us!—to learn to use models to answer questions. When we do that it becomes clear that erosion isn’t to blame
for the rise in sea level. And the best part? We don’t need to rely on experts, we can figure it out for ourselves! | {"url":"https://concord.org/blog/the-importance-of-scientifically-literate-citizens/","timestamp":"2024-11-04T05:55:56Z","content_type":"text/html","content_length":"63343","record_id":"<urn:uuid:29b44c7a-0f27-48eb-ae02-1e7cbd6132dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00348.warc.gz"} |
The decorative designs on many native baskets are geometric. Different decorative shapes are created either through different methods of weaving the materials that make up the basket or by weaving in
colored grasses over the main basket material (this is also called "false embroidery". There are many possible designs, but the most commonly used designs have been given names by the Tlingit and
other tribes. Most of these designs utilize simple geometric shapes in various combinations and patterns. Furthermore, the designs are often presented as being symmetrical in some way or another.
What Is Symmetry
Many animals have bodies in which the right half is the mirror image of the left half. That is called "reflection symmetry." Some things have reflection symmetry between top and bottom. Footballs and
soup cans, for example, have one shape that is the same for top and bottom, and a different shape that is the same for left and right.
Many Pacific Northwest basket designs display various forms of symmetry. The decorative design on the basket on the right displays reflection symmetry.
Rotational symmetry is when you have a shape and rotate it around a single point. Here are some basket designs that display rotational symmetry.
If you have the same shape reflected in both directions (top-bottom, left-right), it is called four-fold symmetry. Four-fold symmetry is a deep design theme in many Native American cultures. It is
used as an organizing principle for religion, society, and native technology. Many native languages, for example, use base four counting. Teepees are often made with four base poles, each placed in
one of the four directions. Prayers are often offered to "the four winds." Left is a Tlingit design that displays four-fold symmetry.
Coordinate Systems
Since most native baskets are made up of rows and columns of roots or other materials, they make up a grid, like a Cartesian coordinate system. A Cartesian graph can be used to visualize basket
decorative designs very well. It is also very easy to observe how symmetry works in the basket designs when we use a Cartesian graph to analyze them. Furthermore, we can use a Cartesian graph to
create our own symmetrical patterns, whether they demonstrate reflection symmetry across one axis, across two axes, or rotational symmetry. Using the Basket Weaver software, you can apply a Cartesian
graph to create patterns commonly seen in native basket designs, or create your own original designs. | {"url":"https://csdt.org/culture/northwestbasketweaver/symmetry.html","timestamp":"2024-11-09T00:00:52Z","content_type":"text/html","content_length":"10508","record_id":"<urn:uuid:fdc38c35-dab2-4878-b302-915884ac3944>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00770.warc.gz"} |
Texas Go Math Kindergarten Unit 4 Assessment Answer Key
Refer to our Texas Go Math Kindergarten Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Kindergarten Unit 4 Assessment Answer Key.
Texas Go Math Kindergarten Unit 4 Assessment Answer Key
Question 1.
Cylinder is colored in green
cube in red, sphere in blue and cone in purple.
Concepts and Skills
Question 2.
Rectangle as 4 sides and 4 vertices
triangle has 3 sides and 3 vertices.
Question 3.
Cube has flat surfaces
sphere has curved surfaces.
Question 4.
A square has 4 sides and 4 vertices
all sides are equal in the square.
1. Use blue to color the sphere. Use green to color the cylinder. Use red to color the cube. Use purple to color the cone. TEKS K.6.B 2. Draw a line from the word side to a side of each shape. Draw a
line from the word vertex
to a vertex, or corner, of each shape. TEKS K.6.D 3. Circle the words that describe a cube. TEKS K.6.B 4. I have four square vertices and four straight sides. What am I? Draw the shape. TEKS K6.D
Question 5.
The given shape is a circle
so, marked the circle.
Question 6.
A triangle has 3 sides and 3 vertices
so, marked the triangle.
Question 7.
A cylinder has 2 flat surfaces and one curved surface.
A cone has one flat surface and a curved surface.
Question 8.
A triangle has 3 sides and 3 vertices
so, marked the number 3.
5. Mark under the shape that matches the shape at the beginning of the row. TEKS K.6.A 6. Mark under the shape that is a triangle. TEKS K.6.A 7. Mark beside the number that shows how many flat
surfaces the cylinder has. TEKS K.6.B 8. Mark beside the number that shows how many sides the triangle has. TEKS K.6.D
Texas Test Prep
Question 9.
A square has a flat surface
and a cube has 6 sides
Question 10.
In first group there are cylinder and a cube
in the second group there is a sphere and a cube
so, marked the first cube.
Question 11.
In a first group there is a cylinder and a cube
in the second group there is cube and a sphere
so, marked the first group.
Question 12.
In first group the blue crayon is longer than the green crayon
in second group the green crayon is longer than the blue crayon
so, marked the second group.
9. Mark under the shape that is flat. TEKS K.6.D 10. Mark under the set that shows a cube and a cylinder. TEKS K.6.B 11. Mark under the set that shows a cylinder and a cube. TEKS K.6.B 12. Mark under
the set that shows the green crayon is longer than the blue crayon. TEKS K.7.B
Performance Task
Performance Task This task will assess the child’s understanding of two-dimensional shapes.
Leave a Comment
You must be logged in to post a comment. | {"url":"https://gomathanswerkey.com/texas-go-math-kindergarten-unit-4-assessment-answer-key/","timestamp":"2024-11-05T03:39:19Z","content_type":"text/html","content_length":"248962","record_id":"<urn:uuid:a103cb16-7590-4516-b22d-b85d41a2f737>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00554.warc.gz"} |
Multiplication Partial Products Worksheets
Math, specifically multiplication, creates the foundation of many academic self-controls and real-world applications. Yet, for many learners, mastering multiplication can position a difficulty. To
resolve this difficulty, instructors and parents have welcomed a powerful tool: Multiplication Partial Products Worksheets.
Intro to Multiplication Partial Products Worksheets
Multiplication Partial Products Worksheets
Multiplication Partial Products Worksheets -
Multiplication Partial Products Worksheet by Kristy Simonson 73 1 50 PDF Worksheet to practice using the partial product strategy for multiplying 2 digit by 2 digit numbers I used in my Grade 4 5
Math Multiplication unit Subjects Math Grades 4 th 6 th Types Activities Assessment Math Centers
These Partial product multiplication worksheets and Area model multiplication examples and test are gives to make kids more successful in complex multiplication Another super easy method to multiply
bigger numbers is the box method
Relevance of Multiplication Practice Understanding multiplication is crucial, laying a strong structure for innovative mathematical concepts. Multiplication Partial Products Worksheets provide
structured and targeted technique, cultivating a much deeper comprehension of this basic math operation.
Advancement of Multiplication Partial Products Worksheets
Partial Products And Box Method Mini Anchor Chart Part Of An Interactive Math Journal Math
Partial Products And Box Method Mini Anchor Chart Part Of An Interactive Math Journal Math
ID 1486793 05 10 2021 Country code BS Country Bahamas School subject Math 1061955 Main content Calculate the product using the partial products method of multiplying 1560062 Students will calculate
the product using the partial products method of multiplying 2 3 and 4 digit numbers Loading ad Share Print Worksheet Finish
Overview This pre assessment launches a set of activities that return to the multiplication work started in Unit 2 In the 12 activ ities that follow students will move from building and sketching 2
digit by 1 digit multiplication combinations to using the standard algorithm to multiply up to 3 digit by 2 digit numbers
From traditional pen-and-paper workouts to digitized interactive layouts, Multiplication Partial Products Worksheets have progressed, catering to varied learning styles and choices.
