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Area
Stang
How many Stangs are in a Chadara?
The answer is one Chadara is equal to 0.0034294219269103 Stangs and that means we can also write it as 1 Chadara = 0.0034294219269103 Stangs. Feel free to use our online unit conversion calculator to convert the unit from Chadara to Stang. Just simply enter value 1 in Chadara and see the result in Stang.
By using our Chadara to Stang conversion tool, you know that one Chadara is equivalent to 0.0034294219269103 Stang. Hence, to convert Chadara to Stang, we just need to multiply the number by 0.0034294219269103. We are going to use very simple Chadara to Stang conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Chadara} = 1 \times 0.0034294219269103 = \text{0.0034294219269103 Stangs}$$
What is Chadara Unit of Measure?
Chadar or Chadara is a unit of measurement for area. Chadara is one of the traditional units for land measurement in Indian state of Karnataka and other southern states. One chadara is equal to 100 square feet.
What is Stang Unit of Measure?
Stang is a unit of measurement for area. It is one of the traditional Welsh units for land measurement. One stang is equal to 3240 square yards.
What is the symbol of Stang?
The symbol of Stang is stang. This means you can also write one Stang as 1 stang.
10.0034294219269103
20.0068588438538206
30.010288265780731
40.013717687707641
50.017147109634551
60.020576531561462
70.024005953488372
80.027435375415282
90.030864797342193
100.034294219269103
1000.34294219269103
10003.43
Chadara to Other Units Conversion Table
1 chadara in ankanam is equal to1.39
1 chadara in aana is equal to0.29218407596786
1 chadara in acre is equal to0.0022956820843815
1 chadara in arpent is equal to0.0027173398629323
1 chadara in are is equal to0.09290304
1 chadara in barn is equal to9.290304e+28
1 chadara in bigha [assam] is equal to0.0069444444444444
1 chadara in bigha [west bengal] is equal to0.0069444444444444
1 chadara in bigha [rajasthan] is equal to0.0036730945821855
1 chadara in bigha [bihar] is equal to0.0036737692872888
1 chadara in bigha [gujrat] is equal to0.0057392102846648
1 chadara in bigha [nepal] is equal to0.0013717421124829
1 chadara in bovate is equal to0.0001548384
1 chadara in bunder is equal to0.0009290304
1 chadara in caballeria is equal to0.00002064512
1 chadara in caballeria [cuba] is equal to0.000069227302533532
1 chadara in caballeria [spain] is equal to0.00002322576
1 chadara in carreau is equal to0.00072017860465116
1 chadara in carucate is equal to0.000019115851851852
1 chadara in cawnie is equal to0.0017204266666667
1 chadara in cent is equal to0.22956820843815
1 chadara in centiare is equal to9.29
1 chadara in circular foot is equal to127.32
1 chadara in circular inch is equal to18334.63
1 chadara in cong is equal to0.009290304
1 chadara in cover is equal to0.0034434040029652
1 chadara in cuerda is equal to0.0023639450381679
1 chadara in chatak is equal to2.22
1 chadara in decimal is equal to0.22956820843815
1 chadara in dekare is equal to0.0092903101288176
1 chadara in dismil is equal to0.22956820843815
1 chadara in dhur [tripura] is equal to27.78
1 chadara in dhur [nepal] is equal to0.54869684499314
1 chadara in dunam is equal to0.009290304
1 chadara in drone is equal to0.00036168981481481
1 chadara in fanega is equal to0.0014448373250389
1 chadara in farthingdale is equal to0.0091801422924901
1 chadara in feddan is equal to0.0022288175576519
1 chadara in ganda is equal to0.11574074074074
1 chadara in gaj is equal to11.11
1 chadara in gajam is equal to11.11
1 chadara in guntha is equal to0.091827364554637
1 chadara in ghumaon is equal to0.0022956841138659
1 chadara in ground is equal to0.041666666666667
1 chadara in hacienda is equal to1.0368642857143e-7
1 chadara in hectare is equal to0.0009290304
1 chadara in hide is equal to0.000019115851851852
1 chadara in hout is equal to0.0065366521280684
1 chadara in hundred is equal to1.9115851851852e-7
1 chadara in jerib is equal to0.0045955882352941
1 chadara in jutro is equal to0.0016143013032146
1 chadara in kanal is equal to0.018365472910927
1 chadara in kani is equal to0.005787037037037
1 chadara in kara is equal to0.46296296296296
1 chadara in kappland is equal to0.060224970828471
1 chadara in killa is equal to0.0022956841138659
1 chadara in kranta is equal to1.39
1 chadara in kuli is equal to0.69444444444444
1 chadara in kuncham is equal to0.022956841138659
1 chadara in lecha is equal to0.69444444444444
1 chadara in labor is equal to0.000012959997314953
1 chadara in legua is equal to5.1839989259811e-7
1 chadara in manzana [argentina] is equal to0.0009290304
1 chadara in manzana [costa rica] is equal to0.0013292827545157
1 chadara in marla is equal to0.36730945821855
1 chadara in morgen [germany] is equal to0.0037161216
1 chadara in morgen [south africa] is equal to0.0010844290883623
1 chadara in mu is equal to0.013935455930323
1 chadara in murabba is equal to0.000091827283375259
1 chadara in mutthi is equal to0.74074074074074
1 chadara in ngarn is equal to0.02322576
1 chadara in nali is equal to0.046296296296296
1 chadara in oxgang is equal to0.0001548384
1 chadara in paisa is equal to1.17
1 chadara in perche is equal to0.27173398629323
1 chadara in parappu is equal to0.036730913350104
1 chadara in pyong is equal to2.81
1 chadara in rai is equal to0.00580644
1 chadara in rood is equal to0.0091827364554637
1 chadara in ropani is equal to0.018261504747991
1 chadara in satak is equal to0.22956820843815
1 chadara in section is equal to0.0000035870064279155
1 chadara in sitio is equal to5.16128e-7
1 chadara in square is equal to1
1 chadara in square angstrom is equal to929030400000000000000
1 chadara in square astronomical units is equal to4.1512520173017e-22
1 chadara in square attometer is equal to9.290304e+36
1 chadara in square bicron is equal to9.290304e+24
1 chadara in square centimeter is equal to92903.04
1 chadara in square chain is equal to0.022956747098315
1 chadara in square cubit is equal to44.44
1 chadara in square decimeter is equal to929.03
1 chadara in square dekameter is equal to0.09290304
1 chadara in square digit is equal to25600
1 chadara in square exameter is equal to9.290304e-36
1 chadara in square fathom is equal to2.78
1 chadara in square femtometer is equal to9.290304e+30
1 chadara in square fermi is equal to9.290304e+30
1 chadara in square feet is equal to100
1 chadara in square furlong is equal to0.00022956820843815
1 chadara in square gigameter is equal to9.290304e-18
1 chadara in square hectometer is equal to0.0009290304
1 chadara in square inch is equal to14400
1 chadara in square league is equal to3.9855467952662e-7
1 chadara in square light year is equal to1.0379591129628e-31
1 chadara in square kilometer is equal to0.000009290304
1 chadara in square megameter is equal to9.290304e-12
1 chadara in square meter is equal to9.29
1 chadara in square microinch is equal to14399987296941000
1 chadara in square micrometer is equal to9290304000000
1 chadara in square micromicron is equal to9.290304e+24
1 chadara in square micron is equal to9290304000000
1 chadara in square mil is equal to14400000000
1 chadara in square mile is equal to0.0000035870064279155
1 chadara in square millimeter is equal to9290304
1 chadara in square nanometer is equal to9290304000000000000
1 chadara in square nautical league is equal to3.0095769444276e-7
1 chadara in square nautical mile is equal to0.000002708616860586
1 chadara in square paris foot is equal to88.06
1 chadara in square parsec is equal to9.757278878282e-33
1 chadara in perch is equal to0.36730945821855
1 chadara in square perche is equal to0.18190585456633
1 chadara in square petameter is equal to9.290304e-30
1 chadara in square picometer is equal to9.290304e+24
1 chadara in square pole is equal to0.36730945821855
1 chadara in square rod is equal to0.36730804433647
1 chadara in square terameter is equal to9.290304e-24
1 chadara in square thou is equal to14400000000
1 chadara in square yard is equal to11.11
1 chadara in square yoctometer is equal to9.290304e+48
1 chadara in square yottameter is equal to9.290304e-48
1 chadara in stang is equal to0.0034294219269103
1 chadara in stremma is equal to0.009290304
1 chadara in sarsai is equal to3.31
1 chadara in tarea is equal to0.01477465648855
1 chadara in tatami is equal to5.62
1 chadara in tonde land is equal to0.001684246555475
1 chadara in tsubo is equal to2.81
1 chadara in township is equal to9.9638979356835e-8
1 chadara in tunnland is equal to0.0018819998379386
1 chadara in vaar is equal to11.11
1 chadara in virgate is equal to0.0000774192
1 chadara in veli is equal to0.0011574074074074
1 chadara in pari is equal to0.00091827364554637
1 chadara in sangam is equal to0.0036730945821855
1 chadara in gunta is equal to0.091827364554637
1 chadara in point is equal to0.22957020325687
1 chadara in lourak is equal to0.0018365472910927
1 chadara in loukhai is equal to0.007346189164371
1 chadara in loushal is equal to0.014692378328742
1 chadara in tong is equal to0.029384756657484
1 chadara in kuzhi is equal to0.69444444444444
1 chadara in veesam is equal to11.11
1 chadara in lacham is equal to0.036730913350104
1 chadara in katha [nepal] is equal to0.027434842249657
1 chadara in katha [assam] is equal to0.034722222222222
1 chadara in katha [bihar] is equal to0.073475385745775
1 chadara in dhur [bihar] is equal to1.47
1 chadara in dhurki is equal to29.39
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority.
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# Finite dimensional subspace if a normed vector space is closed using equivalence of norms
I have shown any norms on a finite dimensional real vector space are equivalent then the question asks why would this imply every finite-dimensional subspace of normed vector space is closed. (Closed in the sense that it is toplogically closed, its complement is an open subset.)
I understand that equivalent norms yield the same notion of convergence however I have very few ideas on where to start. I have seen a few posts showing the subspace is complete instead, but I do not think it is in the spirit of this problem.
How should I proceed? Many thanks in advance!
• Think of completeness. Aug 20 '20 at 18:19
• @DanielFischer Thank you! Just confirming, are you referring to showing the normed vector space (the Ambient space) is complete first and then using complete subset of a complete space is closed? Aug 20 '20 at 20:21
• The ambient space need not be complete. But a complete subset of a metric space is closed, whether the ambient space is complete or not. Aug 20 '20 at 20:23
I know that if $$X$$ is normed space over some field $$\mathbb{F}$$ and finite-dimensional with dimension $$n$$, so you can prove $$X$$ is isomorphic to $$\mathbb{F}^{n}$$ with the euclidean norm. $$[1]$$
A collorary of above result is that if $$X$$ be a finite dimensional vector space with norms $$||\cdot||_{1}$$ and $$||\cdot||_{2}$$. Then $$||\cdot||_{1}$$ and $$||\cdot||_{2}$$ are equivalent.
Now, if you can prove that result $$[1]$$ then you have that any finite dimensional subspace of normed linear space is closed.
• I think you need result $[1]$ to prove the equivalence of the norms. In that case both results that I mentioned are colloraries of $[1]$. Aug 20 '20 at 16:58
• @JustWandering How do you showed that two norms in finite-$\mathbb{R}$-vector space dimensional are equivalent? Aug 20 '20 at 17:03
• I used the fact that they are both equivalent to $d_1$ norm and that required some tricks in compactness and topological properties such that a continuous function on a compact set attains its bounds etc. Aug 20 '20 at 17:07
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# Calculating birthday probabilities with R instead of math
Probability math is hard. Use brute force simulation to find the probability that a household has a cluster of birthdays.
r
tidyverse
simulations
probability
Author
Published
Friday, May 3, 2024
Even though I’ve been teaching R and statistical programming since 2017, and despite the fact that I do all sorts of heavily quantitative research, I’m really really bad at probability math.
The last time I truly had to do set theory and probability math was in my first PhD-level stats class in 2012. The professor cancelled classes after the first month and gave us all of October to re-teach ourselves calculus and probability theory (thank you Sal Khan), and then the rest of the class was pretty much all about pure set theory stuff. It was… not fun.
But I learned a valuable secret power from the class. During the final couple weeks of the course, the professor mentioned in passing that it’s possible to skip most of this probability math and instead use simulations to get the same answers. That one throwaway comment changed my whole approach to doing anything based on probabilities.
## Why simulate?
In one problem set from November 20121, we had to answer this question using both actual probability math and R simulation:
1 I wrote this in a .Rnw file! R Markdown wasn’t even a thing yet!
An urn contains 10 red balls, 10 blue balls, and 20 green balls. If 5 balls are selected at random without replacement, what is the probability that at least 1 ball of each color will be selected?
### ew probability math
We can find this probability by finding the probability of not selecting one or more of the colors in the draw and subtracting it from 1. We need to find the probability of selecting no red balls, no blue balls, and no green balls, and then subtract the probability of the overlapping situations (i.e. no red or blue balls, no red or green balls, and no blue or green balls).
To do this, we can use n-choose-k notation from combinatorics to represent the number of choices from a pool of possible combinations. This notation looks like this:
$\dbinom{n}{k} = \frac{n!}{k! (n-k)!}$
If we’re selecting 5 balls from a pool of 40, we can say “40 choose 5”, or $$\binom{40}{5}$$. To calculate that, we get this gross mess:
\begin{aligned} \dbinom{40}{5} &= \frac{40!}{5! (40-5)!} \\[10pt] &= \frac{40!}{5!\ 35!} \\[10pt] &= \frac{40 \times 39 \times 38 \times 37 \times 36}{5!} \qquad \text{(cancel out the 35!)} \\[10pt] &= \frac{78,960,960}{120} \\[10pt] &= 658,008 \end{aligned}
Or we can do it with R:
choose(40, 5)
## [1] 658008
So with this binomial choose notation, we can calculate the official formal probability of drawing at least one red, blue, and green ball from this urn:
\begin{aligned} \text{Pr}(\text{at least one red, blue, and green}) &= 1 - \frac{\text{Ways to get no red or no blue or no green}}{\text{Ways to get 5 balls from 40}} \\[10pt] &= 1 - \dfrac{ \begin{array}{@{}c@{}} (\text{Ways to get no red}) + (\text{Ways to get no blue}) + (\text{Ways to get no green}) - \\ (\text{Ways to get no red or blue}) - (\text{Ways to get no red or green}) - (\text{Ways to get no blue or green}) \end{array} }{\text{Ways to get 5 balls from 40}} \\[10pt] &= 1 - \dfrac{ \dbinom{30}{5} + \dbinom{30}{5} + \dbinom{20}{5} - \dbinom{20}{5} - \dbinom{10}{5} - \dbinom{10}{5} }{ \dbinom{40}{5} } \end{aligned}
If we really really wanted, we could then calculate all of that by hand, but ew.
We can just use R instead:
# Ways to draw 5 balls without getting a specific color
no_red <- choose(30, 5)
no_blue <- choose(30, 5)
no_green <- choose(20, 5)
# Ways to draw 5 balls without getting two specific colors
no_red_blue <- choose(20, 5)
no_red_green <- choose(10, 5)
no_blue_green <- choose(10, 5)
# Ways to draw 5 balls in general
total_ways <- choose(40, 5)
# Probability of drawing at least 1 of each color
prob_real <- 1 - (no_red + no_blue + no_green - no_red_blue - no_red_green - no_blue_green) / total_ways
prob_real
## [1] 0.5676
Great. There’s a 56.76% chance of drawing at least one of each color. We have an answer, but this was really hard, and I could only do it because I dug up my old problem sets from 2012.
### yay brute force simulation
I really don’t like formal probability math. Fortunately there’s a way I find a heck of a lot easier to use. Brute force simulation.
Instead of figuring out all these weird n-choose-k probabilities, we’ll use the power of computers to literally draw from a hypothetical urn over and over and over again until we come to the right answer.
Here’s one way to do it:2
2 Again, this is 2012-era R code; nowadays I’d forgo the loop and use something like purrr::map() or sapply().
# Make this randomness consistent
set.seed(12345)
# Make an urn with balls in it
urn <- c(rep('red', 10), rep('blue', 10), rep('green', 20))
# How many times we'll draw from the urn
simulations <- 100000
count <- 0
for (i in 1:simulations) {
# Pick 5 balls from the urn
draw <- sample(urn, 5)
# See if there's a red, blue, and green; if so, record it
if ('red' %in% draw && 'blue' %in% draw && 'green' %in% draw) {
count <- count + 1
}
}
# Find the simulated probability
prob_simulated <- count / simulations
prob_simulated
## [1] 0.5681
Sweet. The simulation spat out 0.5681, which is shockingly close to 0.5676. If we boosted the number of simulations from 100,000 to something even higher,3 we’d eventually converge on the true answer.
3 Going up to 2,000,000 got me to 0.5676.
I use this simulation-based approach to anything mathy as much as I can. Personally, I find it far more intuitive to re-create the data generating process rather than think in set theory and combinatorics. In my program evaluation class, we do an in-class activity with the dice game Zilch where we figure out the probability of scoring something in a given dice roll. Instead of finding real probabilities, we just simulate thousands of dice rolls and mark if something was rolled. We essentially recreate the exact data generating process.
This approach is also the core of modern Bayesian statistics. Calculating complex integrals to find posterior distributions is too hard, so we can use Markov Chain Monte Carlo (MCMC) processes bounce around the plausible space for a posterior distribution until they settle on a stable value.
## Birthday probabilities
A couple days ago, I came across this post on Bluesky:
This is neat because it’s also the case in my household. We have “birthday season” from May to November, and have a dearth of birthdays from November to May. They’re all clustered in half the year. I’d never thought about how unlikely that was.
There’s probably some formal probability math that can answer Karl’s question precisely, but that’s hard. So instead, I decided to figure this out with my old friend—brute force simulation.
The data generating process is a little more complicated than just drawing balls from urns, and the cyclical nature of calendars adds an extra wrinkle to simulating everything, but it’s doable (and fun!), so I figured I’d share the details of the simulation process here. And with the recent release of {ggplot2} 3.5 and its new coord_radial() and legend placement settings and point-based text sizing and absolute plot-based positioning, I figured I’d make some pretty plots along the way.
Let’s load some libraries, make a custom theme, and get started!
library(tidyverse)
library(ggtext)
library(patchwork)
clrs <- MetBrewer::met.brewer("Demuth")
# Custom ggplot theme to make pretty plots
# Get the font at https://fonts.google.com/specimen/Montserrat
theme_calendar <- function() {
theme_minimal(base_family = "Montserrat") +
theme(
axis.text.y = element_blank(),
axis.title = element_blank(),
panel.grid.minor = element_blank(),
plot.title = element_text(face = "bold", hjust = 0.5),
plot.subtitle = element_text(hjust = 0.5)
)
}
update_geom_defaults("text", list(family = "Montserrat"))
## Visualizing birthday distributions and spans
### All birthdays within 6-month span
First, let’s work with a hypothetical household with four people in it with birthdays on January 4, March 10, April 28, and May 21. We’ll plot these on a radial plot and add a 6-month span starting at the first birthday:
# All these happen within a 6-month span
birthdays_yes <- c(
ymd("2024-01-04"),
ymd("2024-03-10"),
ymd("2024-04-28"),
ymd("2024-05-21")
)
tibble(x = birthdays_yes) |>
ggplot(aes(x = x, y = "")) +
annotate(
geom = "segment",
x = birthdays_yes[1], xend = birthdays_yes[1] + months(6), y = "",
linewidth = 3, color = clrs[5]) +
geom_point(size = 5, fill = clrs[10], color = "white", pch = 21) +
annotate(
"text",
label = "Yep", fontface = "bold",
x = I(0.5), y = I(0),
size = 14, size.unit = "pt"
) +
scale_x_date(
date_breaks = "1 month", date_labels = "%B",
limits = c(ymd("2024-01-01"), ymd("2024-12-31")),
expand = expansion(0, 0)
) +
scale_y_discrete(expand = expansion(add = c(0, 1))) +
theme_calendar()
The four birthdays all fit comfortably within the 6-month span. Neat.
### All birthdays within 6-month span, but tricky
Next, let’s change the May birthday to December 1. These four birthdays still all fit within a 6-month span, but it’s trickier to see because the calendar year resets in the middle. Earlier, we plotted the yellow span with annotate(), but if we do that now, it breaks and we get a warning. We can’t draw a line segment from December 1 to six months later:
birthdays_yes_but_tricky <- c(
ymd("2024-01-04"),
ymd("2024-03-10"),
ymd("2024-04-28"),
ymd("2024-12-01")
)
tibble(x = birthdays_yes_but_tricky) |>
ggplot(aes(x = x, y = "")) +
annotate(
geom = "segment",
x = birthdays_yes_but_tricky[4],
xend = birthdays_yes_but_tricky[4] + months(6),
y = "",
linewidth = 3, color = clrs[5]) +
geom_point(size = 5, fill = clrs[10], color = "white", pch = 21) +
annotate(
"text",
label = "Yep\n(but broken)", fontface = "bold",
x = I(0.5), y = I(0),
size = 14, size.unit = "pt"
) +
scale_x_date(
date_breaks = "1 month", date_labels = "%B",
limits = c(ymd("2024-01-01"), ymd("2024-12-31")),
expand = expansion(0, 0)
) +
scale_y_discrete(expand = expansion(add = c(0, 1))) +
theme_calendar()
## Warning: Removed 1 row containing missing values or values outside the scale range (geom_segment()).
Instead, we can draw two line segments—one from December 1 to December 31, and one from January 1 to whatever six months from December 1 is. Since this plot represents all of 2024, we’ll force the continued time after January 1 to also be in 2024 (even though it’s technically 2025). Here I colored the segments a little differently to highlight the fact that they’re two separate lines:
tibble(x = birthdays_yes_but_tricky) |>
ggplot(aes(x = x, y = "")) +
annotate(
geom = "segment",
x = birthdays_yes_but_tricky[4],
xend = ymd("2024-12-31"),
y = "",
linewidth = 3, color = clrs[5]) +
annotate(
geom = "segment",
x = ymd("2024-01-01"),
xend = (ymd("2024-01-01") + months(6)) -
(ymd("2024-12-31") - birthdays_yes_but_tricky[4]),
y = "",
linewidth = 3, color = clrs[4]) +
geom_point(size = 5, fill = clrs[10], color = "white", pch = 21) +
annotate(
"text",
label = "Yep\n(but tricky)", fontface = "bold",
x = I(0.5), y = I(0),
size = 14, size.unit = "pt"
) +
scale_x_date(
date_breaks = "1 month", date_labels = "%B",
limits = c(ymd("2024-01-01"), ymd("2024-12-31")),
expand = expansion(0, 0)
) +
scale_y_discrete(expand = expansion(add = c(0, 1))) +
theme_calendar()
Writing two separate annotate() layers feels repetitive, though, and it’s easy to make mistakes. So instead, we can make a little helper function that will create a data frame with the start and end date of a six-month span. If the span crosses December 31, it returns two spans; if not, it returns one span:
calc_date_arc <- function(start_date) {
days_till_end <- ymd("2024-12-31") - start_date
if (days_till_end >= months(6)) {
x <- start_date
xend <- start_date + months(6)
} else {
x <- c(start_date, ymd("2024-01-01"))
xend <- c(
start_date + days_till_end,
(ymd("2024-01-01") + months(6)) - days_till_end
)
}
return(tibble(x = x, xend = xend))
}
Let’s make sure it works. Six months from March 15 is September 15, which doesn’t cross into a new year, so we get just one start and end date:
calc_date_arc(ymd("2024-03-15"))
## # A tibble: 1 × 2
## x xend
## <date> <date>
## 1 2024-03-15 2024-09-15
Six months from November 15 is sometime in May, which means we do cross into a new year. We thus get two spans: (1) a segment from November to the end of December, and (2) a segment from January to May:
calc_date_arc(ymd("2024-11-15"))
## # A tibble: 2 × 2
## x xend
## <date> <date>
## 1 2024-11-15 2024-12-31
## 2 2024-01-01 2024-05-16
Plotting with this function is a lot easier, since it returns a data frame. We don’t need to worry about using annotate() anymore and can instead map the x and xend aesthetics from the data to the plot:
tibble(x = birthdays_yes_but_tricky) |>
ggplot(aes(x = x, y = "")) +
geom_segment(
data = calc_date_arc(birthdays_yes_but_tricky[4]), aes(xend = xend),
linewidth = 3, color = clrs[5]
) +
geom_point(size = 5, fill = clrs[10], color = "white", pch = 21) +
annotate(
"text",
label = "Yep\n(but tricky)", fontface = "bold",
x = I(0.5), y = I(0),
size = 14, size.unit = "pt"
) +
scale_x_date(
date_breaks = "1 month", date_labels = "%B",
limits = c(ymd("2024-01-01"), ymd("2024-12-31")),
expand = expansion(0, 0)
) +
scale_y_discrete(expand = expansion(add = c(0, 1))) +
theme_calendar()
### Some birthdays outside a 6-month span
Let’s change the set of birthdays again so that one of them falls outside the six-month window. Regardless of where we start the span, we can’t collect all the points within a continuous six-month period:
birthdays_no <- c(
ymd("2024-01-04"),
ymd("2024-03-10"),
ymd("2024-04-28"),
ymd("2024-09-21")
)
p1 <- tibble(x = birthdays_no) |>
ggplot(aes(x = x, y = "")) +
geom_segment(
data = calc_date_arc(birthdays_no[1]), aes(xend = xend),
linewidth = 3, color = clrs[3]
) +
geom_point(size = 5, fill = clrs[10], color = "white", pch = 21) +
annotate(
"text",
label = "Nope", fontface = "bold",
x = I(0.5), y = I(0),
size = 14, size.unit = "pt"
) +
scale_x_date(
date_breaks = "1 month", date_labels = "%B",
limits = c(ymd("2024-01-01"), ymd("2024-12-31")),
expand = expansion(0, 0)
) +
scale_y_discrete(expand = expansion(add = c(0, 1))) +
theme_calendar()
p2 <- tibble(x = birthdays_no) |>
ggplot(aes(x = x, y = "")) +
geom_segment(
data = calc_date_arc(birthdays_no[4]), aes(xend = xend),
linewidth = 3, color = clrs[3]
) +
geom_point(size = 5, fill = clrs[10], color = "white", pch = 21) +
annotate(
"text",
label = "Still nope", fontface = "bold",
x = I(0.5), y = I(0),
size = 14, size.unit = "pt"
) +
scale_x_date(
date_breaks = "1 month", date_labels = "%B",
limits = c(ymd("2024-01-01"), ymd("2024-12-31")),
expand = expansion(0, 0)
) +
scale_y_discrete(expand = expansion(add = c(0, 1))) +
theme_calendar()
(p1 | plot_spacer() | p2) +
plot_layout(widths = c(0.45, 0.1, 0.45))
## Simulation time!
So far, we’ve been working with a hypothetical household of 4, with arbitrarily chosen birthdays. For the simulation, we’ll need to work with randomly selected birthdays for households of varying sizes.
### Counting simulated birthdays and measuring spans
But before we build the full simulation, we need to build a way to programmatically detect if a set of dates fit within a six-month span, which—as we saw with the plotting—is surprisingly tricky because of the possible change in year.
If we didn’t need to contend with a change in year, we could convert all the birthdays to their corresponding day of the year, sort them, find the difference between the first and the last, and see if it’s less than 183 days (366/2; we’re working with a leap year). This would work great:
birthdays_yes <- c(
ymd("2024-01-04"),
ymd("2024-03-10"),
ymd("2024-04-28"),
ymd("2024-05-21")
)
birthdays_sorted <- sort(yday(birthdays_yes))
birthdays_sorted
## [1] 4 70 119 142
max(birthdays_sorted) - min(birthdays_sorted)
## [1] 138
But time is a circle. If we look at a set of birthdays that crosses a new year, we can’t just look at max - min. Also, there’s no guarantee that the six-month span will go from the first to the last; it could go from the last to the first.
birthdays_yes_but_tricky <- c(
ymd("2024-01-04"),
ymd("2024-03-10"),
ymd("2024-04-28"),
ymd("2024-12-01")
)
birthdays_sorted <- sort(yday(birthdays_yes_but_tricky))
birthdays_sorted
## [1] 4 70 119 336
max(birthdays_sorted) - min(birthdays_sorted)
## [1] 332
So we need to do something a little trickier to account for the year looping over. There are several potential ways to do this, and there’s no one right way. Here’s the approach I settled on.
We need to check the span between each possible window of dates. In a household of 4, this means finding the distance between the 4th (or last) date and the 1st date, which we did with max - min. But we also need to check on the distance between the 1st date in the next cycle (i.e. 2025) and the 2nd date (in 2024), and so on, or this:
• Distance between 4th and 1st:
• ymd("2024-12-01") - ymd("2024-01-04") or 332 days
• Distance between (1st + 1 year) and 2nd:
• ymd("2025-01-04") - ymd("2024-03-10") or 300 days
• Distance between (2nd + 1 year) and 3rd:
• ymd("2025-03-10") - ymd("2024-04-28") or 316 days
• Distance between (3rd + 1 year) and 4th:
• ymd("2025-04-28") - ymd("2024-12-01") or 148 days
That last one, from December 1 to April 28, is less than 180 days, which means that the dates fit within a six-month span. We saw this in the plot earlier too—if we start the span in December, it more than covers the remaining birthdays.
One easy way to look at dates in the next year is to double up the vector of birthday days-of-the-year, adding 366 to the first set, like this:
birthdays_doubled <- c(
sort(yday(birthdays_yes_but_tricky)),
sort(yday(birthdays_yes_but_tricky)) + 366
)
birthdays_doubled
## [1] 4 70 119 336 370 436 485 702
The first four represent the regular real birthdays; the next four are the same values, just shifted up a year (so 4 and 370 are both January 1, etc.)
With this vector, we can now find differences between dates that cross years more easily:4
4 Some of these differences aren’t the same as before (317 instead of 361; 149 instead of 148). This is because 2024 is a leap year and 2025 is not, and {lubridate} accounts for that. By adding 366, we’re pretending 2025 is also a leap year. But that’s okay, because we want to pretend that February 29 happens each year.
birthdays_doubled[4] - birthdays_doubled[1]
## [1] 332
birthdays_doubled[5] - birthdays_doubled[2]
## [1] 300
birthdays_doubled[6] - birthdays_doubled[3]
## [1] 317
birthdays_doubled[7] - birthdays_doubled[4]
## [1] 149
Or instead of repeating lots of lines like that, we can auto-increment the different indices:
n <- length(birthdays_yes_but_tricky)
map_dbl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x])
## [1] 332 300 317 149
If any of those values are less than 183, the birthdays fit in a six-month span:
any(map_lgl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x] <= 183))
## [1] TRUE
Let’s check it with the other two test sets of birthdays. Here’s the easy set of birthdays without any cross-year loops:
birthdays_yes <- c(
ymd("2024-01-04"),
ymd("2024-03-10"),
ymd("2024-04-28"),
ymd("2024-05-21")
)
birthdays_doubled <- c(
sort(yday(birthdays_yes)),
sort(yday(birthdays_yes)) + 366
)
map_dbl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x])
## [1] 138 300 317 343
any(map_lgl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x] <= 183))
## [1] TRUE
And here’s the set we know doesn’t fit:
birthdays_no <- c(
ymd("2024-01-04"),
ymd("2024-03-10"),
ymd("2024-04-28"),
ymd("2024-09-21")
)
birthdays_doubled <- c(
sort(yday(birthdays_no)),
sort(yday(birthdays_no)) + 366
)
map_dbl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x])
## [1] 261 300 317 220
any(map_lgl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x] <= 183))
## [1] FALSE
It works!
### Actual simulation
Now that we have the ability to check if any set of dates fits within six months, we can generalize this to any household size. To make life a little easier, we’ll stop working with days of the year for. Leap years are tricky, and the results changed a little bit above if we added 366 or 365 to the repeated years. So instead, we’ll think about 360° in a circle—circles don’t suddenly have 361° every four years or anything weird like that.
In this simulation, we’ll generate n random numbers between 0 and 360 (where n is the household size we’re interested in). We’ll then do the doubling and sorting thing and check to see if the distance between any of the 4-number spans is less than 180.
simulate_prob <- function(n, num_simulations = 1000) {
results <- map_lgl(1:num_simulations, ~{
birthdays <- runif(n, 0, 360)
birthdays_doubled <- sort(c(birthdays, birthdays + 360))
any(map_lgl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x] <= 180))
})
mean(results)
}
Here’s the probability of seeing all the birthdays in a six-month span in a household of 4:
withr::with_seed(1234, {
simulate_prob(4)
})
## [1] 0.522
What about a household of 6, like in Karl’s original post?
withr::with_seed(1234, {
simulate_prob(6)
})
## [1] 0.189
We can get this more precise and consistent by boosting the number of simulations:
withr::with_seed(1234, {
simulate_prob(6, 50000)
})
## [1] 0.1866
Now that this function is working, we can use it to simulate a bunch of possible household sizes, like from 2 to 10:
simulated_households <- tibble(household_size = 2:10) |>
mutate(prob_in_arc = map_dbl(household_size, ~simulate_prob(.x, 10000))) |>
mutate(nice_prob = scales::label_percent(accuracy = 0.1)(prob_in_arc))
ggplot(simulated_households, aes(x = factor(household_size), y = prob_in_arc)) +
geom_pointrange(aes(ymin = 0, ymax = prob_in_arc), color = clrs[2]) +
geom_text(aes(label = nice_prob), nudge_y = 0.07, size = 8, size.unit = "pt") +
scale_y_continuous(labels = scales::label_percent()) +
labs(
x = "Household size",
y = "Probability",
title = "Probability that all birthdays occur within \na single six-month span across household size",
caption = "10,000 simulations"
) +
theme_minimal(base_family = "Montserrat") +
theme(
panel.grid.minor = element_blank(),
panel.grid.major.x = element_blank(),
plot.caption = element_text(hjust = 0, color = "grey50"),
axis.title.x = element_text(hjust = 0),
axis.title.y = element_text(hjust = 1)
)
## But birthdays aren’t uniformly distributed!
We just answered Karl’s original question: “Suppose n points are uniformly distributed on a circle. What is the probability that they belong to a connected half circle”. There’s probably some official mathy combinatorial way to build a real formula to describe this pattern, but that’s too hard. Simulation gets us there.
### Uneven birthday disributions
But in real life, birthdays aren’t actually normally distributed. There are some fascinating patterns in days of birth. Instead of drawing birthdays from a uniform distribution where every day is equally likely, let’s draw from the actual distribution. There’s no official probability-math way to do this—the only way to do this kind of calculation is with simulation.
The CDC and the Social Security Administration track the counts of daily births in the Unitd States. In 2016, FiveThirtyEight reported a story about patterns in daily birthrate frequencies and they posted their CSV files on GitHub, so we’ll load their data and figure out daily probabilities of birthdays.
births_1994_1999 <- read_csv(
"https://raw.githubusercontent.com/fivethirtyeight/data/master/births/US_births_1994-2003_CDC_NCHS.csv"
) |>
# Ignore anything after 2000
filter(year < 2000)
"https://raw.githubusercontent.com/fivethirtyeight/data/master/births/US_births_2000-2014_SSA.csv"
)
births_combined <- bind_rows(births_1994_1999, births_2000_2014) |>
mutate(
full_date = make_date(year = 2024, month = month, day = date_of_month),
day_of_year = yday(full_date)
) |>
mutate(
month_cateogrical = month(full_date, label = TRUE, abbr = FALSE)
)
glimpse(births_combined)
## Rows: 7,670
## Columns: 8
## $year <dbl> 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 1994, 19… ##$ month <dbl> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4,…
## $date_of_month <dbl> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,… ##$ day_of_week <dbl> 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5,…
## $births <dbl> 8096, 7772, 10142, 11248, 11053, 11406, 11251, 8653, 7910, 10498, 11706, 11567, 11212, 11570, 8660, 8123, 10567, 11541, 11257, 11682, 11811, 8833, 8310, 11125, 11981, 11514, 11702, 11666, 8988, 8096, 10765, 11755, 11483, 11523, 11677, 8991, 8309, 10984, 12152, 11515, 1162… ##$ full_date <date> 2024-01-01, 2024-01-02, 2024-01-03, 2024-01-04, 2024-01-05, 2024-01-06, 2024-01-07, 2024-01-08, 2024-01-09, 2024-01-10, 2024-01-11, 2024-01-12, 2024-01-13, 2024-01-14, 2024-01-15, 2024-01-16, 2024-01-17, 2024-01-18, 2024-01-19, 2024-01-20, 2024-01-21, 2024-01-22, 2024-01…
## $day_of_year <dbl> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 7… ##$ month_cateogrical <ord> January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, January, Ja…
But first, just because this is one of my favorite graphs ever, let’s visualize the data! Here’s a heatmap showing the daily average births for 366 days:
avg_births_month_day <- births_combined |>
group_by(month_cateogrical, date_of_month) %>%
summarize(avg_births = mean(births))
ggplot(
avg_births_month_day,
aes(x = factor(date_of_month), y = fct_rev(month_cateogrical), fill = avg_births)
) +
geom_tile() +
scale_fill_viridis_c(
option = "rocket", labels = scales::label_comma(),
guide = guide_colorbar(barwidth = 20, barheight = 0.5, position = "bottom")
) +
labs(
x = NULL, y = NULL,
title = "Average births per day",
subtitle = "1994–2014",
fill = "Average births"
) +
coord_equal() +
theme_minimal(base_family = "Montserrat") +
theme(
legend.justification.bottom = "left",
legend.title.position = "top",
panel.grid = element_blank(),
axis.title.x = element_text(hjust = 0)
)
There are some really fascinating stories here!
• Nobody wants to have babies during Christmas or New Year’s. Christmas Day, Christmas Eve, and New Year’s Day seem to have the lowest average births.
• New Year’s Eve, Halloween, July 4, April 1,5 and the whole week of Thanksgiving6 also have really low averages.
• The 13th of every month has slightly fewer births than average—the column at the 13th is really obvious here.
• The days with the highest average counts are in mid-September, from the 9th to the 20th—except for September 11.
5 No one wants joke babies?
6 American Thanksgiving is the fourth Thursday of November, so the exact day of the month moves around each year
With this data, we can calculate the daily probability of having a birthday:
prob_per_day <- births_combined |>
group_by(day_of_year) |>
summarize(total = sum(births)) |>
mutate(prob = total / sum(total)) |>
mutate(full_date = ymd("2024-01-01") + days(day_of_year - 1)) |>
mutate(yearless_date = format(full_date, "%B %d"))
glimpse(prob_per_day)
## Rows: 366
## Columns: 5
## $day_of_year <dbl> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 7… ##$ total <dbl> 164362, 196481, 228259, 232354, 230826, 229776, 230233, 223766, 224116, 232521, 231414, 230499, 223803, 231343, 222374, 224055, 230038, 229465, 225387, 228138, 228190, 225098, 229163, 233021, 230897, 228394, 228012, 228365, 222831, 226669, 229472, 230364, 230984, 228433, 2298…
## $prob <dbl> 0.0019176, 0.0022923, 0.0026631, 0.0027108, 0.0026930, 0.0026808, 0.0026861, 0.0026107, 0.0026147, 0.0027128, 0.0026999, 0.0026892, 0.0026111, 0.0026991, 0.0025944, 0.0026140, 0.0026838, 0.0026771, 0.0026296, 0.0026617, 0.0026623, 0.0026262, 0.0026736, 0.0027186, 0.0026938, 0… ##$ full_date <date> 2024-01-01, 2024-01-02, 2024-01-03, 2024-01-04, 2024-01-05, 2024-01-06, 2024-01-07, 2024-01-08, 2024-01-09, 2024-01-10, 2024-01-11, 2024-01-12, 2024-01-13, 2024-01-14, 2024-01-15, 2024-01-16, 2024-01-17, 2024-01-18, 2024-01-19, 2024-01-20, 2024-01-21, 2024-01-22, 2024-01-23,…
## $yearless_date <chr> "January 01", "January 02", "January 03", "January 04", "January 05", "January 06", "January 07", "January 08", "January 09", "January 10", "January 11", "January 12", "January 13", "January 14", "January 15", "January 16", "January 17", "January 18", "January 19", "January 2… Here are the 5 most common days:7 7 There’s me on September 19. prob_per_day |> select(yearless_date, prob) |> slice_max(order_by = prob, n = 5) ## # A tibble: 5 × 2 ## yearless_date prob ## <chr> <dbl> ## 1 September 09 0.00302 ## 2 September 19 0.00301 ## 3 September 12 0.00301 ## 4 September 17 0.00299 ## 5 September 10 0.00299 And the 5 least probable days—Leap Day, Christmas Day, New Year’s Day, Christmas Eve, and July 4th: prob_per_day |> select(yearless_date, prob) |> slice_min(order_by = prob, n = 5) ## # A tibble: 5 × 2 ## yearless_date prob ## <chr> <dbl> ## 1 February 29 0.000613 ## 2 December 25 0.00162 ## 3 January 01 0.00192 ## 4 December 24 0.00199 ## 5 July 04 0.00216 ### Using actual birthday probabilities Instead of drawing random numbers between 0 and 360 from a uniform distribution, we can draw day-of-the-year numbers. This is easy with sample(). Here’s a random 4-person household: withr::with_seed(1234, { sample(1:366, size = 4, replace = TRUE) }) ## [1] 284 336 101 111 That gives us a uniform probability distribution—all the numbers between 1 and 366 are equally likely. sample() has a prob argument that we can use to feed a vector of probabilities: withr::with_seed(1234, { sample( prob_per_day$day_of_year,
size = 4,
replace = TRUE,
prob = prob_per_day$prob) }) ## [1] 284 101 111 133 These days of the year now match the actual distribution of birthdays in the United States. If we simulated thousands of birthdays, we’d get more in September, fewer on the 13th of each month, and far fewer around Thanksgiving and Christmas. We can now update our simulation to use this more realistic distribution of birthdays: simulate_prob_real <- function(n, num_simulations = 1000) { results <- map_lgl(1:num_simulations, ~{ birthdays <- sample( prob_per_day$day_of_year,
size = n,
replace = TRUE,
prob = prob_per_day\$prob
)
birthdays_doubled <- sort(c(birthdays, birthdays + 366))
any(map_lgl(1:n, ~ birthdays_doubled[.x + n - 1] - birthdays_doubled[.x] <= (366 / 2)))
})
mean(results)
}
Here’s the probability of having all the birthdays within the same six months for a household of 4:
withr::with_seed(1234, {
simulate_prob_real(4)
})
## [1] 0.495
And 6:
withr::with_seed(1234, {
simulate_prob_real(6)
})
## [1] 0.202
And here’s the probability across different household sizes:
sims_real <- tibble(household_size = 2:10) |>
mutate(prob_in_arc = map_dbl(household_size, ~simulate_prob_real(.x, 10000))) |>
mutate(nice_prob = scales::label_percent(accuracy = 0.1)(prob_in_arc))
ggplot(sims_real, aes(x = factor(household_size), y = prob_in_arc)) +
geom_pointrange(aes(ymin = 0, ymax = prob_in_arc), color = clrs[9]) +
geom_text(aes(label = nice_prob), nudge_y = 0.07, size = 8, size.unit = "pt") +
scale_y_continuous(labels = scales::label_percent()) +
labs(
x = "Household size",
y = "Probability",
title = "Probability that all birthdays occur within a\nsingle 6-month span across household size",
subtitle = "Based on average daily birth probabilities from 1994–2014",
caption = "10,000 simulations; daily probabilities from the CDC and SSA"
) +
theme_minimal(base_family = "Montserrat") +
theme(
panel.grid.minor = element_blank(),
panel.grid.major.x = element_blank(),
plot.caption = element_text(hjust = 0, color = "grey50"),
plot.subtitle = element_text(hjust = 0, color = "grey50"),
axis.title.x = element_text(hjust = 0),
axis.title.y = element_text(hjust = 1)
)
In the end, these are all roughly the same as the uniform birthday distribution, but it feels more accurate since the probabilities are based on real-life frequencies.