Sorts Of Multiplication Partial Products Worksheets
Fundamental Multiplication Sheets Simple exercises focusing on multiplication tables, assisting learners build a strong arithmetic base.
Word Problem Worksheets
Real-life circumstances integrated into troubles, boosting vital reasoning and application abilities.
Timed Multiplication Drills Tests developed to boost rate and precision, assisting in rapid mental mathematics.
Benefits of Using Multiplication Partial Products Worksheets
Multiply Using Partial Products 4th Grade Worksheets
Multiply Using Partial Products 4th Grade Worksheets
Free Printable Math Worksheets Partial Product Multiplication Box Generate as many math problem worksheets as you need Simply choose your options click print and you re good to go Practice Sheet Type
Partial Product Print Ctrl P 1 Page Generate and print maths practice sheets Split complex triple digit problems into simple boxes
Partial Products Method 1 The partial product method allows students to find the quotient of multidigit factors by using place value concepts rather than relying on a series of memorized steps In
this worksheet students learn how to use the partial product method then use it to solve nine multi digit multiplication problems Designed for
Boosted Mathematical Abilities
Regular technique sharpens multiplication efficiency, improving total math abilities.
Improved Problem-Solving Abilities
Word troubles in worksheets create analytical thinking and technique application.
Self-Paced Discovering Advantages
Worksheets suit specific discovering speeds, fostering a comfy and versatile understanding setting.
Exactly How to Create Engaging Multiplication Partial Products Worksheets
Incorporating Visuals and Colors Vivid visuals and shades capture focus, making worksheets aesthetically appealing and involving.
Consisting Of Real-Life Circumstances
Associating multiplication to daily scenarios adds relevance and usefulness to workouts.
Customizing Worksheets to Various Ability Levels Tailoring worksheets based on differing efficiency degrees guarantees inclusive learning. Interactive and Online Multiplication Resources Digital
Multiplication Equipment and Games Technology-based sources offer interactive understanding experiences, making multiplication interesting and enjoyable. Interactive Web Sites and Apps On the
internet platforms supply diverse and accessible multiplication technique, supplementing standard worksheets. Personalizing Worksheets for Different Discovering Styles Visual Students Visual help and
layouts help comprehension for learners inclined toward visual learning. Auditory Learners Spoken multiplication problems or mnemonics accommodate students who comprehend principles through acoustic
methods. Kinesthetic Learners Hands-on tasks and manipulatives support kinesthetic students in recognizing multiplication. Tips for Effective Implementation in Learning Consistency in Practice
Regular method strengthens multiplication skills, promoting retention and fluency. Balancing Repetition and Range A mix of recurring exercises and diverse issue layouts keeps passion and
understanding. Offering Constructive Feedback Responses aids in identifying areas of enhancement, urging ongoing progression. Obstacles in Multiplication Method and Solutions Motivation and
Interaction Difficulties Tedious drills can cause disinterest; ingenious strategies can reignite motivation. Getting Rid Of Anxiety of Math Negative understandings around math can hinder development;
developing a favorable learning environment is essential. Effect of Multiplication Partial Products Worksheets on Academic Efficiency Studies and Study Searchings For Research indicates a positive
correlation in between constant worksheet usage and boosted math efficiency.
Multiplication Partial Products Worksheets emerge as flexible tools, fostering mathematical efficiency in learners while suiting varied understanding styles. From fundamental drills to interactive on
the internet sources, these worksheets not only boost multiplication abilities however likewise promote important thinking and analytical capabilities.
Multiply Using Partial Products 4Th Grade Worksheets Db excel
Partial Products Worksheets 4Th Grade Go Math Multiplication Db excel
Check more of Multiplication Partial Products Worksheets below
Grade 5 multiplication worksheets
Double Digit Multiplication Partial Product Method Shyla Acquarelli Library Formative
Multiply Using Partial Products Worksheets Worksheets Master
Partial Products Multiplication Worksheets Free Printable
Multiplication With Partial Products Worksheets
Partial Product Multiplication Lesson Classroom Caboodle Partial Product multiplication
Box method multiplication worksheets PDF Partial product
These Partial product multiplication worksheets and Area model multiplication examples and test are gives to make kids more successful in complex multiplication Another super easy method to multiply
bigger numbers is the box method
Multiplying two 2 digit numbers using partial products Khan Academy
A year ago we figured out a way to multiply a two digit number times one digit number What we did is we broke up the two digit numbers in terms of its place value so the three here in the tenths
place that s three tens this is seven ones So we view 37 sixes as the same thing as 30 sixes three tens times six plus seven sixes seven times six
These Partial product multiplication worksheets and Area model multiplication examples and test are gives to make kids more successful in complex multiplication Another super easy method to multiply
bigger numbers is the box method
A year ago we figured out a way to multiply a two digit number times one digit number What we did is we broke up the two digit numbers in terms of its place value so the three here in the tenths
place that s three tens this is seven ones So we view 37 sixes as the same thing as 30 sixes three tens times six plus seven sixes seven times six
Partial Products Multiplication Worksheets Free Printable
Double Digit Multiplication Partial Product Method Shyla Acquarelli Library Formative
Multiplication With Partial Products Worksheets
Partial Product Multiplication Lesson Classroom Caboodle Partial Product multiplication
Box multiplication partial products method 2 digit by 1 digit worksheet 4 Partial Product
Partial Products Anchor Chart Partial products Everyday Math Math Anchor Charts
Partial Products Anchor Chart Partial products Everyday Math Math Anchor Charts
partial products worksheets Free Product multiplication Printable Interactive Math Journals
Frequently Asked Questions (Frequently Asked Questions).
Are Multiplication Partial Products Worksheets appropriate for all age groups?
Yes, worksheets can be customized to different age and skill levels, making them versatile for various students.
How usually should students practice utilizing Multiplication Partial Products Worksheets?
Regular method is crucial. Routine sessions, ideally a couple of times a week, can yield substantial renovation.
Can worksheets alone enhance mathematics abilities?
Worksheets are a beneficial device however should be supplemented with diverse knowing methods for extensive ability growth.
Exist on-line systems using complimentary Multiplication Partial Products Worksheets?
Yes, several academic sites supply open door to a wide variety of Multiplication Partial Products Worksheets.
How can parents support their kids's multiplication practice in your home?
Motivating regular technique, giving assistance, and creating a positive discovering atmosphere are helpful steps. | {"url":"https://crown-darts.com/en/multiplication-partial-products-worksheets.html","timestamp":"2024-11-05T06:39:07Z","content_type":"text/html","content_length":"29818","record_id":"<urn:uuid:5c3efa4f-eaca-4f5d-978a-356a96cc805a>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00246.warc.gz"} |
Topological Data Analysis
2. Topology
What do we mean with topology? Wikipedia describes topology as follows:
In mathematics, topology … is concerned with the properties of a geometric object that are preserved under continuous deformations, such as stretching, twisting, crumpling, and bending; that is,
without closing holes, opening holes, tearing, gluing, or passing through itself.
What does this even mean?
Coordinate invariance
Topology considers how things are connected, not the actual coordinates. The ellipses below are all topologically equal; it does not matter if they are rotated or not.
Deformation invariance
The geometric shape of an object can be deformed by stretching or bending it, as long as we don’t create new holes or remove existing ones. The letters A written in two different fonts below are
topologically the same. The letter B is different because we cannot deform a letter A to create a letter B without creating a new hole.
Compressed representation
Consider a circle consisting of an infinite number of points. We can represent this same circle with one triangle consisting of only 3 points; they are topologically equivalent.
Image source: Ayasdi white paper "Deep Dive: Topological Data Analysis"
A running joke tells of a topologist who is not able to distinguish a mug from a donut. Indeed, these are equivalent. If you start making the bottom of the mug thicker and thicker, you’ll be working
towards a donut (torus) shape. Both have only 1 hole: the handle, and the hole in the donut.
2.1. Topology vs geometry vs algebra
Consider the idea of a "circle".
In geometry, we talk about its curvature, width, rotational symmetry, etc.
In topology, we look at the circle as being made from flexible material: we can stretch and deform it, but should not poke hole in it or break it.
In algebra, we consider the circle as a collection of points and a rule to connect them. Here, we have a notion of "nearness" of points. This will become important later when we talk about "algebraic
To summarise: a donut and a mug have a different geometry but the same topology, whereas a sieve and a plate have a similar geometry but a different topology (i.e. number of holes).
The image below shows topological equivalence across a wide number of shapes. It shows that a disk is equivalent to a singular point or a hollow sphere with a hole in it. Also, a number 8 is
equivalent to two circles connected with a line, etc.
Source: Singh G et al (2008) Topological analysis of population activity in visual cortex. Journal of Vision, 8(11), 1-18
2.2. From data to topology
As we’re working with datapoints (often in high-dimensional space) and topology (in terms of torus shapes, circles, etc), how do we marry these two together?
In topological data analysis, we convert the datapoints into a network (actually simplicial complexes, but we’ll go into that later). There are different ways for doing this, depending on whether it
is for visualisation purposes (mapper), or for analysis (persistent homology).
We’ll look further in both of them in the following sections. Things that we are particularly interested in, are flares and holes. | {"url":"https://vda-lab.gitlab.io/topological-data-analysis/_topology.html","timestamp":"2024-11-06T05:46:34Z","content_type":"text/html","content_length":"39021","record_id":"<urn:uuid:6c363b90-aa95-4e50-93d3-8edb05905566>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00706.warc.gz"} |
Two types A and B are a model of the concept ImplicitInteroperable, if there is a superior type, such that binary arithmetic operations involving A and B result in this type. This type is
CGAL::Coercion_traits<A,B>::Type. In case types are RealEmbeddable this also implies that mixed compare operators are available.
The type CGAL::Coercion_traits<A,B>::Type is required to be implicit constructible from A and B.
In this case CGAL::Coercion_traits<A,B>::Are_implicit_interoperable is CGAL::Tag_true.
See also | {"url":"https://doc.cgal.org/5.5.2/Algebraic_foundations/classImplicitInteroperable.html","timestamp":"2024-11-08T17:48:12Z","content_type":"application/xhtml+xml","content_length":"10720","record_id":"<urn:uuid:095076dc-57d7-43b3-a84d-00124ab0d4b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00359.warc.gz"} |
Antennas September 2003
Antennas September 2003
Antenna installation
In september 2003, during the IARU-Region-1 VHF contest, we used nine antennas on three portable towers. Two of them carried four DK7ZB 5-element yagis vertically stacked. On the third tower was a
single Cushcraft 17-element yagi. The left picture shows all antennas together and the right picture shows one of our yagi group in detail. The vertical distance between two yagis is 1.6 m.
The DK7ZB 5-Element Yagi
│ Element... │Reflector│Radiador│Director1│Director 2│Director 3│
│ ...length │ 1022 mm│ 960 mm│ 910 mm│ 915 mm│ 882 mm│
│...diameter │ 10 mm│ 10 mm│ 10 mm│ 10 mm│ 10 mm│
│...position │ 0 mm│ 410 mm│ 815 mm│ 1475 mm│ 1970 mm│
We build this Yagi according to a proposal by DK7ZB. Please look at http://www.qsl.net/dk7zb for the details. The most important features are: 2 m boomlength, 9 dBd gain, front-to-back ratio > 20 dB,
50° horizontal beam width, 60° vertical beam width. The feedpoint impedance of the driven element is 28 Ohms. This impedance is transformed to 50 Ohms by a quater-wavelength matching section build of
two 75 Ohm coax lines in parallel. The following pictures shows the connection box (open and closed) of our yagi together with the quater-wavelength matching section.
Our group of four DK7ZB 5-Element Yagis
In order to increase the gain over that obtainable from one DK7ZB Yagi we stacked four yagis. The increase in gain is due to the reduction in vertical beamwidth. The gain of the yagi group is about
15 dBd. One method to connect four yagis is using a four way power divider. All lines between the power divider and the yagis must have the same length. This results in an inappropriate layout.
That's why we looked for another solution.
The following figure shows how we solved this problem. The yagis are connected two by two by three 50 Ohm coax lines. The first line connects Yagi 1 and Yagi 2. The second line connects Yagi 3 and
Yagi 4. A third line is used to connect both groups. The impedance at the middle of line 1 and 2 is 25 Ohms. This impedance is transformed to (100 Ohms + 100 Ohms)/2 = 50 Ohms by the third line. Line
1 and 2 must have the same length. Line 3 is a quater-wavelength matching section.
Such a matching line is easy to build. We used a piece of 50 Ohm coaxial line with N-plugs at both ends and a N-socket in the middle of the line. The connection to the N-socket is protected by a
small plastic box.
Connection of two Yagi groups
I order to run both Yagi groups together at one Transceiver we build a two way power divider.
The power divider's inner and outer diameter was calculated by means of the RF design software "AppCad". This software is available free of charge by Agilent Technologies at http://
This is a picture of our homemade two way power divider. | {"url":"https://www.dk0a.de/cnt/index.php?page=antennas-september-2003","timestamp":"2024-11-14T21:03:59Z","content_type":"application/xhtml+xml","content_length":"29692","record_id":"<urn:uuid:603cf307-741c-444f-a830-f5cafa97331f>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00256.warc.gz"} |
A rectangular building is 4 meters by 6 meters. A dog is tied to a rope that is 10 meters long, and the other end is tied to the midpoint of one of the short sides of the building. Find the total
area of the region that the dog can reach (not including the inside of the building), in square meters. Plz don't send me the link of the simlilar problem, since I don't understand it.
UniCorns555 Feb 17, 2024
ok ill make thise easy to understand so:
we can find the area the dog can reach on one side then multiply by 2 to get the total area. Imagine wrapping a string around a sqaure. If we start wrapping the 10m string around the building towards
the right then the string will form a quarter circle till the side of the building and the pivot will change to the corner of the building and the radius that makes the next semicircle will be \(10-2
=8\). Like wise this will continue to the side of the building the the pivot will again change to be the far corner of the building. Then the radius will change again to \(8-6=2\). Finally there will
be another quarter circle with radius 2. There is no overlap as the other side will copy the same area due to symmetry.