But most importantly, we didn’t have to do any math to get the right answer. Brute force simulation techniques got us there.
## Citation
BibTeX citation:
@online{heiss2024,
author = {Heiss, Andrew},
title = {Calculating Birthday Probabilities with {R} Instead of Math},
date = {2024-05-03},
url = {https://www.andrewheiss.com/blog/2024/05/03/birthday-spans-simulation-sans-math},
doi = {10.59350/r419r-zqj73},
langid = {en}
}
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https://gmatclub.com/forum/three-business-partners-q-r-and-s-agree-to-divide-their-167212.html?kudos=1
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# Three business partners, Q, R, and S, agree to divide their
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Three business partners, Q, R, and S, agree to divide their [#permalink]
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The Official Guide For GMAT® Quantitative Review, 2ND Edition
Three business partners, Q, R, and S, agree to divide their total profit for a certain year in the ratios 2:5:8, respectively. If Q's share was $4,000, what was the total profit of the business partners for the year? (A)$26,000
(B) $30,000 (C)$52,000
(D) $60,000 (E)$300,000
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Re: Three business partners, Q, R, and S, agree to divide their [#permalink]
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07 Feb 2014, 08:11
4
IMO - B
If Q's share is $4,000 and Q get 2x, then 2x = 4000, x = 2000. If R gets 5, then 5x = 10000. If S gets 8, then 8x = 16000 Q + R + S = 4000 + 10000 + 16000 = 30000 _________________ GGG (Gym / GMAT / Girl) -- Be Serious Its your duty to post OA afterwards; some one must be waiting for that... Manager Joined: 11 Jan 2014 Posts: 85 Concentration: Finance, Statistics GMAT Date: 03-04-2014 GPA: 3.77 WE: Analyst (Retail Banking) Re: Three business partners, Q, R, and S, agree to divide their [#permalink] ### Show Tags 07 Feb 2014, 09:40 4 I'd say (B), too. 2:5:8 -> 2x+5x+8x = 15x. Since Q's share is 4,000. 2x=4,000 -> x=2,000. 15*(2,000) = 30,000. Math Expert Joined: 02 Sep 2009 Posts: 61211 Re: Three business partners, Q, R, and S, agree to divide their [#permalink] ### Show Tags 08 Feb 2014, 03:47 1 SOLUTION Three business partners, Q, R, and S, agree to divide their total profit for a certain year in the ratios 2:5:8, respectively. If Q's share was$4,000, what was the total profit of the business partners for the year?
(A) $26,000 (B)$30,000
(C) $52,000 (D)$60,000
(E) $300,000 The total profit was divided in the ratio Q:R:S = 2:5:8, hence Q's share was 2/(2 + 5 + 8) = 2/15 of the total profit. Q's share was$4,000, thus 2/15*{total} = 4,000 --> {total} = 30,000.
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Re: Three business partners, Q, R, and S, agree to divide their [#permalink]
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07 Feb 2014, 04:44
SOLUTION
Three business partners, Q, R, and S, agree to divide their total profit for a certain year in the ratios 2:5:8, respectively. If Q's share was $4,000, what was the total profit of the business partners for the year? (A)$26,000
(B) $30,000 (C)$52,000
(D) $60,000 (E)$300,000
The total profit was divided in the ratio Q:R:S = 2:5:8, hence Q's share was 2/(2 + 5 + 8) = 2/15 of the total profit. Q's share was $4,000, thus 2/15*{total} = 4,000 --> {total} = 30,000. Answer: B. _________________ Intern Joined: 18 Sep 2016 Posts: 43 Re: Three business partners, Q, R, and S, agree to divide their [#permalink] ### Show Tags 28 Sep 2016, 02:07 [quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition Three business partners, Q, R, and S, agree to divide their total profit for a certain year in the ratios 2:5:8, respectively. If Q's share was$4,000, what was the total profit of the business partners for the year?
(A) $26,000 (B)$30,000
(C) $52,000 (D)$60,000
(E) \$300,000
total profit is equal to the sum of the ratios * a variable (x) ======> = $$2x+5x+8x= (2+5+8)*x$$ =15x
if Q's share is 4000 ====> 2x=4000====> x=2000
therefore, from the firstequation, total profit= $$2000*15=30000$$
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# What is the exponent key ? - Canon F-710 Calculator
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## Related Questions:
### How do i enter an exponent on the
It depends on the base of the power function.
For exponential function use [SHIFT][LN] (e^x) followed by the exponent value. If exponent is negative use the change sign +/- key before exponent.
For a general power, type in the base (number), then press the [Y^x] key followed by the exponent.
For a power of 10, use the [Exp] key. Careful here. For example to enter 1.6x10^(-19), type in 1.6, press the [Exp] key, then the change sign key +/- and the exponent. Press [=]
1 . 6 [Exp] +/- 19 [=]
Sep 28, 2014 | Sharp EL-531WBBK 12-Digit 272-Functions...
### How do i do exponent 4
The key you are looking for looks like y raised to the exponent x. It is below the "hyp" key. To determine an exponent, enter you base, hit the y to the x key and enter the exponent.
For example, 3 to the power of 4, enter 3, hit the y to the x key, and enter 4 and hit the = key.
Good luck.
Paul
May 26, 2014 | Sharp el-531x scientific calculator
### How to use it for exponents?
To enter exponents (1, 2, 3, -3..) use the general power key marked with Y^x or X^y or ^. Enter the base press the general power key then the exponent. If exponent is negative try enclosing it in parentheses. use the change sign key (-) to enter the negative sign before the exponent.
To enter power of 10, you can use the shortcut key marked Exp (the 10 must not appear when the Exp is used to enter exponents for power of 10).
Feb 24, 2013 | Sharp Office Equipment & Supplies
### Where is the exponent key on the 531x
If you mean the key to raise a number to an exponent, it's the key marked "y^x", two keys above the "7" key. If you mean the key to enter an exponent of ten, it's the key marked "Exp", just above the "7" key.
Feb 18, 2013 | Sharp el-531x scientific calculator
### Finding exponents key
To enter exponents of an arbitrary base, use the key that is marked [Y^x]
To enter exponent of 10 press the [EE] key. In scientific notation display, the exponent of the power of 10 is displayed to the right of the mantissa.
Apr 08, 2012 | Texas Instruments TI-30XA Calculator
### My exponent key does not revert to regular script after i'm done with the exponent. it stays as superscript.
Press the right-arrow key to tell the calculator you're done with the exponent. Otherwise the calculator thinks you're typing in a very complicated exponent.
Nov 16, 2010 | Texas Instruments TI-84 Plus Calculator
### Where is the exponent key on Casio fx-115es calculator?
To raise 10 to an exponent, use the 10^x function (SHIFT log). To raise e to an exponent, use the e^x function (SHIFT ln). To raise an arbitrary value to an exponent, use x^y (just above the sin key). To enter an exponent of ten, use the x10^x key to the right of the decimal point.
Oct 27, 2010 | Casio FX-115ES Scientific Calculator
### How do you put in the number in scientific notation on the TI 34 II calculator?
Type in the mantissa, then the EE key (for Enter Exponent) above the 7 key, then the exponent. If it's a negative exponent, press the (-) key to the right of the decimal point before entering the exponent.
Sep 03, 2010 | Texas Instruments TI-34II Explorer Plus...
### Decimals in the exponents for TI-30Xa
Hi,
You have a key labeled [Y to the x], use it for all the powers or roots that do not have a dedicated key or key sequence. Let the key be [^]
Ex: 13.675 to the power 1.33 is entered as 13.675 [^] (1.33) [=].
The result is 32.41778634
Ex: (-64)[^](2/3) [=] gives 16 as result. The (-) is the change sign key
Ex: (2.87655)^(-6.778) is entered as 2.87655 [^] ( (-) 6.778 ) [=] 0.0007758
The (-) is the change sign (or the negative of), not the regular minus. When you have a negative exponent or a complicated exponent, it is safer to enclose the whole exponent inside parentheses.
Hope it helps.
Thank you for using FixYa
Nov 21, 2009 | Texas Instruments TI-30XA Calculator
### Exponent Sign
To write 50 exponent in math mode, type your number then hit the key that has (x with the solid white exponent key). It's between the (x squared key) and the (log key) then you can enter the number 50 for your exponent. You can also hit the (x squared key) delete the 2 exponent and type in the exponent you want.
Jul 04, 2009 | Casio FX-115ES Scientific Calculator
## Open Questions:
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# Resources tagged with: Addition & subtraction
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### Nice or Nasty for Two
##### Age 7 to 14 Challenge Level:
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your opponent.
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Here is a chance to play a version of the classic Countdown Game.
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Got It game for an adult and child. How can you play so that you know you will always win?
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##### Age 7 to 11 Challenge Level:
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Choose a symbol to put into the number sentence.
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Add or subtract the two numbers on the spinners and try to complete a row of three. Are there some numbers that are good to aim for?
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##### Age 7 to 14
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
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##### Age 7 to 11 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Odds and Threes
##### Age 7 to 11 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
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##### Age 5 to 11 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
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##### Age 5 to 7 Challenge Level:
In your bank, you have three types of coins. The number of spots shows how much they are worth. Can you choose coins to exchange with the groups given to make the same total?
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##### Age 5 to 11 Challenge Level:
This is an adding game for two players.
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##### Age 7 to 11 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
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##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
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##### Age 7 to 11 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
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##### Age 5 to 7 Challenge Level:
Are these domino games fair? Can you explain why or why not?
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Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
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A game for 2 or more players. Practise your addition and subtraction with the aid of a game board and some dried peas!
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##### Age 7 to 11 Challenge Level:
Arrange three 1s, three 2s and three 3s in this square so that every row, column and diagonal adds to the same total.
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##### Age 5 to 7 Challenge Level:
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Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
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This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
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Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
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##### Age 7 to 11 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Being Resourceful - Primary Number
##### Age 5 to 11 Challenge Level:
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### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Number Tracks
##### Age 7 to 11 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
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##### Age 5 to 7 Challenge Level:
How many starfish could there be on the beach, and how many children, if I can see 28 arms?
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##### Age 5 to 7 Challenge Level:
Use these head, body and leg pieces to make Robot Monsters which are different heights.
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##### Age 5 to 7 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
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##### Age 7 to 11 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
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Sam got into an elevator. He went down five floors, up six floors, down seven floors, then got out on the second floor. On what floor did he get on?
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Can you draw a continuous line through 16 numbers on this grid so that the total of the numbers you pass through is as high as possible?
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##### Age 5 to 7 Challenge Level:
Move from the START to the FINISH by moving across or down to the next square. Can you find a route to make these totals?
### Zargon Glasses
##### Age 7 to 11 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Twizzle's Journey
##### Age 5 to 7 Challenge Level:
Twizzle, a female giraffe, needs transporting to another zoo. Which route will give the fastest journey?
### Criss Cross Quiz
##### Age 7 to 11 Challenge Level:
A game for 2 players. Practises subtraction or other maths operations knowledge.
### Zios and Zepts
##### Age 7 to 11 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Scoring with Dice
##### Age 7 to 11 Challenge Level:
I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
### Magic Triangle
##### Age 7 to 11 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Number Differences
##### Age 7 to 11 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Pouring the Punch Drink
##### Age 7 to 11 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Diagonal Sums
##### Age 7 to 11 Challenge Level:
In this 100 square, look at the green square which contains the numbers 2, 3, 12 and 13. What is the sum of the numbers that are diagonally opposite each other? What do you notice?
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CC-MAIN-2020-05
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https://web2.0calc.es/members/melody/?answerpage=2768
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crawl-data/CC-MAIN-2019-39/segments/1568514574018.53/warc/CC-MAIN-20190920113425-20190920135425-00521.warc.gz
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# Melody
Nombre de usuario Melody Puntuación 103689 Stats Preguntas 859 Respuestas 27307
+7
1054
19
+103689
Melody 8 may. 2019
+9
371
4
+103689
### Congratulations CPhill on clearing the 100,000 hurdle !
Fuera tema
Melody 4 may. 2019
+13
941
14
+103689
Melody 9 mar. 2019
+6
490
7
+103689
### Happy Birthday ALAN
Melody 26 feb. 2019
+6
1174
12
+103689
Melody 14 feb. 2019
+14
1765
3
+103689
### If a question is ticked that does not mean you cannot continue it.
Melody 27 ene. 2019
+10
689
19
+103689
### Happy Birthday CPhill!
Melody 19 dic. 2018
+10
600
7
+103689
Melody 1 dic. 2018
+7
742
7
+103689
### Try something new - Expansion challenge.
Melody 18 sept. 2018
+8
1092
1
+103689
### Interesting puzzle clip
Fuera tema
Melody 18 may. 2018
+17
2049
20
+103689
### Should you consider anything before you answer a question?
Melody 13 may. 2018
+19
1
1187
6
+103689
Melody 7 feb. 2018
+22
1071
6
+103689
### Happy Birthday Heureka
Melody 1 feb. 2018
+15
1039
2
+103689
### A trig exact ratio memory trick.
Melody 23 ene. 2018
+12
1082
13
+103689
Melody 13 ene. 2018
#6
+103689
+1
7 jul. 2019
#2
+103689
+2
Here is the graph
Using the graph can you see what the answer is?
You also need to be able to work these out without graphing first (it is good to graph afterwards for checking you answer)
X^2 is always going to be positive.
x can't be 1
Do you know how to use calculus Tom?
If so it is easy just to differentiate and then set y'=0 to find the stationary points etc.
7 jul. 2019
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https://www.tutoringhour.com/worksheets/logarithms/quotient-rule/
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# Evaluating Logarithms Using Quotient Rule Worksheets
Have you been sweating over subtracting logs with the same base? Our free worksheets on evaluating logarithms using quotient rule let you slack off while you still increase your uptake of the topic. Each of our pdf worksheets presents 8 log expressions involving subtraction operation. The quotient property of logarithms states that we can find the difference between two logs by dividing their arguments. Writing this as a formula, we have: log A - log B = log (A/B), where A and B are the arguments. Apply the quotient rule, condense the expression to a single logarithm, convert it to exponential form, and obtain the value of each expression.
Our printable evaluating logs using quotient rule worksheets are useful for high school students.
CCSS: HSF-BF
You are here: Pre-Algebra >> Logarithms >> Evaluate >> Quotient Rule
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CC-MAIN-2022-27
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http://plus2net.com/sql_tutorial/sql_where.php
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# SQL WHERE Command
``SELECT * FROM table_name WHERE condition``
WHERE clause is used to collect the records from the table based on some condition specified. More than one condition can be added to the where clause by using various logical expressions like AND, OR, < ( less than ), greater than (> ) etc. Logical expressions plays important role in returning the desire records. Let us start with some examples.
We are interested in the marks of Fourth class students. You can just follow the table creation process and the select query section discussed here. We will use the same table to work with where clause.
Here is the table with few rows.
idnameclassmarksex
1John DeoFour75female
2Max RuinThree90male
3ArnoldThree55male
4Krish StarFour60female
5John MikeFour60female
6Alex JohnFour55male
To get only Class Four records
``SELECT * FROM student WHERE class='Four'``
Here is the result
idnameclassmarksex
1John DeoFour75female
4Krish StarFour60female
5John MikeFour60female
6Alex JohnFour55male
10Big JohnFour55female
15Tade RowFour93male
16GimmyFour93male
21Babby JohnFour69female
31Marry ToeeyFour93male
This will return all the records from the table name=student of class=Four. This is what we require to get all the records of fourth standard students.
## WHERE with AND
Now let us add little more requirement to this and go for all the records of students of fourth standard who have scored mark more than 70.
``SELECT * FROM student WHERE class='Four' AND mark >70``
idnameclassmarksex
1John DeoFour75female
15Tade RowFour93male
16GimmyFour93male
31Marry ToeeyFour93male
We have added one more condition in where clause with a AND combination. This query will return all student records of Fourth class who have scored more than 70. There are different logical combinations using which can work on different type of fields depending on the required conditions. We can see all the non numeric fields we have to use quotes and for numeric fields we need to use quotes. Now let us try some more commands and see what result we will get.
## WHERE with BETWEEN
``SELECT * FROM `student` WHERE mark BETWEEN 60 and 70``
idnameclassmarksex
4Krish StarFour60female
5John MikeFour60female
20JacklyNine65female
21Babby JohnFour69female
34Gain ToeSeven69male
This will return all the records of the student table that have scored marks between 60 and 70.
## Using LIKE with WHERE
``SELECT * FROM `student` WHERE name LIKE '%John%'``
idnameclassmarksex
1John DeoFour75female
5John MikeFour60female
6Alex JohnFour55male
7My John RobFifth78male
10Big JohnFour55female
21Babby JohnFour69female
This search will return us all the records for which inside the name field 'John' is used. John name can be a part of the field. This will return records having name = John Deo or John Mike and even names like Alex John.
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Your Rating
▼ More on getting records from table with different combinations of commands
Raju 11-04-2013 Can we use Where condition linking more that one table?
wale 23-02-2014 I have 12 tables with the same number of fields and field names. I want to sum one of those fields that have numerical values in all tables. How can I accomplish the task.
Ivan 13-09-2014 Anyone can help me how to resolve this issue?? I am getting a type mismatch with the below syntax... I dont know how to resolve it.. adoCompName.RecordSource = "SELECT * FROM Tbl_Comp_Dtl WHERE CompName = ' * " & Text1.Text & " '" adoCompName.Refresh
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https://www.mersenneforum.org/showpost.php?s=2e796618b7360fa2c6ff887f47848c51&p=317755&postcount=2
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Thread: Not smooth enough numbers View Single Post
2012-11-09, 21:31 #2 bsquared "Ben" Feb 2007 41×83 Posts You need to solve the congruence t^2 = N mod p. Then the solutions to (x + sqrt(N))^2 - N = 0 mod p are x = +/-t - b mod p. Then you sieve the progressions x + p, x + 2p, ... up to some bound for each solution x1, x2. Also your smoothness bound is *way* too high. A smoothness bound of 100-200 or so would be appropriate here, with a factor base of 25 or so primes. I suggest you find and read Scott Contini's thesis on the quadratic sieve which explains a lot of this stuff in pretty good detail.
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https://number.academy/151679
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# Number 151679 facts
The odd number 151,679 is spelled 🔊, and written in words: one hundred and fifty-one thousand, six hundred and seventy-nine. The ordinal number 151679th is said 🔊 and written as: one hundred and fifty-one thousand, six hundred and seventy-ninth. Color #151679. The meaning of the number 151679 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 151679. What is 151679 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 151679.
## Interesting facts about the number 151679
### Asteroids
• (151679) 2003 AE20 is asteroid number 151679. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, ETS in Socorro on 1/5/2003.
## What is 151,679 in other units
The decimal (Arabic) number 151679 converted to a Roman number is (C)(L)MDCLXXIX. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
151679 seconds equals to 1 day, 18 hours, 7 minutes, 59 seconds
151679 minutes equals to 3 months, 3 weeks, 7 hours, 59 minutes
### Codes and images of the number 151679
Number 151679 morse code: .---- ..... .---- -.... --... ----.
Sign language for number 151679:
Number 151679 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 151679 is not a prime number. The closest prime numbers are 151673, 151681.
#### Factorization and factors (dividers)
The prime factors of 151679 are 11 * 13789
The factors of 151679 are 1, 11, 13789, 151679.
Total factors 4.
Sum of factors 165480 (13801).
#### Powers
The second power of 1516792 is 23.006.519.041.
The third power of 1516793 is 3.489.605.801.619.839.
#### Roots
The square root √151679 is 389,459882.
The cube root of 3151679 is 53,330438.
#### Logarithms
The natural logarithm of No. ln 151679 = loge 151679 = 11,929522.
The logarithm to base 10 of No. log10 151679 = 5,180925.
The Napierian logarithm of No. log1/e 151679 = -11,929522.
### Trigonometric functions
The cosine of 151679 is -0,972536.
The sine of 151679 is 0,232753.
The tangent of 151679 is -0,239326.
## Number 151679 in Computer Science
Code typeCode value
PIN 151679 It's recommended that you use 151679 as your password or PIN.
151679 Number of bytes148.1KB
CSS Color
#151679 hexadecimal to red, green and blue (RGB) (21, 22, 121)
Unix timeUnix time 151679 is equal to Friday Jan. 2, 1970, 6:07:59 p.m. GMT
IPv4, IPv6Number 151679 internet address in dotted format v4 0.2.80.127, v6 ::2:507f
151679 Decimal = 100101000001111111 Binary
151679 Decimal = 21201001202 Ternary
151679 Decimal = 450177 Octal
151679 Decimal = 2507F Hexadecimal (0x2507f hex)
151679 BASE64MTUxNjc5
151679 SHA16c4f8f8ee8742095855b83342a69a2c45bdcbe3c
151679 SHA224c9cab5eb488045d018bb5ff8d356060b02c289a7de3ba7c4a8ccb1bf
151679 SHA25645d6e5511a50596b4d23d5932c102456d1510d903a9cecd1212e841d07abe977
151679 SHA384295c4af755b7ff27f278c529df3d5803be94b749f61905fe21bff07bd4876585b3bce32d17eede44853c9104ee2fa609
More SHA codes related to the number 151679 ...
If you know something interesting about the 151679 number that you did not find on this page, do not hesitate to write us here.
## Numerology 151679
### Character frequency in the number 151679
Character (importance) frequency for numerology.
Character: Frequency: 1 2 5 1 6 1 7 1 9 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 151679, the numbers 1+5+1+6+7+9 = 2+9 = 1+1 = 2 are added and the meaning of the number 2 is sought.
## № 151,679 in other languages
How to say or write the number one hundred and fifty-one thousand, six hundred and seventy-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 151.679) ciento cincuenta y uno mil seiscientos setenta y nueve German: 🔊 (Nummer 151.679) einhunderteinundfünfzigtausendsechshundertneunundsiebzig French: 🔊 (nombre 151 679) cent cinquante et un mille six cent soixante-dix-neuf Portuguese: 🔊 (número 151 679) cento e cinquenta e um mil, seiscentos e setenta e nove Hindi: 🔊 (संख्या 151 679) एक लाख, इक्यावन हज़ार, छः सौ, उन्यासी Chinese: 🔊 (数 151 679) 十五万一千六百七十九 Arabian: 🔊 (عدد 151,679) مائة و واحد و خمسون ألفاً و ستمائة و تسعة و سبعون Czech: 🔊 (číslo 151 679) sto padesát jedna tisíc šestset sedmdesát devět Korean: 🔊 (번호 151,679) 십오만 천육백칠십구 Danish: 🔊 (nummer 151 679) ethundrede og enoghalvtredstusindsekshundrede og nioghalvfjerds Hebrew: (מספר 151,679) מאה חמישים ואחד אלף שש מאות שבעים ותשע Dutch: 🔊 (nummer 151 679) honderdeenenvijftigduizendzeshonderdnegenenzeventig Japanese: 🔊 (数 151,679) 十五万千六百七十九 Indonesian: 🔊 (jumlah 151.679) seratus lima puluh satu ribu enam ratus tujuh puluh sembilan Italian: 🔊 (numero 151 679) centocinquantunomilaseicentosettantanove Norwegian: 🔊 (nummer 151 679) en hundre og femtien tusen seks hundre og syttini Polish: 🔊 (liczba 151 679) sto pięćdziesiąt jeden tysięcy sześćset siedemdziesiąt dziewięć Russian: 🔊 (номер 151 679) сто пятьдесят одна тысяча шестьсот семьдесят девять Turkish: 🔊 (numara 151,679) yüzellibinaltıyüzyetmişdokuz Thai: 🔊 (จำนวน 151 679) หนึ่งแสนห้าหมื่นหนึ่งพันหกร้อยเจ็ดสิบเก้า Ukrainian: 🔊 (номер 151 679) сто п'ятдесят одна тисяча шістсот сімдесят дев'ять Vietnamese: 🔊 (con số 151.679) một trăm năm mươi mốt nghìn sáu trăm bảy mươi chín Other languages ...
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| 3.15625
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https://www.backyardchickens.com/threads/of-splash-from-these-hens.250467/
| 1,685,650,750,000,000,000
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# % of splash from these hens
#### Dipsy Doodle Doo
##### ODD BIRD
13 Years
Hi!
Anyone know how to figure the percentage of splash chicks from this:
There are 9 blue hens, 2 splash hens, 4 black hens with a blue rooster(s).
I know splash can only come from breeding blue x blue (50% blue, 25% splash, 25% black)
or
blue x splash (50% blue, 50% splash).
Maybe someone smarter than me can figure the chances of getting a splash chick from a days' worth of eggs from those 'odds' (assuming they were all laying at a given time).
Thanks!
Lisa
If all the hens laid in one day and if all were fertile, looks like you would have the chance of ending up with 3 to 4 splash out of the mix.
I would also agree with the above post of 3-4 chances of splash from a days worth of eggs.
Thank you both so much!
Someone asked me and I couldn't wrap my brain around a possible number.
Lisa
It looks like 21.6% of your chicks should be splash, assuming all hens are laying equally. One egg from each hen should yield 3.25 splash chicks.
36.6% should be blue
41.6% should be black
(ps, I love math, and genetics!)
Thanks Bailey, do you mind if I use your figures?
I know it only is really applicable per 1000 eggs or so, but saves asking.
I'm about to move the black hens out (to an undisclosed location).
Could you calculate the percentage of splash chicks from just the 9 blue girls and 2 splash girls with blue roo.
Lisa
Quote:
I realize you are using only blue roos, but I wanted to make sure you know that Splash X Splash = 100% Splash
Without the Black, you would get...
29.5% Splash
50% Blue
20.5% Black
If you are shooting for the most splashes, then certainly take the black hens out of the mix. Splash to splash should give all splash. I don't have any actual experience, but I think I'm pretty clear on how it all works.
Feel free to use my numbers. I can't promise this is how it will actually play out, it's just statistics.
Are these for the blue Ameraucanas I see on your page? If so, you breed me a nice Splash roo, with no crow, so I can put him over my Cuckoo Marans hen to make sex linked Olive Eggers with blue plumage!
Maybe I'll send my girls on a little vacation to get knocked up!
It's one thing to estimate percentages & another thing to deal with reality. Those percentage estimates would be accurate if you were talking about hatching 10,000 chicks. For smaller hatches they very well may not hold true just because the sample size is too small. When I was raising Andalusians I would sometimes get a hatch with no splash birds or no black birds. It's equally possible to hatch a small group of eggs from your flock & get all splash.
Thanks, yes, someone asked me about possible percentages from my Am' pens.
"...I know it only is really applicable per 1000 eggs or so..."
Lisa
| 700
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## g++ tle, c++ac球原因
Posted by Alextokc at 2017-05-18 12:45:06 on Problem 3126
```代码:
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <deque>
#include <queue>
#include <utility>
#include <iterator>
using namespace std;
const int Maxn = 10005;
int T;
bool isprime[Maxn];
bool vis[Maxn];
typedef pair < string, int> node;
string a, b;
inline void prime_sieve(int n) {
memset(isprime, 1, sizeof isprime);
int i, j;
isprime[0] = 0, isprime[1] = 0;
for (i = 2; i <= n; ++i) {
if (isprime[i])
for (j = i + i; j <= n; j += i)
isprime[j] = 0;
}
}
inline int strtonum(string x) {
int num = 0, i;
if (x[0] == '-') {
for (i = 1; i < x.length(); ++i)
num = num * 10 + (x[i] - '0');
return -num;
}
for (i = 0; i < x.length(); ++i)
num = num * 10 + (x[i] - '0');
return num;
}
inline bool check(int num) {
if (num == strtonum(b)|| (num & 1 && isprime[num] && vis[num] == 0 && num >= 0 && num <= 9999))
return true;
return false;
}
inline int bfs() {
queue < node > Q;
node x;
x.first = a, x.second = 0;
Q.push(x);
vis[strtonum(x.first)] = 1;
int i, j, numberx;
node next;
while (!Q.empty()) {
x = Q.front(); Q.pop();
if (x.first == b) return x.second;
for (i = 0; i < 4; ++i) {
if (i == 0) {
for (j = 1; j <= 9; ++j) {
next = x;
next.first[0] = (j + '0');
numberx = strtonum(next.first);
if (check(numberx)) {
vis[numberx] = 1;
next.second = x.second + 1;
Q.push(next);
}
}
} else if (i == 1) {
for (j = 0; j <= 9; ++j) {
next = x;
next.first[1] = (j + '0');
numberx = strtonum(next.first);
if (check(numberx)) {
vis[numberx] = 1;
next.second = x.second + 1;
Q.push(next);
}
}
} else if (i == 2) {
for (j = 0; j <= 9; ++j) {
next = x;
next.first[2] = (j + '0');
numberx = strtonum(next.first);
if (check(numberx)) {
vis[numberx] = 1;
next.second = x.second + 1;
Q.push(next);
}
}
} else if (i == 3) {
for (j = 1; j <= 9; j += 2) {
next = x;
next.first[3] = (j + '0');
numberx = strtonum(next.first);
if (check(numberx)) {
vis[numberx] = 1;
next.second = x.second + 1;
Q.push(next);
}
}
}
}
}
return -1;
}
int main() {
ios::sync_with_stdio(false);
prime_sieve(Maxn);
cin >> T;
while (T--) {
memset(vis, 0, sizeof vis);
cin >> a >> b;
int X;
X = bfs();
if (X == -1) cout << "Impossible" << endl;
else cout << X << endl;
}
return 0;
}
Followed by:
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| 3.421875
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CC-MAIN-2017-22
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longest
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en
| 0.191411
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https://web2.0calc.com/questions/what-is-5-75--2-1-8
| 1,591,323,466,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00096.warc.gz
| 582,299,478
| 6,141
|
+0
# what is 5.75 + (-2 1/8)=
0
599
3
what is 5.75 + (-2 1/8)=
Apr 22, 2014
#3
+109524
+5
You could just do it on the calclulator
$${\mathtt{5.75}}{\mathtt{\,-\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right) = {\mathtt{3.625}}$$
.
Apr 24, 2014
#1
+142
0
paranthese first!
I assume that you have -2&1/8 (-2.125)
So a + multiplicated by a - = a minus.
So 5.75-2.125 = 3.625 (or in fraction 3-5/8)
Apr 23, 2014
#2
+142
+5
paranthese first!
I assume that you have -2&1/8 (-2.125)
So a + multiplicated by a - = a minus.
So 5.75-2.125 = 3.625 (or in fraction 3-5/8)
Apr 23, 2014
#3
+109524
+5
$${\mathtt{5.75}}{\mathtt{\,-\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right) = {\mathtt{3.625}}$$
| 377
| 794
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{[ promptMessage ]}
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HW04-solutions
# HW04-solutions - le(chl528 HW04 seckin(57195 This print-out...
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le (chl528) – HW04 – seckin – (57195) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the x -intercept of the tangent line at the point P ( 4 , 3) on the graph of f when lim h 0 f ( 4 + h ) 3 h = 5 . 1. x -intercept = 23 5 2. x -intercept = 23 3. x -intercept = 23 5 correct 4. x -intercept = 23 5. x -intercept = 17 5 6. x -intercept = 17 Explanation: Since lim h 0 f ( 4 + h ) f ( 4) h is the slope of the tangent line at P ( 4 , 3) , an equation for this tangent line is y 3 = 5( x + 4) , i.e. , y = 5 x + 23 . Consequently, the x -intercept = 23 5 . 002 10.0 points Consider the slope of the given curve at the five points shown. C E A B D List the five slopes in decreasing order. 1. E, B, C, D, A 2. E, C, A, D, B 3. B, D, E, C, A 4. D, C, E, A, B 5. D, B, C, A, E 6. B, D, A, C, E correct Explanation: The order will be the one from most positive slope to most negative slope. Inspection of the graph shows that this is B, D, A, C, E . 003 10.0 points Find an equation for the tangent line at the point P ( 2 ,f ( 2)) on the graph of f when f is defined by f ( x ) = 3 x 2 + x + 3 . 1. y + 11 x + 9 = 0 correct 2. y 11 x + 9 = 0 3. y 11 x 9 = 0 4. y + 11 x + 15 = 0
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le (chl528) – HW04 – seckin – (57195) 2 5. y 11 x + 15 = 0 6. y + 11 x 15 = 0 Explanation: The slope of the tangent line at P ( 2 ,f ( 2)) is the value of the limit lim h 0 f ( 2 + h ) f ( 2) h . Now f ( 2 + h ) = 3( 2 + h ) 2 + ( 2 + h ) + 3 = 3 h 2 11 h + 13 , while f ( 2) = 13. Thus f ( 2 + h ) f ( 2) h = 3 h 2 11 h h = 3 h 11 . As h approaches 0, therefore, f ( 2 + h ) f ( 2) h → − 11 . Consequently, by the point-slope formula, an equation for the tangent line at P ( 2 ,f ( 2)) is y 13 = 11( x + 2) , i.e. , y + 11 x + 9 = 0 . 004 (part 1 of 3) 10.0 points A Calculus student leaves the RLM build- ing and walks in a straight line to the PCL Li- brary. His distance (in multiples of 40 yards) from RLM after t minutes is given by the graph -1 0 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 2 4 6 8 10 t distance i) What is his speed after 3 minutes, and in what direction is he heading at that time? 1. away from RLM at 45 yds/min 2. towards RLM at 30 yds/min 3. away from RLM at 30 yds/min 4. towards RLM at 60 yds/min 5. away from RLM at 60 yds/min correct Explanation: The graph is linear and has positive slope on [2 , 4], so the speed of the student at time t = 3 coincides with the slope of the line on [2 , 4]. Hence speed = 40 · 8 5 4 2 = 60 yds/min . As the distance from RLM is increasing on [2 , 4] the student is thus moving away from the RLM. 005 (part 2 of 3) 10.0 points ii) What is his speed after 9 minutes, and in what direction is he heading at that time? 1. towards RLM at 40 yds/min 2. away from RLM at 20 yds/min .
le (chl528) – HW04 – seckin – (57195) 3 3. away from RLM at 5 yds/min .
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http://math.stackexchange.com/questions/248023/non-isomorphic-induced-representations-from-the-same-representation-of-a-subgro
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# Non-Isomorphic induced representations (from the same representation of a subgroup)
I believe that it is true that if we have a group $G$, and two copies $H_1$, $H_2$ of some group $H$ as subgroups of $G$, we can fix a representation $V$ of $H$ and have the situation:
$$\operatorname{Ind}_{H_{1}} V \ncong \operatorname{Ind}_{H_{2}} V$$
I believe my example is Klein-4 in $S_4$, you can take a normal copy generated by products of disjoint $2$-cycles $(12)(34)$, etc., and a copy $\{ (12)(34),(12),(34), 1 \}$ (or do $\mathbb{Z}_3 \times \mathbb{Z}_3$ in $S_9$ )
Anyway, if $H_1$ and $H_2$ are conjugate subgroups, then the induced representations will be isomorphic as can be seen by the character formula
$1/|H| \sum_{x} V(x^{-1}gx)$ = character of $\operatorname{Ind}_{H_{1}} V$ = character of $\operatorname{Ind}_{H_{2}} V$
Is this correct? Thank you
-
So you're looking at the case of ordinary representations? – Curufin Nov 30 '12 at 14:15
Yes, if $H_{1}$ and $H_{2}$ are conjugate, you will get isomorphic modules (whihever ring you work over). If $H_{1}$ and $H_{2}$ are not conjugate, then you need not get the same character for the induced module, even in the complex case. The example you gave for $S_{4}$ is perfectly valid- if $H_{1}$ is normal, then the induced character will vanish outside $H_{1},$ while if $H_{2}$ is not normal, that need not be the case (it is not in your example).
-
Thank you for the answer, it is what I thought. I wanted to ask here for a "fact-checking", in case I did something wrong in my proof. The reason why I picked those two versions of klein-4 is that one is normal and the other isn't ( when you unpack the character formula, all the terms in the normal case will remain, while the terms in the non-normal case will be 0, when conjugation sends you outside the subgroup ). – Elliot Nov 30 '12 at 15:34
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## Cannibals and missionaries help!!!
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### Cannibals and missionaries help!!!
Hello,
I am fairly new to lisp and need help with what seems to be a well known lisp problem. For one of my classes we are doing the cannibals and missionaries problem where there are 3 of both and 1 boat that can hold 2 people. The objective is to get all members across a river but if the missionaries ever outnumber the cannibals then the cannibals will be converted. We need to write a program to solve this problem, then have it solve it with a breadth or depth first search and a best-first search. The code below is what I have but I cannot get it to work please any help would be appreciated.