\(2(\frac{2^2\pi}{4}+\frac{8^2\pi}{4}+\frac{10^2\pi}{4})=\boxed{84\pi\hspace{2mm}\text{ m }^2}\)
EnormousBighead Feb 17, 2024 | {"url":"https://web2.0calc.com/questions/help-due-tmrw_1","timestamp":"2024-11-08T17:57:46Z","content_type":"text/html","content_length":"21757","record_id":"<urn:uuid:1b5a4d79-95be-467d-b923-e4e1e0bd791f>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00024.warc.gz"} |
An In-Depth Guide to Calculate Mass Percentages: Formula, Tips, and Real Life Applications - The Explanation Express
An In-Depth Guide to Calculate Mass Percentages: Formula, Tips, and Real Life Applications
I. Introduction
Mass percentages are an important concept in chemistry that involves calculating the amount of a particular substance present in a given sample. It is a way of expressing a component’s relative
amount to the total mass of the system. In this article, we will explore the various methods of calculating mass percentages and understand why it is crucial in laboratory work and real-world
II. An In-Depth Guide to Calculating Mass Percentages
Mass percentage is a measure of the relative amount of a substance dissolved in a solution or present in a mixture. To calculate the mass percentage, we divide the mass of the solute by the mass of
the solution and multiply by 100. The mathematical formula for calculating mass percentages is:
Mass percentage = (mass of solute / mass of solution) x 100
We use Mass percentage to understand the composition of solutions. For instance, the mass percentage of ethanol in water can give an idea about the purity of the mixture. The mass percentage can also
be used to calculate the strength of acids and bases, which is crucial in laboratory work. Let’s take a look at some examples to further understand the concept of mass percentages in chemistry.
III. Quick and Easy Methods for Finding Mass Percentages
While the above formula is the most common way of calculating mass percentages, there are quicker ways to find these values. These shortcut formulas save time and can be used for simple solutions.
The most common shortcut formula for calculating mass percentage is:
Mass percentage = (the number of grams of solute / total number of grams in solution) * 100
Another method for calculating mass percentages is the use of conversion factors. In this method, we consider the relationship between two units and use it to calculate the mass percentage. For
instance, to calculate the mass percentage of a substance when we know the concentration in moles, we can use the mole-to-gram conversion factor. By multiplying the mole concentration by the molar
mass of the solute, we can find the mass percentage.
IV. Tips and Tricks for Accurately Calculating Mass Percentages
Calculating accurate mass percentages requires precision and attention to detail. Here are some tips and tricks for accurate calculations:
Using a digital balance for accurate measurements: When calculating mass percentages, the accuracy of measurements is essential. We need to use equipment that can provide precise measurements. A
digital balance is a reliable and accurate tool that should always be used in the laboratory setting.
Recording measurements in significant figures: Significant figures are digits that carry meaning in a measurement. It requires identifying which digits are significant and which are not. Recording
and utilizing significant figures in mass percentage calculations will help prevent rounding errors and increase accuracy.
Using proper units for measurements: To calculate mass percentages, all measurements should be in the same units. For instance, if one measurement is in grams, all other measurements should also be
in grams. This way, we can avoid confusion and ensure the accuracy of our calculations.
V. Understanding the Importance of Mass Percentages in Chemistry
Mass percentages play a crucial role in chemistry and laboratory work. Here are some examples of its significance:
Uses of mass percentages in the laboratory: Mass percentages are crucial when it comes to identifying and characterizing chemical reactions and mixtures. It can also help determine the composition
and structure of materials used in various industries.
Types of chemical reactions where mass percentages are important: Mass percentages are essential in all types of chemical reactions, from acid-base reactions to precipitation reactions. Accurately
measuring and calculating mass percentages can help determine the limiting reagent to carry out a reaction to completion.
Relation of mass percentages to other chemical concepts: Mass percentages are closely related to other concepts in chemistry, such as atomic and molecular weights. It is an essential aspect of
stoichiometry calculations and is therefore relevant in chemical reactions.
VI. Step-by-Step Guide to Calculating Mass Percentages
Calculating mass percentage involves several steps. Here is a detailed guide to help you understand every phase of the process:
Step 1: Weigh the sample to find the mass of the solvent or the mixture.
Step 2: Add the solute to the sample and weigh again to find the total mass of the solution.
Step 3: Calculate the mass of the solute by subtracting the mass of the solvent from the total mass of the solution.
Step 4: Use the mathematical formula to calculate mass percentage.
Step 5: Record the calculated mass percentage, its units, and significant figures.
VII. Common Mistakes to Avoid When Calculating Mass Percentages
Common mistakes while calculating mass percentages can lead to inaccurate readings. Here are some strategies to avoid them:
Incorrectly interpreting a problem. In some instances, students may misinterpret a problem, leading to the wrong calculation. One way to avoid this is to read the problem statement carefully and
understand what is being asked before starting the calculations.
Making rounding errors: Rounding off numbers too early in the calculations can result in an incorrect answer. Always perform all calculations before rounding to the proper number of significant
Forget to convert units: Units should be consistent and in the same units when making calculations. One of the most common mistakes is to use different units, which can lead to confusion and
incorrect calculations. Always double-check all units are in the same form before calculation.
VIII. How to Apply Mass Percentages in Real-Life Situations
Mass percentages are not just important in laboratory settings; they also have practical applications in everyday life. Here are some real-life situations where we use mass percentages:
Examples of everyday situations where mass percentages are used: One of the most common instances is when preparing food or drinks. The solution of sugar in water, for instance, can be made to a
particular mass percentage, which will affect the taste. In the medical sector, mass percentages are used to calculate medication doses based on a patient’s weight and height.
Practical uses of mass percentages in various industries: Mass percentages are important in the manufacturing of various products. For example, the percentage of metals in alloys and the purity of
substances such as gasoline can directly impact product quality.
Use of mass percentages in analytical chemistry: In analytical chemistry, mass percentages are used to calculate results and ensure the accuracy and quality of the analysis. It helps to understand
the composition of a sample, identify impurities, and characterize the compounds present.
IX. Conclusion
Calculating mass percentages is an essential aspect of laboratory work and has practical applications in various industries. In this article, we have explored the different methods of calculating
mass percentages, as well as tips and tricks for accurate measurements. By understanding the importance of mass percentages, we can better comprehend various chemical reactions and make informed
decisions regarding chemical mixtures. With the information and knowledge gained from this article, you can now easily calculate mass percentages for various substances and understand their | {"url":"https://www.branchor.com/how-to-calculate-mass-percent/","timestamp":"2024-11-12T06:45:13Z","content_type":"text/html","content_length":"46650","record_id":"<urn:uuid:07f3f312-efd8-4a07-8925-76345861c814>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00039.warc.gz"} |
EPiQC Researchers Simulate 61-Bit Quantum Computer With Data Compression - Department of Computer Science
UChicago CS News•Jan 13, 2020
EPiQC Researchers Simulate 61-Bit Quantum Computer With Data Compression
Researchers at the University of Chicago and Argonne National Laboratory significantly reduced the gap between quantum computers and simulators by using data compression and a large supercomputer to
conduct a 61-qubit simulations of a quantum search algorithm.
Figure 1. Overview of simulation with data compression
When trying to debug quantum hardware and software by using a quantum simulator, every quantum bit (qubit) counts. Every simulated qubit closer to physical machine sizes halves the gap in compute
power between the simulation and the physical hardware. However, the memory requirement of full-state simulation grows exponentially with the number of simulated qubits, and this limits the size of
simulations that can be run.
Researchers at the University of Chicago and Argonne National Laboratory significantly reduced this gap by using data compression techniques to fit a 61-qubit simulation of Grover’s quantum search
algorithm on a large supercomputer with 0.4 percent error. Other quantum algorithms were also simulated with substantially more qubits and quantum gates than previous efforts.
Classical simulation of quantum circuits is crucial for better understanding the operations and behaviors of quantum computation. However, today’s practical full-state simulation limit is 48 qubits,
because the number of quantum state amplitudes required for the full simulation increases exponentially with the number of qubits, making physical memory the limiting factor. Given n qubits,
scientists need 2^n amplitudes to describe the quantum system.
There are already several existing techniques that trade execution time for memory space. For different purposes, people choose different simulation techniques. This work provides one more option in
the set of tools to scale quantum circuit simulation, applying lossless and lossy data compression techniques to the state vectors.
Figure 1 shows an overview of our simulation design. The Message Passing Interface (MPI) is used to execute the simulation in parallel. Assuming we simulate n-qubit systems and have r ranks in total,
the state vector is divided equally on r ranks, and each partial state vector is divided into nb blocks on each rank. Each block is stored in a compressed format in the memory.
Figure 2 shows the amplitude distribution in two different benchmarks. “If the state amplitude distribution is uniform, we can easily get a high compression ratio with the lossless compression
algorithm,” said researcher Xin-Chuan Wu. “If we cannot get a nice compression ratio, our simulation procedure will adopt error-bounded lossy compression to trade simulation accuracy for compression
Figure 2. Value changes of quantum circuit simulation data. (a) The data value changes within a range. (b) The data exhibit a high spikiness and variance such that lossless compressors cannot work
The entire full-state simulation framework with data compression leverages MPI to communicate between computation nodes. The simulation was performed on the Theta supercomputer at Argonne National
Laboratory. Theta consists of 4,392 nodes, each node containing a 64-core Intel®Xeon PhiTM processor 7230 with 16 gigabytes of high-bandwidth in-package memory (MCDRAM) and 192 GB of DDR4 RAM.
The full paper, “Full-State Quantum Circuit Simulation by Using Data Compression,” was published by The International Conference for High Performance Computing, Networking, Storage, and Analysis
(SC’19). The UChicago research group is from the EPiQC (Enabling Practical-scale Quantum Computation) collaboration, an NSF Expedition in Computing, under grants CCF-1730449. EPiQC aims to bridge the
gap from existing theoretical algorithms to practical quantum computing architectures on near-term devices. | {"url":"https://computerscience.uchicago.edu/news/quantum-compression/","timestamp":"2024-11-05T09:53:04Z","content_type":"text/html","content_length":"129425","record_id":"<urn:uuid:3addbeb1-7d37-48ad-a090-2997989e1a3e>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00014.warc.gz"} |
Understanding Roman Numerals 2019: Conversion, Meaning, and Uses
Roman numerals have been in use for a number of centuries, and their usage may still be seen today, particularly for aesthetic and symbolic reasons. In this piece, we will explore the history of
Roman numerals as well as their practical applications, with a particular emphasis on the year 2019.
The year 2019 is represented by the Roman number MMXIX. In order to translate any number into Roman numerals, we must first get familiar with the fundamental symbols and the meanings behind them. The
values that are represented by Roman numerals are written out using letters from the Latin alphabet. The following is a list of the fundamental symbols and their respective values:
I = 1 V = 5 X = 10 L = 50 C = 100 D = 500 M = 1000
In order to express the year 2019 using Roman numerals, we must first convert it into its place values, which are 2000, 10, and 9. We are going to work our way down from the greatest place value to
the lowest one. Hence, the number 2000 is denoted by the letter M, the number 10 by the letter X, and the number 9 by the letters IX (which means 1 less than 10). As a result, the year 2019 is
abbreviated as MMXIX.
In the past, Roman numerals were used often for a variety of purposes, including counting and record-keeping. They were used to a great extent by the ancient Romans, and this practice was maintained
all the way through the Middle Ages and into the Renaissance. They offer a feeling of heritage and timelessness, which is why they are used so often in clocks, on buildings, and in the credits of
movies nowadays.
For the purpose of identifying and interpreting Roman numerals in a variety of circumstances, it is vital to have a working knowledge of the system. Having the ability to convert numbers into Roman
numerals is a vital talent that may be put to use for a variety of different reasons. When the year 2019 is broken down into its place values and the fundamental symbols and their values are used, we
are able to see that the year 2019 is represented as MMXIX when written in Roman numerals.
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Full-Text Articles in Mathematics
Operations On Submodules With The Multiplicative And Quotient Properties, Jiekai Pang
Operations On Submodules With The Multiplicative And Quotient Properties, Jiekai Pang
Mathematics & Statistics ETDs
Inspired by the works of Petro, Epstein, Vassilev, and Morre, this thesis aims to study the generalized definitions of the semiprime operation, weakly prime operation, and standard closure operation
on rings, that is, the multiplicative operation, weakly multiplicative operation, and standardly multiplicative operation on submodules. Then, we will use Matlis duality to induce the dual notions of
these definitions on submodules of Matlis-dualizable Artinian modules. In order to understand the dual notion of the standardly multiplicative operation, that is, the standardly quotient operation,
we will classify the operations on the injective hull of residue field of the ring K[[x,y]]/(xy) which …
Shevtsov: Teaching Modeling To First-Year Life Science Students: The Ucsc Experience, Martin H. Weissman
Shevtsov: Teaching Modeling To First-Year Life Science Students: The Ucsc Experience, Martin H. Weissman
Annual Symposium on Biomathematics and Ecology Education and Research
No abstract provided.
Modeling Opioid Addiction In Hand Surgery Patients, Eli Goldwyn, Grace Bowman, Kathryn Montovan, Julie Blackwood
Modeling Opioid Addiction In Hand Surgery Patients, Eli Goldwyn, Grace Bowman, Kathryn Montovan, Julie Blackwood
Annual Symposium on Biomathematics and Ecology Education and Research
No abstract provided.
Using Academic Librarians And The Academic Library: Survey Results From Mathematics Faculty In United States And Canada, Elizabeth Novosel, Daniel G. Kipnis
Using Academic Librarians And The Academic Library: Survey Results From Mathematics Faculty In United States And Canada, Elizabeth Novosel, Daniel G. Kipnis
Libraries Scholarship
Presented at STEM Librarian South Conference, November 6, 2024
Despite our diligent outreach and relationship-building efforts, certain groups of faculty within STEM disciplines remain hesitant to engage with academic librarians and utilize library resources.
This challenge raises a critical question: How can librarians effectively support a department when faculty members are unresponsive?
Our research team, composed of three frustrated mathematics librarians and one mathematics faculty member, recognized the need for a fresh approach. Mathematics, a department known for its
independence, has received limited attention in scholarly Library and Information Science (LIS) literature. To address this gap, we embarked on a …
Book V Of The Mathematical Collection Of Pappus Of Alexandria, Translated By John B. Little, Pappus Of Alexandria, John B. Little
Book V Of The Mathematical Collection Of Pappus Of Alexandria, Translated By John B. Little, Pappus Of Alexandria, John B. Little
Holy Cross Bookshelf
John B. Little is the translator.
Book V of the Mathematical Collection is addressed to a certain Megethion, about whom we know nothing else. From the context he may have been a student or patron of Pappus in Alexandria. In a heading
at the start, Pappus says that the general theme will be comparisons between different geometric figures. The overall structure brings interesting relations and connections to the fore. The book
opens with a very well-known and charming discussion of how the importance of such comparisons can be seen by considering the structures built by non-human creatures such as bees. …
Algebraic Structures On Parallelizable Manifolds, Sergey Grigorian
Algebraic Structures On Parallelizable Manifolds, Sergey Grigorian
School of Mathematical and Statistical Sciences Faculty Publications and Presentations
In this paper we explore algebraic and geometric structures that arise on parallelizable manifolds. Given a parallelizable manifold L , there exists a global trivialization of the tangent bundle,
which defines a map ρ p : l ⟶ T p L for each point p ∈ L , where l is some vector space. This allows us to define a particular class of vector fields, known as fundamental vector fields, that
correspond to each element of l . Furthermore, flows of these vector fields give rise to a product between elements of l and L , which in turn induces …
Pricing Variance Swaps For The Discrete Bn-S Model, Semere Gebresilasie
Pricing Variance Swaps For The Discrete Bn-S Model, Semere Gebresilasie
Journal of Stochastic Analysis
No abstract provided.