Code: Select all
`;; Name: ;; Project Name: Cannibals and Missionaries;; Description : To safely transport all members across the river;; to the other side(defun solve-cm (state goal) (path state goal nil));; Define the start state(setf *start* '(3 3 0 0 start-bank));; Define the goal state(setf *goal* '(0 0 3 3 goal-bank));; Safe returns nil if a state is not safe; it returns the state unchanged if it is safe.(defun safe (state) (cond ((or (> (missionaries-at-start state)(cannibals-at-start state)) (> (missionaries-at-goal state)(cannibals-at-goal state)) (> (cannibals-at-start state) 0) (< (cannibals-at-start state) 3) nil)) (t( state)) ) );; The recursive path algorithm searches the space in a depth first fashion.(defun path (state goal been-list) (cond ((null state) nil) ((equal state goal) (reverse (cons state been-list))) ((not (member state been-list :test #'equal)) (or (path (move-two-cannibals state) goal (cons state been-list)) (path (move-two-missionaries state) goal (cons state been-list)) (path (move-one-cannibal state) goal (cons state been-list)) (path (move-one-missionary state) goal (cons state been-list)) (path (move-one-of-each state) goal (cons state been-list))))));; These functions define legal moves in the state space. The take;; a state as argument, and return the state produced by that operation.(defun move-two-cannibals(state) (cond ((equal (boat-position state) 'start-bank) (safe (make-state (- (cannibals-at-start state) 2) (missionaries-at-start state) (+ (cannibals-at-goal state) 2) (missionaries-at-goal state) (opposite (boat-poition state)) ))) ((equal (boat-position state) 'goal-bank) (safe (make-state (+ (cannibals-at-start state) 2) (missionaries-at-start state) (- (cannibals-at-goal state) 2) (missionaries-at-goal state) (opposite (boat-poition state)) )))))(defun move-one-cannibal(state) (cond ((equal (boat-position state) 'start-bank) (safe (make-state (- (cannibals-at-start state) 1) (missionaries-at-start state) (+ (cannibals-at-goal state) 1) (missionaries-at-goal state) (opposite (boat-poition state)) ))) ((equal (boat-position state) 'goal-bank) (safe (make-state (+ (cannibals-at-start state) 1) (missionaries-at-start state) (- (cannibals-at-goal state) 1) (missionaries-at-goal state) (opposite (boat-poition state)) )))))(defun move-two-missionaries(state) (cond ((equal (boat-position state) 'start-bank) (safe (make-state (cannibals-at-start state) (- (missionaries-at-start state) 2) (cannibals-at-goal state) (+ (missionaries-at-goal state) 2) (opposite (boat-poition state)) ))) ((equal (boat-position state) 'goal-bank) (safe (make-state (cannibals-at-start state) (+ (missionaries-at-start state) 2) (cannibals-at-goal state) (- (missionaries-at-goal state) 2) (opposite (boat-poition state)) )))))(defun move-one-missionary(state) (cond ((equal (boat-position state) 'start-bank) (safe (make-state (cannibals-at-start state) (- (missionaries-at-start state) 1) (cannibals-at-goal state) (+ (missionaries-at-goal state) 1) (opposite (boat-poition state)) ))) ((equal (boat-position state) 'goal-bank) (safe (make-state (cannibals-at-start state) (+ (missionaries-at-start state) 1) (cannibals-at-goal state) (- (missionaries-at-goal state) 1) (opposite (boat-poition state)) )))))(defun move-one-of-each(state) (cond ((equal (boat-position state) 'start-bank) (safe (make-state (- (cannibals-at-start state) 1) (- (missionaries-at-start state) 1) (+ (cannibals-at-goal state) 1) (+ (missionaries-at-goal state) 1) (opposite (boat-poition state)) ))) ((equal (boat-position state) 'goal-bank) (safe (make-state (+ (cannibals-at-start state) 1) (+ (missionaries-at-start state) 1) (- (cannibals-at-goal state) 1) (- (missionaries-at-goal state) 1) (opposite (boat-poition state)) )))))(defun opposite (boat-position) (cond ((equal boat-position 'start-bank) 'goal-bank) ((equal boat-position 'goal-bank) 'start-bank)));; These functions define states of the world as an abstract data type.(defun make-state (cs ms cg mg boat-position) (list cs ms cg mg boat-position))(defun cannibals-at-start ( state ) (first state))(defun missionaries-at-start ( state ) (second state))(defun cannibals-at-goal ( state ) (third state))(defun missionaries-at-goal ( state ) (fourth state))(defun boat-position ( state ) (fifth state))`
Diesel298
Posts: 1
Joined: Fri Nov 01, 2013 6:21 am
### Re: Cannibals and missionaries help!!!
In your 'safe' procedure, you have 'nil' as part of the first predicate of the 'cond' (and serves no purpose in this position). It should be the evaluated consequent.
In other words, one of the closing parentheses is in the wrong place.
saulgoode
Posts: 45
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https://mathpickle.com/project/prime-number-catacombs/
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#### Prime Number Catacombs
##### (William Paulsen, 2000)
This is absolutely one of my favourite pedagogic discoveries from recreational mathematics. It is engaging, evocative and curiosity inducing. I now prefer to start students groups off at different numbers. They must try to meet. Good starting numbers are: 2, 11, 17, 31, 37, and 79.
Here is a pdf file – the last three sheets are suitable for printing. The first are suitable for presentation.
PS. Warning: This problem requires students to learn binary! Most curricula have this appearing years after students learn about prime numbers.
Nothing must be advanced in a positive manner. The mind of the pupil is to be the principal operator; it must instruct, convince, and confute itself; and when it arrives at some important truth or result, it must be through its own powers. It ought not even to perceive that it has been guided thither.
F. J. Grund
Introductory Discourse and Lectures, Boston, 1830
#### Standards for Mathematical Practice
MathPickle puzzle and game designs engage a wide spectrum of student abilities while targeting the following Standards for Mathematical Practice:
##### MP1 Toughen up!
Students develop grit and resiliency in the face of nasty, thorny problems. It is the most sought after skill for our students.
##### MP2 Think abstractly!
Students take problems and reformat them mathematically. This is helpful because mathematics lets them use powerful operations like addition.
##### MP3 Work together!
Students discuss their strategies to collaboratively solve a problem and identify missteps in a failed solution. MathPickle recommends pairing up students for all its puzzles.
##### MP4 Model reality!
Students create a model that mimics the real world. Discoveries made by manipulating the model often hint at something in the real world.
##### MP5 Know the tools.
Students master the tools at their fingertips - whether it's a pencil or an online app.
##### MP6 Be precise!
Students learn to communicate using precise terminology. MathPickle encourages students not only to use the precise terms of others, but to invent and rigorously define their own terms.
##### MP7 Be observant!
Students learn to identify patterns. This is one of the things that the human brain does very well. We sometimes even identify patterns that don't really exist 😉
##### MP8 Be lazy!?!
Students learn to seek for shortcuts. Why would you want to add the numbers one through a hundred if you can find an easier way to do it?
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https://www.yuvajobs.com/download-drdo-placement-papers/drdo-aptitude-general-1-738.html
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Search Jobs
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# Aptitude - General 1 by DRDO
Details of Aptitude - General 1 by DRDO conducted by DRDO for job interview.
Share Us With Others
Model for ECE:
Direct Theory question:
1. A triangular-square wave generator uses
* A sine wave oscillator and a comparator
* An integrator and a comparator
* A diffrentiator and comparator
* A sine wave oscillator and a clipper
2. An amplifier has two identical cascaded stages. Each stage has a BW of 20KHz. The overall BW
shall be
* 10KHz
* 12.9KHz
* 20KHz
* 28.3HHz
Direct numerical problem:
1. A parabolic dish antenna has a diameter of 1m. the maximum possible ideal gain of the antenna at a wavelength 3.14cm is
* 20 dB
* 30 dB
* 40 dB
* 50 dB
2. The frequency deviation produced in a VHF carrier by a signal of 100Hz is 50KHz. The frequency modulation is
Application based theoretical or numerical problem:Which one of the pulses has the same form in time domain as well as in frequency domain
Rectangular
Exponenetial
Triangular
Gaussian
Good voice reproduction via PCM requires 128 quantisation levels. If the BW of voice channel is 4KHz then data rate is
256 Kbps
128 Kbps
56 Kbps
28 Kbps
Match the following:
1. List 1 List2
A. Thyristor 1. Second break down
B. MOSFET 2. Large on-state drop
C. IGBT 3. Small on state drop
D. BJT 4. Slow device
Codes:
(a) A-4, B-3, C-2, D-1 (b) A-1, B-2, C-3, D-4
(c) A-4, B-2, C-3, D-1 (d) A-1, B-3, C-2, D-4
List 1 List2
A. Free and forced response 1. Discrete time systems
B. Z transform 2. Dirichlet conditions
C. Probability theory 3. Non-homogenous differential equation
d. Fourier series 4. Random process
Codes:
(a) A-1, B-3, C-2, D-4 (b) A-3, B-1, C-2, D-4
(c) A-1, B-3, C-4, D-2 (d) A-3, B-1, C-4, D-2
Assertion and reasoning type:
The following item consists of two statements labeled as Assertion A and the other as Reason R. Use the following codes:
(a) Both A&R are individually true and R is the correct explanation of A
(b) Both A & R are individually true and R is not the correct explanation of A
(c) A is true and R is false
(d) A is false and R is true
AEvery materials has a different value of energy band gap except metals which have no band gap
R The energy band gap is decided by the equilibrium lattice constant, which is different in different materials.
AOp-amp are commonly used in instrumentation
R Op-amps do not load the circuit due to their very high input impedance
Problem based on circuits and graphs:
Some examples:
For circuits: HW rectifier, voltmeter, RC or RLC or RL transients, feedback systems, clipper, clamper, transistor biasing,555 circits, logic gates, counters, bridges and wattmeters.
Graphs: Frequency response H(s),open loop gain from root loci, Nyquist plot, transfer functions etc
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http://wallstreetpit.com/85587-predicting-gdp-with-arima-forecasts/
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Predicting GDP with ARIMA Forecasts
Is the U.S. economy headed for a new recession? The risk is clearly elevated these days, in part because the euro crisis rolls on. The sluggish growth rate in the U.S. isn’t helping either. But with ongoing job growth, albeit at a slow rate, it’s not yet clear that we’ve reached a tipping point. Given all the mixed signals, however, forecasting, is unusually tough at the moment. In fact, it’s never easy, but it’s necessary just the same. But how to proceed? The possibilities are endless, but one useful way to begin is with so-called autoregressive integrated moving averages (ARIMA). It sounds rather intimidating, but the basic calculation is straightforward and it’s easily performed in a spreadsheet, which helps explain why ARIMA models are so popular in econometrics. A more compelling reason are a number of studies that report that ARIMA models have a history of making relatively accurate forecasts compared with more sophisticated competition.
As a simple example of the power of ARIMA forecasting, let’s consider what this statistical tool is telling us about the next quarterly change in nominal GDP for the U.S. Making a few reasonable assumptions (discussed below), a basic ARIMA forecasting model predicts that fourth quarter nominal GDP will rise 4.7% at a seasonally adjusted annual rate. For comparison, that’s slightly lower than the government’s initial estimate of 5.0% growth for the third quarter. (Remember, we’re talking here of nominal GDP growth vs. real growth, which strips out inflation. Real GDP is the more popularly quoted series.)
For comparison, let’s compare my ARIMA forecast with GDP predictions via the Survey of Professional Forecasters (SPF). SPF data is available on the Philadelphia Fed’s web site and is reported in nominal terms, thus my focus on nominal GDP. To cut to the chase, ARIMA has a history of dispatching superior forecasts compared with SPF. To be precise, I’m comparing ARIMA forecasts with the mean quarter-ahead prediction of economists surveyed quarterly in the SPF reports.
Ok, let’s take a closer look at the details by reviewing a few basic ARIMA concepts. Keep in mind that in the interest of brevity I’m glossing over the details. For a complete discussion of ARIMA, an introductory econometrics text will suffice. One of many examples: Peter Kennedy’s A Guide to Econometrics. Meanwhile, the main point with ARIMA forecasting is that it’s a tool for using a time series’ history to make a forecast. Yes, it’s naïve, but the fact that ARIMA’s errors tend to be low relative to many if not most other forecasting techniques makes this approach worthwhile. It’s not a crystal ball, of course, and so ARIMA forecasts should be considered in context with other prediction techniques.
The first step is taking the data series (in this case the historical quarterly nominal GDP numbers) and regressing them against a lagged set of the same data. For the analysis here, I’m regressing each quarterly GDP number against 1) the previous quarter; 2) two quarters previous; and 3) four quarters previous. Next, I ran a multi-regression analysis on this set of lagged data to estimate the parameters, which tell us how to weight each lagged variable in the formula that spits out the forecast. To check the accuracy of the parameter estimates, I also ran a maximum likelihood procedure. (As a quick aside, all of this analysis can be easily done in Excel, although more sophisticated software packages are available, such as Matlab and EViews.)
ARIMA’s forecasts are naïve, of course, but based on history it does fairly well compared with SPF. The in-sample forecasting errors (residuals, as statisticians call them) for ARIMA’s average deviation from the actual reported GDP number is a slight 0.0049% since 1970. That’s a mere fraction of SPF’s 3.06% residual over the past 30 years. There are several additional error tests we can run, but the simple evaluations above offer a general sense of how a naïve ARIMA model can provide competitive forecasts vs. the expectations of professional economists.
Alas, like all econometric techniques that look backward as a means of looking ahead there’s the risk that sharp and sudden turning points in the trend will surprise an ARIMA model. That’s clear by looking at recent history, as shown in the chart below. Note that when the actual level of nominal GDP turned down in 2008 as the Great Recession unfolded, ARIMA’s forecasting error rate increased. But ARIMA fared no worse than the mean SPF predictions. In fact, you can argue that ARIMA did slightly better than SPF, as implied by tallying up the errors for each during 2008 and comparing one to another.
Errors are inevitable in forecasting. The goal is to keep them to a minimum, a task that ARIMA does quite well. A certain amount of trial and error is building and adjusting ARIMA models is critical. In fact, the errors can help improve ARIMA forecasts. By modeling the errors and incorporating their history into ARIMA’s regressions, there’s a possibility that we can reduce the error in out-of-sample forecasts going forward. That’s because if you design an ARIMA model correctly, the errors should be randomly distributed around a mean of zero. In other words, errors through time should cancel each other out. In that case, the error terms can be useful for enhancing ARIMA’s forecasting powers, i.e., reduce the magnitude of the errors.
Yes, we still need an understanding of economic theory to adjust, interpret and design ARIMA models for predicting GDP and other economic and financial data series. But considering the simplicity and relative reliability of this econometric technique, ARIMA forecasting is a no-brainer. At the very least, it provides a good benchmark for evaluating other forecasts.
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# Circumference Of A Circle Formula
Circumference Of A Circle Formula: A circle consists of all points in a plane that are a given distance, called the radius, from a given point called the center.
#### A segment or line can intersect a circle in several ways.
A segment with endpoints that are the center of the circle and a point of the circle is a radius.
A segment with endpoints that lie on the circle is a chord.
A chord that contains the circle’s center is a diameter.
The circumference of a circle is the distance around the circle.
For a circumference of C units and a diameter of d units or a radius of r units,
C = π.d or C = 2πr
Example: Find the circumference of the circle to the nearest hundredth.
Answer: C = 2πr ⇒ Circumference formula
⇒ 2π(13)
≈ 81.68 (Use a calculator)
The circumference is about 81.68 centimeters.
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https://maplesoft.com/support/help/maple/view.aspx?path=Units/Simple/frem
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frem - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
Units[Simple]
frem
floating-point remainder function in the Simple Units environment
Calling Sequence frem(x1, x2)
Parameters
x1, x2 - expressions
Description
• In the Simple Units environment, any verifications of valid dimensions ignore so-called Unit annotations.
• In the Simple Units environment, the frem procedure overrides the top-level frem procedures. The units for the arguments need to have the same dimension, and the result is given a unit corresponding to that dimension.
• More precisely, if x1 and x2 have the same unit, then the result has that unit. If they have different units of the same dimension, then the result has the default unit for that dimension (as set by the UseUnit or UseSystem commands).
• Every command in the Simple Units environment that needs to determine whether an expression is valid or not, does so using the Units:-TestDimensions command.
Examples
Notes:
– To enter a unit in 2-D Math input, select the unit from the appropriate Units palette. If the unit you want is not there, select $\mathrm{unit}$ and then enter the unit.
– When you edit a unit, double brackets appear around it.
> $\mathrm{with}\left(\mathrm{Units}\left[\mathrm{Simple}\right]\right):$
> $\mathrm{frem}\left(\frac{3}{2}\mathrm{Unit}\left('m'\right),1.49\mathrm{Unit}\left('m'\right)\right)$
${0.010000000}{}⟦{m}⟧$ (1)
> $\mathrm{frem}\left(\frac{3}{2}\mathrm{Unit}\left('\mathrm{ft}'\right),1.49\mathrm{Unit}\left('\mathrm{ft}'\right)\right)$
${0.010000000}{}⟦{\mathrm{ft}}⟧$ (2)
> $\mathrm{frem}\left(\frac{3}{2}\mathrm{Unit}\left('m'\right),1.49\mathrm{Unit}\left('\mathrm{ft}'\right)\right)$
${0.137544000}{}⟦{m}⟧$ (3)
Compatibility
• The Units[Simple][frem] command was introduced in Maple 2021.
• For more information on Maple 2021 changes, see Updates in Maple 2021.
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https://byjus.com/question-answer/how-do-we-know-that-when-do-we-have-to-use-dot-product-and-cross/
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Question
# How do we know that when do we have to use dot product and cross product in a problem?
Solution
## Dot product serves the purpose of finding what component of a vector is the direction of some other vector. Or if we know the vectors them it can serve the purpose of finding the angle between them. While Cross product is used when we want to find a new vector perpendicular to two known vectors as in the case of find the normal to a plane.Or when we want to find the area of a triangle or a parallelogram given that we know two adjacent sides in vector form. We can also use a combination of them as we do to find the volume of a parallelopipe.Physics
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https://bytes.com/topic/access/answers/857540-subform-footer-totals-problem
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# Subform Footer Totals Problem
100+ P: 166 Happy Friday! I was wondering if you can have more than one total in the footer. I ask this because when I set up one of my totals in the footer (lets call it total1), it works (slowly but it works)...but when I set up the other one (total2), I get an error for the total1 value and an error for the total2 value. I have a function, GetYield, that returns a calculation based on the material chosen (matType). Public Function GetYield(matType As Integer, BatchWeight As Double, MatGrav As Double) As Double 'returns the yield of each material 'DM_yield' calculation Select Case matType 'Cement, Coarse, Fine, Pigment Case 1, 2, 3, 4 GetYield = BatchWeight / (MatGrav * 62.4) 'Chemicals Case 5 GetYield = (BatchWeight / 128) * (10 / 62.4) Case Else GetYield = 0 End Select End Function I call this function in a query, which is the control source for the subform: SELECT MixSample.DM_Mix, MixSample.DM_MaterialNo, MixSample.matTypeID, MixSample.materialID, MixSample.matBatchWeight, MatType.matType, Material.material, Material.materialGrav, GetYield([MixSample].[matTypeID],[matBatchWeight],[materialGrav]) AS DMYield, MixSample.pigPercent FROM MixDesign INNER JOIN (Material INNER JOIN (MixSample INNER JOIN MatType ON MixSample.matTypeID = MatType.matTypeID) ON Material.materialID = MixSample.materialID) ON MixDesign.DM_Mix = MixSample.DM_Mix; I have a bound text box on my subform whose control source is the calculated field in the query, called DMYield. In my footer, I have an unbound text box, txtSumDMyield, that is supposed to get the total sum of the DMyield field (named DM_yield) in the form. the control source of the txtSumDMYield field is: =Sum([DMyield]), which only sometimes returns a value. Currently, it is not returning any thing... When I tried to include an unbound text box which also calculated the sum of another field on the form, total1 stopped working and gave me a #error, along with the total2 value equaling #error. total2's sontrol source was similar to total1, =Sum([DM_MixCost]). The unbound text box for DM_MixCost value is also determined by a function. The function returns correct values, but when it comes to getting a total sum for them, I get an error. Any ideas how I can get BOTH totals working? Thank you Dec 5 '08 #1
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# What 5 numbers multiply together to make 700?
Updated: 10/19/2022
Wiki User
5y ago
As a product of its prime factors: 2*2*5*5*7 = 700
Wiki User
5y ago
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http://www.codeproject.com/Articles/4522/Introduction-to-SSE-Programming?fid=16168&df=90&mpp=10&noise=1&prof=True&sort=Position&view=None&spc=None&select=1129427&fr=12&PageFlow=FixedWidth
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11,502,547 members (66,227 online)
# Introduction to SSE Programming
, 10 Jul 2003 CPOL 430.1K 7.6K 121
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An article describes programming floating-point calculations using Streaming SIMD Extensions
## Introduction
The Intel Streaming SIMD Extensions technology enhance the performance of floating-point operations. Visual Studio .NET 2003 supports a set of SSE Intrinsics which allow the use of SSE instructions directly from C++ code, without writing the Assembly instructions. MSDN SSE topics [2] may be confusing for the programmers who are not familiar with the SSE Assembly progamming. However, reading the Intel Software manuals [1] together with MSDN gives the opportunity to understand the basics of SSE programming.
SIMD is a single-instruction, multiple-data (SIMD) execution model. Consider the following programming task: computing of the square root of each element in a long floating-point array. The algorithm for this task may be written by such way:
for each f in array
f = sqrt(f)
Let's be more specific:
for each f in array
{
load f to the floating-point register
calculate the square root
write the result from the register to memory
}
Processor with the Intel SSE support have eight 128-bit registers, each of which may contain 4 single-precision floating-point numbers. SSE is a set of instructions which allow to load the floating-point numbers to 128-bit registers, perform the arithmetic and logical operations with them and write the result back to memory. Using SSE technology, algorithms may be written as:
for each 4 members in array
{
load 4 members to the SSE register
calculate 4 square roots in one operation
write the result from the register to memory
}
The C++ programmer writing a program using SSE Intrinsics doesn't care about registers. He has a 128-byte __m128 type and a set of functions to perform the arithmetic and logical operations. It's up to the C++ compiler to decide which SSE register to use and to make code optimizations. SSE technology may be used when some operation is done with each element of a long floating-point arrays.
## SSE Programming Details
### Include Files
All SSE instructions and __m128 data type are defined in xmmintrin.h file:
#include <xmmintrin.h>
Since SSE instructions are compiler intrinsics and not functions, there are no lib-files.
### Data Alignment
Each float array processed by SSE instructions should have 16 byte alignment. A static array is declared using the __declspec(align(16)) keyword:
__declspec(align(16)) float m_fArray[ARRAY_SIZE];
Dynamic array should be allocated using new _aligned_malloc function:
m_fArray = (float*) _aligned_malloc(ARRAY_SIZE * sizeof(float), 16);
Array allocated by the _aligned_malloc function is released using the _aligned_free function:
_aligned_free(m_fArray);
### __m128 Data Type
Variables of this type are used as SSE instructions operands. They should not be accessed directly. Variables of type _m128 are automatically aligned on 16-byte boundaries.
### Detection of SSE Support
SSE instructions may be used if they are supported by the processor. The Visual C++ CPUID sample [4] shows how to detect support of the SSE, MMX and other processor features. It is done using the cpuid Assembly command. See details in this sample and in the Intel Software manuals [1].
## SSETest Demo Project
SSETest project is a dialog-based application which makes the following calculation with three float arrays:
fResult[i] = sqrt( fSource1[i]*fSource1[i] + fSource2[i]*fSource2[i] ) + 0.5
i = 0, 1, 2 ... ARRAY_SIZE-1
ARRAY_SIZE is defined as 30000. Source arrays are filled using sin and cos functions. The Waterfall chart control written by Kris Jearakul [3] is used to show the source arrays and the result of calculations. Calculation time (ms) is shown in the dialog. Calculation may be done using one of three possible ways:
• C++ code;
• C++ code with SSE Intrinsics;
• Inline Assembly with SSE instructions.
C++ function:
void CSSETestDlg::ComputeArrayCPlusPlus(
float* pArray1, // [in] first source array
float* pArray2, // [in] second source array
float* pResult, // [out] result array
int nSize) // [in] size of all arrays
{
int i;
float* pSource1 = pArray1;
float* pSource2 = pArray2;
float* pDest = pResult;
for ( i = 0; i < nSize; i++ )
{
*pDest = (float)sqrt((*pSource1) * (*pSource1) + (*pSource2)
* (*pSource2)) + 0.5f;
pSource1++;
pSource2++;
pDest++;
}
}
Now let's rewrite this function using the SSE Instrinsics. To find the required SSE Instrinsics I use the following way:
• Find Assembly SSE instruction in Intel Software manuals [1]. First I look for this instruction in Volume 1, Chapter 9, and after this find the detailed Description in Volume 2. This description contains also appropriate C++ Intrinsic name.
• Search for SSE Intrinsic name in the MSDN Library.
Some SSE Intrinsics are composite and cannot be found by this way. They should be found directly in the MSDN Library (descriptions are very short but readable). The results of such search may be shown in the following table:
Required Function Assembly Instruction SSE Intrinsic Assign float value to 4 components of 128-bit value movss + shufps _mm_set_ps1 (composite) Multiply 4 float components of 2 128-bit values mulps _mm_mul_ps Add 4 float components of 2 128-bit values addps _mm_add_ps Compute the square root of 4 float components in 128-bit values sqrtps _mm_sqrt_ps
C++ function with SSE Intrinsics:
void CSSETestDlg::ComputeArrayCPlusPlusSSE(
float* pArray1, // [in] first source array
float* pArray2, // [in] second source array
float* pResult, // [out] result array
int nSize) // [in] size of all arrays
{
int nLoop = nSize/ 4;
__m128 m1, m2, m3, m4;
__m128* pSrc1 = (__m128*) pArray1;
__m128* pSrc2 = (__m128*) pArray2;
__m128* pDest = (__m128*) pResult;
__m128 m0_5 = _mm_set_ps1(0.5f); // m0_5[0, 1, 2, 3] = 0.5
for ( int i = 0; i < nLoop; i++ )
{
m1 = _mm_mul_ps(*pSrc1, *pSrc1); // m1 = *pSrc1 * *pSrc1
m2 = _mm_mul_ps(*pSrc2, *pSrc2); // m2 = *pSrc2 * *pSrc2
m3 = _mm_add_ps(m1, m2); // m3 = m1 + m2
m4 = _mm_sqrt_ps(m3); // m4 = sqrt(m3)
*pDest = _mm_add_ps(m4, m0_5); // *pDest = m4 + 0.5
pSrc1++;
pSrc2++;
pDest++;
}
}
This doesn't show the function using inline Assembly. Anyone who is interested may read it in the demo project. Calculation times on my computer:
• C++ code - 26 ms
• C++ with SSE Intrinsics - 9 ms
• Inline Assembly with SSE instructions - 9 ms
Execution time should be estimated in the Release configuration, with compiler optimizations.
## SSESample Demo Project
SSESample project is a dialog-based application which makes the following calculation with float array:
fResult[i] = sqrt(fSource[i]*2.8)
i = 0, 1, 2 ... ARRAY_SIZE-1
The program also calculates the minimum and maximum values in the result array. ARRAY_SIZE is defined as 100000. Result array is shown in the listbox. Calculation time (ms) for each way is shown in the dialog:
• C++ code - 6 ms on my computer;
• C++ code with SSE Intrinsics - 3 ms;
• Inline Assembly with SSE instructions - 2 ms.
Assembly code performs better because of intensive using of the SSX registers. However, usually C++ code with SSE Intrinsics performs like Assembly code or better, because it is difficult to write an Assembly code which runs faster than optimized code generated by C++ compiler.
C++ function:
// Input: m_fInitialArray
// Output: m_fResultArray, m_fMin, m_fMax
void CSSESampleDlg::OnBnClickedButtonCplusplus()
{
m_fMin = FLT_MAX;
m_fMax = FLT_MIN;
int i;
for ( i = 0; i < ARRAY_SIZE; i++ )
{
m_fResultArray[i] = sqrt(m_fInitialArray[i] * 2.8f);
if ( m_fResultArray[i] < m_fMin )
m_fMin = m_fResultArray[i];
if ( m_fResultArray[i] > m_fMax )
m_fMax = m_fResultArray[i];
}
}
C++ function with SSE Intrinsics:
// Input: m_fInitialArray
// Output: m_fResultArray, m_fMin, m_fMax
void CSSESampleDlg::OnBnClickedButtonSseC()
{
__m128 coeff = _mm_set_ps1(2.8f); // coeff[0, 1, 2, 3] = 2.8
__m128 tmp;
__m128 min128 = _mm_set_ps1(FLT_MAX); // min128[0, 1, 2, 3] = FLT_MAX
__m128 max128 = _mm_set_ps1(FLT_MIN); // max128[0, 1, 2, 3] = FLT_MIN
__m128* pSource = (__m128*) m_fInitialArray;
__m128* pDest = (__m128*) m_fResultArray;
for ( int i = 0; i < ARRAY_SIZE/4; i++ )
{
tmp = _mm_mul_ps(*pSource, coeff); // tmp = *pSource * coeff
*pDest = _mm_sqrt_ps(tmp); // *pDest = sqrt(tmp)
min128 = _mm_min_ps(*pDest, min128);
max128 = _mm_max_ps(*pDest, max128);
pSource++;
pDest++;
}
// extract minimum and maximum values from min128 and max128
union u
{
__m128 m;
float f[4];
} x;
x.m = min128;
m_fMin = min(x.f[0], min(x.f[1], min(x.f[2], x.f[3])));
x.m = max128;
m_fMax = max(x.f[0], max(x.f[1], max(x.f[2], x.f[3])));
}
## Sources
1. Intel Software manuals.
2. MSDN, Streaming SIMD Extensions (SSE). http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vclang/html/vcrefstreamingsimdextensions.asp
3. Waterfall chart control written by Kris Jearakul. http://www.codeguru.com/controls/Waterfall.shtml
4. Microsoft Visual C++ CPUID sample. http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vcsample/html/vcsamcpuiddeterminecpucapabilities.asp
5. Matt Pietrek. Under The Hood. February 1998 issue of Microsoft Systems Journal. http://www.microsoft.com/msj/0298/hood0298.aspx
## About the Author
Software Developer
Israel
No Biography provided
## Comments and Discussions
FirstPrev Next
A question lei_ma200318-Apr-07 22:59 lei_ma2003 18-Apr-07 22:59
Question shaihnc22-Jul-05 10:41 shaihnc 22-Jul-05 10:41
Re: Question punkbuster17-Jan-06 18:55 punkbuster 17-Jan-06 18:55
A question Sachini M7-Jun-05 16:39 Sachini M 7-Jun-05 16:39
Excellent article! I'm very new to this topic and have a question. When using SSE, does the number of iterations of each loop always have to be a multiple of 4? Lets say you need to do a check (if statement inside the loop) at every iteration, is there a way to use SSE? or is there any use using it? Thanks in advance! Regards, Sachini
Re: A question Alex Fr8-Jun-05 2:43 Alex Fr 8-Jun-05 2:43
Re: A question punkbuster16-Jan-06 18:46 punkbuster 16-Jan-06 18:46
array memory alignment ? not_happy011-May-05 20:08 not_happy0 11-May-05 20:08
Re: array memory alignment ? Alex Fr12-May-05 3:07 Alex Fr 12-May-05 3:07
performance loss using SSE David St. Hilaire3-Dec-04 8:18 David St. Hilaire 3-Dec-04 8:18
Re: performance loss using SSE Alex Farber3-Dec-04 8:33 Alex Farber 3-Dec-04 8:33
Re: performance loss using SSE David St. Hilaire3-Dec-04 9:38 David St. Hilaire 3-Dec-04 9:38
Re: performance loss using SSE Alex Farber3-Dec-04 21:30 Alex Farber 3-Dec-04 21:30
AMD support Jens froslev-nielsen1-Dec-04 1:20 Jens froslev-nielsen 1-Dec-04 1:20
q: movaps vs. movups yoaz4-Nov-04 8:31 yoaz 4-Nov-04 8:31
Re: q: movaps vs. movups Alex Farber5-Nov-04 2:35 Alex Farber 5-Nov-04 2:35
Re: q: movaps vs. movups yoaz5-Nov-04 2:54 yoaz 5-Nov-04 2:54
Excelent! + a question yoaz19-Sep-04 23:52 yoaz 19-Sep-04 23:52
Re: Excelent! + a question Alex Farber20-Sep-04 0:39 Alex Farber 20-Sep-04 0:39
Re: Excelent! + a question yoaz20-Sep-04 1:54 yoaz 20-Sep-04 1:54
Is this a VC 2003 compiler BUG ? leandrobecker10-Jun-04 4:51 leandrobecker 10-Jun-04 4:51
Intel compiler Lars Schouw18-Apr-04 21:06 Lars Schouw 18-Apr-04 21:06
Re: Intel compiler Alex Farber18-Apr-04 21:21 Alex Farber 18-Apr-04 21:21
How about double data type? mrskyok16-Feb-04 0:28 mrskyok 16-Feb-04 0:28
Re: How about double data type? Alex Farber16-Feb-04 0:35 Alex Farber 16-Feb-04 0:35
Re: How about double data type? nutty9-Feb-05 3:56 nutty 9-Feb-05 3:56
Re: How about double data type? doug6553611-Aug-08 0:50 doug65536 11-Aug-08 0:50
Very good! Vincent Leong773-Aug-03 20:02 Vincent Leong77 3-Aug-03 20:02
how to use SSE under Linux? EagleCalifornia9-Sep-03 1:43 EagleCalifornia 9-Sep-03 1:43
Re: how to use SSE under Linux? Alex Farber9-Sep-03 2:18 Alex Farber 9-Sep-03 2:18
Re: how to use SSE under Linux? gnuLNX23-Sep-03 3:59 gnuLNX 23-Sep-03 3:59
Re: how to use SSE under Linux? Christophe Avoinne18-Oct-03 2:39 Christophe Avoinne 18-Oct-03 2:39
Re: how to use SSE under Linux? gnuLNX20-Oct-03 1:53 gnuLNX 20-Oct-03 1:53
Re: how to use SSE under Linux? PSuade20-Oct-03 8:40 PSuade 20-Oct-03 8:40
Re: how to use SSE under Linux? gnuLNX20-Oct-03 10:57 gnuLNX 20-Oct-03 10:57
Re: how to use SSE under Linux? PSuade21-Oct-03 9:40 PSuade 21-Oct-03 9:40
Re: how to use SSE under Linux? gnuLNX22-Oct-03 2:25 gnuLNX 22-Oct-03 2:25
Re: how to use SSE under Linux? PSuade22-Oct-03 4:22 PSuade 22-Oct-03 4:22
SSE2 Examples... godot_gildor31-Jul-03 6:52 godot_gildor 31-Jul-03 6:52
Re: SSE2 Examples... Alex Farber2-Aug-03 20:06 Alex Farber 2-Aug-03 20:06
SSE2 Examples... godot_gildor31-Jul-03 6:51 godot_gildor 31-Jul-03 6:51
Visual C++ 6.0 Processor Pack lipunov@hotmail.com21-Jul-03 15:06 lipunov@hotmail.com 21-Jul-03 15:06
Re: Visual C++ 6.0 Processor Pack Alex Farber21-Jul-03 20:46 Alex Farber 21-Jul-03 20:46
Great! Greg S.17-Jul-03 6:07 Greg S. 17-Jul-03 6:07
Excellent; well written, in depth Anonymous11-Jul-03 20:28 Anonymous 11-Jul-03 20:28
Thank you! Robert Buldoc11-Jul-03 12:43 Robert Buldoc 11-Jul-03 12:43
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A072236 Numbers of primes between successive primorials. 1
1, 2, 7, 36, 297, 2905, 39083, 603698, 11637502, 288086265, 7728273214, 251460107734, 9155428058351, 353182833587645, 15035130777416118 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS The sum of this series is pi(x). LINKS Tomás Oliveira e Silva, Tables of values of pi(x) and of pi2(x) [From Donovan Johnson, Apr 25 2010] EXAMPLE There are 3 primes less than 6, 7 primes between 6 and 30 and 36 primes between 30 and 210. MATHEMATICA Table[ PrimePi[ Product[Prime[i], {i, 1, n}]] - PrimePi[ Product[ Prime[i], {i, 1, n - 1}]], {n, 1, 12}] Join[{1}, Differences[PrimePi/@Rest[FoldList[Times, 1, Prime[Range[12]]]]]] (* Harvey P. Dale, Mar 16 2012 *) (* Mathematica's implementation of PrimePi will not work for the 13th primorial because it's too large *) CROSSREFS Cf. A002110 & A000849. Sequence in context: A012712 A012363 A012717 * A007474 A002724 A292206 Adjacent sequences: A072233 A072234 A072235 * A072237 A072238 A072239 KEYWORD hard,nonn AUTHOR Stephan Wagler (stephanwagler(AT)aol.com), Jul 05 2002 EXTENSIONS Edited by Robert G. Wilson v, Jul 08 2002 a(13)-a(14) from Donovan Johnson, Apr 25 2010 STATUS approved
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Last modified January 28 03:49 EST 2020. Contains 331317 sequences. (Running on oeis4.)
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# CBSE Class 12 Mathematics Question Paper 2017: Delhi
CBSE Class 12 Maths 2017 board exam question paper (Set 1, Delhi is available here). With this article, students can download the complete question paper in PDF format. With the help of this paper students can easily understand the level of questions which are asked in CBSE Class 12 Maths board exams.
Updated: Aug 9, 2017 10:00 IST
CBSE Class 12 Board Exam 2017: Maths Paper Analysis
Some randomly selected questions from CBSE Class 12th Maths Question Paper 2017:
Question: If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A–1) = (det A)k.
Question: If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Question: Show that all the diagonal elements of a skew symmetric matrix are zero.
Question: The volume of a sphere is increasing at the rate of 3 cubic centimeter per second. Find the rate of increase of its surface area, when the radius is 2 cm.
Question: Prove that if E and F are independent events, then the events E and F' are also independent.
Question: A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs. 100 and that on a bracelet is Rs. 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit?
It is being given that at least one of each must be produced.
Question:
If xm yn = (x + y)m + n, prove that d2y/dx2 = 0.
Question:
Prove that x2y2 = c (x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2) dx = (y3 – 3x2y) dy, where C is a parameter.
Question:
Often it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. Do you also agree that the value of truthfulness leads to more respect in the society?
Question: If the sum of lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is π/3
Download CBSE Class CBSE Class 12th Maths Question Paper 2017 (Set 1, Delhi) in PDF format
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# 2.8 decimal to fraction
2.8 decimal to fraction keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website
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2.8 decimal
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We found at least 10 Websites Listing below when search with 2.8 decimal to fraction on Search Engine
### 2.8 decimal to fraction
Coolconversion.com DA: 18 PA: 50 MOZ Rank: 68
• 65 rows · 2.8 = 14 5 = 24 5 as a fraction
• To convert the decimal 2.8 to a fraction, just follow these steps: Step 1: Write down the number as a fraction of one: 2.8 = 2.8 1
• Step 2: Multiply both top and bottom by 10 for every number after the decimal point: As we have 1 numbers after the decimal point, we multiply both numerator and denominator by 10
### 2.8 decimal to fraction
Fractioncalculator.pro DA: 22 PA: 45 MOZ Rank: 68
• Here is the answer to the question: 2.8 decimal to fraction or what is 2.8 as a fraction
• Use the decimal to fraction converter/calculator below to write any decimal number as a fraction
• Enter a decimal value: Ex.: 0.625, 0.75, .875, etc.
### Convert 2.8 to a fraction
Fractioncalculator.pro DA: 22 PA: 46 MOZ Rank: 70
• Here is the answer to the question: Convert 2.8 to a fraction or what is 2.8 as a fraction
• Use the decimal to fraction converter/calculator below to write any decimal number as a fraction
• Enter a decimal value: Ex.: 0.625, 0.75, .875, etc.
### What is 2.8 as a fraction
Visualfractions.com DA: 19 PA: 50 MOZ Rank: 72
• In the case of 28 and 10, the greatest common divisor is 2
• This means that to simplify the fraction we can divide by the numerator and the denominator by 2 and we get: 28/2
• And there you have it! In just a few short steps we have figured out what 2.8 is as a fraction.
### What is 2.8 as a fraction
Thefractioncalculator.com DA: 25 PA: 50 MOZ Rank: 79
• 2.8 / 1 To get rid of the decimal point in the numerator, we count the numbers after the decimal in 2.8, and multiply the numerator and denominator by 10 if it is 1 number, 100 if it is 2 numbers, 1000 if it is 3 numbers, and so on
• Therefore, in this case we multiply the numerator and denominator by 10 to get the following fraction: 28 / 10
### 2.8 as a fraction in simplest form
Calculator.name DA: 15 PA: 18 MOZ Rank: 38
• To convert 2.8 to fraction, follow these steps: First write down the decimal number divided by 1 like this: 2.8 / 1
• As we have 1 digits after the decimal point in the numerator, we need to multiply both the numerator and denominator by 10 1 = 10, so that there is no decimal point in the numerator
• 2.8 × 10 / 1 × 10 = 28 / 10
• Since the greatest common factor of 28 and 10 is 2, we can
### Decimal to Fraction Calculator
Calculatorsoup.com DA: 22 PA: 50 MOZ Rank: 78
• Rewrite the decimal number as a fraction with 1 in the denominator$1.625 = \frac{1.625}{1}$Multiply to remove 3 decimal places
• Here, you multiply top and bottom by 10 3 = 1000$\frac{1.625}{1}\times \frac{1000}{1000}= \frac{1625}{1000}$Find the Greatest Common Factor (GCF) of 1625 and 1000, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 125
### Online Fraction Calculator
Mycarpentry.com DA: 19 PA: 32 MOZ Rank: 58
• The online fraction calculator calculates the fraction value of any decimal number.You can select 16ths, 32nds, 64ths, or 100ths precision values
• This calculator that converts decimal values to fractions is very useful on projects where a tape measure is being used.
### 2/8 inches to decimal
Decimal.info DA: 12 PA: 50 MOZ Rank: 70
• To convert the fractional 2/8 inches to decimal, we simply divide the numerator by the denominator
• Here is the math to illustrate: 2 / 8 = 0.25 inches Thus, 2/8 inches to decimal is 0.25
• Moving on, note that there are 12 inches in a foot
• Therefore, to convert 2/8 inches to feet, you simply divide the above decimal answer by 12 like so: 0.25
### 2.8 inches in fraction
Fractioncalculator.pro DA: 22 PA: 44 MOZ Rank: 75
• Here is the answer to the question: 2.8 inches in fraction or what is 2.8 as a fraction
• Use the decimal to fraction converter/calculator below to write any decimal number as a fraction
• Enter a decimal value: Ex.: 0.625, 0.75, .875, etc.
### What is 2/8 as a decimal
Thefractioncalculator.com DA: 25 PA: 49 MOZ Rank: 84
• To convert 2/8 so you can write it as a decimal, simply divide the numerator by the denominator, like this: = 2/8
• Therefore, the solution to 2/8 as a decimal is as follows: 2/8 = 0.25
• To reiterate, just remember that when you want a fraction like 2/8 as a decimal, simply see the fraction bar as a division sign and solve the
### 2.8 Repeating as a Fraction
Calculationcalculator.com DA: 25 PA: 28 MOZ Rank: 64
• How to write 2.8 Repeating as a Fraction? Convert a repeating decimal number as a simplified fraction or mixed fraction
• Just enter the Decimal value into the input box and then press calculate button, the system will automatically calculate the Fraction value.
### 2.8 as a mixed number
Fractioncalculator.pro DA: 22 PA: 43 MOZ Rank: 77
• Welcome! Here is the answer to the question: 2.8 as a mixed number or what is 2.8 as a fraction
• Use the decimal to fraction converter/calculator below to …
### What is 2/8 as a decimal
Visualfractions.com DA: 19 PA: 50 MOZ Rank: 82
• That's literally all there is to it! 2/8 as a decimal is 0.25
• I wish I had more to tell you about converting a fraction into a decimal but it really is that simple and there's nothing more to say about it
• If you want to practice, grab yourself a pen and a pad and try to calculate some fractions to decimal format yourself.
### Fraction 2/8 to Decimal Number Equivalent
Getcalc.com DA: 11 PA: 31 MOZ Rank: 56
The fraction number = 2/8 step 2 Write it as a decimal 2/8 = 0.25 0.25 is the decimal representation for 2/8 For Percentage Conversion: step 1 To represent 0.25 in percentage, write 0.25 as a fraction Fraction = 0.25/1 step 2 multiply 100 to both numerator and denominator (0.25 x 100)/(1 x 100) = 25/100 25% is the percentage representation for 2/8
### Inch Fraction Calculator
Inchcalculator.com DA: 22 PA: 26 MOZ Rank: 63
• The chart below can be used to easily find the correct fraction for your decimal measurement, or vice-versa
• Find decimal equivalents in 1 ⁄ 64 ” increments, including 1 ⁄ 2 “, 1 ⁄ 4 “, 1 ⁄ 8 “, and 1 ⁄ 16 “, and 1 ⁄ 32
• The chart also shows hints on the markings sizes found on a tape measure or ruler.
### Inches to Fraction Calculator
Omnicalculator.com DA: 22 PA: 30 MOZ Rank: 68
• Transforming a distance in its decimal form to its fraction inches is almost the same as converting any decimal to a regular fraction.Almost
• Principally, we have to find the ratio of two numbers, the numerator and the denominator
• The only difference is that the denominator should be to the power of 2: 2, 4, 8, 16, etc.
• So, how to convert decimal to inch fraction?
### Convert to a Decimal 2.8% Mathway
Mathway.com DA: 15 PA: 35 MOZ Rank: 67
• Convert the percentage to a fraction by placing the expression over 100 100
• Percentage means 'out of 100 100 '
• Convert the decimal number to a fraction by shifting the decimal point in both the numerator and denominator.
### Convert Decimal to Fractions How to convert Decimals to
Byjus.com DA: 9 PA: 35 MOZ Rank: 62
• Solve problems based on conversion of decimal to fraction and fraction to decimal are given here
• These examples of decimal to fraction conversions will help you in better understanding of the concept
• Question 1: Find the fraction form of the decimal 0.7
• Solution: Given, decimal number 0.7, we need to find the fraction for 0.7.
### 0.028 or 2.8% as a Fraction
Getcalc.com DA: 11 PA: 33 MOZ Rank: 63
• Input parameters and values: The decimal number = 0.028 step 2 Write it as a fraction 0.028/1 step 3 Multiply 1000 to both numerator and denominator (0.028 x 1000)/(1 x 1000) = 28/1000 It can be written as 2.8% = 2.8/100 or 28/1000 step 4 To simplify 28/1000 to its lowest terms, find LCM (Least Common Multiple) for 28 and 1000
### What is 2.8 repeating as a fraction
Numbermaniacs.com DA: 17 PA: 50 MOZ Rank: 87
• 2.8 is a repeating decimal number and you want to convert it to a fraction or mixed number
• When you say 2.8 repeating, you mean that the 1 is repeating
• Here is the question formulated in mathematical terms with the vinculum line above the decimal number that is repeating
### Decimal to Fraction Calculator
Rapidtables.com DA: 19 PA: 40 MOZ Rank: 80
• How to convert decimal to fraction Conversion stages
• Write the decimal fraction as a fraction of the digits to the right of the decimal period (numerator) and a power of 10 (denominator)
• Find the greatest common divisor (gcd) of the numerator and the denominator
• Reduce the fraction by dividing the numerator and the denominator with the gcd
### Is 2/8 a terminating decimal
Decimal.info DA: 12 PA: 50 MOZ Rank: 84
• We want to know if the decimal number you get when you divide the fraction 2/8 (2 ÷ 8) is terminating or non-terminating
• Here are the steps to determine if 2/8 is a terminating decimal number: 1) Find the denominator of 2/8 in its lowest form
• The greatest common factor (GCF) of 2 and 8 is 2.
### Express 2.8 as a fraction
Answers.everydaycalculation.com DA: 31 PA: 16 MOZ Rank: 70
• Steps to convert decimal into fraction
• Multiply both numerator and denominator by 10 for every number after the decimal point
### Fraction 1 2/8 as decimal
Fractioncalculator.pro DA: 22 PA: 46 MOZ Rank: 92
• Fraction to decimal explained: To convert this mixed number (fraction) to a decimal just follow these two steps: Step 1: divide numerator (2) by the denominator (8): 2 ÷ 8 = 0.25
• Step 2: add this value to the the integer part: 1 + 0.25 = 1.25
### Fraction to Decimal Calculator
Rapidtables.com DA: 19 PA: 40 MOZ Rank: 84
• How to convert fraction to decimal Method #1
• Expand the denominator to be a power of 10
• 3/5 is expanded to 6/10 by multiplying the numerator by 2 and denominator by 2:
### Equivalent fractions for 2/8
Coolconversion.com DA: 18 PA: 33 MOZ Rank: 77
• To find an equivalent fraction to 2 8, or to any other fraction, you just need to multiply (or divide, if the fraction is not yet reduced), both the numerator and the denominator of the given fraction by any non-zero natural number
• For example: By dividing the original fraction by 2, we get: 2 ÷ 2 8 ÷ 2 = 1 4.
### Decimal to Fraction Calculator
Mathsisfun.com DA: 18 PA: 42 MOZ Rank: 87
• Shows the steps to convert a decimal number to its equivalent fraction
• Converting Decimals to Fractions Greatest Common Factor …
### What Is 38 2/8 As A Decimal
Rebab.net DA: 9 PA: 18 MOZ Rank: 55
Enter fraction:= Convert× ResetSwapDecimal result:Calculation:Decimal to fraction converter How to convert fraction to decimalMethod #1Expand the denominator to be a power of 10, Example #13/5 is expanded to 6/10 by multiplying the numerator by 2 and denominator by 2:3=3×2=6=0
### What is 2.8 as a fraction
Study.com DA: 9 PA: 46 MOZ Rank: 84
• When you say 2.8 out loud, you hear: ''2 and 8 tenths''
• This can also be written as a mixed number of 2 8/10.
### What is 2 8/19 as a decimal
Coolconversion.com DA: 18 PA: 50 MOZ Rank: 98
• 2 8/19 is equal to 2.4210526315789 in decimal form
• Use our fraction to decimal calculator to convert any fraction to a decimal and to know if it is a terminating or a recurring (repeating) decimal.
### What is 6 2/8 as a decimal
Visualfractions.com DA: 19 PA: 50 MOZ Rank: 100
• It's very common when learning about fractions to want to know how convert a mixed fraction like 6 2/8 into a decimal
• In this step-by-step guide, we'll show you how to turn any fraction into a decimal really easily
• Let's take a look! Before we get started in the fraction to decimal conversion, let's go over some very quick fraction basics.
### Fraction 2 8/58 decimal equivalent
Coolconversion.com DA: 18 PA: 50 MOZ Rank: 100
• 2 8/58 is equal to 2.1379310344828 in decimal form
• Use our fraction to decimal calculator to convert any fraction to a decimal and to know if it is a terminating or a recurring (repeating) decimal.
### What is 2.8 percent as a fraction
Math.answers.com DA: 16 PA: 36 MOZ Rank: 85
1) Convert 2.8% to decimal form: 2.8% = .028 2) Convert the deicmal to a fraction by putting a 1 under it: .028/1 3) Remove the decimal and add as many numbers that are on the right of the decimal to the right of the 1: .028/1 becomes 028/1000 (also written 28/1000) 4 Reduce the fraction to its lowest term: 28/1000 = 7/250
### What is 4 2/8 as a decimal
Visualfractions.com DA: 19 PA: 50 MOZ Rank: 16
• So the answer is that 4 2/8 as a decimal is 4.25
• And that is is all there is to converting 4 2/8 to a decimal
• We convert it to an improper fraction which, in this case, is 34/8 and then we divide the new numerator (34) by the denominator to get our answer.
### What is 2/8 as a percentage
Visualfractions.com DA: 19 PA: 50 MOZ Rank: 15
• Now we can see that our fraction is 25/100, which means that 2/8 as a percentage is 25%
• We can also work this out in a simpler way by first converting the fraction 2/8 to a decimal.To do that, we simply divide the numerator by the denominator:
### Fraction to Decimal Conversion
Math2.org DA: 9 PA: 35 MOZ Rank: 80
• Fraction to Decimal Conversion Tables Important Note: any span of numbers that is underlined signifies that those numbes are repeated
• For example, 0.09 signifies 0.090909
• Only fractions in lowest terms are listed
• For instance, to find 2/8, first simplify it to 1/4 then search for it in the table below.
### Fraction to Percent Calculator
Calculatorsoup.com DA: 22 PA: 50 MOZ Rank: 13
• So converting a fraction such as 1/4 to a decimal means you need to solve the math: 1 divided by 4
• Multiply by 100 to convert decimal number to percent
• You can reduce a fraction before converting to a decimal but it's not necessary because the answer will be the same.
### Recurring Decimal To Fraction Calculator
Calcul.com DA: 14 PA: 46 MOZ Rank: 98
• For example, since 2 / 3 = 0.666666666, to express the fraction 2 / 3 in the decimal system, we require an infinity of 6s
• Such decimals are referred to as __recurring (or repeating) decimals__
• ##Recurring decimal to fraction## Every recurring decimal has a representation as a fraction.
### Fraction to Percent (Converting Fractional Numbers to
Byjus.com DA: 9 PA: 27 MOZ Rank: 75
• Examples of Fraction to Percent
• Example 1: Convert 3/4 to a percent
• Solution: Step 1: Convert the fraction 3/4 into decimal Step 2: 3/4 = 0.75 Step 3: Multiply the decimal by 100: 0.75 × 100 = 75% Therefore, the solution is 75%
• Example 2: Convert 3/16 to percent
• Solution: Step 1: Convert the fraction 3/16 into decimal Step 2: 3/16 =0.1875
### What is 26 2/8 in decimal form
Decimal.info DA: 12 PA: 50 MOZ Rank: 10
• To get 26 2/8 in decimal form, we basically convert the mixed number to a fraction and then we divide the numerator of the fraction by the denominator of the fraction
• Here are the detailed math steps we use to convert 26 2/8 mixed number to decimal form: Step 1: Multiply the whole number by the denominator: 26 × 8 = 208
### Fractions rounded to 1/4, 1/8, 1/16, 1/32, 1/64, 1/128
Dmcritchie.mvps.org DA: 19 PA: 19 MOZ Rank: 79
• Fractions rounded to 1/4, 1/8, 1/16, 1/32, 1/64, 1/128
• The formatted display of fractional measurements in Excel leaves a lot to be desired
• This example and the formulas shown below were generously provided by Bernie Deitrick as a follow-up to a posting 2000-08-04 in the microsoft.public.excel.misc newsgroup.
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MT1Solutions - Econ 100A Spring 2004 Midterm 1 Solutions A1...
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Econ 100A Spring 2004 Midterm 1 Solutions A1. 17 Points (a) [8 points] Neither analyst is correct. When a tax is placed on producers, they will raise the price they charge consumers (P’ to Pc); thus, consumers are harmed by a tax on producers. When a tax is placed on consumers, they will be willing to pay less to producers than before; thus, producers are harmed by a tax on consumers. Grade Guide: [5 points for correct answers with no/incorrect explanation],[4 points for incorrect answer with a partially correct explanation] (b) [9 points] Graphical Method: When a tax (t) is placed on this market, the price consumers pay rises from P* to P C while the price producers receive remains at P*. From the graph, t = P C – P*. Using the basic formula, Tax incidence on consumers = p = P C – P* = P C P* = 1 ∆τ t P C – P* Alternative Method: Applying the Elasticity form of the tax incidence on consumers formula, η = η = 1 η ε η η is the elasticity of supply, and ε = 0 is the elasticity of demand. Grade Guide: [6 points for correct answer with no/incorrect explanation], [4 points for incorrect answer with correct graph or formula], [2 points for only identifying ε = 0, or that consumers pay more of tax], [-1 point for not reducing or a wrong ε ]
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Solution: A2 Figure 1 shows the world market in the absence of US subsidies Q P p * Q d world Q s world q * US +q * BF= q * world q * US q * BF Q s US Q s BF Figure 2 shows the world market with US subsidies Q P p * Q d world Q s world q US +q BF= q world q US q BF Q s US Q s BF Q s US -- subsidy Q s world -- subsidy p Initially, the world price is determined by world demand and world supply, where world supply is simply the horizontal sum of US and Burkina Faso supply. This price is denoted by p * . World quantity is q * world , US supply is q * US , and Burkina Faso supply is q * BF . Once a subsidy is introduced for US suppliers, the US supply curve shifts to the right. This increases world supply and drives the world price down. While US supply increases to q’ US , Burkina Faso’s supply decreases falls to q’ BF . US suppliers benefit from the subsidy because their output increases and they receive more (price plus subsidy) than before. Burkina Faso suppliers are harmed – their supply decreases and they receive a lower price. World consumers of cotton benefit from the lower prices. US taxpayers are harmed – they have to foot the bill for the subsidy.
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Shapes
Rectangles
Square (as special rectangle)
Triangles
as half rectangles
as set of angles
ANGLES
The full details are in the Nepal Constitution, Schedule 1
SCHEDULE-1
(Relating to Article 6)
NATIONAL FLAG
(A) Method of Making the Shape inside the Border
1) On the lower portion of a crimson cloth draw a line AB of the required length from left to right.
2) From A draw a line AC perpendicular to AB making AC equal to AB plus one third AB. From AC mark off D making line AD equal to line AB. Join BD.
3) From BD mark off E making BE equal to AB.
4) Touching E draw a line FG, starting from the point F on line AC, parallel to AB to the right hand-side. Mark off FG equal to AB.
5) Join CG.
(B) Method of Making the Moon
6) From AB mark off AH making AH equal to one-fourth of line AB and starting from H draw a line HI parallel to line AC touching line CG at point I.
7) Bisect CF at J and draw a line JK parallel to AB touching CG at point K.
8) Let L be the point where lines JK and HI cut one another.
9) Join JG.
10) Let M be the point where line JG and HI cut one another.
11) With centre M and with a distance shortest from M to BD mark off N on the lower portion of line HI.
12) Touching M and starting from O, a point on AC, draw a line from left to right parallel to AB.
13) With centre L and radius LN draw a semi-circle on the lower portion and let P and Q be the points where it touches the line OM respectively.
14) With centre M and radius MQ draw a semi-circle on the lower portion touching P and Q.
15) With centre N and radius NM draw an arc touching PNQ at R and S. Join RS. Let T be the point where RS and HI cut one another.
16) With centre T and radius TS draw a semi-circle on the upper portion of PNQ touching it at two points. 91
17) With centre T and radius TM draw an arc on the upper portion of PNQ touching at two points.
18) Eight equal and similar triangles of the moon are to be made in the space lying inside the semi-circle of No. (16) and outside the arc of
No. (17) of this Schedule.
(C) Method of Making the Sun
19) Bisect line AF at U, and draw a line UV parallel to AB line touching line BE at V.
20) With centre W, the point where HI and UV cut one another and radius MN draw a circle.
21) With centre W and radius LN draw a circle.
22) Twelve equal and similar triangles of the sun are to be made in the space enclosed by the circles of No. (20) and No. (21) with the two
apexes of two triangles touching line HI.
(D) Method of Making the Border
23) The width of the border will be equal to the width TN. This will be of deep blue colour and will be provided on all the sides of the flag.
However, on the five angles of the flag the external angles will be equal to the internal angles.
24) The above mentioned border will be provided if the flag is to be used with a rope. On the other hand, if it is to be hoisted on a pole, the hole on the border on the side AC can be extended according to requirements.
Explanation: - The lines HI, RS, FE, ED, JG, OQ, JK and UV are imaginary. Similarly, the external and internal circles of the sun and the
other arcs except the crescent moon are also imaginary. These are not shown on the flag.
+
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http://www.ck12.org/geometry/Congruent-Angles-and-Angle-Bisectors/exerciseint/Use-the-Side-Splitter-Theorem-part-8/
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Typesetting math: 100%
# Congruent Angles and Angle Bisectors
## Bisectors split the angle into two equal halves.
Estimated9 minsto complete
%
Progress
Practice Congruent Angles and Angle Bisectors
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated9 minsto complete
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Use the Side Splitter Theorem (part 8)
Teacher Contributed
Use the Side Splitting Theorem to solve for the value of $y$$y$.
A
B
E
C
D
18
$3y-32$
$4y+6$
6
qid: 100307
Reviews
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https://www.wyzant.com/resources/answers/44691/division_of_polynomials_x_3_7_x_2
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0
# Division of polynomials: x^3-7/x-2
This is a division of polynomials. It is supposed to be divided by using the division algorithm. Please help me!
### 1 Answer by Expert Tutors
4.8 4.8 (219 lesson ratings) (219)
0
There are a couple of ways to attack this, but I will choose my way.
Let's rewrite the expression as:
(x^3 - 8)/(x - 2) + 1/(x - 2)
This is equal to our original expression.
When factoring a difference of two cubes you get this:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
x^3 - 8 = (x - 2)(x^2 +2x + 4)
Now we have:
(x - 2)(x^2 + 2x + 4)/(x - 2) + 1/(x - 2)
You can cancel out the x-2 term in the numerator & denominator. You get:
x^2 + 2x + 4 + 1/(x-2)
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http://openstudy.com/updates/4fd5823ee4b04bec7f16b89f
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FatimaISWA 3 years ago Find the greatest common factor (GCF): 2x2 - 40x
1. jim_thompson5910
What is the GCF of 2 and -40?
2. jim_thompson5910
Any ideas?
3. FatimaISWA
the GCF of 2 is 18,36,and 90. i'm not sure of -40
4. jim_thompson5910
Its simply 2, I'm not sure where you're getting 18 36 and 90..
5. jim_thompson5910
now what is the GCF of x^2 and x?
6. WONDEMU
the GCF of x^2 and x is x
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https://eudml.org/doc/37951
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# On the distributive radical of an Archimedean lattice-ordered group
• Volume: 59, Issue: 3, page 687-693
• ISSN: 0011-4642
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## Abstract
top
Let $G$ be an Archimedean $\ell$-group. We denote by ${G}^{d}$ and ${R}_{D}\left(G\right)$ the divisible hull of $G$ and the distributive radical of $G$, respectively. In the present note we prove the relation ${\left({R}_{D}\left(G\right)\right)}^{d}={R}_{D}\left({G}^{d}\right)$. As an application, we show that if $G$ is Archimedean, then it is completely distributive if and only if it can be regularly embedded into a completely distributive vector lattice.
## How to cite
top
Jakubík, Ján. "On the distributive radical of an Archimedean lattice-ordered group." Czechoslovak Mathematical Journal 59.3 (2009): 687-693. <http://eudml.org/doc/37951>.
@article{Jakubík2009,
abstract = {Let $G$ be an Archimedean $\ell$-group. We denote by $G^d$ and $R_D(G)$ the divisible hull of $G$ and the distributive radical of $G$, respectively. In the present note we prove the relation $(R_D(G))^d=R_D(G^d)$. As an application, we show that if $G$ is Archimedean, then it is completely distributive if and only if it can be regularly embedded into a completely distributive vector lattice.},
author = {Jakubík, Ján},
journal = {Czechoslovak Mathematical Journal},
keywords = {Archimedean $\ell$-group; divisible hull; distributive radical; complete distributivity; Archimedean -group; divisible hull; distributive radical; complete distributivity},
language = {eng},
number = {3},
pages = {687-693},
publisher = {Institute of Mathematics, Academy of Sciences of the Czech Republic},
title = {On the distributive radical of an Archimedean lattice-ordered group},
url = {http://eudml.org/doc/37951},
volume = {59},
year = {2009},
}
TY - JOUR
AU - Jakubík, Ján
TI - On the distributive radical of an Archimedean lattice-ordered group
JO - Czechoslovak Mathematical Journal
PY - 2009
PB - Institute of Mathematics, Academy of Sciences of the Czech Republic
VL - 59
IS - 3
SP - 687
EP - 693
AB - Let $G$ be an Archimedean $\ell$-group. We denote by $G^d$ and $R_D(G)$ the divisible hull of $G$ and the distributive radical of $G$, respectively. In the present note we prove the relation $(R_D(G))^d=R_D(G^d)$. As an application, we show that if $G$ is Archimedean, then it is completely distributive if and only if it can be regularly embedded into a completely distributive vector lattice.
LA - eng
KW - Archimedean $\ell$-group; divisible hull; distributive radical; complete distributivity; Archimedean -group; divisible hull; distributive radical; complete distributivity
UR - http://eudml.org/doc/37951
ER -
## References
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1. Birkhoff, G., Lattice Theory, Revised Edition Providence (1948). (1948) Zbl0033.10103MR0029876
2. Byrd, R. D., Lloyd, J. T., 10.1007/BF01136029, Math. Z. 101 (1967), 123-130. (1967) Zbl0178.02902MR0218284DOI10.1007/BF01136029
3. Darnel, M. R., Theory of Lattice-Ordered Groups, M. Dekker, Inc. New York-Basel- Hong Kong (1995). (1995) Zbl0810.06016MR1304052
4. Jakubík, J., Representation and extension of $\ell$-groups, Czech. Math. J. 13 (1963), 267-283 Russian. (1963) MR0171865
5. Jakubík, J., Distributivity in lattice ordered groups, Czech. Math. J. 22 (1972), 108-125. (1972) MR0325487
6. Jakubík, J., 10.1023/A:1013781300217, Czech. Math. J. 51 (2001), 889-896. (2001) MR1864049DOI10.1023/A:1013781300217
7. Lapellere, M. A., Valente, A., Embedding of Archimedean $\ell$-groups in Riesz spaces, Atti Sem. Mat. Fis. Univ. Modena 46 (1998), 249-254. (1998) MR1628633
8. Sikorski, R., Boolean Algebras, Second Edition Springer Verlag Berlin (1964). (1964) Zbl0123.01303MR0126393
## NotesEmbed?
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http://domainconverters.com/frequency/yoctohertz-conversion/
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# Yoctohertz (yHz) a Frequency Measurement Unit and Conversion Chart
Domainconverters > Frequency Conversions > yoctohertz(yHz) Conversion
## Q: What is a Yoctohertz ?
It is non-SI multiple of the frequency unit hertz.
## Q: How much is 1 Yoctohertz ?
It is 1.00E-24 times hertz .
Symbol : yHz
So 1 yoctohertz = 1.00E-24 hertz.
Example : 23 yoctohertz = 1.00E-24 X 23 Hertz.
Or we can say, 23yHz = 2.3E-23 Hz.
## Yoctohertz Conversion Table and Chart
Use this conversion chart to compare one yoctohertz with other frequency measurement units.
1.0E-6 aHz1.0E-22 cHz
1.0E-23 dHz1.0E-25 daHz
1.0E-42 EHz1.0E-9 fHz
1.0E-33 GHz1.0E-24 Hz
1.0E-27 kHz1.0E-30 MHz
1.0E-18 µHz1.0E-21 mHz
1.0E-15 nHz1.0E-39 PHz
1.0E-12 pHz1.0E-36 THz
1 yHz1.0E-48 YHz
0.001 zHz1.0E-45 ZHz
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https://www.reddit.com/r/iamverysmart/comments/2b6gdx/good_lord_this_guys_really_smart/
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×
This is an archived post. You won't be able to vote or comment.
[–] 469 points470 points (27 children)
Does he actually understand general relativity?
That's the first thing people notice about you? Dude, you need to work on your small talk
[–][deleted] (12 children)
[deleted]
[–] 130 points131 points (10 children)
That's special relativity.
God I'm just as bad as that guy :/
[–] 48 points49 points (8 children)
What's the difference between general and special relativity? You seem to be REALLY smart! No, really, I'm curious.
[–] 33 points34 points (7 children)
I've only ever done a course on SR so I am hardly an expert. As far as I know SR deals only with inertial reference frames (i.e. ones not accelerating), which is the special case, whereas GR deals also with acceleration/gravity, i.e. the general case.
[–] 41 points42 points (6 children)
You're right, but there is something more to it. The central idea of general relativity can be said to be "matter tells space how to curve, space tells matter how to move."
SR deals with moving objects and how they see each other. It's more or less a direct consequence of Maxwell's laws, if you follow them to their logical conclusion.
GR is a geometric model of the universe, that uses the curvature of a 4-dimensional spacetime due to energy concentrations to explain why we see gravitational effects.
I think I heard it said somewhere that the discovery of SR was more or less inevitable around the time Einstein wrote about it, but if he hadn't gone on to figure out GR, we probably wouldn't know about it yet today.
[–] 63 points64 points (2 children)
Do you actually understand general relativity? Did you just explain the concept in a way that I understand it?? Good lord, you're REALLY smart!
[–] 47 points48 points (1 child)
Didn't use elucidate in a sentence 0/10
[–] 0 points1 point (0 children)
10/10 worst comment ever
[–] 3 points4 points (2 children)
I've heard that, but I'd argue that GR does arise from somewhat similar mathematics to SR, i.e. popularisation of the idea that non-Euclidean spaces are actually a viable way of describing the universe. As far as I understand it (and my GR is a bit shaky), SR relies on relaxing the metric requirement of Euclidean space, while the space in which GR is set arises from relaxing the affine requirement.
I'm doing it, aren't I? Sorry about that.
[–] 2 points3 points (1 child)
I think we were all doing it, to be honest.
[–] 2 points3 points (0 children)
Yeah. But GR is really not something I should be talking about. I've researched Euclidean space plenty, but that's pretty much where my expertise ends.
[–] 6 points7 points (0 children)
Whoops. I'm in the wrong here. I suck.
[–] 11 points12 points (0 children)
"...cool...so, did you catch the race last n-"
"I AM NO REDNECK."
[–] 132 points133 points (2 children)
He has no place for small talk. Only primitives speak with small talk
[–] 48 points49 points (1 child)
Smart minds discuss ideas average minds discuss events and weak minds discuss people, that's why he always makes sure to tell people he understands general relatively so they know he's snart
[–] 1 point2 points (0 children)
*smrt
[–] 138 points139 points (0 children)
Why is this fat nerd telling me about Einsteins theories? Christ, what an aspie.
[–] 11 points12 points (4 children)
What he doesn't understand when they think that is that it is obvious that he doesn't know a thing about it but continues to explain it anyway.
[–] 4 points5 points (0 children)
My friends brother in law does this. He thinks he's great with understanding physics and has never taken a physics class.
[–] 4 points5 points (2 children)
"I UNDERSTAND GENERAL RELATIVITY"
"What's a manifold?"
"Uh...."
[–] 13 points14 points (0 children)
Shit man, I'm a physics Ph.D. student and I don't even know the answer to that.
[–] 2 points3 points (0 children)
The number of people who mistakenly think they understand general relativity is much, much greater than the number of people who understand general relativity.
[–] 2 points3 points (1 child)
What did you read? The general relativity thing is clearly #5.
[–] 1 point2 points (0 children)
Seriously.
[–] 1 point2 points (1 child)
How do you explain that you understand general relativity in casual conversation?
"You see the alternating tensor to the mu prime subbed x is equal to lambda to the... so then you take the scalar product..."
Seems like some quality small talk.
[–] 2 points3 points (0 children)
" Hi, my name is Bob and I look even more appealing behind a gravitational lens. "
[–] 0 points1 point (0 children)
This is kind of funny because there's fairly recent PRL papers where people just go like "There exists a solution to GR that predicts this, but we won't work out the specifics because those equations are a bugger to work with" (with slightly less casual words). Understand. General Relativity. At least with quantum mechanics we can work out the equations sometimes...
[–] 228 points229 points (8 children)
7) "God this guy's an arrogant twat"
[–] 78 points79 points (7 children)
8) "I've lost all faith in the human race. I think I'll go home and mix ammonia with bleach and take a real good whiff."
[–] 39 points40 points (5 children)
That's how you make crystals
[–] 25 points26 points (1 child)
it reminds me of a banana muffin recipe i saw :^)
[–] 24 points25 points (0 children)
[–] 15 points16 points (1 child)
(for anyone unaware, this is a reference to a lethal "prank". Don't do it!)
[–] 3 points4 points (0 children)
But chlorine gas is such a good painkiller !
[–] 0 points1 point (0 children)
You forgot the pennies, that's the most important part.
[–] 3 points4 points (0 children)
Bleach and alcohol are my favorite way to make scented candles.
[–][deleted] (35 children)
[deleted]
[–] 184 points185 points (0 children)
elucidating intensifies
[–] 43 points44 points (23 children)
Good god do I want to see that.
[–][deleted] (22 children)
[deleted]
[–] 53 points54 points (21 children)
Who am I?
I am a genius. Unfortunately, I am also uneducated. I have no time, no finances, no credit, and no psychological capacity for college. Feel free to ask for specifics, my life is (mostly) and open book. Point is, I was unable to finish college, and I can't go back now. So I do what I can to learn as much as I can about everything, every day.
Good lord.
[–][deleted] (9 children)
[deleted]
[–] 22 points23 points (2 children)
| Diet: Mostly other
Is "other" just Doritos and M'Dew?
[–] 26 points27 points (1 child)
Oh my god I am dying over "M'Dew."
"M'Dew." sipping intensifies
[–] 4 points5 points (0 children)
Me too. Death toll mounting.
[–] 16 points17 points (5 children)
Where are you guys finding more? I'm not smart enough to elucidate the source of the new material.
[–] 16 points17 points (3 children)
Take one of the longer sentences in the post and put it in double quotes, then let Google point you to enlightenment.
[–] 23 points24 points (2 children)
Wow, that was surprisingly depressing.
[–] 10 points11 points (1 child)
Yeah, I'm kind of sad now.
[–] 0 points1 point (0 children)
Unsave. Shit. I need to scrub my eyes.
[–] 2 points3 points (0 children)
If you google any of his elegant elucidations, his OkCupid account comes up
[–] 11 points12 points (4 children)
I actually used to be like that. Then, you know, got a job, went back to college, now I can feel permanently inferior because everyone I compare myself to is an established academic.
[–] 1 point2 points (3 children)
The part where he brags about not having had a vacation since 2004 is the worst. Broke as I am, travel is still a serious priority for me.
[–] 0 points1 point (2 children)
Yeah, like, when I was like that I was mostly useless by travelling around too much to do anything productive. Part of the reason to finish university for me was to be able to go to international conferences and find jobs in random countries! At my current company I'd get 7 weeks of vacation a year if I stayed that long but I'm moving for grad school.
[–] 1 point2 points (1 child)
There's a pretty wide area between traveling so much that it's a detriment to your life and not taking a vacation for a decade.
[–] 0 points1 point (0 children)
Yeah, plus it's perfectly possible to integrate an actual career and travel.
[–] 3 points4 points (0 children)
He's from my home town... Embarrassing.
[–] 4 points5 points (0 children)
I am also uneducated.
But he totes understands general relativity.
[–] 0 points1 point (0 children)
I'll say this in his defense. He knows what he wants. He's not hiding his character (flaws and all) behind anything.
The downside is that he's going to repulse the vast majority of people who see his profile. But the people who, by some quirk of nature, find the profile endearing? He's got a great shot with them.
[–] -3 points-2 points (0 children)
I am a genius
Ok...
(mostly) and open book
A genius that can't use simple, first grade grammar.
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Blur out the necessary bits and it should be fine, I think.
[–] 5 points6 points (1 child)
Black bar over the eyes should be fine.
[–] 10 points11 points (0 children)
It's more traditional to put the black bar over the genitals.
Wait, what sub am I in?
[–] 4 points5 points (0 children)
This cannot be real surely? The 'halloween' photo says it all.
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a pierced fedora?
"Tipping a pierced fedora" does sound like some accessible book on the mysteries of the physics of space though.
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Read the first few points and I'm not surprised in the slightest.
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genius dropped out of university ok then
[–] 60 points61 points (18 children)
Those confident enough in their mental abilities let other people compliment their intelligence.
Only those who are insecure feel the need to bring it up themselves.
[–] 37 points38 points (10 children)
Very true. My buddy's dad told us growing up "if you really are something, you don't need to tell anyone."
[–] 87 points88 points (4 children)
"Real gangsta ass niggas don't flex nuts Cause real gangsta ass niggas know they got em"
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I don't understand this reference
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Game of Thrones - "Any man who must say, "I am the king" is no true king. "
[–] 2 points3 points (0 children)
Gotcha! Never watched the show. I suspect it's a fairly common lesson in shows and such.
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That's awful advice. I spent two years literally invisible because I was waiting for someone else to bring it up.
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The rule of status: If you need to tell people you have it, you don't. Obama never has to stamp his foot and shout 'I am the President!'
[–] 2 points3 points (2 children)
Well Lincoln had to once or twice.
[–] 1 point2 points (1 child)
In fairness, one of those times was stomping his boot on Jefferson Davis's face during their overtime brawl to decide who won the Civil War.
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He was just setting up for the Great Emancipator though. Doesn't count.
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Tywin Lannister agrees.
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[–] 50 points51 points (2 children)
"The word is 'elucidate.'"
"Can you use it in a sentence?"
"Sure. 'Elucidate' is hard to spell."
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Good lord, you're really smart.
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"I don't know what elucidate means"
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Oh yeah, this fat redneck NASCAR fan just totally schooled me on Einstein's theories of gravitation.
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one of those 'autistic' kids...their vision is based on movement
Hey come on, man. There's no need to bring autism into this. Don't be a dick.
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Hey come on, man. There's no need to bring dicks into this. Don't be a shit.
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[deleted]
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Hey come on, man. There's no need to bring niggerfaggotjewretards into this. Don't be a stanfordtree.
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[deleted]
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Lol
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At first I thought sooner or later that guy would get put in his place by someone that is actually intelligent. Then I realized he probably never interacts with other human beings outside the Internet.
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He dropped out of college
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Obviously by "dropped out" you mean he voluntarily left because it was too simple and he decided self teaching would be just as good (and when his potential employer asks for proof of knowledge, he proudly presents a portfolio of Facebook statuses).
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He was more interested in REAL learning (Wikipedia binges and Youtube videos) than being indoctrinated by the man.
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Does he actually understand general relativity?
No, he doesn't. You understand general relativity in the way I understand I car engine - I know the basic principles on which one works. Would I be able to build one? To do something with it? No, so the knowledge is mostly useless.
[–] 10 points11 points (0 children)
I built a general relativity once. It was kind of hard, but I figured it out.
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No, so the knowledge is mostly useless.
I'm a physics Ph.D. student, and I disagree completely. There is nothing useless about knowing the ideas behind relativity, because it is so contrary to common sense and simple observations of the world we live in, that just understanding that the world is actually drastically different is a huge step towards appreciating how little you know. You don't have to be able to do anything math with it for it to change the way you think about the universe. In that way, I think the math part of the relativity is the "mostly useless" part, in that the only people that really need to know it are a very specific subset of academic researchers and people who work at space agencies.
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Yes, mostly useless. It may change your thinking a little bit but unless you're involved with it for real it has very little effect on your day to day life.
My knowledge of an engine may be able to help me diagnose what is wrong if something gets broken, but I could not do much beyond that.
Relativity is one of those things people like to pretend to know to look smart.
[–] 2 points3 points (3 children)
Exactly. He probably has just read the "rubber-sheet" analogy and thinks he's the bee's knees. I know several physicists at my uni who don't truly understand GR (because it isn't their field, obviously, but the point still stands)
[–] 2 points3 points (2 children)
I'm pretty sure the general relativity guys I know still don't truly understand it. I think I have some biases there (as a quantum person, I just see it as really fucking complicated equations), but still. It's not completely intuitive or observable in everyday life.
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Yeah. I picked up a book on it from the uni library, just out of interest, and opened it to a random page. I noped outta there pretty ricky-tick
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I took out a quantum field theory book from the library. Have had it on loan for ages. Every few months I open it again and understand a bit more. It's nice!
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useless for the average person unless they're showing off. anything that relies on GPS has to account for relativity.
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"Dude like if you are accelerating you can't tell the difference from that and being in a gravitational field! I just explained general relativity!"
Jesus H Christ I hope this guy is trolling.
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Welllll... You described one of the observations that led to general relativity, but to actually explain it you'd at least need to explain spacetime curvature, and the effects of the mutual influence between spacetime and energy.
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You didn't explain that in a way that I can understand. Good lord, you're really not as smart as this guy.
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Don't make excuses. Now we all know that you are not very smart.
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Until you use "elucidate" properly in a sentence I'm afraid I don't care what you say!
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"I have an elucidate on Skype tonight."
(I'm assuming e-luci-date means an electronic date with a girl named Luci.)
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This sub should be renamed "spot the kid with aspergers"
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How much do you have to use "elucidate" for it to define you.
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that scalated in an expected way
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Anyone else read that in the voice of the Simpsons' Comic Book Guy?
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Yes.
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Why do mediocre people latch on to the theory of relativity like it's the defining characteristic of smart people? Anyone can understand it if they bother to spend a few minutes reading.
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seriously. they always talk about how they understand "e=mc2". yeah, so did everyone in my high school physics class.
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Oh wow, those letters and numbers stand for something! I can now bring this up in conversation all the time.
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the sad thing is when they claim to "understand it", it's usually them just know what each letter represents and maybe some vague understanding of energy and mass being related.
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No they can't? People spend years in college studying to understand the theory of relativity. I don't really et how you could say that just a few minutes would give even close to understanding of the topic.
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This has to be a joke, right? Right?
[–] 3 points4 points (2 children)
I'm not sure why but I get the vibe that their favorite show is The Big Bang Theory.
[–] 2 points3 points (1 child)
No, they probably call TBBT "nerd blackface" or some other victim thing.
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Well, that is a mostly accurate, if extreme, way to put it.
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What website is this from?
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It looks like okcupid.
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Good Lord!
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# What was the date 4 months ago?
## Saturday December 23, 2023
4 months before today was 23 Dec 2023, a Saturday. Subtracting 4 months in the past is usually just counting from April; however, longer calculations will push us from 2024 into 2023 or further. Even if the calculation remains within the year, I usually take note of the 7 days left in the end of April to prep for an additional calculation before 2024-04-23 04:50:43.662543. But that’s all there is to it - understand the month, if 4 months will change the year, season, or fiscal calendar, and subtract your dates.
## How we calculated 4 months before today
All of our day calculators are measured and QA'd by our engineer. Read more about the Git process here. But here's how 4 months ago gets calculated on each visit:
1. We started with date inputs: used current day of 23 Apr, then set the calculation - 4 months, and factored in the year 2024
2. Noted your current time of year: 4 months in April will keep us in April and we don’t need to worry about adjusting months or years.
3. Counted backwards months from current day: date - 23 Apr, factoring in the 7 days left in April to calculate Saturday December 23, 2023
4. Did NOT factor in workdays: In this calculation, we kept weekend. See below for just workdays or the fiscal calendar
### Tips when solving for December 23
• Current date: 23 Apr
• Day of the week: Saturday
• New Date: Saturday December 23, 2023
• New Date Day of the week: Saturday
• December marks the end of the year, so account for all the out of office notifications when counting backwards.
• This calculation crosses at least one month. Remeber, this will change our day of the week.
• The solution crosses into a different year.
• Your date will be a weekend. Consider if you only want workdays in your calculation.
## Ways to calculate 4 months ago
1. Calculate it: Start with a time ago calculator. 4 months is easiest solved on a calculator. For ours, we've already factored in the 7 days in April + all number of days in each month and the number of days in . Simply add your months and choose the length of time, then click "calculate". This calculation does not factor in workdays or holidays (see below!).
2. Use April's calendar: Begin by identifying 23 Apr on a calendar, note that it’s Saturday, and the total days in March (trust me, you’ll need this for smaller calculations) and days until last year (double trust me, you'll need this for larger calculations). From there, count backwards 4 times months by months, subtracting months from until your remainder of months is 0.
3. Use excel: For more complex months calculations or if you h8 our site (kidding), I use Excel functions like =TODAY()-4 to get or =WORKDAY(TODAY()), -4, cell:cell) for working months.
## Working months in 4 calendar months
4 months is Saturday December 23, 2023 or could be if you only want workdays. This calculation takes 4 months and only subtracts by the number of workdays in a week. Remember, removing the weekend from our calculation will drastically change our original Saturday December 23, 2023 date.
Work months Solution
Monday
Tuesday
Wednesday
Thursday
Friday
4 months back
Saturday
December 23
Sunday
## The past 4 months is equivalent to:
Counting back from today is Saturday December 23, 2023 using a full calendar, and is also 2976 hours ago and 97.81% of the year.