A Phenomenological Study Of Persistence For Black Women In Mathematics Doctoral Programs, Shushannah Marie Smith
A Phenomenological Study Of Persistence For Black Women In Mathematics Doctoral Programs, Shushannah Marie Smith
Doctoral Dissertations and Projects
The purpose of this phenomenological study was to describe the doctoral persistence experiences of Black women mathematicians. Guided by McGee’s fragile and robust mathematical identity framework,
the research design followed Moustakas’ qualitative transcendental phenomenological approach. The researcher explored the contributing factors to the phenomenon guided by the central research
question, “What are the doctoral persistence experiences of Black women mathematicians?” The study involved 11 Black women participants who hold or were within one year of holding a doctoral degree
in mathematics and were selected to ensure diversity and representativeness. Data collection encompassed three methods: a mathematical identity development timeline, one-on-one …
Uniqueness Of Maximum Points Of A Sequence Of Functions Arising From An Adapted Algorithm For ‘The Secretary Problem’, Boning Wang, Giang Vu Thanh Nguyen
Uniqueness Of Maximum Points Of A Sequence Of Functions Arising From An Adapted Algorithm For ‘The Secretary Problem’, Boning Wang, Giang Vu Thanh Nguyen
OUR Journal: ODU Undergraduate Research Journal
This paper is aimed at a sequence of functions that is extended from an adaptive algorithm of the classical ‘secretary problem’. It was proved in Nguyen et al. (2024) the uniqueness of maximizers of
a function sequence that represents the expected score of an element in a ‘candidate’ sequence. This function sequence is indeed considered as a special case of an extended function sequence that
corresponds to the case α = 0. More specifically, we are motivated to prove the uniqueness of maximizers for this extended function sequence in the case α = 1. Nevertheless, the corresponding proof
is rather …
Scaled Global Operators And Fueter Variables On Non-Zero Scaled Hypercomplex Numbers, Daniel Alpay, Ilwoo Choo, Mihaela Vajiac
Scaled Global Operators And Fueter Variables On Non-Zero Scaled Hypercomplex Numbers, Daniel Alpay, Ilwoo Choo, Mihaela Vajiac
Mathematics, Physics, and Computer Science Faculty Articles and Research
In this paper we describe the rise of global operators in the scaled quaternionic case, an important extension from the quaternionic case to the family of scaled hypercomplex numbers H[t], t ∈ R^∗ ,
of which the H[−1] = H is the space of quaternions and H[1] is the space of split quaternions.We also describe the scaled Fueter-type variables associated to these operators, developing a coherent
theory in this field. We use these types of variables to build different types of function spaces on H[t]. Counterparts of the Hardy space and of the …
Counting The Classes Of Projectively-Equivalent Pentagons On Finite Projective Planes Of Prime Order, Maxwell Hosler
Counting The Classes Of Projectively-Equivalent Pentagons On Finite Projective Planes Of Prime Order, Maxwell Hosler
Rose-Hulman Undergraduate Mathematics Journal
In this paper, we examine the number of equivalence classes of pentagons on finite projective planes of prime order under projective transformations. We are interested in those pentagons in general
position, meaning that no three vertices are collinear. We consider those planes which can be constructed from finite fields of prime order, and use algebraic techniques to characterize them by their
symmetries. We are able to construct a unique representative for each pentagon class with nontrivial symmetries. We can then leverage this fact to count classes of pentagons in general. We discover
that there are (1/10)((p+3)(p-3)+4 …
Nano Topology And Decision Making In Medical Applications, Samir Mukhtar, Mohamed Shokry, Manar Omran
Nano Topology And Decision Making In Medical Applications, Samir Mukhtar, Mohamed Shokry, Manar Omran
Journal of Engineering Research
Nano Topology is one of the essential topics that receive special attention from some athletes in the field of General Topology, Operations Research, and Computer Science, because it has a vital role
in the generalizing most of the various mathematical concepts. Recently, many efforts have been made to study many types of Nano Topology, as the previous studies lacked real applications in
Engineering, Medicine, Pharmacy, and Social Sciences. In this paper, we present some different applications of these studies. The paper is divided into two parts: Firstly, we study the theory of The
Nano Topology and investigate its relation with …
New Operation Defined Over Dual-Hesitant Fuzzy Set And Its Application In Diagnostics In Medicine, Manar Mohamed Omran, Reham Abdel-Aziz Abo-Khadra
New Operation Defined Over Dual-Hesitant Fuzzy Set And Its Application In Diagnostics In Medicine, Manar Mohamed Omran, Reham Abdel-Aziz Abo-Khadra
Journal of Engineering Research
In recent decades, several types of sets, such as fuzzy sets, interval-valued fuzzy sets, intuitionistic fuzzy sets, interval-valued intuitionistic fuzzy sets, type 2 fuzzy sets, type n fuzzy sets,
and hesitant fuzzy sets, have been introduced and investigated widely. In this paper, we propose dual hesitant fuzzy sets (DHFSs), which encompass fuzzy sets, intuitionistic fuzzy sets, hesitant
fuzzy sets, and fuzzy multi-sets as special cases. Then we investigate the basic operations and properties of DHFSs. We also discuss the relationships among the sets mentioned above, and then propose
an extension principle of DHFSs. Additionally, we give an example to illustrate …
Decision-Making In Diagnosing Heart Failure Problems Using Dual Hesitant Fuzzy Sets, Manar Mohamed Omran, Reham Abdel-Aziz Abo-Khadra
Decision-Making In Diagnosing Heart Failure Problems Using Dual Hesitant Fuzzy Sets, Manar Mohamed Omran, Reham Abdel-Aziz Abo-Khadra
Journal of Engineering Research
In recent decades, several types of sets, such as fuzzy sets, interval-valued fuzzy sets, intuitionistic fuzzy sets, interval-valued intuitionistic fuzzy sets, type 2 fuzzy sets, type n fuzzy sets,
and hesitant fuzzy sets, have been introduced and investigated widely. In this paper, we propose dual hesitant fuzzy sets (DHFSs), which encompass fuzzy sets, intuitionistic fuzzy sets, hesitant
fuzzy sets, and fuzzy multi-sets as special cases. Then we investigate the basic operations and properties of DHFSs. We also discuss the relationships among the sets mentioned above, and then propose
an extension principle of DHFSs. Additionally, we give an example to illustrate …
Discrete Time Series Forecasting Of Hive Weight, In-Hive Temperature, And Hive Entrance Traffic In Non-Invasive Monitoring Of Managed Honey Bee Colonies: Part I, Vladimir A. Kulyukin, Daniel Coster,
Aleksey V. Kulyukin, William Meikle, Milagra Weiss
Discrete Time Series Forecasting Of Hive Weight, In-Hive Temperature, And Hive Entrance Traffic In Non-Invasive Monitoring Of Managed Honey Bee Colonies: Part I, Vladimir A. Kulyukin, Daniel Coster,
Aleksey V. Kulyukin, William Meikle, Milagra Weiss
Computer Science Faculty and Staff Publications
From June to October, 2022, we recorded the weight, the internal temperature, and the hive entrance video traffic of ten managed honey bee (Apis mellifera) colonies at a research apiary of the Carl
Hayden Bee Research Center in Tucson, AZ, USA. The weight and temperature were recorded every five minutes around the clock. The 30 s videos were recorded every five minutes daily from 7:00 to 20:55.