Did you know?
Saturday Saturday December 23, 2023 was the 357 day of the year. At that time, it was 97.81% through 2023.
## In 4 months, the average person Spent...
• 26635.2 hours Sleeping
• 3541.44 hours Eating and drinking
• 5803.2 hours Household activities
• 1726.08 hours Housework
• 1904.64 hours Food preparation and cleanup
• 595.2 hours Lawn and garden care
• 10416.0 hours Working and work-related activities
• 9582.72 hours Working
• 15683.52 hours Leisure and sports
• 8511.36 hours Watching television
## What happened on December 23 (4 months ago) over the years?
### On December 23:
• 1951 National Football League Championship, LA Memorial Coliseum: Los Angeles Rams beat Cleveland Browns, 24-17; first coast-to-coast televised NFL title game
• 1951 National Football League Championship, LA Memorial Coliseum: Los Angeles Rams beat Cleveland Browns, 24-17; first coast-to-coast televised NFL title game
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Science Forums
# The Final Piece Of The Puzzle!
## Recommended Posts
In such a case, I can write [imath]\Phi(\vec{x})[/imath] in a very simple form consistent with standard notation:
$\Phi(\vec{x})=-\frac{\kappa M}{r}$
where [imath]\kappa[/imath] is the proper proportionality constant to yield the correct potential generated by the mass M.
I was just reading the OP and it's getting late and I had a question, so figured it would be a good place to stop for the night.
If your wave equation propagates in 4 dimensional Euclidean space (an impression I got from the special relativity thread) then the vector field causing the refractive index, n, would presumably be an inverse (n-1)th conservative force (F $\propto$ 1/r3). That force, I would think, would be the derivative of Phi making Phi proportional to 1/r2 (in contrast to normal Newtonian gravitational potential) so I'm not quite following how or why you deduced Phi $\propto$ 1/r.
Are ds, c, and the geodesic in four dimensions? If so, how have I gone wrong?
Thank you,
~modest
• Replies 129
• Created
#### Popular Posts
Well Anssi, in spite of the fact that I will be essentially out of contact for the next couple of months, following your recommendation, I decided to post this. Let's see what happens. If you have a
I would say that the "supposed position of a defined object" can be called "an event".A collision between a virtual particle and a fermion can be seen as "an event", as can the simultaneous change in
Yeah, that's the problem. Whenever it comes to complex algebra, mistakes are easy to make.The problem is that my work has not gone through that process and there are minor insignificant errors all ov
Coldcreation, I really wish I knew why you feel compelled to post to this thread. You obviously have not the first hint of what I am talking about.
There's another erroneous presumption I guess we should add to the list.
The reason why I feel compelled to post to this thread is simply that it intrigues me why someone, with the background you have, would see the need to overhaul an aspect of a theory—Einstein's concept of gravity as a curved spacetime phenomenon—by replacing it with a concept (Euclidean space) that has shown not to be tenable. Indeed, it is precisely the concept of gravity as curvature that did away with the notions erroneous within that which it replaced (Newtonian gravitation), and the idea that space was Euclidean, independent of time, and absolute (in the sense that there existed a unique gods-eye reference frame).
Besides quantum theory there have been few theories produced since the beginning of modern science that have a better track record, i.e., there is excellent corroboration with empirical observations and experimental data related to phenomena such as gravitational time dilation and frequency shift, light deflection and gravitational time delay, orbital effects and the relativity of direction (precession of apsides, orbital decay, geodetic precession and frame-dragging). Nearly being the key word, it seems that a modification, if at all, rather than a complete overhaul, would be more appropriate, in order to clear up the discrepancies with GR and that which is observed and theorized to be operational on meso- and microscopic scales. What makes you think that gravity is not due to a curved spacetime manifold?
That “almost” is there for honesty. There exist aspects of modern physics which I have not yet proved are indeed implied by my fundamental equation. It is not an easy equation to solve. [...] and finally, (as the last piece of the puzzle of identifying constraints implied by my equation with supposed conclusions of modern physics) showing that Einstein's theory of General Relativity is essentially an approximate solution to my fundamental equation. I have not yet achieved “complete corroboration with all empirical evidence” but I have certainly deduced enough to convince me that modern physics is a tautology (as it is quite clear that my fundamental equation is certainly a tautology). As I said, it is not exactly an easy equation to solve. [...] Yes, and I will make a prediction: there probably exists other empirical evidence which can be shown to be approximate solutions to my fundamental equation.
I was referring to your statements:
"The fact that my result is not exactly the same as that obtained from Einstein's theory is not too troubling.
And:
"It appears (at least to me at the moment) that the effect of that extra term in my solution is to make the gravitational field appear to be slightly stronger than estimated via Einstein's field theory. If that conclusion is correct, then it could also explain the “dark matter” problem."
If there are indeed differences that you describe perhaps you should center your predictions around those.
For starters you could attempt to predict rotational curves without cold dark matter for galaxies more accurately than GR with CDM...
That would be one giant leap for mankind.
Of course you do because you have absolute faith in the validity of the standard “by guess and by golly” approach used by modern physics.
Actually I do not. I've been known on occasion to display a dislike for certain mainstream (and certain less mainstream) physical theories (e.g., inflation, with its false vacuum, strings with its 21 dimensions, branes, even the big bang theory with its CDM, DE and repulsive force: lambda).
For example, Einstein's general relativity tells us many things about the physical universe, 'reality' (and is thus not a religion).
Yeah, I know! That is exactly what you are convinced of and you would like the rest of us to believe. I say it is no more than another consequence of internal self consistency of the argument (but as I get something a little different, either I have made a deductive error, which is certainly possible, or Einstein's theory is not internally consistent). The fact that no one has succeeded in presenting a quantized version of that theory is somewhat indicative of its failure to be internally consistent.
Each is entitled to 'believe' that which pleases him or her. As it stands, GR is in empirical agreement with observations That is what makes it compelling (nothing to do with belief).
Again, if you can make a prediction to within a more accurate decimal place, please do so.
Your tau axis seems more akin to some belief system. You did after all write of it; "tau is a complete fabrication of our imagination"
Your hypothesis is not solely that "an explanation must be internally self consistent!" There is more involved than just that.
You clearly have never read the proof as, if you had, you could explicitly point out the other hypotheses.
Maybe there's a better word than hypothesis: contentions, or presumptions for example. One example is that you feel Euclidean space is more consistent a schema than curved spacetime. Your definition, or interpretation, of time is another. The idea that gravity is a pseudo-force is yet another. Not to mention tau again. These are all part of your hypothesis, or contention(s).
As I have commented earlier, you seem to have utterly no comprehension of what puzzles are under discussion. And, as an aside, if you were at all aware of my proof you would clearly comprehend the nature of that tau axis. [snip]
That would be a good place to start clearing up what people find confusing about your proof. To what extent do you feel tau is relevant to the natural world. How can the concept of tau be tested? Why do you need to include tau in your equations? What would be the consequences if it were to be removed?
If you like, you can hyperlink to a thread where you've discussed tau already. But that explanation clearly belongs in this thread.
PS. Your OP would have been easier to follow had you started with an abstract, and ended with a conclusion (at least for those who have not sifted through all your other threads).
CC
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Thank you modest for taking the trouble to think about what I have said.
If your wave equation propagates in 4 dimensional Euclidean space (an impression I got from the special relativity thread) then the vector field causing the refractive index, n, would presumably be an inverse (n-1)th conservative force (F $\propto$ 1/r3). That force, I would think, would be the derivative of Phi making Phi proportional to 1/r2 (in contrast to normal Newtonian gravitational potential) so I'm not quite following how or why you deduced Phi $\propto$ 1/r.
Your assertion concerning the radial form of a conservative force is mathematically true in a Euclidean geometry in the Newtonian picture as it is intimately related to the way the surface area changes with respect to the radius. The effect is easily seen in a simplified notion of photon exchange forces.
Exchange forces are attributed to momentum exchanges mediated by the exchange of virtual particles. The force (i.e., that change in momentum) is directly related to the probability that a given virtual particle will be exchanged and that probability is in turn proportional to the cross section of the interaction as seen by the interacting bodies: i.e., the same spacial density of interacting virtual particles from a given source, as seen from a more distant object, declines as the area of the encompassing sphere increases. That area is proportional to r squared and therefore the magnitude of the force becomes inverse to r squared.
And you are correct. In a four dimensional Euclidean geometry in the Newtonian picture a conservative force would be proportional to the inverse of r cubed, yielding a potential proportional to the inverse of r squared. However, that is not what we have here. In this case, both interacting bodies are momentum quantized in the tau direction. The whole circumstance yields utterly no variation in the physical probability density of these interacting bodies (or the density of the virtual exchange particles) in the tau direction. This situation essentially projects out the tau dimension and all forces and potentials come back to the three dimensional dynamics (except for the subtle consequences of that momentum in the tau direction: i.e., what we call mass).
Are ds, c, and the geodesic in four dimensions? If so, how have I gone wrong?
Yes, ds and c must both be evaluated in the four dimensional space as they are exactly that subtle consequence of momentum in the tau direction of which I spoke. Mass, the quantization of momentum in the tau direction, and likewise the tau component of its path and the tau component of its velocity do not change in the tau direction thus the existence of the tau dimension can not be ignored with respect to these terms.
I hope that makes things a little clearer. As I have commented elsewhere, I am currently working on a post of my proof of my fundamental equation and I think that post will clear up a lot of these kinds of questions.
The reason why I feel compelled to post to this thread is simply that it intrigues me why someone, with the background you have, would see the need to overhaul an aspect of a theory—Einstein's concept of gravity as a curved spacetime phenomenon—by replacing it with a concept (Euclidean space) that has shown not to be tenable.
Again you demonstrate that you do not have even the slightest idea as to what I am doing. I have no compulsion to overhaul any aspect of Einstein's theory; my only purpose is to endeavor to find the consequences of my proof. Since you have utterly no idea as to why my fundamental equation must be valid, your comments are totally off the subject. I won't put you on my ignore list because you appear to have a little education; however, I don't think I will respond to your posts until I have a little more evidence that you are trying to understand this stuff.
Have fun -- Dick
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To CC,
One of the points DD's consistently made in this thread and in others is that his system and proofs show that physics is a logical tautology deriving from his fundamental equation. As far as I can tell, DD's perspective holds that this is a sort of disproof of physics as anything other than a mental construction of the human kind having little to do with the machinations of physical world beyond our interpretation of it.
Seems, DD has different aims in this respect than your average scientist. He's out to disprove the independence of physics and scientific method as it stands from philosophical and logical concerns by showing that it is in fact equivalent to any other axiomatic system we're familiar with. As such, it would constitute a logical argument against science as independent of the concerns of philosophy.
"There is no such thing as philosophy-free science; there is only science whose philosophical baggage is taken on board without examination." —Daniel Dennett, Darwin's Dangerous Idea, 1995.
DD, do correct me if I'm wrong in my interpretation of your argument, but this is the gist I get from what I've read so far of your work.
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Well Mclean, I am impressed. I was kind of unimpressed by the character “KickAssClown”, but your recent posts have led me to believe there are possibilities there.
Your post number 21 to this thread seems to indicate a rather intelligent approach. I have no real argument with that post at all. All I have is a few subtle adjustments to your perspective.
One of the points DD's consistently made in this thread and in others is that his system and proofs show that physics is a logical tautology deriving from his fundamental equation.
Absolutely and incontrovertibly true.
As far as I can tell, DD's perspective holds that this is a sort of disproof of physics as anything other than a mental construction of the human kind having little to do with the machinations of physical world beyond our interpretation of it.
Essentially true; however, there are a number of conclusions which might be drawn from that statement that really kind of misrepresent the situation.
Seems, DD has different aims in this respect than your average scientist. He's out to disprove the independence of physics and scientific method as it stands from philosophical and logical concerns by showing that it is in fact equivalent to any other axiomatic system we're familiar with. As such, it would constitute a logical argument against science as independent of the concerns of philosophy.
Again an essentially correct statement; however, it seems to imply an attitude which is somewhat askew of what I actually have in mind. To quote post #14 I made to this thread
I discovered the proof of my fundamental equation when I was still a graduate student back in the sixties but it seemed pretty worthless because I couldn't solve the equation. I discovered the first solution around 1983 and after some work, I attempted to publish my proof about twenty years ago. I don't think it made it past any of the referees. The physicists said it was philosophy, the philosophers said it was mathematics and the mathematicians said it was physics.
I have since come to the conclusion that it is indeed philosophy, that is why I am posting to the “Philosophy of Science” forum. Certainly it has absolutely nothing to say about mathematics as nothing in my work yields anything new to the field of mathematics. And, essentially, my work has nothing to say about physics (except perhaps the fact that Einstein's theory of General Relativity has some problems) as, for all practical circumstances, it essentially confirms most all of modern physics.
Just as an aside with regard to the confirmation of modern physics, Newtons orbital calculations essentially (except for a few exceptions) confirmed the charts prepared through the “cycle and epicycle” theory of Claudius Ptolomy (which, today, is clearly seen as no more than a mathematical mechanism for cataloging cosmic positions). Newton admitted the possibility of error in his calculations; however, as it turned out the differences were errors on the part of the astronomers citing the actual data. Likewise, Einstein's theory could still be correct as it is entirely possible that I have made an error in my algebra. On the other hand, perhaps Einstein is wrong. Time will tell.
The reason I bring this up is the fact that I really have no complaint with physics. Considering the problems they have solved and the foundations they have to work with, they have done an excellent job. It is all based upon the presumption that the “by guess and by golly” attack is the only attack available to them.
My real complaint is with the field of philosophy. It is the philosophers who have dropped the ball here. They have taken the issues underlying the fundamental questions of interest and lathered them over with gobs and gobs of esoteric bullshit. It is time that philosophers became a little more exact in their analyses. I think I have kind of thrown out the ball there but, except for Anssi, have seen no rational reaction.
Philosophy was once considered the Queen of all scientific investigations. But no more. Today philosophy is considered to be a field totally concerned only with stirring bullshit and that is a sad report on their efforts over the last three thousand years.
"There is no such thing as philosophy-free science; there is only science whose philosophical baggage is taken on board without examination." —Daniel Dennett, Darwin's Dangerous Idea, 1995.
I couldn't put it any better myself (though I don't think I would call evolution a “dangerous idea”).
For what it is worth, consider yourself corrected. :confused:
Have fun -- Dick
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In this case, both interacting bodies are momentum quantized in the tau direction. The whole circumstance yields utterly no variation in the physical probability density of these interacting bodies (or the density of the virtual exchange particles) in the tau direction. This situation essentially projects out the tau dimension and all forces and potentials come back to the three dimensional dynamics (except for the subtle consequences of that momentum in the tau direction: i.e., what we call mass).
That makes sense. I think I had read previously, but forgot, the consequences of uncertainty in tau being infinite. Thank you for the explanation. I'll pick up where I left off when the girlfriend eases up on the whip (metaphorically speaking :confused:)
~modest
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I have a comment and a question about tau dimension.
I have since come to the conclusion that it [the Fundamental Equation] is indeed philosophy, that is why I am posting to the “Philosophy of Science” forum
Well, this is exactly what I have been trying to explain to DD for over one year now. And, I have presented why his Fundamental Equation is important to philosophy. When DD indicated a few posts ago that he agrees with the comment that ....physics is a logical tautology deriving from his fundamental equation...I have suggested that the reason is because his Fundamental Equation is derived from "tautology itself"---that is, from the philosophic Law of Identity, which was presented by Aristotle > 2000 years ago as the ultimate Fundamental Equation A = A ! This is why I agree with DD that his Fundamental Equation has great philosophic importance. It shows how A = A (which is a statement about ontology) can be transformed into an isomorphic equation [the DD Fundamental Equation] that is a statement about epistemology (explanation itself). Imo, this is the final piece of the "philosophic puzzle"--the DD Fundamental Equation completes the thinking of Aristotle about the "mathematical" relationship between existence and knowledge--it puts philosophy on a sound mathematical basis. But of course--only if one accepts the Law of Identity as a valid premise.
==
My question is about the tau dimension used by DD.
Now, DD indicates that "mass" is derived from his tau dimension--yet he also indicated (I do believe) that this dimension is "abstract". So, I recall a similar situation with use of the Schroedinger Equation as relates to shell structure in the atomic nucleus for nucleons. This model predicts (using the Schroedinger Equation) the interactions of independent nucleons as if they are being "acted on" by an energy potential (V) that is an abstract dimension related to a central harmonic energy well. That is, the "mass" of the nucleons is a function of this interaction with an abstract concept.
So my question--is this what DD is saying with the relationship between "mass" and abstract "tau" ? Is his claim about mass & tau the same philosophic relationship as between the mass of a nucleon and abstract central energy well within the nucleus as a whole as predicted by quantum mechanics via use of Schroedinger Equation ? If yes, then what DD is claiming makes perfect sense to me. If no, please explain why.
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Hi, I had not noticed you had posted this thing (Didn't pick up on the title I guess
I'll try to get around to start a walk through soon, probably this weekend.
But first one comment. It would be very nice if the discussion about the underlying issues would be held here;
Simply because it will make it easier to follow and backtrack this thread if the responses are more strictly about the derivation of general relativity.
It is of course okay to make small comments, but long conversations about the underlying issues just make it time consuming for me to find information about the actual derivation later.
Following that note, I'll post a little comment to Coldcreation to that thread.
-Anssi
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To CC,
DD's perspective holds that this is a sort of disproof of physics as anything other than a mental construction of the human kind having little to do with the machinations of physical world beyond our interpretation of it.
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I agree that indeed that is the DD's conclusion. However, it is a rather dangerous conclusion. DD negates experience. All that exists is sets and logic. DD concludes that physics is nothing but entertainment. in other words, physics is math which is entertainment of the mind. Only sets and relationships between sets exist, only math exist. Experience may exist but we know nothing of it.
His central premise is we know nothing other than that which we create in our mind. While enticing, this is rather illogical imo. Because the very sets and logic that give rise to mathematics are the product of experience of ourselves. If we do not trust ourselves, our experience, then we can not even begin to trust the sets. Only nothing would be valid.
I find DD's equation scientific, but his conclusions dangerous because they negate experience.
From the bottom up, DD's equation is scientific as a synthesis; not an analysis. It is no different than mathematically synthesizing any other set of sums into a single expression. To that end I agree with DD that it is falsifiable as a mathematical expression. As has been posted by others, if there exists one valid theory inconsistent with DD's equation, then the equation is falsifiable.
From the top down, his analysis of the equation is purely logical. He sets up some definitions and analyzes those to come up with the equation. To the extent that the specified elements are elements of DD's space, they appear valid--no different then any other valid set.
But what of this equation? In essence, so what? Of course we invented math. We certainly did not mine it in a cave.
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Okay, let's get to it...
Relativity is the mathematical transformation between two different geometric coordinate systems. Back in Newton's day, such transformations were quite straight forward as Euclidean coordinate systems were assumed applicable to reality. If the origin of a Euclidean coordinate system (coordinate system “b”) was at point (x0,y0,z0) in the original coordinate system (coordinate system “a”) then any point in coordinate system “a”, say point (xa,ya,za) was simply represented by the point (xa-x0,ya-y0,za-z0) in coordinate system “b”.
This transformation was exactly the same even if the point being referred to as (x0,y0,z0) was moving in any arbitrary manner.
I.e. even if the coordinate system b was moving inside the coordinate system a in an arbitrary manner.
The difference between my system and Euclid's original system is that tau axis. Momentum quantization along that tau axis (mass) introduces some very subtle consequences when it comes to physical measurements such that the simple transformation above does not yield the measurements as taken by a person at rest in the moving system: i.e., special relativity is a necessary part of such a transformation in order to compensate for the effects of momentum quantization along the tau axis.
Since you mention momentum quantization along the tau axis, I stopped to think about that a bit. I am not sure why was it quantized instead of being a continuous variable. I suppose it had got something to do with the uncertainty in the position of tau being infinite (since the momentum is constant/known?), but my understanding is quite shaky when it comes to the details of how the uncertaintly principle plays out here.
Also, I suspect this issue related to the use of dirac constant in the definition of mass operator [imath]-i\frac{\hbar}{c}\frac{\partial}{\partial \tau}[/imath]...?
Nevertheless, yes I remember how special relativistic transformation came into play.
.
.
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The question then arises, how is it that Einstein's theory appears to circumvent Maupertuis' proof? The answer revolves around the principal of “least action” he invented as a means of calculating paths consistent with Newtonian physics (essentially minimizing the energy with respect to the path). Maupertuis showed that the problem was a consequence of the fact that, when it came to gravitational paths, different velocities led to different paths: i.e., two different objects behavior could not be reduced to geodesic motion in the same reference frame, something which must be true in the proper inertial frame. The inclusion of time in Einstein's space-time continuum allows this critical variation to be achieved; however, it turns out that this is exactly that same issue which creates the critical problems when it comes to “quantizing the gravitational field”. Thus it is that there are very real problems bringing quantum mechanics into Einstein's General Relativity theory. To date all attempts, that I am aware of, have resulted in failure.
Yup, thank you about that whole explanation. It might be interesting to take a quick look at that Maupertuis' proof but I did not find it... Nevertheless, I have pretty good suspicion about how the inclusion of time axis and its relativistic transformation can get around proofs that rely on newtonian definition of time.
I don't have very good idea about the problems with the traditional attempts of gravity quantization. Wikipedia only vaguely mentions problems with re-normalization.
Compare this to my presentation which is totally consistent with quantum mechanics from the very beginning. Beyond that, in my presentation, mass is defined to be momentum in the tau direction of a four dimensional Euclidean geometry. As a consequence, that hypothetical (see as unexaminable) tau dimension can be simply scaled to make the velocity of every elemental entity through that four dimensional geometry look exactly the same.
i.e. whatever velocity they are "missing" in the [imath]x,y,z[/imath]-axes, is to be attributed to their velocity along the [imath]\tau[/imath]-axis...
That final fact totally circumvents Maupertuis' proof. It seems certainly reasonable to once again look at the consequences of general transformations and perhaps find that “non-inertial” geometry which yields gravity as a pseudo force. So let us proceed to examine some aspects of that circumstance.
To begin with, my fundamental equation does indeed require a specific frame of reference: that frame of reference being at rest with respect to the entire universe. In that particular frame, (remember, that frame is a standard Euclidean frame) any object (any collection of fundamental elements forming a stable pattern) which can be seen as essentially not interacting with the rest of the universe (i.e., those interactions may be ignored and we are looking at a “free” object) will obey Newton's laws of motion in the absence of a force (it will not accelerate in any way). That is, the only forces which appear in such a circumstance are those pseudo forces we want to examine in this analysis: i.e., apparent forces which are entirely due to the fact that our coordinate system is not at rest with respect to the universe.
True, we have created a situation which obviously circumvents Maupertuis' proof...
Hmmm, well not very obvious to me, as I don't know how his proof worked exactly...
I'm guessing what you are referring to is that, since Maupertuis' proof had something to do with the inability to "reduce two different objects' behaviour to geodesic motion in the same reference frame", this presentation form with a [imath]\tau[/imath]-axis (that gets "projected out"), will allow two different objects with different velocities in [imath]x,y,z[/imath] space to have the same total velocities when the velocity along [imath]\tau[/imath] is included.
Not really sure yet how that will circumvent the proof.
I understand though, that in the standard GR presentation form, the objects in gravitational free fall are following geodesic paths in the general relativistic coordinate system.
but, since energy is now not a function of velocity,
...i.e. not a function of the total velocity in the [imath]x,y,z,\tau[/imath]-space...? Are you saying that because, for instance, an object is considered to gain (kinetic) energy when it gains velocity in the [imath]x,y,z[/imath] directions, while its total velocity remains unchanged?
we have also made the original formation of his principal of action into an unusable procedure (his relationship related velocities and we now have no relationship to minimize);
...since velocities never change in the [imath]x,y,z,\tau[/imath]-space...?
however, there is another attack (actually the same attack but somewhat subtly different). If gravity is to be a mere pseudo force, we can use the fact that our model must reproduce the exactly the same classical pseudo forces produced by Newton mechanics. This is true as these forces are no more than a direct consequence of expressing the path in a non-inertial frame or, in my case, a reference frame not at rest with respect to the universe.
Yup.
It is interesting to look at centrifugal force as a well understood Newtonian pseudo force. From the perspective of an observer at the center of the rotation, a string exerting a force equal to the centrifugal force will appear to maintain the object under the influence of that pseudo force at rest: i.e., a rock at the end of a string swinging in a circle will appear to be at rest in a reference frame rotating with that rock. We can then see the object as a test probe into the force field describing that specific pseudo force. Since my total interest is in explaining gravity as a pseudo force, I want to examine this force as seen from the perspective of being m times the negative gradient of a gravitational potential (which is the typical way of defining a gravitational potential). In this case [imath]\vec{F}=-m\vec{\nabla}\Phi [/imath] where [imath]\Phi(\vec{x})[/imath] is the gravitational potential.
Had a little adventure in Wikipedia about the definition of gravitational potential etc, and yes that seems to make perfect sense to me.
I gathered that the "gradient of a gravitational potential" is essentially the radial derivative of the function describing gravitational potential... I.e. the "slope" of a graph describing that function.
Also I gathered that the definition of gravitational potential is the "potential energy" PER unit mass, so that's why you are including the "m" there in your equation (I was getting a bit confused at first).
So in other words, [imath]\vec{F}=-m\vec{\nabla}\Phi [/imath] refers to the strength of the "fictional force" that affects an object with given mass in a given location in a gravitational field. (Just thought I'll say that out loud to benefit other people who are starting with almsot zero physics knowledge, like me :D)
Any freshman physics text will provide an excellent derivation of centrifugal force. The result is the quite simple form, [imath]\;\vec{F}=m\omega^2\hat{r}\;[/imath] where omega is the angular velocity and [imath]\hat{r}[/imath] is a unit vector in the radial direction.
I'll take that on faith for now.
It follows that the analogous gravitational representation implies that the required [imath]\Phi(\vec{x})[/imath] (which, by the way, must by symmetry be a radial function) must obey the relationship
$-\frac{\partial}{\partial r}\Phi®=r\omega^2$.
I.e. the radial derivative, or the "slope" of the gravitational potential, is related to the angular velocity somehow?
Hmmm, you must be referring to a situation where some object is orbiting the center of the gravitational field, so then its angular velocity and the distance from the center are related to the strength of the associated "fictional force", which is keeping it in stable orbit.
Intuitively, that makes sense, but I am not entirely sure how you got that exact expression (specifically, the "r" on the right side).
I mean, I understand that [imath]\vec{\nabla}\Phi = \frac{\partial}{\partial r}\Phi®[/imath] since its a radial function, and I guess we are essentially associating gravitational force with a centrifugal force capable of negating it...?
$-m\frac{\partial}{\partial r}\Phi® = m\omega^2\hat{r}$
...I guess... :shrug: ?
And you have removed the "m" from both sides, but I am not sure how the "r" finds its way to the right side the way it did. I mean, I understand the associated fictional force is a function of the radius, but how do we know the [imath]\omega^2[/imath] is simply multiplied by the radius?
I'll pause here as I am getting the feeling I may well be interpreting you wrong already...
-Anssi
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Hi Anssi, I am sorry I hadn't let you know about the post. There is another post you should be aware of, I am speaking of “Laying out the representation to be solved”. I would appreciate any comments you might wish to make on the clarity of that post. I don't think you will have any problems understanding what I said, but, if you do, we can try and clear them up.
Meanwhile, I will attempt to clarify this thread.
It would be very nice if the discussion about the underlying issues would be held here;
Simply because it will make it easier to follow and backtrack this thread if the responses are more strictly about the derivation of general relativity.
I wouldn't argue with you but I suspect you will not find many takers. There are a lot of people here who's sole purpose is to create confusion by making it difficult to backtrack threads. Rade long ago made his intentions quite clear and at the moment I am seriously suspicious of lawcat. I am considering placing lawcat on my ignore list. If he keeps up the kind of comments he has been making I will do so. I didn't see the “ignore list” as a proper response to ignorance but I am beginning to realize that intentional ignorance is a bothersome issue which should be ignored.
Since you mention momentum quantization along the tau axis, I stopped to think about that a bit. I am not sure why was it quantized instead of being a continuous variable.
It is a direct consequence of the manner and purpose with which tau was introduced. That issue will show up (hopefully more clearly) in my further posts regarding my proof of my fundamental equation. For the moment, let's not worry about it as discussion at this point will probably generate nothing but further confusion. Just take it as a proved aspect of tau.
Also, I suspect this issue related to the use of dirac constant in the definition of mass operator [imath]-i\frac{\hbar}{c}\frac{\partial}{\partial \tau}[/imath]...?
That issue you seem to have somewhat backwards; however, again I suggest you not worry about it for the moment. It will be clarified in the actual proof which I am currently trying to restate.
Yup, thank you about that whole explanation. It might be interesting to take a quick look at that Maupertuis' proof but I did not find it... Nevertheless, I have pretty good suspicion about how the inclusion of time axis and its relativistic transformation can get around proofs that rely on newtonian definition of time.
As I said, Maupertuis' proof has to do with the fact that objects with different velocities follow different paths in a gravitational field. This is totally counter to the proposition that “pseudo forces” create paths which are mirror images of the path of the non-inertial frame. That idea means that the path certainly can not depend upon the velocity of the object; not in a Euclidean geometry anyway.
i.e. whatever velocity they are "missing" in the [imath]x,y,z[/imath]-axes, is to be attributed to their velocity along the [imath]\tau[/imath]-axis...
Absolutely correct!
Hmmm, well not very obvious to me, as I don't know how his proof worked exactly...
Well, if everything in the universe is traveling through my four dimensional Euclidean geometry at exactly the same velocity, then velocity differences don't exist and the actual paths can once more be mirror images of the path of the non-inertial frame. Path dependence on the velocity of the objects vanishes absolutely.
Are you saying that because, for instance, an object is considered to gain (kinetic) energy when it gains velocity in the [imath]x,y,z[/imath] directions, while its total velocity remains unchanged?
Not really. The issue here is that nothing can depend upon the velocity if the velocity of every object in the universe is the same.
...since velocities never change in the [imath]x,y,z,\tau[/imath]-space...?
Exactly!
I gathered that the "gradient of a gravitational potential" is essentially the radial derivative of the function describing gravitational potential... I.e. the "slope" of a graph describing that function.
That is correct.
So in other words, [imath]\vec{F}=-m\vec{\nabla}\Phi [/imath] refers to the strength of the "fictional force" that affects an object with given mass in a given location in a gravitational field. (Just thought I'll say that out loud to benefit other people who are starting with almsot zero physics knowledge, like me :D)
I think you have that pretty straight.
I.e. the radial derivative, or the "slope" of the gravitational potential, is related to the angular velocity somehow?
As I stated, centrifugal force is given by
$\vec{F}=mr \omega^2\hat{r}$
(Ah, Anssi you have once again caught an error in a post of mine. I omitted that “r” in there. I have edited the original post to correct the error. I thank you once again. You are clearly the only person who reads my stuff carefully.) Sorry you took it on faith to be correct. You shouldn't do that with my stuff; I do certainly make errors.
At any rate, the force due to the gravitational potential is given by [imath]\vec{F}=-m\vec{\nabla}\Phi[/imath]. Setting these two forces to be identical (together with the fact that [imath]\vec{\nabla}\Phi=\frac{\partial}{\partial r}\Phi® \hat{r}[/imath] because there is no change in the potential outside a change in r) and, setting these two forces to be identical one has immediately the fact that
$r\omega^2=-\frac{\partial}{\partial r}\Phi®$
which I am sure you would have accepted had it not been for my gross error.
Hmmm, you must be referring to a situation where some object is orbiting the center of the gravitational field, so then its angular velocity and the distance from the center are related to the strength of the associated "fictional force", which is keeping it in stable orbit.
No, I am talking about an object restrained by a string to the center of the coordinate system. The coordinate system is not an inertial system (or, in my case, at rest with respect to the universe) but is rather rotating with an angular velocity of omega. Because I do not know the coordinate system is rotating, I presume the force is caused by a gravitational potential. Thus, what I am calculating is the consequence of the rotation and then attributing it to gravitational effects; essentially casting the supposed gravitational force as a "pseudo" force.
Intuitively, that makes sense, but I am not entirely sure how you got that exact expression (specifically, the "r" on the right side).
Yeah, I omitted that r in my specification of the centrifugal force.
I'll pause here as I am getting the feeling I may well be interpreting you wrong already...
Nope, you are doing beautifully. As per usual, you just caught me in a serious error and I thank you for that.
Know that I appreciate you beyond belief.
Thanks -- Dick
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Hi Anssi, I am sorry I hadn't let you know about the post. There is another post you should be aware of, I am speaking of “Laying out the representation to be solved”. I would appreciate any comments you might wish to make on the clarity of that post. I don't think you will have any problems understanding what I said, but, if you do, we can try and clear them up.
Okay, yeah, I saw it yesterday, but didn't read it through yet.
As I said, Maupertuis' proof has to do with the fact that objects with different velocities follow different paths in a gravitational field. This is totally counter to the proposition that “pseudo forces” create paths which are mirror images of the path of the non-inertial frame. That idea means that the path certainly can not depend upon the velocity of the object; not in a Euclidean geometry anyway.
Hmm, there's something I'm definitely missing here... :I
I suppose you are referring to, for instance, a cannon ball being shot at different velocities, and consequently following different trajectories... I'm just thinking that, since only the rate of descent would be attributed to the fictional force, the different trajectories shouldn't be a problem, at least not very obviously so.
I'm also thinking that with a coriolis effect the path obviously depends on the velocity of the object too, i.e. the final path is not exactly a mirror image of the path of the non-inertial frame... Hmmm, unfortunately I just can't find any information about that Maupertuis' proof, I don't know why is it so hard to find. I'm just finding all sort of material about his expidition to lapland :D
Well, if everything in the universe is traveling through my four dimensional Euclidean geometry at exactly the same velocity, then velocity differences don't exist and the actual paths can once more be mirror images of the path of the non-inertial frame. Path dependence on the velocity of the objects vanishes absolutely.
No comments, as I don't yet even understand the problem exposed by Maupertuis... :P
(Ah, Anssi you have once again caught an error in a post of mine. I omitted that “r” in there. I have edited the original post to correct the error...
Oh okay, now [imath]r\omega^2=-\frac{\partial}{\partial r}\Phi®[/imath] makes sense to me :D
I guess I should have expected to see an "r" somewhere in there, but I had already absorbed too much information at one sitting, and was starting to feel cloudy :)
No, I am talking about an object restrained by a string to the center of the coordinate system. The coordinate system is not an inertial system (or, in my case, at rest with respect to the universe) but is rather rotating with an angular velocity of omega. Because I do not know the coordinate system is rotating, I presume the force is caused by a gravitational potential. Thus, what I am calculating is the consequence of the rotation and then attributing it to gravitational effects; essentially casting the supposed gravitational force as a "pseudo" force.
Yup.
Back to OP;
It follows that the analogous gravitational representation implies that the required [imath]\Phi(\vec{x})[/imath] (which, by the way, must by symmetry be a radial function) must obey the relationship
$-\frac{\partial}{\partial r}\Phi®=r\omega^2$.
This fact directly implies that [imath]-\Phi=\frac{1}{2}(r\omega)^2=\frac{1}{2}|\vec{v}|^2[/imath]
At first I did not realize at all how you got to that expression (so unfamiliar with math), but after toying around with wolfram alpha a bit, I realized that [imath]\frac{\partial}{\partial r}\frac{1}{2}(r\omega)^2 = r\omega^2[/imath], so that's why you are saying [imath]-\Phi=\frac{1}{2}(r\omega)^2[/imath].
Then, regarding [imath]\frac{1}{2}(r\omega)^2=\frac{1}{2}|\vec{v}|^2[/imath] we are in unbelievable luck; my eyes just accidentally landed on a wikipedia text mentioning that the alternative to [imath]F = m\omega^2r[/imath] is [imath]F = \frac{mv^2}{r}[/imath]
So that implies;
$\frac{v^2}{r} = r\omega^2$
i.e.
$v^2 = (r\omega)^2$
Hence; $\frac{1}{2}(r\omega)^2=\frac{1}{2}|\vec{v}|^2$
And even I can't believe I figured that out, like I said, complete accident since I don't know the derivation of the centrifugal force :D hehe
Anyway, looks valid to me.
or, multiplying by [imath]\frac{2}{c^2}[/imath], one can conclude that
$\frac{2}{c^2}\Phi®=-\left(\frac{|\vec{v}|}{c}\right)^2$.
Yup.
In other words, the gravitational potential (as seen in the frame where the object being observed appears to be at rest) seems to be directly related to the actual velocity as seen from the correct frame (the frame at rest with the universe).
Wait a minute... About the "velocity as seen from the correct frame"; are we now talking about a situation where an object is orbiting the center of a gravitational field? I.e, the frame where the object being observed appears to be rest, is essentially a rotating frame?
Or a situation where an object is basically just sitting on the ground?
Or both?
This result is very interesting. As the observed object is actually moving in the correct frame, we should expect a clock (or any temporal physical process moving with that object) to proceed in accordance with special relativity. This implies that the correct relativistic transformation of the instantaneous time differentials should be given by
$dt'=dt\sqrt{1-\left(\frac{|\vec{v}|}{c}\right)^2}\equiv dt\sqrt{1+\frac{2\Phi}{c^2}}$
Indeed. I realize that is essentially the standard lorentz factor, but for the benefit of lurkers, in this analysis that relationship was derived in the thread about special relativity, and the same conclusion can be seen in the end of the little animation related to the analysis:
YouTube - Presentation of a moving clock http://www.youtube.com/watch?v=-UDrWCIgmTk
$Cycles_m = Cycles_r \sqrt{1 - sin^2(\theta)}$
I.e, if you backtrack from there a bit;
$Cycles_m = Cycles_r \sqrt{1 - \left (\frac{v}{c} \right )^2}$
which happens to be exactly the standard gravitational red shift.
Looked at the wikipedia page for gravitational red shift and gravitational time dilation, and yes looks very familiar, albeit I did not find that exact expression.
This implies that any geometry which yields gravity as a pseudo force must also yield the standard gravitational red shift...
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......i.e., instead of seeing the speed of light as slower in a gravitational field we could just as well see the speed as unchanged and the distances as increased. After all, once time is defined, distances are reckoned via the speed of light. Though that satisfies the original goal expressed above, the idea of refraction (the speed of light being slowed in a gravitational field) is a much simpler expression of the solution. It is certainly most convenient method of finding the proper geodesics. In fact, there is a very simple view of the situation which will yield exactly that result.
Got all the way to that paragraph and did not have problems with it. I'll have a rest here again.
-Anssi
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$dt'=dt\sqrt{1-\left(\frac{|\vec{v}|}{c}\right)^2}\equiv dt\sqrt{1+\frac{2\Phi}{c^2}}$
which happens to be exactly the standard gravitational red shift.
Looked at the wikipedia page for gravitational red shift and gravitational time dilation, and yes looks very familiar, albeit I did not find that exact expression.
Substitute $\Phi = -GM/r$ (from the definition of gravitational potential) and that's probably how wiki has it expressed. It is sometimes expressed as a function of potential in the literature. Eq. 4.4:
Eq. 17.18:
~modest
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Hmm, there's something I'm definitely missing here... :I
I suppose you are referring to, for instance, a cannon ball being shot at different velocities, and consequently following different trajectories... I'm just thinking that, since only the rate of descent would be attributed to the fictional force, the different trajectories shouldn't be a problem, at least not very obviously so.
I'm also thinking that with a coriolis effect the path obviously depends on the velocity of the object too, i.e. the final path is not exactly a mirror image of the path of the non-inertial frame... Hmmm, unfortunately I just can't find any information about that Maupertuis' proof, I don't know why is it so hard to find. I'm just finding all sort of material about his expidition to lapland :D
Sorry about that. I also googled Maupertuis and found no reference whatsoever to the proof. The central problem here is that I studied physics so long ago that I suspect many things were given quite different spins back then. It could also be that my approach might be colored by idiosyncrasies of of individual professors. Whatever the cause, I have been out of contact with the academy for many many years. I seem to remember Maupertuis getting credit back then.
In my original paper, I had referenced that issue to page 6 of the 1965 edition of “Introduction to General Relativity” by Adler, Bazin and Schiffer.
The following is an exact quote of what is presented on page 6 of the 1965 edition.
Why has Einstein's idea of geometrizing the gravitational field of force not been conceived before? To answer this question let us look at the most geometrical of all variational principles of mechanics, namely, the principle of Maupertuis. In its simplest form it states the following: Let a particle move in a field of force with the potential V(x,y,z). If it travels from a point [imath]P_1[/imath] to a point [imath]P_2[/imath] with the varying velocity v, its trajectory is that actual curve which yields a stationary value for the action integral [imath]\int^{P_2}_{P_1}vds[/imath] among all paths connecting [imath]P_1[/imath] and [imath]P_2[/imath] which can be run through with the same constant energy [imath]E=\frac{1}{2}mv^2 +V[/imath] of the particle. We may express this principal in the obvious variational formula
$\delta\int^{P_2}_{P_1}\left(\frac{2}{m}(E-V)\right)ds=0$
In the case of V=0, we obtain the rectilinear motion asserted by the law of inertia. In the case of a nonvanishing potential V(x,y,z), we can introduce a metric based on the line element
$dl^2=\frac{2}{m}[E-V(x,y,z)](dx^2_1+dx^2_2+dx^2_3)$
and formulate the trajectory condition as
$\delta\int^{P_2}_{P_1}dl=0$
In the new differential geometry with this line element dl, the trajectory would indeed be a geodesic. But observe that, for different particles in the same field and with different energies E, the geometry would have to be a different one, which is impossible. This fact precluded a geometrization of dynamics.
We can see the same difficulty from the following consideration. Suppose that the gravitational field of the sun creates a non-Euclidean geometry and that the planets have to move along the geodesics of this geometry. It is well known that, if we prescribe a point in space and a direction through this point, there exists exactly one geodesic passing through the point with the prescribed direction. On the other hand, two particles in a gravitational field fired from the same point in the same direction will move along the same trajectory only if their initial velocities are equal. Thus only one projectile could at most follow the corresponding geodesic. Indeed geometry deals with the space variables and directions, but velocity is a concept involving time, and it is the initial velocity which enters into the determination of a trajectory. In the theory of special relativity Einstein had shown that space and time variables are inextricably connected and transform among each other under Lorentz transformations. A reduction of gravitational theory to geodesic motion in an appropriate geometry could be carried out only in the four-dimensional space-time continuum of relativity theory. That this is indeed possible is the main thesis of this book. That a reduction of the theory of gravitation to geometry was hardly possible before the special theory of relativity should be clear from the preceding considerations.
Note that the whole issue boils down to “different velocities” for different objects.
Note that with regard to the centrifugal and Coriolis forces, there is a simple Euclidean transformation which yields a straight line path for a free particle: i.e., the transformation to a non-rotating system. The issue with gravity is that they were searching for a transformation to a non-Euclidean system. Since the orbits return upon themselves the geometry simply can not be Euclidean because, in a Euclidean system, straight lines can not yield closed paths. In other words to find a rational result, they had to examine much more complex systems.
And even I can't believe I figured that out, like I said, complete accident since I don't know the derivation of the centrifugal force :D hehe
I think you are beginning to learn some mathematics.
Regarding centrifugal force the issue is actually quite simple. Most people measure angles in degrees. Physicists tend to measure angles in “radians” (it makes a lot of angular mathematics easy). By definition, instead of 360 degrees in a circle, there are [imath]2\pi[/imath] radians. That makes arc lengths easy to specify. If the angle [imath]\theta[/imath] is measured in radians the arc length is just [imath]r\theta[/imath]. Angular velocity is generally represented by [imath]\frac{d\theta}{dt}=\omega[/imath]. Thus the velocity along the circle of that rock out there on the end of the string is given by [imath]r\omega[/imath]. Note that the speed of the rock does not change; what changes is its direction. The change in velocity is always perpendicular to the path; it is towards the center, the direction the string is pulling. So the change in velocity (the acceleration) is the component of its velocity parallel to the string after time dt. Well the distance the rock has moved along the path is [imath]vdt=r\omega dt[/imath] so the change in velocity in the time dt is [imath]dv=(vdt)d\theta=r\omega d\theta[/imath] (draw a picture, the two triangles, radius against distance moved and velocity against velocity change, involved here are similar). Divide that by dt and one has the acceleration [imath]\frac{dv}{dt}=r\omega \frac{d\theta}{dt}= r\omega^2[/imath].
Wait a minute... About the "velocity as seen from the correct frame"; are we now talking about a situation where an object is orbiting the center of a gravitational field?
No, I am still talking about the centrifugal force, the rock on a string. Seen from the rotating frame (where it is at rest) there is an apparent force holding it out there against the string. I am just calling that force a “gravitational force” because, looking at it from my rotating frame, I have no idea what is causing it to pull on that string. The “correct frame” is the one which is not rotating.
Looked at the wikipedia page for gravitational red shift and gravitational time dilation, and yes looks very familiar, albeit I did not find that exact expression.
I will plead senility on that one. I don't know exactly where I got the expression (that was a lot of years ago). I will go with modest on the validity of the expression; see his second link (equation 17.19).
Got all the way to that paragraph and did not have problems with it. I'll have a rest here again.
Thank you very much for your efforts. I won't post any more of my proof of the fundamental equation until you have proof read the “laying out the representation” post.
Have fun -- Dick
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Just a short reply for now...
In my original paper, I had referenced that issue to page 6 of the 1965 edition of “Introduction to General Relativity” by Adler, Bazin and Schiffer.
The following is an exact quote of what is presented on page 6 of the 1965 edition.
Note that the whole issue boils down to “different velocities” for different objects.
Okay, I didn't understand all of that, but I have a faint idea of it having to do with getting inconsistent energies, and about relativistic time relationships adjusting them correctly... I think.
Note that with regard to the centrifugal and Coriolis forces, there is a simple Euclidean transformation which yields a straight line path for a free particle: i.e., the transformation to a non-rotating system. The issue with gravity is that they were searching for a transformation to a non-Euclidean system. Since the orbits return upon themselves the geometry simply can not be Euclidean because, in a Euclidean system, straight lines can not yield closed paths. In other words to find a rational result, they had to examine much more complex systems.
Right, okay.
Regarding centrifugal force the issue is actually quite simple. Most people measure angles in degrees. Physicists tend to measure angles in “radians” (it makes a lot of angular mathematics easy). By definition, instead of 360 degrees in a circle, there are [imath]2\pi[/imath] radians. That makes arc lengths easy to specify. If the angle [imath]\theta[/imath] is measured in radians the arc length is just [imath]r\theta[/imath]. Angular velocity is generally represented by [imath]\frac{d\theta}{dt}=\omega[/imath]. Thus the velocity along the circle of that rock out there on the end of the string is given by [imath]r\omega[/imath]. Note that the speed of the rock does not change; what changes is its direction. The change in velocity is always perpendicular to the path; it is towards the center, the direction the string is pulling. So the change in velocity (the acceleration) is the component of its velocity parallel to the string after time dt. Well the distance the rock has moved along the path is [imath]vdt=r\omega dt[/imath] so the change in velocity in the time dt is [imath]dv=(vdt)d\theta=r\omega d\theta[/imath] (draw a picture, the two triangles, radius against distance moved and velocity against velocity change, involved here are similar). Divide that by dt and one has the acceleration [imath]\frac{dv}{dt}=r\omega \frac{d\theta}{dt}= r\omega^2[/imath].
Ah, I see, pretty clever :)
No, I am still talking about the centrifugal force, the rock on a string. Seen from the rotating frame (where it is at rest) there is an apparent force holding it out there against the string. I am just calling that force a “gravitational force” because, looking at it from my rotating frame, I have no idea what is causing it to pull on that string. The “correct frame” is the one which is not rotating.
Right, okay.
I will plead senility on that one. I don't know exactly where I got the expression (that was a lot of years ago). I will go with modest on the validity of the expression; see his second link (equation 17.19).
Yes, looks like there it stands in exactly the same form. Thank you for digging that up Modest.
Thank you very much for your efforts. I won't post any more of my proof of the fundamental equation until you have proof read the “laying out the representation” post.
Yup, I'll take a look at it.
-Anssi
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That idea together with differential calculus created an extremely powerful mathematical method of predicting the dynamic behavior of objects (an object being any suitable stable defined collection of information). Since acceleration is the time derivative of velocity (where velocity is the time derivative of position in that “inertial frame”) force can be thought of as the time derivative of momentum (momentum being given by [imath]m\vec{v}[/imath]). If m is a constant, the expression [imath]\vec{F}=\frac{d}{dt}(m \vec{v})[/imath] is identical to [imath]\vec{F}=m\vec{a}[/imath] and if m is allowed to change, that fact simply allows a rather simple mechanism to handle cases where identical forces cause different accelerations. As a consequence, m ends up being little more than a parameter allowing a more versatile definition of force.
So here we are using the equation [imath]\vec{F}=m\vec{a}[/imath] as the definition of force in any inertial frame. That is, any frame that is not accelerating, or equivalently one that has no force acting on it, which is sort of a circular definition so maybe we should avoid it. But doesn’t this bring up the issue that different frames will not agree on the force on a object as they may not agree on length or mass so they wont agree on the acceleration of an object?
Also, what about the possibility that the mass will be a function of t or is it also required by our choice of using an inertial reference frame that the mass is a constant?
That brings up the interesting question, “what does the dynamic behavior look if one is using a non-inertial frame”. Clearly, a non-inertial frame is a frame which is accelerating relative to an inertial frame. It should be clear to the reader that any object at rest in any inertial frame (which then, by definition, has no forces accelerating it) will appear to be accelerating if its position is represented by coordinates in a non-inertial frame representation. It should also be quite clear that it is not really accelerating at all, it is only the reference frame which is actually changing (accelerating). The apparent motion of any force free physical object who's position is being represented via the non-inertial frame will be an exact mirror image of that frames acceleration.
So is there a way to tell what frame we are in, is it as simple as saying that in a non-inertial reference frame we are experiencing a force and the transformation you are talking about is a purely mathematical means of making the force on a object vanish in our explanation.
That is, we are now interested in finding a reference frames where if an object is experiencing a force it no longer has a force acting on it in the new coordinate system. And the rest of the universe is now experiencing a force in the opposite direction that results in the same acceleration.
$dt'=dt\sqrt{1-\left(\frac{|\vec{v}|}{c}\right)^2}\equiv dt\sqrt{1+\frac{2\Phi}{c^2}}$
which happens to be exactly the standard gravitational red shift. This implies that any geometry which yields gravity as a pseudo force must also yield the standard gravitational red shift; or, alternately, gravitational red shift is not really a valid test of Einstein's general theory of relativity. This really isn't very enlightening as the gravitational red shift can be shown to be required by conservation of energy, but it does nonetheless imply that the above analysis is valid.
That is, the only thing that a different force will imply is the actual form and value of [imath]\Phi[/imath] which is the equivalent velocity of an object. That is, it is the velocity that the object will have in a reference frame where the force on the object vanishes that will make the Lorenz transformation the correct transformation.
But I don’t understand how this results in red shift of a object as it looks like all that you have done is calculate the change of a clock in an accelerating reference frame (the left side of the equation) in comparison to what a clock in an inertial reference frame will measure (the right side of the equation).
More importantly, the above suggests an attack towards determining the geometry which will yield gravity as a pseudo force in our four dimensional Euclidean geometry. I have already shown how static structures appear as three dimensional objects in this geometry so let us examine what is commonly called “a gravitational well”. The gravitational well consists of a vertical hole where there is a gravitational field in the vertical direction. If an experimenter in a gravitational well sets up a clock via a light pulse traveling back and forth between two horizontally displaced mirrors, since we can establish horizontal measure (simple vertical lines carry those measures to different heights in the hole) and his clock must run slow, we must see the apparent velocity of light to be
$c'=c\sqrt{1+\frac{2\Phi}{c^2}}$
So do we conclude that the passage of the clock in the gravitational well is running slower because just like in your example of centrifugal force the object must act as though it is moving in comparison to a inertial reference frame. That is, the Lorenz transformation is valid we just have to solve for the required function to substitute in for the velocity of the object in it’s inertial frame.
Won’t this result though, in the conclusion that the length of his clock is longer then our clock rather then a faster speed of light since we have defined length and speed to be interdependent of each other and so it would make sense to measure his clock by means of how long light takes to move from one of the mirrors to the other. As a result we would conclude that the lines that are extended up from the mirrors are not straight but are bent lines.
Or is this the very problem with general reactivity, it assumes that the speed of light is constant and as a result must generate a geometry where the lines are bent? And since the geometry that you are using is a Euclidean geometry those lines must be straight lines so we have no way to conclude that the speed of light is the same in a gravity well and must in fact conclude that it is slower. But if this is the case what will someone in a gravity well see when looking at a clock that is not in the gravity well? Will they conclude that our clock is running faster then one not in the gravity well.
Also, I don’t understand how you arrived at the above equation but I suspect that it is an issue of velocity rather then length being scaled by the Lorenz transformation.
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With regard to the issue of refraction, my fundamental equation is a wave equation with Dirac delta function interactions. Clearly, in the absence of interactions, the probability wave representing an event will proceed at a fixed velocity. Any specific delta function interaction can be seen as an impact changing the direction and energy of that probability wave. What is important here is the fact that interaction will depend upon the distance between the two elements connected by that delta function interaction: i.e., the hypothetical element (which must be a boson) must carry the momentum and energy being transfered and the transfer must be consistent with the Heisenberg uncertainty principal: i.e., the uncertainty in momentum is directly related to the uncertainty in position. This implies that the further apart the interacting fermions are, the less momentum transfered must be (see virtual particle exchange).
I'm struggling here a bit.
First I am not exactly sure what do you mean by a probability wave representing an event. I was aligned to think of this in terms of probability waves representing the positions of defined objects. Do you just see them as essentially the same thing, or do you refer to something like a probability wave representing something like a collision between virtual particle and a fermion?
Anyway, I read that page about the virtual particle exchange, did not understand everything about it, but I picked up some idea of what you are talking about. I did not pick up though, why the Heisenberg uncertainty principle implies that the further apart the interaction fermions are, the less momentum transfer must be... I guess it has got something to do with how the wave functions interfere, but I am completely unfamiliar with that subject and didn't manage to figure it out... :P
Any physical object (any structure stable enough to be thought of as an object) must have internal forces maintaining that structure. Any interaction with another distant object must be via the virtual particle exchange I just commented about. Thus it is that one would expect the fundamental element of that physical object interacting via that delta function would have its momentum altered, not the whole object; however, that alteration would create a discrepancy in the structure of the object under discussion. Since that object must have internal forces maintaining its structure, it is to be expected that those internal interactions (which are also mediated by delta function exchange forces) will bring the trajectory that interacting fundamental element essentially back to its original path (at least on average).
Thus it is that the path of that fundamental element can be seen as crooked as compared to its path in the absence of that distant object. Of issue is the fact that, if the influence of the distant object is ignored, the influenced element will inexplicitly appear to be proceeding at a slightly slower velocity than it would if the distant object didn't exist. What is important here is that this effect decreases as the distance from the distant object increases. That means that the net effect is to yield a very slight change in the speed of the elements which make up that object as one moves across the object. The net effect of such an interaction is to refract the wave function of the object under examination.
Right I see, the point is that those sub-elements of the object that are closer to the distant object, are slowed down slightly more than those sub-elements that are further away, so the total average path would be expected to curve... I understand the analogy to refraction now.
So I suppose when you refer to the speed of an element, you must be referring to its speed in [imath]x,y,z,\tau[/imath]-space. I.e. it is because every element must move at the same fixed speed, that more wiggling means slower speed towards the average direction of the entire object.
I spent quite a while thinking about this, and yeah, it seems valid.
If the distant object and the object under observation are not moving with respect to one another (they are moving parallel to one another in the tau direction), the net effect of that refraction is to curve the paths of the two objects towards one another: i.e., there will be an apparent attraction between them.
Yup.
It is also evident that, since the mass of the source object (the source of these bosons external to the object of interest) is proportional to the total momentum of that object, one should expect the apparent density (as seen from the object of interest) should be proportional to its mass: i.e., one should expect the exchange forces to be proportional to mass.
Right... When you say "total momentum", you are essentially referring to the number and density of the individual elements moving along [imath]\tau[/imath] (at fixed speed).
Clearly the interaction just discussed arises from differential effects in the basic interactions thus it will amount to a force considerably less than the underlying force standing behind that differential effect. Thus it is that the two forces I have already discussed (the forces due to massless boson exchange: shown to yield electromagnetic effects and the forces due to massive boson exchange: shown to yield fundamental nuclear forces) will end up being split into four forces.
I may be forgetting something, but where and when the massive boson exchange/nuclear forces were brought up...?
Differential effects will yield a correction to both basic forces which correspond quite well with the forces observed in nature. The differential effect on massless boson exchange yields what appears to be a very weak gravitational force (weak when compared to the underlying electromagnetic effects) and the differential effect on massive boson exchange yields what appears to be a very weak nuclear force (weak compared to the underlying nuclear force).
Right, that makes perfect sense to me, if there's a force mediated by boson exchange, there must be a corresponding differential effect via refraction, due to the fixed speed of the elements...
What is interesting is that the “weak nuclear force” can be shown to violate parity symmetry whereas the “weak electrical force” (gravity) does not. This is a direct consequence of the fact that the nuclear exchange bosons are massive.
That is very interesting... Another very unexpected feature of modern physics turning out to be a very expected feature of the symmetry requirements. Should probably discuss that issue little bit at some point as well.
Apart from this analysis, does any satisfying explanation exist as to why weak nuclear force violates parity symmetry?
I'll have a pause here again, but first I still have one more comment. When I'm thinking about the differential effect caused by boson exchange forces, I can't help but think of the unexplained behaviour of foucault pendulum during a solar eclipse.
Decrypting the Eclipse - NASA Science
Current theories of gravity don't explain the Allais effect, and AFAIK the best attempt to explain it so far is that the shadow of the moon causes air to cool down and its density to increase, which causes a gravitational effect. That apparently causes too small gravitational effect to actually explain the Allais effect though (I don't have the skills to work it out, but certainly that explanation sounds dubious at best).
At any rate, someone more skilled than me could probably work out whether this paradigm explains Allais effect directly as a gravitational effect. I guess the moon should be expected to affect the virtual particle exchange between the sun and the pendulum one way or another. I can't work out what sort of interference one might expect between all the relevant bodies, but I'm just thinking if this explains the Allais effect as a gravitational effect, that would be remarkable. And another first. :shrug:
-Anssi
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Cody
Problem 2130. Orthonormal matrix
Solution 2158824
Submitted on 13 Mar 2020 by HH
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
a=[1 0 0 ;0 1 0;0 0 1] tf = 1; assert(isequal(ortho_normal(a),tf))
a = 1 0 0 0 1 0 0 0 1
2 Pass
a=[0 0 1 0;0 1 0 0;1 0 0 0;0 0 0 1] tf = 1; assert(isequal(ortho_normal(a),tf))
a = 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1
3 Pass
a=eye(20) k=a; p=a; a(:,1)=k(:,end); a(:,end)=k(:,1); a(:,5)=p(:,9); a(:,9)=p(:,5) tf = 1; assert(isequal(ortho_normal(a),tf))
a = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 a = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4 Pass
a=[1 0 1 0;0 1 0 0;1 0 0 0;0 0 0 1] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1
5 Pass
a=[0 0 2 0;0 1 0 0;1 0 -1 0;0 0 0 1] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 0 0 2 0 0 1 0 0 1 0 -1 0 0 0 0 1
6 Pass
a=[1 2 3 ;1 1 5;3 0 1] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 1 2 3 1 1 5 3 0 1
7 Pass
a =[1 0 0 ;0 0 0; 0 0 0] tf = 0; assert(isequal(ortho_normal(a),tf))
a = 1 0 0 0 0 0 0 0 0
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# 1233: Relativity
(Redirected from Relativity)
Relativity Title text: It's commonly believed that Lorentz contraction makes objects appear flatter along the direction of travel. However, this ignores light travel times. In fact, a fast-moving butt would appear rotated toward the observer but not substantially distorted. Shakira was right.
## Explanation
Gedankedank is a humorous portmanteau of Gedankenexperiment (German for "thought experiment") and badonkadonk (slang term for an attractive, round butt).
Albert Einstein often used thought experiments to explore scientific hypotheses too impractical or impossible to actually perform, in order to examine their consequences. Moving close to the speed of light, c, is nigh-impossible with existing technology; and according to Einstein's theory of special relativity accelerating a mass exactly to c is impossible. Einstein is well known not to have cared about his appearance (e.g. (WARNING: TVTropes link) his uncombed hair, the tongue photo, etc.) so it is unlikely that he pondered how relativistic velocity would affect the appearance of his butt.
Lorentz contraction is a consequence of special relativity, whereby objects contract in the direction of travel. An object traveling exactly at c (rather than near light speed) would be squished completely flat. It would also have infinite mass, infinite energy, and experience no passage of time until slowed by an outside force.
The appearance of rotation while moving close to the speed of light is known as Terrell rotation. The title text then connects this rotation to the Shakira song "Hips Don't Lie".
## Transcript
[We see a head and shoulders view of Einstein. He looks pensive.]
[Einstein's "line" is in a thought bubble.]
Einstein: If I were traveling at the speed of light, my butt would look awesome.
Einstein was famed for his Gedankedank.
# Discussion
Should be "Gedankengang" (Plural, it's "Gedanke" + "n"). Means: chain/train/line of thought(s) 91.66.205.94 05:52, 3 July 2013 (UTC)
Interesting idea, because it matches the text more closely, but he really wasn't that famous for his "chain of thought" - more famous for his Gedankenexperimente, for example, chasing a light beam (leading to Special Relativity) or the thought experiments that lead to the EPR paradox. --196.35.92.54 09:52, 3 July 2013 (UTC)
Anyway, "Gedankedank" does not really make sense, it's not a word. 130.60.152.125
Might be refering to "Badonkadonk", even though "Gedankedank" isnt a german word. 178.26.88.31
That's what I think: He's taken "Gedankenexperiment", which sounds funny in English, and taken the "Gedanken" part and reduplicated it to match the sound of "badonkadonk", without worrying about whether it makes sense in German. --196.35.92.54 12:22, 3 July 2013 (UTC)
"chain of thought" might be a too literal translation. I think "Gedankengang" is a bit broader than just meaning "chain of thought". Its literal translation would be "go of the thoughts" or "way the thoughts are going". If Randall has done some research before creating the word, then this is also a possiblity. In German one could really say, that Albert Einstein is famous for his "Gedankenexperimente", "Gedanken" or "Gedankengänge". Maybe we should offer both explanations? --212.255.32.112 13:35, 6 July 2013 (UTC)
Also, from reading http://forums.xkcd.com/viewtopic.php?f=7&t=103423, in last week's "what if" there was:
20 meters per second is about how fast an average person with a good arm could throw a bouncy ball. Therefore, to determine the result of an impact, we can make use of what Einstein called a gedankenexperiment, or "thought experiment"
196.35.92.54 10:02, 3 July 2013 (UTC)
I'm not sure what the "Lorentz contraction" in the title text is referring to, should this be added to the description? 96.228.23.74 14:06, 3 July 2013 (UTC)
Side note: Logical conclusion -- Shakira's hips must like big butts. JamesCurran (talk) 15:30, 3 July 2013 (UTC)
" it is unlikely that he pondered how relativistic velocity would affect the appearance of his butt. " This, ladies and gentlemen, is why i read this! Funnier than xkcd sometimes... Way too many times, sometimes... 189.5.99.20 03:43, 4 July 2013 (UTC)
I just realized the number of this comic. Friday will be comic 1234. I doubt there will be a joke about it since I think only 404 and 1000 had any self-reference. 76.106.251.87 14:37, 4 July 2013 (UTC)
Also, as part of a discussion for THIS comic, there was a bit of magnificent basterdry on Randall's part in seeding within his What If all the information needed to get the joke. Almost anyone that read it would get an immediate chuckle out of it. For me, I had even forgotten the source of where I had learned what "gedankenexperiment" meant. 76.106.251.87 05:20, 5 July 2013 (UTC)
AIUI, Einstein shouldn't notice any change himself - well unless he's suddenly also Mr. Fantastic... the MARVEL Mr. Fantastic of the Fantastic Four, and only his lower body is travelling that fast, and his head is in a different frame, where the relative velocities between the two frames is ~c. (But that's stupid. If he was Mr. Fantastic, he'd just fix his butt. Or.. is this superfast motion how Mr. Fantastic gets his powers (and Mystique)? Does that mean Quicksilver and Flash can also exhibit these shape-shifting powers?). Anyway, coming back to the point - I didn't get what the title text states... so I employed some Google-fu and found this: http://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/Flash/ContractInvisible.html I think you'd only see the edge of the disk though, and not a ring... given the camera is a 2 dimensional thing in the plane of the disk. Reason I mention that is, I'm still unsure how this works for a 3D object where the object is moving across your FOV. Anyone have any ideas? 220.224.246.97 10:45, 5 July 2013 (UTC)
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# How Many Milliliters Are There in a Pound?
There are 473.18 milliliters in 1 pound. One milliliter is equal to roughly 0.002 pound. To convert from milliliters to pounds, one must first convert from milliliters to liquid ounces, and then convert from ounces to pounds. There are roughly 0.034 ounces in 1 milliliter.
A milliliter is a metric unit of capacity equal to 1/1000 of a liter and about 0.06 cubic inch. Milliliters are frequently used to measure small amounts of liquids such as water. A pound is a unit of weight or mass equal to 16 ounces or 7,000 grains. Pounds are frequently used to measure solid goods.
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# correct
1. ### What is the correct formula to calculate the angle of vector A = 0i + 1j
I know the angle is 90 degrees. and the formula to calculate the angle of vector A is inverse tan of (Ay / Ax), but Excel returns an error message because Ax is zero. Thanks.
2. ### d'alembert worked example but is trig correct?
Trying to follow another D'Alembert problem (see attached). Can't figure the step where they wrote \delta W_G = -mgR \sin \phi \delta \phi, shouldn't it be \delta W_G = -mgR (1 - \cos \phi) \delta \phi since we are measuring against the vertical force of gravity?
3. ### Is this correct 2 (Different topic)
While waiting for a responce in my earlier post, I looked theough my project one more time and realized that it had anoter big problem. Again I found a formula I'd like to verify, this time with velocity. Velocity^2 = 2 x acceleration x distance Then take the square root of v^2 to get...
4. ### Is this correct?
My formula for a ball rolling down a ramp: (Weight)sin(angle) = a (Weight)cos(angle) = b (Friction) times b = c a - c = d d divided by mass = e e + (5 x 9.8 x sin(angle) ) / 2 = acceleration Is this correct?
5. ### Finding correct gear ratio
I'm trying to find a formula that will allow me to predict what rear end gear I would need to run at a particular oval race track. For instance, Say we run at Daytona, a 2.5 mile track, in a 3400lb car with 850 hp engine turning 8200rpms and tire circumference of 85in and a 1:1 final...
6. ### Correct?
Sorry to bug you guys again. Wanted to make sure I am doing this correctly: A kangaroo jumps vertically to a height of 2.7m. How long was he in the air before returning to earth? I know y0 = 0, y1 = 2.7m and a = -9.80m/s^2 Find velocity with: v^2 = v0^2 + 2a (y1 - y0) v^2 =...
7. ### Is this magnetic field equation correct in material as well?
Or is it only valid in vacuum condition? Segment of straight wire carrying current I (at distance r) dB = u.Idl x r / 4pi . r^2 (I've came across it on a book about Maxwell's Equations.) Does correction have to be made for polarization or magnetization of the material medium? If so, how?
8. ### Are these statements correct?
Hi Everyone: I am writing an essay and would appreciate it if you would check the accuracy of these statements, and let me know how you would correct them, if they weren't accurate. Thanks! Physicists know that the different states of matter are defined at a minute level. A rock is made up...
9. ### Help me understand correct interpretation of Gauss' Magnetic Law
I have been looking at http://www.maxwells-equations.com and they had a different interpretation to what I expected to Gauss' magnetic law. I know this is the divergence operator but for the case of a bar magnet inside a sphere (completely) I imagined that it was the number of magnetic...
10. ### is this the correct way to find impedance
load absorbs 75583 W at a lagging power factor of pf= 0.85. If the current flowing through the load is ix= 0.100A-rms , determine the impedance of the load. VA = 75583 / .85 = 88921 VA = I^2 Z Z = VA / I^2 Z = 88931 / 0.1^2 8.89 x 10^6 ohms
11. ### Is this correct?
Women weighs 100 lb-f on her home planet where acc due to gravity is 35 ft/s^2. How much will she weigh on a planet where the acc due to gravity is 25 ft/s^2?? Using w=mg 100 lb-f = m ( 35 ft/s^2) m = 2.287 lb-f* ft/s^2 So now you use the same formula w= 2.287 lb-f * ft/s^s *(25...
12. ### Problem getting the correct induced current (solution included)
I DO see why the correct solution is correct. What I do not see is why my version is incorrect. Here they are: Correct way: http://i.imgur.com/3SPhU.jpg My way: http://i.imgur.com/RWTbe.jpg Alternatively, you can download them as I am attaching them to this post as well. Basically, I...
13. ### Do I have the correct acceleration??
When a parachut opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915N and the drag force has a magnitude of 1027N. The mass of the sky diver is 93.4kg...
14. ### Am I correct in setting up this ODE?
The motion of various shaped objects that bob in a pool of water can be modeled by a second order differential equation derived from F = ma. The forces acting on the object are: 1)force due to gravity, 2)a frictional force by water, 3)and a buoyant force based on Archimedes'...
15. ### Is it correct [potential difference between two spheres (mathematical)]
Greetings everyone!! Consider two conducting spheres of radius R carrying equal and opposite charge Q. let the distance between their surfaces be d. We will take the distance from surface of the sphere carrying positive charge. Now the electric field at a point at distance x between the spheres...
16. ### Did I do this correct F=ma to find Vi???
Need help on this. A car whose mass is 2600 kg can produce an unbalanced braking force of 2765N. Calculate the car's initial speed in order to have a stopping distance of 3036.3m. Vi=? So... m=2600kg F=2765N s=3036.3m Vf=0 So I found acceleration of a=F/m = 2765/2600 = 1.06346m/s^2 So I have to...
17. ### Finding correct angle for equal tension
Finding equal tension A 4m length of steel girder is hung horizontally by steel cable each end, the left end is f1 the right end is f2. The steel girder has a mass of 20 kg, 1m from f1 there is a 80 N force hanging down. F2 cable is positioned 30 degree's from the horizontal with a force in the...
18. ### Are my answers correct?
An object starts from rest w/ a constant acceleration of 8m/s^2 along straight line. a)find the speed at the end of 5 sec.------40sec b)the ave. spped for 5 sec interval------20m/s A stone is thrown vertically upward w/ an initial velocity of 49m/s a)How long does it take to reach the highest...
19. ### Are these two kinetic equations correct for the following two problems?
Question 1: Skier travels down straight ramp with constant acceleration of 8.0 ms. Skiers velocity after 5 seconds is 40ms-1 During the time of 5 seconds, what distance is travelled? (this is what i did) V1 = intial velocity V2 = final velocity d = V1 x t + 1/2 x a x t² 0 x 5 + 1/2 x...
20. ### Can someone pls correct me if im wrong thnks :)
You are standing by the side of the street motionless when a car is traveling at a speed of 20m/s passes you. It continues traveling at this speed for 4.0 secs. Then the driver hits the brakes and, with a constant acceleration, comes to rest; traveling 25m while breaking. After 2 secs at rest...
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# Month: July 2021
## Equivalent Definitions of e
(A new question of the week) It is not unusual for mathematicians to define a concept in multiple ways, which can be proved to be equivalent. One definition may lead to a theorem, which another presentation uses as the definition, from which the original definition can be proved as a theorem. Here, in yet another …
## Finding a Function Value Recursively
(A new question of the week) May was a particularly good month for interesting questions! Here is one requiring us to find one value of a function, based on an unusual property: If $$a+b=2^x$$, then $$f(a)+f(b)=x^2$$. The problem turned out to be not as hard as it looked, yet the function itself is quite interesting …
## Separable Differential Equations
(A new question of the week) We received a couple different questions recently about solving differential equations by separation of variables, and why the method is valid. We’ll start with a direct question about it, and then look at an attempt at an alternate perspective using differentials.
## Long Division with Zero, Revisited
(A new question of the week) One of our first posts, in 2018, was about zeros in long division. But we still get many questions about this issue, and it’s time to dig in deeper. We’ll look here at two of them, answering the twin questions, “When do you put a zero in the quotient …
## Two Integration Puzzlers
Two recent questions (that came to us within two hours) dealt with apparent contradictions in integration. The first seems to give a result of zero that is clearly wrong; the second seems to give two different results for the same integral.
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# hw4 - BE 110 Homework Assignment#4 CONTINUUM MECHANICS Fall...
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BE 110 CONTINUUM MECHANICS Fall 2009 Homework Assignment #4 Due Thursday 10/12/09, 3:30 PM before class No late submissions will be accepted. Practice Problems . 1. Find the resultant force and moment distribution in the foot of a diving athlete with weight W standing upright on a platform just before the jump. The athlete is standing on the edge of the platform supported only by the toes without support on the heels. The length of the foot from the heel to the base of the toes is L. 2. Let τ ij be the stress tensor. What is the physical interpretation of the following quantities: (a) τ 33 (b) τ 13 (c) -(1/3) τ ij δ ij (d) τ ij i where δ ij is the Kronecker symbol and i is a unit vector. 3. The stress components τ ij have been determined to be 3 x 2 + 4 xy - 8 y 2 ax 2 - 6 xy + by 2 0 ax 2 - 6 xy + by 2 2 x 2 + 2 xy - 3 y 2 0 0 0 0 The body force components are assumed to be zero. Under what conditions is this stress field in equilibrium? Sketch in graphical form the stress components in the x-y plane.
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# Python program to calculate the average score
Posted by Editorial Staff | Updated on | .
Today we are going to share a Python program to calculate the average score. If you are a python beginner and want to start learning the python programming, then keep your close attention in this …
# Python program to check alphabet or not
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Today we are going to share a Python program to check alphabet or not. If you are a python beginner and want to start learning the python programming, then keep your close attention in this …
# Python program to print an identity matrix
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Today we are going to share a Python program to print an identity matrix. If you are a python beginner and want to start learning the python programming, then keep your close attention in this …
# Python program to print duplicates number from a given list
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Today we are going to share a Program to print duplicates number from a given list. If you are a python beginner and want to start learning the python programming, then keep your close attention …
# Python program to find Hash of a file
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Today we are going to share a Python program to find Hash of a file. If you are a python beginner and want to start learning the python programming, then keep your close attention in …
# Python program to map two lists into a dictionary
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Today we are going to share a Python program to map two lists into a dictionary. If you are a python beginner and want to start learning the python programming, then keep your close attention …
# Python program to calculate the average of numbers in a given list
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Today we are going to share a Python program to calculate the average of numbers in a given list. If you are a python beginner and want to start learning the python programming, then keep …
# Python program to reverse a string using recursion
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# Python program to find the intersection of two lists
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# Python program to find the Union of two lists
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# Python program to check if a number is a perfect number
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# Python program to print table of a given number
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# Python program to convert kilometers to miles
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We are going to share a Python program to convert kilometers to miles. If you are a python beginner and want to start learning the python programming, then keep your close attention in this tutorial …
# Python program to find ASCII value of character
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# Python program to reverse a string using recursion
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We are going to share a Python program to reverse a string using recursion. If you are a python beginner and want to start learning the python programming, then keep your close attention in this …
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# How can I obtain segmented linear regressions with a priori breakpoints?
I need to explain this in excruciating detail because I don't have the basics of statistics to explain in a more succinct way. Asking here in SO because I am looking for a python solution, but might go to stats.SE if more appropriate.
I have downhole well data, it might be a bit like this:
``````Rt T
0.0000 15.0000
4.0054 15.4523
25.1858 16.0761
27.9998 16.2013
35.7259 16.5914
39.0769 16.8777
45.1805 17.3545
45.6717 17.3877
48.3419 17.5307
51.5661 17.7079
64.1578 18.4177
66.8280 18.5750
111.1613 19.8261
114.2518 19.9731
121.8681 20.4074
146.0591 21.2622
148.8134 21.4117
164.6219 22.1776
176.5220 23.4835
177.9578 23.6738
180.8773 23.9973
187.1846 24.4976
210.5131 25.7585
211.4830 26.0231
230.2598 28.5495
262.3549 30.8602
266.2318 31.3067
303.3181 37.3183
329.4067 39.2858
335.0262 39.4731
337.8323 39.6756
343.1142 39.9271
352.2322 40.6634
367.8386 42.3641
380.0900 43.9158
388.5412 44.1891
390.4162 44.3563
395.6409 44.5837
``````
(the Rt variable can be considered a proxy for depth, and T is temperature). I also have 'a priori' data giving me the temperature at Rt=0 and, not shown, some markers that i can use as breakpoints, guides to breakpoints, or at least compare to any discovered breakpoints.
The linear relationship of these two variables is in some depth intervals affected by some processes. A simple linear regression is
`q, T0, r_value, p_value, std_err = stats.linregress(Rt, T)`
and looks like this, where you can see the deviations clearly, and the poor fit for T0 (which should be 15):
I want to be able to perform a series of linear regressions (joining at ends of each segment), but I want to do it: (a) by NOT specifying the number or locations of breaks, (b) by specifying the number and location of breaks, and (c) calculate the coefficients for each segment
I think I can do (b) and (c) by just splitting the data up and doing each bit separately with a bit of care, but I don't know about (a), and wonder if there's a way someone knows this can be done more simply.
I have seen this: http://stats.stackexchange.com/a/20210/9311, and I think MARS might be a good way to deal with it, but that's just because it looks good; I don't really understand it. I tried it with my data in a blind cut'n'paste way and have the output below, but again, I don't understand it:
-
The short answer is that I solved my problem using R to create a linear regression model, and then used the `segmented` package to generate the piecewise linear regression from the linear model. I was able to specify the expected number of breakpoints (or knots) `n` as shown below using `psi=NA` and `K=n`.
R version 3.0.1 (2013-05-16)
Platform: x86_64-pc-linux-gnu (64-bit)
``````# example data:
bullard <- structure(list(Rt = c(5.1861, 10.5266, 11.6688, 19.2345, 59.2882,
68.6889, 320.6442, 340.4545, 479.3034, 482.6092, 484.048, 485.7009,
486.4204, 488.1337, 489.5725, 491.2254, 492.3676, 493.2297, 494.3719,
495.2339, 496.3762, 499.6819, 500.253, 501.1151, 504.5417, 505.4038,
507.6278, 508.4899, 509.6321, 522.1321, 524.4165, 527.0027, 529.2871,
531.8733, 533.0155, 544.6534, 547.9592, 551.4075, 553.0604, 556.9397,
558.5926, 561.1788, 562.321, 563.1831, 563.7542, 565.0473, 566.1895,
572.801, 573.9432, 575.6674, 576.2385, 577.1006, 586.2382, 587.5313,
589.2446, 590.1067, 593.4125, 594.5547, 595.8478, 596.99, 598.7141,
599.8563, 600.2873, 603.1429, 604.0049, 604.576, 605.8691, 607.0113,
610.0286, 614.0263, 617.3321, 624.7564, 626.4805, 628.1334, 630.9889,
631.851, 636.4198, 638.0727, 638.5038, 639.646, 644.8184, 647.1028,
647.9649, 649.1071, 649.5381, 650.6803, 651.5424, 652.6846, 654.3375,
656.0508, 658.2059, 659.9193, 661.2124, 662.3546, 664.0787, 664.6498,
665.9429, 682.4782, 731.3561, 734.6619, 778.1154, 787.2919, 803.9261,
814.335, 848.1552, 898.2568, 912.6188, 924.6932, 940.9083), Tem = c(12.7813,
12.9341, 12.9163, 14.6367, 15.6235, 15.9454, 27.7281, 28.4951,
34.7237, 34.8028, 34.8841, 34.9175, 34.9618, 35.087, 35.1581,
35.204, 35.2824, 35.3751, 35.4615, 35.5567, 35.6494, 35.7464,
35.8007, 35.8951, 36.2097, 36.3225, 36.4435, 36.5458, 36.6758,
38.5766, 38.8014, 39.1435, 39.3543, 39.6769, 39.786, 41.0773,
41.155, 41.4648, 41.5047, 41.8333, 41.8819, 42.111, 42.1904,
42.2751, 42.3316, 42.4573, 42.5571, 42.7591, 42.8758, 43.0994,
43.1605, 43.2751, 44.3113, 44.502, 44.704, 44.8372, 44.9648,
45.104, 45.3173, 45.4562, 45.7358, 45.8809, 45.9543, 46.3093,
46.4571, 46.5263, 46.7352, 46.8716, 47.3605, 47.8788, 48.0124,
48.9564, 49.2635, 49.3216, 49.6884, 49.8318, 50.3981, 50.4609,
50.5309, 50.6636, 51.4257, 51.6715, 51.7854, 51.9082, 51.9701,
52.0924, 52.2088, 52.3334, 52.3839, 52.5518, 52.844, 53.0192,
53.1816, 53.2734, 53.5312, 53.5609, 53.6907, 55.2449, 57.8091,
57.8523, 59.6843, 60.0675, 60.8166, 61.3004, 63.2003, 66.456,
67.4, 68.2014, 69.3065)), .Names = c("Rt", "Tem"), class = "data.frame", row.names = c(NA,
-109L))
library(segmented) # Version: segmented_0.2-9.4
# create a linear model
out.lm <- lm(Tem ~ Rt, data = bullard)
# Set X breakpoints: Set psi=NA and K=n:
o <- segmented(out.lm, seg.Z=~Rt, psi=NA, control=seg.control(display=FALSE, K=3))
slope(o) # defaults to confidence level of 0.95 (conf.level=0.95)
# Trickery for placing text labels
r <- o\$rangeZ[, 1]
est.psi <- o\$psi[, 2]
v <- sort(c(r, est.psi))
xCoord <- rowMeans(cbind(v[-length(v)], v[-1]))
Z <- o\$model[, o\$nameUV\$Z]
id <- sapply(xCoord, function(x) which.min(abs(x - Z)))
yCoord <- broken.line(o)[id]
# create the segmented plot, add linear regression for comparison, and text labels
plot(o, lwd=2, col=2:6, main="Segmented regression", res=TRUE)
abline(out.lm, col="red", lwd=1, lty=2) # dashed line for linear regression
text(xCoord, yCoord,
labels=formatC(slope(o)[[1]][, 1] * 1000, digits=1, format="f"),
pos = 4, cex = 1.3)
``````
-
What you want is technically called spline interpolation, particularly order-1 spline interpolation (which would join straight line segments; order-2 joins parabolas, etc).