We curated the collected data into a dataset of 758,703 records (208,760–weight; 322,570–temperature; 155,373–video). A principal objective of Part I of our investigation was to use the curated
dataset to investigate …
(Si13-07) Optimal Range For Value Of Two-Person Zero-Sum Game Models With Uncertain Payoffs, Sana Afreen, Ajay Kumar Bhurjee
(Si13-07) Optimal Range For Value Of Two-Person Zero-Sum Game Models With Uncertain Payoffs, Sana Afreen, Ajay Kumar Bhurjee
Applications and Applied Mathematics: An International Journal (AAM)
Game theory deals with the decision-making of individuals in conflicting situations with known payoffs. However, these payoffs are imprecisely known, which means they have uncertainty due to
vagueness in the data set of most real-world problems. Therefore, we consider a two-person zero-sum game model on a larger scale where the payoffs are imprecise and lie within a closed interval. We
define the pure and mixed strategy as well as value for the game models. The proposed method computes the optimal range for the value of the game model using interval analysis. To derive some
important results, we establish some lemmas …
(Si13-06) Analysis Of Some Unified Integral Equations Of Fredholm Type Associated With Multivariable Incomplete H And I-Functions, Rahul Sharma, Vinod Gill, Naresh Kumar, Kanak Modi, Yudhveer Singh
(Si13-06) Analysis Of Some Unified Integral Equations Of Fredholm Type Associated With Multivariable Incomplete H And I-Functions, Rahul Sharma, Vinod Gill, Naresh Kumar, Kanak Modi, Yudhveer Singh
Applications and Applied Mathematics: An International Journal (AAM)
In this research paper, we examine various effective methods for addressing the problem of solving Fredholm-type integral equations. Our investigation commences by applying the principles of
fractional calculus theory. We employ series representations and products of multivariable incomplete H-functions and multivariable incomplete I-functions to solve these integrals. The outcomes
derived from our analysis possess a general nature and hold the potential to yield numerous results.
Two Is Enough, But Three (Or More) Is Better: In Ai And Beyond, Olga Kosheleva, Vladik Kreinovich, Victor L. Timchenko, Yury P. Kondratenko
Two Is Enough, But Three (Or More) Is Better: In Ai And Beyond, Olga Kosheleva, Vladik Kreinovich, Victor L. Timchenko, Yury P. Kondratenko
Departmental Technical Reports (CS)
At present, the most successful AI technique is deep learning -- the use of neural networks that consist of multiple layers. Interestingly, it is well known that neural networks with two data
processing layers are sufficient -- in the sense that they can approximate any function with any given accuracy. Because of this, until reasonably recently, researchers and practitioners used such
networks. However, recently it turned out, somewhat unexpectedly, that using three or more data processing layers -- i.e., using what is called deep learning -- makes the neural networks much more
efficient. In this paper, on numerous examples from …
The P -Adic Schrödinger Equation And The Two-Slit Experiment In Quantum Mechanics, Wilson A. Zuniga-Galindo
The P -Adic Schrödinger Equation And The Two-Slit Experiment In Quantum Mechanics, Wilson A. Zuniga-Galindo
School of Mathematical and Statistical Sciences Faculty Publications and Presentations
p-Adic quantum mechanics is constructed from the Dirac-von Neumann axioms identifying quantum states with square-integrable functions on the N-dimensional p-adic space, Q_{p}^{N}. This choice is
equivalent to the hypothesis of the discreteness of the space. The time is assumed to be a real variable. p-Adic quantum mechanics is the response to the question: what happens with the standard
quantum mechanics if the space has a discrete nature? The time evolution of a quantum state is controlled by a nonlocal Schrödinger equation obtained from a p-adic heat equation by a temporal Wick
rotation. This p-adic heat equation describes a particle performing …
Relative Equilibria Of Pinwheel Point Mass Systems In A Planar Gravitational Field, Ritwik Gaur
Relative Equilibria Of Pinwheel Point Mass Systems In A Planar Gravitational Field, Ritwik Gaur
Rose-Hulman Undergraduate Mathematics Journal
In this paper, we consider a planar case of the full two-body problem (F2BP) where one body is a pinwheel (four point masses connected via two perpendicular massless rods) and the other is a point
mass. Relative equilibria (RE) are defined to be ordered pairs (r,θ) such that there exists a rotating reference frame under which the two bodies are in equilibrium when the distance between the
point mass and the center of the pinwheel is r and the angle of the pinwheel within its orbit is θ. We prove that relative equilibria exist for …
A Preliminary Fuzzy Inference System For Predicting Atmospheric Ozone In An Intermountain Basin, John R. Lawson, Seth N. Lyman
A Preliminary Fuzzy Inference System For Predicting Atmospheric Ozone In An Intermountain Basin, John R. Lawson, Seth N. Lyman
Mathematics and Statistics Faculty Publications
High concentrations of ozone in the Uinta Basin, Utah, can occur after sufficient snowfall and a strong atmospheric anticyclone creates a persistent cold pool that traps emissions from oil and gas
operations, where sustained photolysis of the precursors builds ozone to unhealthy concentrations. The basin's winter-ozone system is well understood by domain experts and supported by archives of
atmospheric observations. Rules of the system can be formulated in natural language ("sufficient snowfall and high pressure leads to high ozone"), lending itself to analysis with a fuzzy-logic
inference system. This method encodes human expertise as machine intelligence in a more prescribed …
Fractional Derivative-Based Analysis Of The Heat Transfer Properties Of Fluid Flow Over A Contracting Permeable Infinite-Length Cylinder, Anas Saeb Alhasan
Fractional Derivative-Based Analysis Of The Heat Transfer Properties Of Fluid Flow Over A Contracting Permeable Infinite-Length Cylinder, Anas Saeb Alhasan
Thesis/ Dissertation Defenses
Understanding the complex interplay between the contracting behavior of the cylinder and the fluid flow dynamics has implications for the design of porous structures for heat exchange and filtration
systems. In this study, we investigate the dynamics and thermal behavior of fluid flow past a contracting permeable infinite cylinder. First, we developed a mathematical model based on the
Navier-Stokes equations to describe the fluid dynamics around the contracting permeable infinite cylinder. A new simple well-behaved definition of fractional derivative called conformable fractional
derivative introduced by authors Khalil et al. is employed to generalize the PDE’s of momentum and energy. The …
Hierarchical Neural Networks, P-Adic Pdes, And Applications To Image Processing, Wilson A. Zuniga-Galindo, B. A. Zambrano-Luna, Baboucarr Dibba
Hierarchical Neural Networks, P-Adic Pdes, And Applications To Image Processing, Wilson A. Zuniga-Galindo, B. A. Zambrano-Luna, Baboucarr Dibba
School of Mathematical and Statistical Sciences Faculty Publications and Presentations
The first goal of this article is to introduce a new type of p-adic reaction–diffusion cellular neural network with delay. We study the stability of these networks and provide numerical simulations
of their responses. The second goal is to provide a quick review of the state of the art of p-adic cellular neural networks and their applications to image processing.
Q-Rational Functions And Interpolation With Complete Nevanlinna–Pick Kernels, Daniel Alpay, Paula Cerejeiras, Uwe Kaehler, Baruch Schneider
Q-Rational Functions And Interpolation With Complete Nevanlinna–Pick Kernels, Daniel Alpay, Paula Cerejeiras, Uwe Kaehler, Baruch Schneider
Mathematics, Physics, and Computer Science Faculty Articles and Research
In this paper we introduce the concept of matrix-valued q-rational functions. In comparison to the classical case, we give different characterizations with principal emphasis on realizations and
discuss algebraic manipulations. We also study the concept of Schur multipliers and complete Nevanlinna–Pick kernels in the context of q-deformed reproducing kernel Hilbert spaces and provide first
applications in terms of an interpolation problem using Schur multipliers and complete Nevanlinna–Pick kernels.