There is already a question here on Stack Overflow dealing with Spline Interpolation in Python, which will help you on your question. Here's the link. Post back if you have further questions after trying those tips.
-
I don't understand how I can use the methods in the linked question. I am quite dim when it comes to this, so perhaps you could explain a bit more? I have irregularly-spaced 1D data and I want to perform some linear regression analysis on it; the linked question is about point interpolation of regularly-gridded multivariate data: 2D, 3D?? – a different ben Aug 26 '12 at 6:04
@a different ben: Oops, my bad. A quick Google search revealed that Scipy has spline interpolation. I never used it, but see if the material in this link helps you: docs.scipy.org/doc/scipy/reference/tutorial/interpolate.html – HerrKaputt Aug 26 '12 at 10:21
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https://www.tvokids.com/transcript/114351/school-age/videos/area-triangle
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(music plays)
In a colourful TV studio, Robyn stands next to a wall touch screen showing a triangle and a formula. She is in her mid-twenties, with long blond hair. She wears a blue T-shirt with a print that reads "Home Work Zone" and black trousers.
She says HEY, TVO KIDS.
I'M TEACHER ROBYN.
TODAY, WE'RE GOING TO EXPLORE
HOW TO FIND THE ARE OF A
TRIANGLE USING A FORMULA.
IT'S REALLY EASY, REMEMBER THAT
AREA IS THE AMOUNT OF SPACE
A TWO DIMENSIONAL SHAPE TAKES
UP. HERE'S THE FORMULA:
TO FIND THE AREA OF A TRIANGLE,
MULTIPLY THE BASE BY THE HEIGHT
AND DIVIDE THE WHOLE THING
BY TWO.
THE BASE IS THE BOTTOM
OF THE TRIANGLE,
AND THE HEIGHT
IS HOW FAR UP IT GOES.
She clicks on a top menu and an equilateral triangle with a dotted line in the middle appears.
She continues HERE'S AN EXAMPLE: HERE WE'VE
BEEN GIVEN THE NUMBERS,
BUT SOMETIMES, YOU MAY NEED TO
USE A RULER,
OR OTHER MEASURING DEVICE TO
WE KNOW OUR FORMULA IS BASE
TIMES HEIGHT DIVIDED BY TWO
NOW, ALL WE NEED TO DO IS PLUG
IN THE NUMBERS.
WE ALWAYS WANT TO MAKE SURE THAT
WE'RE USING OUR UNITS
WHEN REPRESENTING THE BASE AND
THE HEIGHT, IN THIS CASE,
THE BASE IS 12 CENTIMETRES
TIMES SEVEN CENTIMETRES.
WHEN YOU'RE FIGURING OUT
THE AREA OF THE TRIANGLE,
MAKE SURE YOU GO STEP BY STEP.
She uncovers the bottom of the screen to show the answer.
She continues NOW WE KNOW THAT THE AREA OF THE
TRIANGLE IS 84 CENTIMETRES
SQUARED DIVIDED BY TWO.
WE SAY CENTIMETRES SQUARED
TWO DIMENSIONS.
FINALLY, WE DIVIDE 84 BY TWO AND
42 CENTIMETRES SQUARED.
LET'S TRY ANOTHER EXAMPLE:
HERE, WE CAN SEE THAT THE UNITS
HAVE CHANGED.
NOW, WE'RE TALKING IN METRES.
HOWEVER, THE FORMULA IS THE SAME
WE KNOW THAT THE BASE OF THE
TRIANGLE IS EIGHT METRES
AND THE HEIGHT IS FIVE METRES.
WE START BY MULTIPLYING EIGHT
METRES BY FIVE METRES.
THAT GIVES US
40 METRES SQUARED.
NOW, WE'LL QUICKLY DIVIDE
THAT BY TWO,
AND WE END WITH THE AREA
OF THE TRIANGLE,
WHICH IS 20 METRES SQUARED.
ONCE YOU HAVE THIS FORMULA,
IT'S REALLY SIMPLE. WAY TO GO,
TVO KIDS. SEE YOU NEXT TIME!
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https://cran.microsoft.com/snapshot/2019-10-31/web/packages/bang/vignettes/bang-hef-vignette.html
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# Conjugate Hierarchical Models
#### 2017-11-20
The bang package simulates from the posterior distributions involved in certain Bayesian models. See the vignette Introducing bang: Bayesian Analysis, No Gibbs for an introduction. In this vignette we consider the Bayesian analysis of certain conjugate hierarchical models. We give only a brief outline of the structure of these models. For a full description see Chapter 5 of Gelman et al. (2014).
Suppose that, for $$j = 1, \ldots, J$$, experiment $$j$$ of $$J$$ experiments yields a data vector $$Y_j$$ and associated parameter vector $$\theta_j$$. Conditional on $$\theta_j$$ the data $$Y_j$$ are assumed to follow independent response distributions from the exponential family of probability distributions. A prior distribution $$\pi(\theta_j \mid \phi)$$ is placed on each of the population parameters $$\theta_1, \ldots, \theta_J$$, where $$\phi$$ is a vector of hyperparameters. For mathematical convenience $$\pi(\theta_j \mid \phi)$$ is selected to be conditionally conjugate, that is, conditionally on $$\phi$$ the posterior distribution of $$\theta_j$$ of the same type as $$\pi(\theta_j \mid \phi)$$.
Use of a conditionally conjugate prior means that it is possible to derive, and simulate from, the marginal posterior density $$\pi(\phi \mid \boldsymbol{\mathbf{y}})$$. The hef function does this using the generalized ratio-of-uniforms method, implemented by the function ru in the rust package (Northrop 2017)). By default a model-specific transformation of the parameter vector $$\phi$$ is used to improve efficiency. See the documentation of hef and the two examples below for details. Simulation from the full posterior density $$\pi(\boldsymbol{\mathbf{\theta}}, \phi \mid \boldsymbol{\mathbf{y}}) = \pi(\boldsymbol{\mathbf{\theta}} \mid \phi, \boldsymbol{\mathbf{y}}) \pi(\phi \mid \boldsymbol{\mathbf{y}})$$ follows directly because conditional conjugacy means that it is simple to simulate from $$\pi(\boldsymbol{\mathbf{\theta}} \mid \phi, \boldsymbol{\mathbf{y}})$$, given the values simulated from $$\pi(\phi \mid \boldsymbol{\mathbf{y}})$$. The simulation is performed in the function hef.
## Beta-binomial model
We consider the example presented in Section 5.3 of Gelman et al. (2014), in which the data (Tarone 1982) in the matrix rat are analysed. These data contain information about an experiment in which, for each of 71 groups of rats, the total number of rats in the group and the numbers of rats who develop a tumor is recorded, so that $$J = 71$$. Conditional on $$\boldsymbol{\mathbf{\theta}} = (\theta_1, \ldots, \theta_J) = (p_1, \ldots, p_J)$$ we assume independent binomial distributions for $$(Y_1, \ldots, Y_J)$$, that is, $$Y_j \sim {\rm binomial}(n_j, p_j)$$. We use the conditionally conjugate priors $$p_j \sim {\rm Beta}(\alpha, \beta)$$, so that $$\phi = (\alpha, \beta)$$.
The conditional conjagacy of the priors means that the marginal posterior of $$(\alpha, \beta)$$ given $$\boldsymbol{\mathbf{y}} = (y_1, \ldots, y_J)$$ can be determined (equation (5.8) of Gelman et al. (2014)) as $\pi(\alpha, \beta \mid \boldsymbol{\mathbf{y}}) \propto \pi(\alpha, \beta) \prod_{j=1}^J \frac{B(\alpha + y_j, \beta + n_j - y_j)}{B(\alpha, \beta)},$ where $$\pi(\alpha, \beta)$$ is the hyperprior density for $$(\alpha, \beta)$$. By default $$\phi = (\alpha, \beta)$$ is transformed prior to sampling using $$(\rho_1, \rho_2)=(\log (\alpha/\beta), \log (\alpha+\beta))$$. The aim of this is to improve efficiency by rotating and scaling the (mode-relocated) conditional posterior density in an attempt to produce near circularity of this density’s contours.
To simulate from the full posterior density $$\pi(\boldsymbol{\mathbf{\theta}}, \alpha, \beta \mid \boldsymbol{\mathbf{y}}) = \pi(\boldsymbol{\mathbf{\theta}} \mid \alpha, \beta, \boldsymbol{\mathbf{y}}) \pi(\alpha, \beta \mid \boldsymbol{\mathbf{y}})$$ we first sample from $$\pi(\alpha, \beta \mid \boldsymbol{\mathbf{y}})$$. Simulation from the conditional posterior distribution of $$\boldsymbol{\mathbf{\theta}}$$ given $$(\alpha, \beta, \boldsymbol{\mathbf{y}})$$ is then straightforward on noting that $\theta_j \mid \alpha, \beta, y_j \sim {\rm beta}(\alpha + y_j, \beta + n_j - y_j)$ and that $$\theta_j, j = 1, \ldots, J$$ are conditionally independent.
The hyperprior for $$(\alpha, \beta)$$ used by default in hef is $$\pi(\alpha, \beta) \propto (\alpha+\beta)^{-5/2}, \alpha>0, \beta>0$$, following Section 5.3 of Gelman et al. (2014). A user-defined prior may be set using set_user_prior.
library(bang)
# Default prior, sampling on (rotated) (log(mean), log(alpha + beta)) scale
rat_res <- hef(model = "beta_binom", data = rat, n = 10000)
plot(rat_res)
plot(rat_res, ru_scale = TRUE)
The plot on the left shows the values sampled from the posterior distribution of $$(\alpha, \beta)$$ with superimposed density contours. On the right is a similar plot displayed on the scale used for sampling, that is, $$(\rho_1, \rho_2)$$.
The following summary is of properties of the generalized ratio-of uniforms algorithm, in particular the probability of acceptance, and summary statistics of the posterior sample of $$(\alpha, \beta)$$.
summary(rat_res)
#> ru bounding box:
#> box vals1 vals2 conv
#> a 1.0000000 0.00000000 0.00000000 0
#> b1minus -0.2382163 -0.40313465 -0.03906170 0
#> b2minus -0.2174510 0.05447431 -0.35297539 0
#> b1plus 0.2231876 0.36718411 -0.06551353 0
#> b2plus 0.2512577 0.05665707 0.44459818 0
#>
#> estimated probability of acceptance:
#> [1] 0.5258729
#>
#> sample summary
#> alpha beta
#> Min. : 0.6152 Min. : 3.883
#> 1st Qu.: 1.7895 1st Qu.:10.675
#> Median : 2.2198 Median :13.329
#> Mean : 2.4051 Mean :14.341
#> 3rd Qu.: 2.8034 3rd Qu.:16.830
#> Max. :14.1474 Max. :80.895
## Gamma-Poisson Model
We perform a fully Bayesian analysis of an empirical Bayesian example presented in Section 4.2 of Gelfand and Smith (1990), who fix the hyperparameter $$\alpha$$ described below at a point estimate derived from the data. The pump dataset (Gaver and O’Muircheartaigh 1987) is a matrix in which each row gives information about one of 10 different pump systems. The first column contains the number of pump failures. The second column contains the length of operating time, in thousands of hours.
pump
#> failures time
#> [1,] 5 94.320
#> [2,] 1 15.720
#> [3,] 5 62.880
#> [4,] 14 125.760
#> [5,] 3 5.240
#> [6,] 19 31.440
#> [7,] 1 1.048
#> [8,] 1 1.048
#> [9,] 4 2.096
#> [10,] 22 10.480
The general setup is similar to the beta-binomial model described above but now the response distribution is taken to be Poisson, the prior distribution is gamma and $$J = 10$$. For $$j = 1, \ldots, J$$ let $$Y_j$$ denote the number of failures and $$e_j$$ the length of operating time for pump system $$j$$. Conditional on $$\boldsymbol{\mathbf{\lambda}} = (\lambda_1, \ldots, \lambda_J)$$ we assume independent Poisson distributions for $$(Y_1, \ldots, Y_J)$$ with means that are proportional to the exposure time $$e_j$$, that is, $$Y_j \sim {\rm Poisson}(e_j \lambda_j)$$. We use the conditionally conjugate priors $$\lambda_j \sim {\rm gamma}(\alpha, \beta)$$, so that $$\phi = (\alpha, \beta)$$. We use the parameterization where $$\beta$$ is a rate parameter, so that $${\rm E}(\lambda_j) = \alpha/\beta$$ a priori.
The marginal posterior of $$(\alpha, \beta)$$ given $$\boldsymbol{\mathbf{y}} = (y_1, \ldots, y_J)$$ can be determined as $\pi(\alpha, \beta \mid \boldsymbol{\mathbf{y}}) \propto \pi(\alpha, \beta) \prod_{j=1}^J \frac{\beta^\alpha\Gamma(\alpha + y_j)}{(\beta+e_j)^{\alpha+y_j} \Gamma(\alpha)},$ where $$\pi(\alpha, \beta)$$ is the hyperprior density for $$(\alpha, \beta)$$. The scale used for sampling is $$(\rho_1, \rho_2)=(\log (\alpha/\beta), \log \beta))$$.
To simulate from the full posterior $$\pi(\boldsymbol{\mathbf{\lambda}}, \alpha, \beta \mid \boldsymbol{\mathbf{y}}) = \pi(\boldsymbol{\mathbf{\lambda}} \mid \alpha, \beta, \boldsymbol{\mathbf{y}}) \pi(\alpha, \beta \mid \boldsymbol{\mathbf{y}})$$ we first sample from $$\pi(\alpha, \beta \mid \boldsymbol{\mathbf{y}})$$ and then note that $\lambda_j \mid \alpha, \beta, y_j \sim {\rm gamma}(\alpha + y_j, \beta + e_j)$ and that $$\lambda_j, j = 1, \ldots, J$$ are conditionally independent.
By default hef takes $$\alpha$$ and $$\beta$$ to be independent gamma random variables a priori. The parameters of these gamma distributions can be set by the user, using the argument hpars or a different prior may be set using set_user_prior.
We produce similar output to the beta-binomial example above.
pump_res <- hef(model = "gamma_pois", data = pump, hpars = c(1, 0.01, 1, 0.01))
plot(pump_res)
plot(pump_res, ru_scale = TRUE)
summary(pump_res)
#> ru bounding box:
#> box vals1 vals2 conv
#> a 1.0000000 0.00000000 0.00000000 0
#> b1minus -0.5174980 -0.91869101 -0.06060116 0
#> b2minus -0.5150835 0.15757254 -0.92429417 0
#> b1plus 0.4124640 0.65433383 -0.11046433 0
#> b2plus 0.4224941 0.08788857 0.67847965 0
#>
#> estimated probability of acceptance:
#> [1] 0.5094244
#>
#> sample summary
#> alpha beta
#> Min. :0.2007 Min. :0.08544
#> 1st Qu.:0.8185 1st Qu.:1.29559
#> Median :1.0598 Median :1.94558
#> Mean :1.1496 Mean :2.16025
#> 3rd Qu.:1.3915 3rd Qu.:2.73560
#> Max. :3.6629 Max. :9.28544
## References
Gaver, Donald P., and I. G. O’Muircheartaigh. 1987. “Robust Empirical Bayes Analyses of Event Rates.” Technometrics 29 (1). [Taylor & Francis, Ltd., American Statistical Association, American Society for Quality]: 1–15. http://www.jstor.org/stable/1269878.
Gelfand, Alan E., and Adrian F. M. Smith. 1990. “Sampling-Based Approaches to Calculating Marginal Densities.” Journal of the American Statistical Association 85 (410). [American Statistical Association, Taylor & Francis, Ltd.]: 398–409. http://www.jstor.org/stable/2289776.
Gelman, A., J. B. Carlin, H. S. Stern, D. B. Dunson, A. Vehtari, and D. B. Rubin. 2014. Bayesian Data Analysis. Third. Florida, USA: Chapman & Hall / CRC. http://www.stat.columbia.edu/~gelman/book/.
Northrop, P. J. 2017. rust: Ratio-of-Uniforms Simulation with Transformation. https://CRAN.R-project.org/package=rust.
Tarone, Robert E. 1982. “The Use of Historical Control Information in Testing for a Trend in Proportions.” Biometrics 38 (1). [Wiley, International Biometric Society]: 215–20. http://www.jstor.org/stable/2530304.
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https://memorylimitexceeded.gitlab.io/leetcode/problems/0211-add-and-search-word---data-structure-design.html
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## # 211. Add and Search Word - Data structure design
Design a data structure that supports the following two operations:
``````void addWord(word)
bool search(word)
search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.
``````
Example:
``````addWord("bad")
search("b..") -> true
``````
Note: You may assume that all words are consist of lowercase letters a-z.
### # Solution
Approach 1: use a trie.
### # Code (Python)
Approach 1:
``````class TrieNode:
def __init__(self):
self.is_word = False
self.children = {}
class WordDictionary:
def __init__(self):
"""
"""
self._trie = TrieNode()
def addWord(self, word: 'str') -> 'None':
"""
Adds a word into the data structure.
"""
node = self._trie
for char in word:
if char not in node.children:
node.children[char] = TrieNode()
node = node.children[char]
node.is_word = True
def search(self, word: 'str') -> 'bool':
"""
Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
"""
return self._search(0, word, self._trie)
def _search(self, index, word, node):
if index == len(word):
return node.is_word
char = word[index]
if char != '.':
if char not in node.children:
return False
return self._search(index + 1, word, node.children[char])
else:
for child_node in node.children.values():
if self._search(index + 1, word, child_node) == True:
return True
return False
``````
### # Code (C++)
Approach 1:
``````class WordDictionary {
private:
struct TrieNode {
TrieNode *children[26];
bool isEnd;
TrieNode() {
for (int i = 0; i < 26; ++i)
{
children[i] = NULL;
}
isEnd = false;
}
};
TrieNode *root;
bool doSearch(string word, int index, TrieNode *start) {
if (start == NULL)
return false;
if (index == word.size())
return start->isEnd;
TrieNode *curr = start;
int tail = 25;
if (word[index] != '.')
{
}
for (int j = head; j <= tail; ++j)
{
if (doSearch(word, index+1, curr->children[j]))
return true;
}
return false;
}
public:
/** Initialize your data structure here. */
WordDictionary() {
root = new TrieNode();
}
/** Adds a word into the data structure. */
TrieNode *curr = root;
for (int i = 0; i < word.size(); ++i)
{
int j = word[i] - 'a';
if (curr->children[j] == NULL)
curr->children[j] = new TrieNode();
curr = curr->children[j];
}
curr->isEnd = true;
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
bool search(string word) {
return doSearch(word, 0, root);
}
};
/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary* obj = new WordDictionary();
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# How Many Mph Is 600Hp?
The 2020 Alpina B7 is capable of hitting 205 mph.
## How fast can a 600 hp car go?
The 599GTO will take 3.1 seconds to get to 60 mph, but the Lambo is the fastest 600-hp car, with a top speed of 207 mph.
## How many mph is 1000HP?
The first car to go over 200 mph was this one. The last run of the car was a demonstration circuit that ran at a slow pace. There is a display at the National Motor Museum.
## How fast can a 700 hp car go?
That’s enough to push the car from 0 to 62 mph in 3.4 seconds and on to 199 mph in just over an hour.
## Is 300 hp a lot for a car?
The sweet spot for drivers is between 200 and 300 horses per hour. If the vehicle is a heavy truck or another large model, be careful.
See also Do You Need To Be In Us To File I-90?
## How fast can a 900 hp car go?
It is able to deliver over 900hp. It goes from zero to 62 mph in less than a second. The top speed is over 200 mph.
## How much force is 1hp?
The amount of work that can be done in a given amount of time is described by the power of the engine. 500 foot-pounds per second is what the constant 1 horsepower equates to. The amount of work required to move a load of 550 pounds over 1 foot in 1 second is known as 1 horsepower.
## How much horsepower does it take to fly?
It takes 1 Mega watt to power 1341 horses. A fully loaded aircraft with two GE 90 to 115B engines can produce 23 Mega watt of power on a cruise flight. 30.843 is the power output of this vehicle.
## How fast does a pontoon go?
A pontoon can go as fast as 25 miles per hour. There are high-speed models that can go up to 50 mph.
## How fast does a 30hp outboard go?
2 people, 26 mph, with normal load, WOT of 5740rpm. The prop is made out of aluminum. The Boston Whaler Super Sport has a 3 blade, 10×13, and is capable of speeds up to 30 mph.
## How fast will a pontoon go with a 40 hp?
A 40hp pontoon boat with an average load of people can go between 10 and 15 mph. The size of the boat, the number of people on it, and the weather are some of the factors that affect it.
## Is 170 hp good?
Is it good to have 170 horse power? A car’s engine can reach between 200 and 300hp. A lot of different model years have a range between 100 and 200hp.
## How much horsepower does a f1 car have?
The power unit of the 1.6L V6 is capable of producing over a thousandhp. If you combine the power with the car’s specifications, you have a vehicle that can reach 400 km per hour.
## Is 400 hp too much?
One of the most important numbers for car enthusiasts is the power of the engine. The type of car has an effect on whether or not 400hp is a lot. For a family sedan, it is more than you would need. 400 is an adequate amount for sports and luxury cars.
## Is 400 hp a lot for a truck?
Semi trucks are very powerful and stay within the 400 to 600hp range. The higher the engine’s power, the more worthwhile it is to consider factors such as weight and fuel economy.
## Is 500 horsepower a lot?
The amount of power in a car is very close to 500. Let me tell you what’s going on. First of all, 500 is nice and round, halfway between zero and 1,000.
## What does 1000hp mean?
Godsmack’s studio album of the same name has a lead single called “1000hp”.
## How was horsepower invented?
James Watt was the inventor of steam engines. The term was used to describe the strength of a person. Watt found that a big horse could pull a load of 150 pounds while walking 2.5 miles per hour.
## How many horsepower does a Lamborghini have?
The 759-hp Lamborghini Aventador is one of the models currently on the market. The Urus has a power output of 641 horsepower. Huracan’s power is 630hp.
## Can we fly in real life?
Humans are not capable of flying. The force of gravity is too much for us to overcome. Birds are able to fly through other parts of the body. They have a light frame and hollow bones that help counteract gravity.
## What would it take to make a human fly?
It is not possible for humans to fly like birds, according to scientists. A bird can fly because it has the right muscles for it’s size. It has a lightweight skeleton with hollow bones, which makes it less heavy on its wings.
## How much can a jet engine lift?
How much can an engine be lifted? 100,000 pounds of thrust can be converted into 210,000 pounds by using each engine. The engine made a new record of 134,300 pounds.
## How fast will a 150 hp Tritoon boat go?
There are two things. How fast will the boat travel? Depending on the size of the boat and the load it is carrying, they can reach speeds of up to 40 miles per hour.
## How fast does a 150 hp boat go?
A 115 to 150-hp engine can produce a reasonable performance for an entry level fiberglass sportboat. A powerboat in this range can reach a top speed of up to 45 mph and a cruising speed of up to 20 mph.
## How fast can a 100 hp boat motor go?
A boat with 100hp can go very fast, and it can keep the average boat at cruising speed. You can expect a boat speed between 21 and 66 miles per hour.
## How do you calculate the speed of a boat?
S is the boat speed and C is the Crouch constant, which is how the speed is written. The above equation is referred to as Crouch’s formula.
## How fast will a Mercury 150 go?
We were able to reach our top speed of 35.1 mph at 6000rpm with a 150-hp Mercury Four Stroke. The best cruise was at 3000rpm with a speed of 14.2 mph.
## Related Posts
#### Do You Get Paid For Being Sacked?
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# Difference between revisions of "BME103 s2013:T900 Group5 L3"
BME 103 Spring 2013 Home
People
Lab Write-Up 1
Lab Write-Up 2
Lab Write-Up 3
Course Logistics For Instructors
Photos
Wiki Editing Help
# OUR TEAM
Name: Cody Gates Camera Operator Name: Alexander Oropel Research and Development Scientist Name: Matt McClintock Protocol Name: StudentRole(s) Name: StudentRole(s)
# LAB 3 WRITE-UP
## Original System: PCR Results
PCR Test Results
Sample Name Ave. INTDEN* Calculated μg/mL Conclusion (pos/neg) Positive Control 5725984.00 11.66 μg/mL N/A Negative Control 2820659.67 4.50 μg/mL N/A Tube Label: 1 Patient ID: 92336 rep 1 4902761.33 9.63 μg/mL pos Tube Label: 2 Patient ID: 92336 rep 2 4957051.00 9.77 μg/mL pos Tube Label: 3 Patient ID: 92336 rep 3 5446934.67 10.93 μg/mL pos Tube Label: 4 Patient ID: 44606 rep 1 2497338.33 3.70 μg/mL neg Tube Label: 5 Patient ID: 44606 rep 2 2202675.67 2.97 μg/mL neg Tube Label: 6 Patient ID: 44606 rep 3 1789569.00 1.95 μg/mL neg
* Ave. INTDEN = Average of ImageJ integrated density values from three Fluorimeter images
Bayesian Statistics
These following conditional statistics are based upon all of the DNA detection system results that were obtained in the PCR lab for 20 hypothetical patients who were diagnosed as either having cancer or not having cancer.
Bayes Theorem is an equation in probability theory and statistics that relates inverse representations of probabilities concerning two events, or rather, it expresses a degree of change when accounting for evidence. Bayes Theorem is represented as follows:
P(A|B) = P(B|A) * P(A) / P(B)
the probability of A given B = (the probability of B given A * the probability of A) / the probability of B
This information will be utilized to determine various probabilities listed below when accounting for the positive/negative values determined by the entire class as well as an outside document listing the actual yes/no cancer diagnosis
Calculation 1: The probability that the sample actually has the cancer DNA sequence, given a positive diagnostic signal.
• A = Cancer-Positive Conclustion = 9/20 = .45
• B = Positive PCR Reactions = 26/60 = .433
• P (B|A) = Positive PCR given cancer Positive conclustion = 11/13 = .846
• P(A|B) = .879=88%
Calculation 2: The probability that the sample actually has a non-cancer DNA sequence, given a negative diagnostic signal.
• A = Cancer negative conclustion = 11/20 = .55
• B = Negative PCR reactions = 17/30 = .567
• P (B|A) = Negative PCR given cancer-negative conclustion = 16/17 = .94
• P(A|B) = .911 = 91%
Calculation 3: The probability that the patient will develop cancer, given a cancer DNA sequence.
• A = "yes" cancer diagnosis = 7/20 = .35
• B = "positive" test conclusion = 9/20 = .45
• P (B|A) = Positive given yes = 6/20 = .3
• P(A|B) = .233 = 23%
Calculation 4: The probability that the patient will not develop cancer, given a non-cancer DNA sequence.
• A = "no" cancer diagnosis = 13/20 = .65
• B = "negative" test conclusion = 11/20 = .55
• P (B|A) = Negative given no = 1/2 = .5
• P(A|B) = .591 = 59%
## New System: Design Strategy
We concluded that a good system "Must Have":
• Results that are easy to determine. This means a clear indication of positive or negative results when compared to the controls. This is integral to the design success because the results must be easy differentiable as to not require re-testing.
• Software that is simple to use. The software for open PCR is incredibly easy to use and a program similar would be ideal. Anything that requires computer coding or computational design is too complicated, so the software must already be made to use. A user should be able to plug in the information he or she wants and get the desired response from the software, no computing needed.
We concluded that we would Want a good system to have:
• Samples that are easily identifiable throughout the experiment. This means that there is no changing of labels or titles for the samples. This is important because it is crucial to keep the samples consistent and not mixed up. When transferring the samples there should be prepared labels to keep them straight.
• Accessibility. This means that anyone can access the materials needed and replicate the findings. For the purpose of DNA amplification no advanced technology is required, the PCR machine is easy for anyone to access.
We concluded that a good system Must Not Have:
• High cost. This means that the system must not be too expensive. This is important because the system should be replicable and useful to everyone.
• inconsistent timing. This was the most annoying problem with the original PCR design. The PCR machine kept changing times, and the group was unsure if the amplification was taking place properly. This also made it impossible to gauge the rest of the experiment timing which delayed the experiment and forced the group to reschedule further testing.
We concluded that a good system Should Avoid:
• High energy consumption. Possible options are a battery or solar power. Not only would this make the experiment accesible at any location, it is a more sustainable option.
• Manual analysis of the images. This was the most tedious process of the experiment. ImageJ was useful in providing a medium for calculating light density, but a program that could do calculations on its own would be ideal.
## New System: Machine/ Device Engineering
SYSTEM DESIGN
KEY FEATURES
We chose to include these new features
• Feature 1 - explanation of how this addresses any of the specifications in the "New System: Design Strategy" section
• Feature 2 - explanation of how this addresses any of the specifications in the "New System: Design Strategy" section
• Etc.
[OR]
We chose keep the devices the same as the original system
• Feature 1 - explanation of how a pre-existing feature addresses any of the specifications in the "New System: Design Strategy" section
• Feature 2 - explanation of how a pre-existing feature addresses any of the specifications in the "New System: Design Strategy" section
• Etc.
INSTRUCTIONS
## New System: Protocols
DESIGN
We chose to include these new approaches/ features
• Feature 1 - explanation of how this addresses any of the specifications in the "New System: Design Strategy" section
• Feature 2 - explanation of how this addresses any of the specifications in the "New System: Design Strategy" section
• Etc.
[OR]
We chose keep the protocols the same as the original system
• Feature 1 - explanation of how a pre-existing feature addresses any of the specifications in the "New System: Design Strategy" section
• Feature 2 - explanation of how a pre-existing feature addresses any of the specifications in the "New System: Design Strategy" section
• Etc.
MATERIALS
PROTOCOLS
• PCR Protocol
1. Step 1
2. Step 2
3. Etc.
• DNA Measurement and Analysis Protocol
1. Step 1
2. Step 2
3. Etc.
## New System: Research and Development
BACKGROUND
Polymerase chain reaction is the process of amplifying a strand of DNA from a DNA template strand. From here the scientist is capable of amplifying any specific gene they choose. In this research we are targeting the single nucleotide polymorphism that is rs1787996, which contains a single nucleotide variation or SNV. The CHEK2 gene is essentially a gene that is capable of coding for susceptibility to breast cancer. The relation to SNP is that it is essentially a variation of the CHEK 2 gene that is present within humans, or Homo sapiens. The cancer-related function of the gene is that it essentially changes the base Thymine to Cytosine, changing the normal allele ATT to ACT, which is the cancer related allele.
DESIGN
Primers for PCR
Amplification of cancer-associated DNA
Cancer allele forward primer: 5' TATGTATGCACTGTAAGAGTT
Cancer allele reverse prime: 5' CTAGGAGAGCTGGTAATTTGG
A disease allele will give a PCR product because the primer associated with the process will identify the sequences that will code for cancer. From there the primer will allow for nucleotide bases to be placed in a reverse sequence from the template DNA. Essentially this will continuously amplify the cancerous DNA gene while the PCR process is in effect.
Our primers address the following design needs
• Design specification 1 - explanation of how an aspect of the primers addresses any of the specifications in the "New System: Design Strategy" section
• Design specification 2 - explanation of how an aspect of the primers addresses any of the specifications in the "New System: Design Strategy" section
• Etc.
## New System: Software
[THIS SECTION IS OPTIONAL. If your team has creative ideas for new software, and new software is a key component included in your new protocols, R&D, or machine design, you may describe it here. You will not receive bonus points, but a solid effort may raise your overall page layout points. If you decide not to propose new software, please delete this entire section, including the ==New System: Software== header.]
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Convert 8 mm to cm
Conversion details
To convert mm to cm use the following formula:
1 mm equals 0.1 cm
So, to convert 8 mm to cm, multiply 0.1 by 8 i.e.,
8 mm = 0.1 * 8 cm = 0.8 cm
For conversion tables, definitions and more information on the mm and cm units scroll down or use the related mm and cm quick access menus located at the top left side of the page.
mm is the symbol for millimeter
cm is the symbol for centimeter
From 0.10 to 4.00 mm, 40 entries
0.1 mm = 0.01 cm
0.2 mm = 0.02 cm
0.3 mm = 0.03 cm
0.4 mm = 0.04 cm
0.5 mm = 0.05 cm
0.6 mm = 0.06 cm
0.7 mm = 0.07 cm
0.8 mm = 0.08 cm
0.9 mm = 0.1 cm
1 mm = 0.1 cm
1.1 mm = 0.11 cm
1.2 mm = 0.12 cm
1.3 mm = 0.13 cm
1.4 mm = 0.14 cm
1.5 mm = 0.15 cm
1.6 mm = 0.16 cm
1.7 mm = 0.17 cm
1.8 mm = 0.18 cm
1.9 mm = 0.19 cm
2 mm = 0.2 cm
2.1 mm = 0.21 cm
2.2 mm = 0.22 cm
2.3 mm = 0.23 cm
2.4 mm = 0.24 cm
2.5 mm = 0.25 cm
2.6 mm = 0.26 cm
2.7 mm = 0.27 cm
2.8 mm = 0.28 cm
2.9 mm = 0.29 cm
3 mm = 0.3 cm
3.1 mm = 0.31 cm
3.2 mm = 0.32 cm
3.3 mm = 0.33 cm
3.4 mm = 0.34 cm
3.5 mm = 0.35 cm
3.6 mm = 0.36 cm
3.7 mm = 0.37 cm
3.8 mm = 0.38 cm
3.9 mm = 0.39 cm
4 mm = 0.4 cm
Click here for a list of all conversion tables of mm to other compatible units.
millimeter
Millimeter is a subdivision of the meter unit. The milli prefix stands for 0.001 therefore, 1 millimeter = 0.001 meter units. Meter is a unit of measurement of length. The definition for meter is the following:
1 millimeter is thousandth (1/1000) of a meter
The symbol for millimeter is mm
centimeter
Centimeter is a subdivision of the meter unit. The centi prefix stands for 0.01 therefore, 1 centimeter = 0.01 meter units. Meter is a unit of measurement of length. The definition for meter is the following:
1 centimeter is 0.01 meters
The symbol for centimeter is cm
Other people are also searching for information on mm conversions.
Following are the most recent questions containing mm. Click on a link to see the corresponding answer.
10pm to mm
600pm to mm
10mm to pm
10cm= mm
5 dm to mm
0,005mm in dm
5dam=__mm
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78 mm = m
512kb in mm
Home | Base units | Units | Conversion tables | Unit conversion calculator
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# Used Camrys Cars depreciate over time, whether made by Honda or, in this case, Toyota. These data…
Used Camrys Cars depreciate over time, whether made by Honda or, in this case, Toyota. These data show the prices of Toyota Camrys listed for sale by individuals in The Philadelphia Inquirer. One column gives the asking price (in thousands of dollars) and a second column gives the age (in years).
(a) Do you expect the resale value of a car to drop by a fixed amount each year?
(b) Fit a linear equation with price as the response and age as the explanatory variable. What do the slope and intercept tell you, if you accept this equation’s description of the pattern in the data?
(c) Plot the residuals from the linear equation on age. Do the residuals suggest a problem with the linear equation?
(d) Fit the equation
Do the residuals from this fit “fix” the problem found in part (c)?
(e) Compare the fitted values from this equation with those from the linear model. Show both in the same scatterplot. In particular, compare what this graph has to say about the effects of increasing age on resale value.
(f) Compare the values of and between these two equations. Give units where appropriate. Does this comparison agree with your impression of the better model? Should these summary statistics be compared?
(g) Interpret the intercept and slope in this equation. (h) Compare the difference in asking price for cars that are 1 and 2 years old to that for cars that are 11 and 12 years old. Use the equation with the log of age as the explanatory variable. Is the difference the same or different?
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# Removing elements that have consecutive duplicates in Python
I was curios about the question: Eliminate consecutive duplicates of list elements, and how it should be implemented in Python.
What I came up with is this:
``````list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
i = 0
while i < len(list)-1:
if list[i] == list[i+1]:
del list[i]
else:
i = i+1
``````
Output:
``````[1, 2, 3, 4, 5, 1, 2]
``````
Which I guess is ok.
So I got curious, and wanted to see if I could delete the elements that had consecutive duplicates and get this output:
``````[2, 3, 5, 1, 2]
``````
For that I did this:
``````list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
i = 0
dupe = False
while i < len(list)-1:
if list[i] == list[i+1]:
del list[i]
dupe = True
elif dupe:
del list[i]
dupe = False
else:
i += 1
``````
But it seems sort of clumsy and not pythonic, do you have any smarter / more elegant / more efficient way to implement this?
``````>>> L = [1,1,1,1,1,1,2,3,4,4,5,1,2]
>>> from itertools import groupby
>>> [x[0] for x in groupby(L)]
[1, 2, 3, 4, 5, 1, 2]
``````
If you wish, you can use map instead of the list comprehension
``````>>> from operator import itemgetter
>>> map(itemgetter(0), groupby(L))
[1, 2, 3, 4, 5, 1, 2]
``````
For the second part
``````>>> [x for x, y in groupby(L) if len(list(y)) < 2]
[2, 3, 5, 1, 2]
``````
If you don't want to create the temporary list just to take the length, you can use sum over a generator expression
``````>>> [x for x, y in groupby(L) if sum(1 for i in y) < 2]
[2, 3, 5, 1, 2]
``````
Oneliner in pure Python
``````[v for i, v in enumerate(your_list) if i == 0 or v != your_list[i-1]]
``````
To Eliminate consecutive duplicates of list elements; as an alternative, you may use `itertools.izip_longest()` with list comprehension as:
``````>>> from itertools import izip_longest
>>> my_list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
>>> [i for i, j in izip_longest(my_list, my_list[1:]) if i!=j]
[1, 2, 3, 4, 5, 1, 2]
``````
## protected by eyllanescJul 28 '18 at 10:03
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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# Cash Earnings Per Share (Cash EPS)
## What is Cash Earnings per Share (Cash EPS)?
Cash earnings per share (cash EPS) is the operating cash flow generated by a company divided by the number of shares outstanding. Cash earnings per share (Cash EPS) is different from traditional earnings per share (EPS), which takes the company’s net income and divides it by the number of shares outstanding. In other words, EPS measures how much of the company’s profit can be allocated to each share of stock, while Cash EPS measures how much cash flow can be allocated to each share of stock.
Cash earnings per share (Cash EPS) is a profitability ratio that compares a company’s cash flow against their volume of shares outstanding. It is unique from the more widely used net profit metric, earnings per share (EPS), which only considers net income per share.
Cash EPS doesn’t take into account any non-cash elements. For example, it doesn’t include depreciation which is part of profit-based earnings per share calculations. Because of this, CEPS can be a more accurate measurement of a company’s financial and operational states. The bigger a company’s CEPS, the better its performance is believed to be over for the particular period considered.
The CEPS ratio comes in handy when comparing trends within a company, and trends among different companies in like industries.
## Cash Earnings Per Share Formula
To calculate cash earnings per share formula you have to sum up net income with non-cash items such as deferred tax, depreciation and amortization and then dividing it by the number of outstanding shares.
`Cash EPS = Operating Cash Flow ÷ Number of Shares Outstanding`
All the data needed to calculate cash EPS can be found in the financial statements and notes to accounts. For the most accurate CEPS calculation, it is crucial to determine all of the non-cash elements in the income statement.
Alternatively, the cash EPS formula can be calculated using EBITDA like this:
`Cash EPS = (Net Income + Depreciation & Amortization × (1 - tax)) ÷ Number of Shares Outstanding`
Both of these formulas can yield slightly different results due to the inclusion of change in working capital in the first formula. Hence, investors more commonly use the first formula. Analysts can find all these items in the financial statements and notes to accounts. It is important to identify the non-cash elements in the Income statement to accurately calculate CashEPS.
Operating Cash Flow is EBITDA plus change in working capital and other non-cash adjustments. This value is stated by the company in the Cash flow statement.
Net Income, Depreciation & Amortization, tax are stated in the Income statement.
Number of shares outstanding could be either basic or diluted and both can be found in the notes to accounts.
Operating cash flow, which is indicated in a company’s cash flow statement, is simply EBITDA with the addition of change in working capital and all other non-cash modifications. The number of shares outstanding may be basic or diluted and are found in the notes to accounts, and all the rest are included in the income statement.
When a company has high cash earnings, it may indicate that they are performing very well. Analysts usually consider the annual growth rate of the ratio over a number of years. As a result, this means a high growth rate over time is a positive sign.
The CEPS of one company may also be compared with those of other companies within the same industry and with similar product varieties. This is one key benefit offered by CEPS over regular EPS. Because CEPS may be adjusted for all non-cash items, largely drawn from management estimates, it enables analysts to compare the values for several companies.
## Earnings Per Share vs. Cash Flow Per Share
A company’s earnings per share is the portion of its profit that is allocated to each outstanding share of common stock. Like cash flow per share, earnings per share serves as an indicator of a company’s profitability. Earnings per share is calculated by dividing a company’s profit, or net income, by the number of outstanding shares.