Euler Archive Spotlight: Ed Sandifer's Influence, Erik Tou
Euler Archive Spotlight: Ed Sandifer's Influence, Erik Tou
The Euler Archive owes its existence to the work and enthusiasm of C. Edward Sandifer. In this issue, we give a history of Ed's work and influence on the creation and growth of the Euler Archive.
Translating Scientific Latin Texts With Artificial Intelligence: The Works Of Euler And Contemporaries, Sylvio R. Bistafa
Translating Scientific Latin Texts With Artificial Intelligence: The Works Of Euler And Contemporaries, Sylvio R. Bistafa
The major hindrance in the study of earlier scientific literature is the availability of Latin translations into modern languages. This is particularly true for the works of Euler who authored about
850 manuscripts and wrote a thousand letters and received back almost two thousand more. The translation of many of these manuscripts, books and letters have been published in various sources over
the last two centuries, but many more have not yet appeared. Fortunately, nowadays, artificial intelligence (AI) translation can be used to circumvent the challenges of translating such substantial
number of texts. To validate this tool, benchmark tests have …
On The Vibration Of Strings: An English Translation Of Leonhard Euler’S `Sur La Vibration Des Cordes' (E140), Reilly R. Fortune
On The Vibration Of Strings: An English Translation Of Leonhard Euler’S `Sur La Vibration Des Cordes' (E140), Reilly R. Fortune
We present an English translation of E140 - 'Sur la Vibration des cordes.'
About The Cases In Which The Formula X^4+Mxxyy+Y^4 Can Be Reduced To A Square, Georg Ehlers
About The Cases In Which The Formula X^4+Mxxyy+Y^4 Can Be Reduced To A Square, Georg Ehlers
Euler continues a previous study on the title Quartic (E696) with a new approach. His starting point here is the observation that, when the title Quartic is solved for m, the resulting fraction
becomes an integer when z=ax^2y^2-(x^2±y^2). He provides many quadratic forms for m that allow special solutions, and tables for |m|≤200. Even though the tables are known today to be incomplete, they
allow an insight into the enormous amount of work that was needed for their compilation.
The interest in the title Quartic predates Euler, and the fact that he resumed the study indicates a …
Ed Sandifer: A Running Mathematician And Mathematical Runner, Rick Cleary
Ed Sandifer: A Running Mathematician And Mathematical Runner, Rick Cleary
Historian of mathematics C. Edward Sandifer was an outstanding marathon runner as well as a first rate mathematician. This note is a review of Prof. Sandifer's athletic successes, and a look at the
attributes that he brought to both his professional and running careers.
How Ed Did It - A Memorial Conference To Honor Ed Sandifer, Lawrence D'Antonio
How Ed Did It - A Memorial Conference To Honor Ed Sandifer, Lawrence D'Antonio
I look back at the virtual conference from February 2023 that was organized to honor the memory of the historian of mathematics Ed Sandifer, who had died in August 2022. The program of the conference
is given at the end of the article. | {"url":"https://network.bepress.com/explore/physical-sciences-and-mathematics/mathematics/","timestamp":"2024-11-03T09:18:57Z","content_type":"text/html","content_length":"109559","record_id":"<urn:uuid:17a3cc51-b542-4f7f-a94b-3e0f8a7c2574>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00148.warc.gz"} |
Doing homework the hard way
In yesterday's
I linked to some
lecture notes
of Vigoda on Valiant's result. Those notes also cite a paper of Zankó. Now every paper has a story but this one is a little more interesting than most.
In my first year as an assistant professor at the University of Chicago, I taught a graduate complexity course where I gave a homework question to show that computing the permanent of a matrix A with
nonnegative integer entries is in #P. Directly constructing a nondeterministic Turing machine such that Perm(A) is the number of accepting computations of M(A) is not too difficult and that was the
approach I was looking for.
In class we had shown that computing the permanent of a zero-one matrix is in #P so Viktoria Zankó decided to reduce my homework question to this problem. She came up with a rather clever reduction
that converted a matrix A to a zero-one matrix B with Perm(A)=Perm(B). This reduction indeed answered my homework question while, unbeknownst to Zankó at the time, answered an open question of
Valiant. Thus Zankó got a publication by solving a homework problem the hard way. | {"url":"https://blog.computationalcomplexity.org/2003/03/doing-homework-hard-way.html?m=1","timestamp":"2024-11-03T16:39:53Z","content_type":"application/xhtml+xml","content_length":"49239","record_id":"<urn:uuid:99786a4e-49b4-4754-9fec-5f9d63a183c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00486.warc.gz"} |
PROPORTIONS OF LENGTHS - Numbers and Proportions - Numbers: Their Tales, Types, and Treasures
Numbers: Their Tales, Types, and Treasures.
Chapter 10: Numbers and Proportions
10.2.PROPORTIONS OF LENGTHS
Let us represent two magnitudes geometrically by lengths a and b. Consider, for example, the two line segments shown at the top of figure 10.3. How can we learn something about the relation between
the two line segments—that is, about their proportion? If possible, we would like to find the natural numbers n and m, such that the proportion a:b could be expressed as n:m.
In order to achieve this, the Pythagoreans devised the following method, which we illustrate in figure 10.3.
Figure 10.3: Determining the proportion of two line segments.
One starts by examining how often the shorter line segment b would fit into the longer one. Obviously, b fits into a twice, and then a short part r[1] of the segment a would be left over. Hence, we
write a = 2 × b + r[1], where r[1] is shorter than b. The next question would be, how often wouldr[1] fit into b? Figure 10.3 shows that, obviously, b = 1 × r[1] + r[2], with r[2] < r[1]. The next
step shows that the remainder r[2] would fit into r[1] twice, with an even smaller remainder r[3]. Finally, r[3] fits exactly twice into r[2]. That is, there is no remainder r[4], or r[4] = 0.
What have we achieved now? Obviously, we found a length r[3] (the last nonvanishing remainder) that fits a whole number of times into all previous lengths, and, therefore, a = 19r[3], and b = 7r[3].
Both line segments a and b can be expressed with the help of the small line segment r[3]; they are both integer multiples of r[3]. The line segment r[3] is a “unit” that allows us to measure a and b
simultaneously and is called the “greatest common measure” of a and b. If r[3] would be taken as the unit of length, then we would have a = 19 and b = 7. We then say that a is to b in the same
relation as 19 to 7. One says the proportion a:b equals 19:7. Today, this proportion would be considered a fraction with the numerical value of 19 divided by 7. In decimal notation this would be
(the bar over the six digits after the point indicates that they are continuously repeated).
Greek philosophers in the fifth century BCE seemed to have held the belief that this method of finding an integer proportion would actually always work for any two quantities and would come to an end
after a finite number of steps. Philosophers Leucippus and Democritus claimed that any extended continuous quantity cannot be divided infinitely. It was the birth of the theory of atomism—that is,
that any division of an extended quantity would finally terminate in atoms, which cannot be further divided. Likewise, the method shown in figure 10.3 would terminate, in the worst case, when the
remainder was the size of an atom and hence indivisible. | {"url":"https://schoolbag.info/mathematics/numbers/85.html","timestamp":"2024-11-06T10:41:32Z","content_type":"text/html","content_length":"12557","record_id":"<urn:uuid:437ee18c-fb79-4ea3-8f2f-7fbdb33cc2c2>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00679.warc.gz"} |