Since depreciation, amortization, one-time expenses, and other irregular expenses are generally subtracted from a company’s net income, the outcome of an earnings per share calculation could be artificially deflated. Additionally, earnings per share may be artificially inflated with income from sources other than cash. Non-cash earnings and income can include sales in which the purchaser acquired the goods or services on credit issued through the selling company, and it may also include the appreciation of any investments or selling of equipment.
Since the cash flow per share takes into consideration a company’s ability to generate cash, it is regarded by some as a more accurate measure of a company’s financial situation than earnings per share. Cash flow per share represents the net cash a firm produces on a per-share basis.
## Cash Earnings Per Share Example Analysis
Investors will always look at the performance of the company by considering operating cash flow relative to the company’s net income. Therefore a higher Cash EPS implies that the company is performing well. This is a metric that analysts look at to ascertain if the company has been growing over the higher and thus a higher growth rate is favorable. Most importantly one can compare Cash EPS of companies operating within the same sector dealing with handling the same product mix. This is the upside of the Cash EPS relative to normal EPS because it is adjusted for non-cash transactions it lets investors and analysts have a comparison of two companies.
Basic EPS is also disposed to accounting manipulation thus making it an unreliable metric for the company’s performance. Management can use this to mask the performance of the company especially considering nowadays stock repurchase programs instead of dividends are becoming famous as a way of returning profits to stockholders. As a result, some executives can increase EPS through the reduction of the outstanding stock and therefore use EPS to hype their compensation programs. This is where Cash EPS becomes important as it can do away some of the issues apparent in accounting manipulation.
If the CEPS of a company have been decreasing then it might be because the company issued more equity stock to get capital for funding expansion plans. Interestingly the plan might increase earnings in the long term or the number of shares might have increased because of the conversion of additional stock options.
## Benefits of Using Cash EPS
• CEPS is less prone to accounting manipulation, which offers a clearer picture of cash flow and real earnings. Added transparency is a sign of good corporate governance.
• CEPS shows investors on a per share basis how much profit each share generates. This helps identify incremental value.
• CEPS is not subject to the same short-term market focus seen with EPS.
## Final Statements
Using Cash Earnings per Share comes with a great advantage over using normal EPS, as the CEPS is specially adjusted for non-cash items – this allowing the analysts to compare their results and numbers with those of different companies.
If a company has high cash earning, then this might mean that the said company has a strong underlying performance. As an analyst, you would look into the annual growth rate of this profitability ratio – over several years, of course – expecting to find a high growth rate, which is something that every company out there wants.
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×
[–]Moves in silence like lasagna 6 points7 points (0 children)
You don't need the def. Try an HP ring
[–] 4 points5 points (2 children)
Exa HP Take this advice from someone who lost his 6/8 knight half an hour after maxing vit
[–]Sutch | Chaos Theory | Nab 2 points3 points (1 child)
It's not very good advice if you just died, is it? :P
[–] 0 points1 point (0 children)
I died because I chose not to use Exa HP
[–][deleted] (1 child)
[deleted]
[–]Pride Rock | USE 0 points1 point (0 children)
Pyra really shouldn't be used unless you have res armor. So good on you. However, if you have a decent mp heal pet you regen the mp faster than you use it. c:
[–][S] 1 point2 points (0 children)
thanks peeps.
[–]6/8 1 point2 points (4 children)
I can completely understand where the other players are coming from, but it is all a matter of preference. I use an exa def, because I am super fucking clumsy, I also carry an exa vit with me. If you run into everything thrown at you, an exa def is better. Mathematically speaking, if you have on exa hp and get hit constantly you will loose more life then if you use an exa def. If you think about it, it will make sense, or I am fucking retarted. Also, willing to hear someone's other reason of why not to use exa def. Seriously I need to know.
[–] 6 points7 points (3 children)
Because enemies still do 15% damage regardless of your def (because of the 85% def cap). A knight has enough Def to reach this cap for most if not all enemies and HP allows you to tank more hits.
EDIT: to sound more concise.
[–]6/8 2 points3 points (2 children)
I completely understand now. I didn't know there was a cap. Now... To buy an exa hp lol
[–]6/8 3 points4 points (1 child)
Edit: ok I did the math.
With the exa def on Giant eyes hit me for 36 With the exa hp on Giant eyes hit me for 44
So mathematically speaking... Exa def: 680/36=18.8888.... I can get 18 hits before dying. Exa hp: 820/44=18.6363.... I can get less hits.
Still very very confused on which to use, because using my Exa hp I loose more hp faster. EDIT:EDIT: If you also factor in the amount of time a maxed vit will take to regen your hp you will actually save time and take just as many shots, if you use an exa def.
[–]6/8 0 points1 point (0 children)
I retract my previous statement. Source? I fucking died hahaha. There goes a 6/8 lmao
[–]Delibob 0 points1 point (0 children)
Personally, Swithing from an Exa Def to an Exa Hp was the best choice. Now, if you use res, use expo, if you use acrop, use nile. The expo gives a decent mp and hp boost, while also taking up the slack that res does in def, while with acrop, you have more than enough def for most things, so the extra speed and dex is nicer than the def and vit.
[–] 0 points1 point (0 children)
Use an HP ring and just a tip for abysses b/c you are maxing vit: stand inside malphas at first and right when he becomes vulnerable, shotgun him with your stun as many times as you can.
[–] 0 points1 point (0 children)
HP all the way.
that way, you can tank more hits.
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sin y dy/ cos^2 y
2. ## Re: anti derivative please
\displaystyle \begin{align*} \int{\frac{\sin{y}}{\cos^2{y}}\,dy} = -\int{\frac{-\sin{y}}{\cos^2{y}}\,dy} \end{align*}
Let \displaystyle \begin{align*} u = \cos{y} \implies du = -\sin{y}\,dy \end{align*} and the integral becomes
\displaystyle \begin{align*} -\int{\frac{1}{u^2}\,du} &= -\int{u^{-2}\,du} \\ &= -\left(\frac{u^{-1}}{-1}\right) + C \\ &= \frac{1}{u} + C \\ &= \frac{1}{\cos{y}} + C \\ &= \sec{y} + C \end{align*}
3. ## Re: anti derivative please
Originally Posted by sluggerbroth
sin y dy/ cos^2 y
ever wonder why you had to learn all those identities ?
$\frac{\sin{y}}{\cos^2{y}} = \frac{1}{\cos{y}} \cdot \frac{\sin{y}}{\cos{y}} = \sec{y} \cdot \tan{y}$
$\int \sec{y} \cdot \tan{y} \, dy = \sec{y} + C$
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The 10 main groups are: 000099, general works; 100199, philosophy and psychology; 200299, religion; 300399, social sciences; 400499, language; 500599, natural sciences and mathematics; 600699, technology; 700799, the arts; 800899, literature and rhetoric; and 900999, history, biography, and geography.
How do you find the Dewey Decimal Classification?
The best place to begin your search is the Library of Congress Online Catalog. When you open a record for a book in the catalog, click on the Full Record tab at the top of the page and look for a field labeled Dewey Class No. If this field is listed, it will give the book’s Dewey classification.
What is meant by Universal Decimal Classification?
The Universal Decimal Classification (UDC) is a bibliographic and library classification representing the systematic arrangement of all branches of human knowledge organized as a coherent system in which knowledge fields are related and inter-linked.
What are the 800s in Dewey Decimal System?
The first three digits indicate the category under which the book is classified. For example, the 500s are for natural sciences, the 700s are for the arts and the 800s are for literature.
What are the 900s in the Dewey Decimal System?
800 – 899 = Literature. 900 – 999 = History & Geography.
How do you read a Dewey Decimal number?
Reading a Dewey Decimal Call Number
1. Books are arranged sequentially. The first number of a Dewey Decimal call number indicates the general class the call number falls within. …
2. Numbers following the first 3 numbers. The numbers define the subject of the book. …
3. The Cutter Number is the next set of numbers.
What is Dewey Decimal System example?
In the Dewey Decimal System, books are filed digit by digit, not by whole number. This means, for example, that our book at 595.789/BROC would come after 595.0123 and before 595.9. In our collection, Biographies do not use the 921 Dewey number, but rather the letter B for biography.
How many categories are in the Dewey Decimal System?
ten Structure and Notation At the broadest level, the DDC is divided into ten main classes, which together cover the entire world of knowledge. Each main class is further divided into ten divisions, and each division into ten sections (not all the numbers for the divisions and sections have been used).
What is 398.2 in the Dewey Decimal System?
According to libraries Dewey Decimal System, which all books are shelved in, 398.2 represents the beginning of the fiction/fantasy section of books. … It’s the Dewey Decimal call number ( on the spine label) of library books. It represents Fantasy and Fairy Tales.
What does DDC mean?
Dewey Decimal Classification, DDC for short, is a system for organising knowledge. In principle, it can be used to classify knowledge in any form, be it text, music, images or other knowledge resources, printed or digital. Subjects are subdivided by means of classes.
What is the Dewey Decimal Classification for books on tigers?
The Dewey Decimal System organizes information into 10 broad areas, which are broken into smaller and smaller topics. Different topics are assigned numbers, known as call numbers. For example, Tigers are given the number 599.756.
What is DDC in library science?
The Dewey Decimal Classification (DDC), colloquially the Dewey Decimal System, is a proprietary library classification system which allows new books to be added to a library in their appropriate location based on subject. It was first published in the United States by Melvil Dewey in 1876.
What is difference between DDC and UDC?
As opposed to DDC, which is used mainly in public libraries, UDC tends to be used in larger, academic, research, national or special libraries.
Is UDC based on DDC?
UDC is a practical scheme based on the demands of pamphlets, reports and periodical literature rather than the framework of a theory. The scheme is based on DDC and claims to be the first Analytico-synthetic classification scheme.
How many Schedules are there in DDC?
At the broadest level, the DDC is divided into ten main classes, which together cover the entire world of knowledge. Each main class is further divided into ten divisions, and each division into ten sections (not all the numbers for the divisions and sections have been used).
What is the difference between 92 and 920 in Dewey Decimal System?
Single biographies are classified with the number 92 and the first three letters of the subject’s last name. If the book is a collective biography, it is classified with the number 920 and the first three letters of the author’s last name.
What is the purpose of the Dewey Decimal System?
The Dewey Decimal system is a classification system used by libraries to arrange books via subject. Each book is issued a shelfmark number, usually found on the spine of the book, and arranged in numerical order.
What makes up the Dewey Decimal call number?
Numbers following the first 3 numbers: The numbers define the subject of the book. The number . 2 is referring to a book about a specific age group and 0973 means the history and description about America. Cutter Number (the next set of numbers): The Cutter Number indicates the author or title of the book.
What is a Dewey Decimal call number?
Like Library of Congress call numbers, Dewey Decimal call numbers group books together by broad topic or classification (number before the decimal), and then use further letters and numbers to group books into more specific topics and subtopics (letters and numbers after the decimal).
How do you get a Dewey decimal number?
Build a DDC number
1. Locate the base number for the item you are classifying. …
2. Do one of the following. …
3. Build the number. …
4. To continue number building, navigate to the next number specified in the add instruction or to another base number. …
5. When the built number is complete and correct, click Save.
What replaced the Dewey Decimal System?
Among libraries shifting away from Dewey, variations on the Book Industry Standards and Communications (BISAC) model are the most prevalent replacements, and while the idea of switching is attractive to many librarians, it is not perfect.
What is the Dewey decimal test?
The Dewey Decimal System is a proprietary library classification system introduced by Melvil Dewey in 1876 as a means of being able to order books by subject. … The way it works is by placing books on the shelf by subject using numbers from 000 to 999 and using decimal points for more detail.
What are the features of Dewey Decimal Classification?
Features of Dewey Decimal System
• 10 Main Classes (first schedule or first summary)
• 100 Divisions (second schedule or second summary)
• 1000 Sections (third schedule or third summary)
What is the Dewey decimal number for sports?
025.431 025.431: The Dewey blog: Sports.
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How do you spell Hannukah?
# proportional
[pruh-pawr-shuh-nl, -pohr-] /prəˈpɔr ʃə nl, -ˈpoʊr-/
1.
having due proportion; corresponding.
2.
being in or characterized by proportion.
3.
of, relating to, or based on proportion; relative.
4.
Mathematics.
1. (of two quantities) having the same or a constant ratio or relation:
The quantities y and x are proportional if y/x = k, where k is the constant of proportionality.
2. (of a first quantity with respect to a second quantity) a constant multiple of:
The quantity y is proportional to x if y = kx, where k is the constant of proportionality.
Origin
1350-1400
1350-1400; Middle English proporcional < Latin prōportiōnālis. See proportion, -al1
Related forms
proportionality, noun
Can be confused
proportional, proportionate.
Synonyms
1. harmonious, comparative, accordant, consonant, proportionate.
Dictionary.com Unabridged
Based on the Random House Dictionary, © Random House, Inc. 2014.
Cite This Source
British Dictionary definitions for inter-proportional
## proportional
/prəˈpɔːʃənəl/
1.
of, involving, or being in proportion
2.
(maths) having or related by a constant ratio
noun
3.
(maths) an unknown term in a proportion: in a/b = c/x, x is the fourth proportional
Derived Forms
proportionality, noun
Collins English Dictionary - Complete & Unabridged 2012 Digital Edition
Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012
Cite This Source
Word Origin and History for inter-proportional
## proportional
late 14c. (implied in proportionally), from Late Latin proportionalis "pertaining to proportions," from proportio (see proportion). Related: Proportionally.
Online Etymology Dictionary, © 2010 Douglas Harper
Cite This Source
### Difficulty index for proportional
Few English speakers likely know this word
### Word Value for inter
5
6
Scrabble Words With Friends
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Next: Time relationships in the Up: DERIVING THE SYSTEM OF Previous: DERIVING THE SYSTEM OF
## Vectorial relationships on the earth's surface
In the triangle (PSG) in Figure (1), the following expression relates the horizontal traveling of the source and geophone rays with the offset:
(3)
The vectors and are related to the table that gives the lateral distance traveled by the ray as a function of the ray parameter and the travel time, . The projection on the earth's surface of a ray path (of parameter p and traveltime t) is the 2-D vector . Thus, relation (3) yields
(4)
where and are the 2-D vectors of the source and geophone coordinates.
A second relation expresses the coordinates of the point of emergence of the zero-offset ray, E. In the triangle (PEM), we have :
(5)
Again, the vector is related to the table of lateral distances, , yielding
(6)
where is the 2-D vector of the emergence point coordinates.
Rayp3d
Figure 1
Three-dimensional view of the source, receiver, and reflection points for 3-D v(z) DMO. The dashed lines represent the ray paths in the earth, and the bold solid lines represent the distances on the surface of the earth.
Next: Time relationships in the Up: DERIVING THE SYSTEM OF Previous: DERIVING THE SYSTEM OF
Stanford Exploration Project
11/17/1997
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# Thread: What are the advantages and disadvantages of a 220V 50 Hz & 110V 60 Hz environment
1. micro ATX User
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## What are the advantages and disadvantages of a 220V 50 Hz & 110V 60 Hz environment
Here in India we have a 220V 50 Hz electricity distribution and there in US and in several other countries exists a 110V 60 Hz electricity distribution.
My questions are -
2.What possibly are or were the practical reasons behind the various countries for opting for a particular environment.
3.Can the two environment be compared with a conclusion - Environmnet X is Superior to environmet Y .
4.Why was the figure 110 and 220 only chosen and amazingly one is just the double of the other in magnitude ( though not in frequency) .Why weren't any other voltages chosen .How was it decided that this particular amplitude of voltage would serve the exact purpose and not any other voltage and then the equipments were designed / manufactured accordingly.
5.Is frequency of AC a function of the voltage viz 60 Hz corresponds to 110V and 50 Hz corresponds to 220Volts .Is there a mathematical relation between the frequency and the voltage , if yes what's the formula.
6.Which of the two systems is more efficient / competent in any way say in monetary terms / cost of design/manufacture and implementation et al .
I am just trying to seek an answer to a question which I have been asking my teachers since I was a kid and till now no body has given me a satisfactory answer.I don't intend to make a derogatory comparison .
Incase this question is beyond the scope / spectrum of the site , then may I please request very humbly for the forum member to give me some relevant links which precisesly and explicitly explain the above facts.
Thanks and regards
2. micro ATX User
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Generally, i would say the 220v system is more efficient, regarding power delivery and losses.
But this comes at the price of better isolation requirements and the higher lethal risk if something is going wrong.
This is caused by the ohms law, (I=U/R) which means that at 220v the lethal current is more easily reached then with 110v.
Frequency is choose for the intended consumer. Not that i can tell you much about the details, but for very big motors (e.g. electric locomotive) a far lower frequency is used (16,70 Hz in DE, CH and AU).
I think that the figures are chosen due to material and physical limits at the beginning of the electric age.
Then they are kept to be compatible.
3. Originally Posted by Big-G
Here in India we have a 220V 50 Hz electricity distribution and there in US and in several other countries exists a 110V 60 Hz electricity distribution.
My questions are -
2.What possibly are or were the practical reasons behind the various countries for opting for a particular environment.
3.Can the two environment be compared with a conclusion - Environmnet X is Superior to environmet Y .
4.Why was the figure 110 and 220 only chosen and amazingly one is just the double of the other in magnitude ( though not in frequency) .Why weren't any other voltages chosen .How was it decided that this particular amplitude of voltage would serve the exact purpose and not any other voltage and then the equipments were designed / manufactured accordingly.
5.Is frequency of AC a function of the voltage viz 60 Hz corresponds to 110V and 50 Hz corresponds to 220Volts .Is there a mathematical relation between the frequency and the voltage , if yes what's the formula.
6.Which of the two systems is more efficient / competent in any way say in monetary terms / cost of design/manufacture and implementation et al .
I am just trying to seek an answer to a question which I have been asking my teachers since I was a kid and till now no body has given me a satisfactory answer.I don't intend to make a derogatory comparison .
Incase this question is beyond the scope / spectrum of the site , then may I please request very humbly for the forum member to give me some relevant links which precisesly and explicitly explain the above facts.
Thanks and regards
http://www.ieee.org/portal/cms_docs_...der/trivia.pdf
4. 220V
* less current for the same power, so less heating (PC power supplies are more efficient at 220V)
* more dangerous
Here is what I wrote for the following Quora question
"American has 120v and UK has 240v. Is there any advantages to 120v and vice versa?"
America has 220v; there is a 220v circuit in most American houses for the high consumption devices (clothes dryer, range etc.)
HOWEVER
There is one terrible disadvantage to the way 110v is often implemented in America, namely through a neutral line that divides the 220v into two circuits of 110v. Some devices in the house are on one circuit, some on the other. Now the problem: if that dividing neutral should ever fail due to a bad connection then some devices can be exposed to more than 110v and destroyed (it is almost paradoxical that the loss of a connection can lead to higher voltage).
To be explicit, have a washing machine on one circuit and a PC on the other; disconnect the dividing neutral and the 220v now goes across the two devices in series and the voltage is now unevenly distributed across them.
The following video illustrates it better than I can
https://www.bing.com/videos/search?F...al&view=detail
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# KiBps to Tibps → CONVERT Kibibytes per Second to Tebibits per Second
expand_more
info 1 KiBps is equal to 0.000000007450580596923828125 Tibps
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
KiBps
## Kibibytes per Second (KiBps) Versus Tebibits per Second (Tibps) - Comparison
Kibibytes per Second and Tebibits per Second are units of digital information used to measure storage capacity and data transfer rate.
Both Kibibytes per Second and Tebibits per Second are the "binary" units. One Kibibyte is equal to 1024 bytes. One Tebibit is equal to 1024^4 bits. There are 134,217,728 Kibibyte in one Tebibit. Find more details on below table.
Kibibytes per Second (KiBps) Tebibits per Second (Tibps)
Kibibytes per Second (KiBps) is a unit of measurement for data transfer bandwidth. It measures the number of Kibibytes that can be transferred in one Second. Tebibits per Second (Tibps) is a unit of measurement for data transfer bandwidth. It measures the number of Tebibits that can be transferred in one Second.
## Kibibytes per Second (KiBps) to Tebibits per Second (Tibps) Conversion - Formula & Steps
The KiBps to Tibps Calculator Tool provides a convenient solution for effortlessly converting data rates from Kibibytes per Second (KiBps) to Tebibits per Second (Tibps). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Kibibyte) and target (Tebibit) data units.
Source Data Unit Target Data Unit
Equal to 1024 bytes
(Binary Unit)
Equal to 1024^4 bits
(Binary Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Kibibyte to Tebibit in a simplified manner.
÷ 1024
÷ 1024
÷ 1024
÷ 1024
x 1024
x 1024
x 1024
x 1024
Based on the provided diagram and steps outlined earlier, the formula for converting the Kibibytes per Second (KiBps) to Tebibits per Second (Tibps) can be expressed as follows:
diamond CONVERSION FORMULA Tibps = KiBps x 8 ÷ 10243
Now, let's apply the aforementioned formula and explore the manual conversion process from Kibibytes per Second (KiBps) to Tebibits per Second (Tibps). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Tebibits per Second = Kibibytes per Second x 8 ÷ 10243
STEP 1
Tebibits per Second = Kibibytes per Second x 8 ÷ (1024x1024x1024)
STEP 2
Tebibits per Second = Kibibytes per Second x 8 ÷ 1073741824
STEP 3
Tebibits per Second = Kibibytes per Second x 0.000000007450580596923828125
Example : By applying the previously mentioned formula and steps, the conversion from 1 Kibibytes per Second (KiBps) to Tebibits per Second (Tibps) can be processed as outlined below.
1. = 1 x 8 ÷ 10243
2. = 1 x 8 ÷ (1024x1024x1024)
3. = 1 x 8 ÷ 1073741824
4. = 1 x 0.000000007450580596923828125
5. = 0.000000007450580596923828125
6. i.e. 1 KiBps is equal to 0.000000007450580596923828125 Tibps.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Kibibytes per Second to Tebibits per Second using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Kibibyte ?
A Kibibyte (KiB) is a binary unit of digital information that is equal to 1024 bytes (or 8,192 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'kibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'kilobyte' (KB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
arrow_downward
#### What is Tebibit ?
A Tebibit (Tib or Tibit) is a binary unit of digital information that is equal to 1,099,511,627,776 bits and is defined by the International Electro technical Commission(IEC). The prefix 'tebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'terabit' (Tb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Kibibytes per Second (KiBps) to Tebibits per Second (Tibps)
Apply the formula as shown below to convert from 1 Kibibytes per Second (KiBps) to Tebibits per Second (Tibps).
A B C
1 Kibibytes per Second (KiBps) Tebibits per Second (Tibps)
2 1 =A2 * 0.000000007450580596923828125
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kibibytes per Second (KiBps) to Tebibits per Second (Tibps) Conversion
You can use below code to convert any value in Kibibytes per Second (KiBps) to Kibibytes per Second (KiBps) in Python.
kibibytesperSecond = int(input("Enter Kibibytes per Second: "))
tebibitsperSecond = kibibytesperSecond * 8 / (1024*1024*1024)
print("{} Kibibytes per Second = {} Tebibits per Second".format(kibibytesperSecond,tebibitsperSecond))
The first line of code will prompt the user to enter the Kibibytes per Second (KiBps) as an input. The value of Tebibits per Second (Tibps) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Tebibits(Tibit) are there in a Kibibyte(KiB)?expand_more
There are 0.000000007450580596923828125 Tebibits in a Kibibyte.
#### What is the formula to convert Kibibyte(KiB) to Tebibit(Tibit)?expand_more
Use the formula Tibit = KiB x 8 / 10243 to convert Kibibyte to Tebibit.
#### How many Kibibytes(KiB) are there in a Tebibit(Tibit)?expand_more
There are 134217728 Kibibytes in a Tebibit.
#### What is the formula to convert Tebibit(Tibit) to Kibibyte(KiB)?expand_more
Use the formula KiB = Tibit x 10243 / 8 to convert Tebibit to Kibibyte.
#### Which is bigger, Tebibit(Tibit) or Kibibyte(KiB)?expand_more
Tebibit is bigger than Kibibyte. One Tebibit contains 134217728 Kibibytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation.
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# X, Y, and Z are all nonzero numbers. Find the value of
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X, Y, and Z are all nonzero numbers. Find the value of [#permalink]
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X, Y, and Z are all nonzero numbers. Find the value of Z=(X-Y)/(2X+Y)
(1) |X|=Y
(2) X=-3
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20 Jul 2003, 08:49
A??
Since all three are non-zero
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20 Jul 2003, 20:13
|x|=y means x=y or -x=y.
In the first one z=o it has to be eliminated due to the terms.
In the second one z=2
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# How to do CSG on 3D triangulated objects, originally stored as a .obj file, using openGL or other libraries?
What I would like to do is to create a crater on a field. To do so, I thought that creating a universal crater object (something cone-shaped) and then subtracting it from the field on a specified point might be a good idea.
So I would specify a "middle point" on the field, take the lowest point of the crater and move the crater by its lowest point onto the middle point of the field and then push it by its size of the crater into the field. Then apply said subtracting operation.
My question is: Are there any techniques/libraries which help me accomplish this? They should be either openGL-compatible or work on .obj file format.
## 1 Answer
Describing a full CSG algorithm in a forum post is something really hard, and probably you will not find a complete answer to what you are asking here, but I can give you some suggestions.
First of all, CSG on polygon meshes is not usually employed on real-time graphics, because the algorithms are still considered to be too much resource-intensive. If you are looking for these kinds of algorithms, I would suggest reading books on computational geometry, of which Mark De Berg's is commonly considered a good introduction to the topic. If you are planning to pre-compute everything, then classic CSG algorithms are the way to go.
However, for just adding craters on a field, CSG seems to be too much overkill to me, and I would suggest some completely different approaches. If you want just a visual result (i.e., your player won't be able to go into the crater), you could use techniques such as z-buffer carving or parallax occlusion mapping, combined with some sort of decal system (preferably, using some deferred-like rendering system). Otherwise, if you need proper collision detection, you will actually need to change your mesh. You can easily apply some vertex modification operators around the center of each crater, such as pushing them down according to a radial-basis function (i.e., spheres, gaussians, etc). To accomplish this in realtime graphics, I would suggest using compute shaders or vertex feedback transforms, instead of computing everything in the CPU. These techniques will even allow you to add craters dynamically. Think of some really nice meteor strikes taking place.
I'm sorry for not being too descriptive, but I hope that some of the keywords that I have put here may help you out on finding a solution to your problem.
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Calculating Fall Clearance with Different Fall Arrest Systems | ABC Academy & GA Articles
## Calculating Fall Clearance with Different Fall Arrest Systems
Date: Wednesday, May 25, 2016 Watch Now Time: 3:00 p.m. (ET) / 2:00 p.m. (CT) / 1:00 p.m. (MT) / 12:00 p.m. (PT) Length: 60 minutes Fee: Free Speaker: Chris Irwin, Global Safety Trainer, Mine Safety Appliances (MSA) Category: Training and Education, AQC
Presented and Sponsored by Mine Safety Appliances (MSA)
WEBINAR DESCRIPTION
Calculating fall clearance - The use of a "Personal Fall Arrest System" is something that, on the surface, doesn't seem to be that
complicated but quite often is either not done at all or done incorrectly. With the broad spectrum of fall arrest equipment
that has come onto the market in recent years, knowing how to calculate fall clearance based on one simple equation simply is no longer feasible.
This Webinar will cover the intricacies of calculating clearance for the main types of fall arrest systems in-use today, showing how the process varies and giving tips on working to simplify the process. The main focus of the presentation will include a discussion on (1) why this topic is important to understand and, then, (2) quickly transition to talking about how to figure out system space requirements for applications such as utilizing energy-absorbing lanyards, self-retracting lifelines, horizontal lifelines, and vertical lifelines.Time will be taken to discuss general rules that can be applied in each of the above-mentioned situations, while directing the listener back to instructions provided with their particular systems. Emphasis will be put on how recent regulations have truly added many "variables" to our calculations.
THIS WEBINAR IS DESIGNED TO:
• Basic strategies for calculating clearance with the two main types of connecting devices--EALs & SRLs--will be discussed
• Explaining fall clearance calculations with simpler systems, further time will be spent discussing applying their use on horizontal and vertical lifelines.
• The learner will select the CORRECT formula for calculating clearance, based on a proposed scenario and equipment being used.
• Time will be taken to discuss how new (1) regulations and (2) technologies have changed how fall clearance is calculated.
• Strategies will be given as general rules for always knowing how to safely calculate clearance based on system types and user needs.
RESOURCES
PowerPoint Presentation(PDF)
Chris Irwin, ASHM, is a global fall protection trainer and safety program developer with MSA, having worked in the development of both fall protection and confined space entry training programs for the company. Having received a Bachelor of Sciences degree in Safety and Health Management, Chris began his career at an OSHA Voluntary Protection Program (VPP) Star manufacturing facility as an Environmental, Health, and Safety specialist before moving on to work as a site safety supervisor for the general contractor of a construction company tasked with building a dual-cycle power plant. Since 2009 Chris has worked full-time as a safety trainer for open enrollment and contractual trainings in everything from basic fall protection, competent person, and train-the-trainer programs to tower, wind turbine, and confined space rescue. Besides training, he has also worked extensively carrying out on-site OSHA-style safety inspections pertaining to all areas of the 1910 and 1926 regulations, consulting with safety managers to establish corporate programs and policies. Additionally, Chris is a certified OSHA 10 and 30-Hour trainer for both General Industry and Construction. Prior to entering the safety field Chris worked as a Spanish teacher in a rural Pennsylvania high school for grades 7 through 12.
Comments are closed for this post, but if you have spotted an error or have additional info that you think should be in this post, feel free to contact us.
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# 经济代写|计量经济学代写Econometrics代考|ECON2271
## 经济代写|计量经济学代写Econometrics代考|Conditional Moment Tests
One important approach to model specification testing that we have not yet discussed is to base tests directly on certain conditions that the error terms of a model should satisfy. Such tests are sometimes called moment specification tests but are more frequently referred to as conditional moment, or CM, tests. They were first suggested by Newey (1985a) and Tauchen (1985) and have been further developed by White (1987), Pagan and Vella (1989), Wooldridge (1991a, 1991b), and others. The basic idea is that if a model is correctly specified, many random quantities which are functions of the error terms should have expectations of zero. The specification of a model sometimes allows a stronger conclusion, according to which such functions of the error terms have zero expectations conditional on some information set – whence the terminology of conditional moment tests.
Since an expectation is often referred to as a moment, the condition that a random quantity has zero expectation is generally referred to as a moment condition. Even if a population moment is zero, its empirical counterpart, which we will call an empirical moment, will (almost) never be so exactly, but it should not be significantly different from zero. Conditional moment tests are based directly on this fact.
Conditional moment tests can be used to test many different aspects of the specification of econometric models. Suppose that the economic or statistical theory behind a given parametrized model says that for each observation $t$ there is some function of the dependent variable $y_t$ and of the model parameters $\boldsymbol{\theta}$, say $m_t\left(y_t, \boldsymbol{\theta}\right)$, of which the expectation is zero when the DGP used to compute the expectation is characterized by $\theta$. Thus, for all $t$ and for all $\theta$,
$$E_{\boldsymbol{\theta}}\left(m_t\left(y_t, \boldsymbol{\theta}\right)\right)=0 .$$
We may think of (16.48) as expressing a moment condition. In general, the functions $m_t$ may also depend on exogenous or predetermined variables.
Even though there is a different function for each observation, it seems reasonable, by analogy with empirical moments, to take the following expression as the empirical counterpart of the moment in condition (16.48):
$$m(\boldsymbol{y}, \hat{\boldsymbol{\theta}}) \equiv \frac{1}{n} \sum_{t=1}^n m_t\left(y_t, \hat{\boldsymbol{\theta}}\right),$$
where $\hat{\boldsymbol{\theta}}$ denotes a vector of estimates of $\boldsymbol{\theta}$. Expression (16.49) is thus a form of empirical moment. A one-degree-of-freedom CM test would be computed by dividing it by an estimate of its standard deviation and would be asymptotically distributed as $N(0,1)$ under suitable regularity conditions. There might well be more than one moment condition, of course, in which case the test statistic could be calculated as a quadratic form in the empirical moments and an estimate of their covariance matrix and would have the chi-squared distribution asymptotically.
## 经济代写|计量经济学代写Econometrics代考|Information Matrix Tests
One important type of conditional moment test is the class of tests called information matrix, or IM, tests. These were originally suggested by White (1982), although the conditional moment interpretation is more recent; see Newey (1985a) and White (1987). The basic idea is very simple. If a model that is estimated by maximum likelihood is correctly specified, the information matrix must be asymptotically equal to minus the Hessian. If it is not correctly specified, that equality will in general not hold, because the proof of the information matrix equality depends crucially on the fact that the joint density of the data is the likelihood function; see Section 8.6.
Consider a statistical model characterized by a loglikelihood function of the form
$$\ell(\boldsymbol{y}, \boldsymbol{\theta})=\sum_{t=1}^n \ell_t\left(y_t, \boldsymbol{\theta}\right),$$
where $\boldsymbol{y}$ denotes an $n$-vector of observations $y_t, t=1, \ldots, n$, on a dependent variable, and $\boldsymbol{\theta}$ denotes a $k$-vector of parameters. As the subscript $t$ indicates, the contribution $\ell_t$ made by observation $t$ to the loglikelihood function may depend on exogenous or predetermined variables that vary across the $n$ observations. The null hypothesis for the IM test is that
$$\operatorname{plim}{n \rightarrow \infty}\left(\frac{1}{n} \sum{t=1}^n\left(\frac{\partial^2 \ell_t(\boldsymbol{\theta})}{\partial \theta_i \partial \theta_j}+\frac{\partial \ell_t(\boldsymbol{\theta})}{\partial \theta_i} \frac{\partial \ell_t(\boldsymbol{\theta})}{\partial \theta_j}\right)\right)=0,$$
for $i=1, \ldots, k$ and $j=1, \ldots, i$. Expression (16.65) is a typical element of the information matrix equality. The first term is an element of the Hessian, and the second is the corresponding element of the outer product of the gradient. Since the number of such terms is $\frac{1}{2} k(k+1)$, the number of degrees of freedom for an IM test is potentially very large.
# 计量经济学代考
## 经济代写|计量经济学代写Econometrics代考|Conditional Moment Tests
$$E_{\boldsymbol{\theta}}\left(m_t\left(y_t, \boldsymbol{\theta}\right)\right)=0 .$$
$$m(\boldsymbol{y}, \hat{\boldsymbol{\theta}}) \equiv \frac{1}{n} \sum_{t=1}^n m_t\left(y_t, \hat{\boldsymbol{\theta}}\right),$$
## 经济代写|计量经济学代写Econometrics代考|Information Matrix Tests
$$\ell(\boldsymbol{y}, \boldsymbol{\theta})=\sum_{t=1}^n \ell_t\left(y_t, \boldsymbol{\theta}\right)$$
$$\operatorname{plim} n \rightarrow \infty\left(\frac{1}{n} \sum t=1^n\left(\frac{\partial^2 \ell_t(\boldsymbol{\theta})}{\partial \theta_i \partial \theta_j}+\frac{\partial \ell_t(\boldsymbol{\theta})}{\partial \theta_i} \frac{\partial \ell_t(\boldsymbol{\theta})}{\partial \theta_j}\right)\right)=0$$
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Math作业代写、数学代写常见问题
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# Area of Sectors and Segments
## Area of parts of a circle.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Area of Sectors and Segments
This concept teaches students how to find the area of sectors and segments of circles.
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## Area of Sectors and Segments: Pacman Area
Area of Sectors and Segments Interactive
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## Area of Sectors and Segments Principles
This video gives more detail about the mathematical principles presented in Area of Sectors and Segments.
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## Area of Sectors and Segments Examples
This video shows how to work step-by-step through one or more of the examples in Area of Sectors and Segments.
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## Area of Sectors and Segments Practice
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## Area of Sectors and Segments Discussion Questions
A list of student-submitted discussion questions for Area of Sectors and Segments.
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## Area of Sectors and Segments Post Read
To stress understanding of a concept by summarizing the main idea and applying that understanding to create visual aids and generate questions and comments using a Concept Matrix.
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## Area of Sectors and Segments Pre Read
To activate prior knowledge, make personal connections, reflect on key concepts, encourage critical thinking, and assess student knowledge on the topic prior to reading using a Quickwrite.
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## Area of Sectors and Segments Stop and Jot Table
Learn new vocabulary words and help remember them by coming up with your own sentences with the new words using a Stop and Jot table.
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## Watch Out for the Weather!
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## Area Of Sectors And Segments
How many angles can you identify in the world around you?
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• Study Guide
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# How much did Earth's gravity alter the course of 2005 YU55?
• 2milehi
#### 2milehi
Just like the title says - with the meteor passing close by the Earth, gravity should alter the meteor's course. So how much was it?
Earth: 5x10^24kg
Aircraft carrier: 7x10^7kg
Separation:300,000,000m
http://www.wsanford.com/~wsanford/calculators/gravity-calculator.html
That wasn't my question. As the meteor passed by Earth there was a varible force that acted on the meteor. This force would alter the trajectory of the meteor. Was it a degree, an arcminute, an arcsecond?
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That wasn't my question. As the meteor passed by Earth there was a varible force that acted on the meteor. This force would alter the trajectory of the meteor. Was it a degree, an arcminute, an arcsecond?
Dave was giving you a way to calculate the sideways force on the object. Do you know how long the object was within the orbit of the moon? You could get a pretty good approximation for how much the trajectory was altered by applying the average gravitational attraction of the Earth on the object in a perpendicular direction for the duration of the time inside the Moon's orbit.
Earth boosted 2005YU55's semimajor axis by about 2 million km. It reduced its inclination from about a half a degree to about a third of a degree.
In this image, the green orbit is 2005 YU55. The interior one is before, the exterior one is after.
Thank you for answering my question. :-)
Actually, the question is quite a bit misguided.
The trajectory remain the same as it ever was - all gravitational perturbations included.
Look at it as an object traveling in a perfectly straight line tnrough some very uneven surface (gravity wells of Earth, Sun, Moon...). The object still rolls/flies in a straght line... But the line is warped.
To continue: Even if you fall back on the classical/Newtonian physics the question still doesn't make sense. "changed" compared to what? To what it would be in the Earth's absence? But... hold on a sec, in Earth's absence that asteroid wouldn't even arrive here in the first place - Earth's gravitational pull is part of the reason that asteroid got where it was anyway.
Tony understood my question. Griz you are not adding anything to my question. I will expand to clear any confusion - what was the change in the orbital characteristics of 2005 YU55 when it recently passed by Earth.
This gif makes a lot more sense when comparing it to
Tony's pic
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Well, perhaps I misunderstood your question. (I'm actually inclined to think so).
But the basic fact remains that the question "how much did the Earth gravity changed the course of that asteroid" is pretty much meaningless. It didn't. In any shape, size or form UNLESS you mean how it shaped the course from the start, to begin with. But then... the question about "change" is once again a bit odd.
Well, perhaps I misunderstood your question. (I'm actually inclined to think so).
But the basic fact remains that the question "how much did the Earth gravity changed the course of that asteroid" is pretty much meaningless. It didn't. In any shape, size or form UNLESS you mean how it shaped the course from the start, to begin with. But then... the question about "change" is once again a bit odd.
Change - like before and after. Tell me that you don't see it with the following image.
The meteor was "sling-shot" when passing by Earth. Space probes use Jupiter to "sling-shot" to add speed and change direction.
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What -I- don't see is how this pretty picture indicates a changed orbit.
What -you- don't see is the fact that orbits are never the same perfect ellipses. Doesn't happen.
So, what your picture shows is just a normsl, pretty ordinary orbit.
Perhaps you should make distinction between the words course/trajectory/orbit?
Earth boosted 2005YU55's semimajor axis by about 2 million km. It reduced its inclination from about a half a degree to about a third of a degree.
In this image, the green orbit is 2005 YU55. The interior one is before, the exterior one is after.
Griz maybe you missed this post. The inside green circle represents the orbital characteristics of the meteor before it was "sling-shot" by Earth. The outer green circle represents the new orbital characteristics of the meteor. The Earth change the direction and the speed of the meteor via gravitational attraction.
It seems you are going out of your way NOT to understand what I am asking/saying.
Grizzled, I'm sure your technically correct, but I think you're simply confusing some people. Suffice it to say I think the OP's question was answered by the picture.
Drakkith,
I can't help it if some people are so easily confused. Perhaps they should study a bit?
I also can't help it if all they want are some pretty pictures which they (for some mysterious reason) insist on posting over and over again.
At least you agree that I am "technically" correct.
Thanks.
Drakkith,
I can't help it if some people are so easily confused. Perhaps they should study a bit?
Why do you think they are here at PF if not to study and learn? EVERYONE is easily confused when they know very little about something. Just because you know more about the subject does not give you a reason to be rude. Instead it might help if you explained what the different terms mean for this particular scenario.
I also can't help it if all they want are some pretty pictures which they (for some mysterious reason) insist on posting over and over again.
There is nothing wrong with pictures and diagrams. They help people to visualize a concept, which is very important for most people when attempting to learn something new.
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Drakkith,
I can't help it if some people are so easily confused.
I also can't help it if all they want are some pretty pictures which they (for some mysterious reason) insist on posting over and over again.
If pretty pictures answer the question asked then pretty pictures are the thing.
At least you agree that I am "technically" correct.
I am "technically" correct when I mention that Mercury precesses. The issue, of course, is whether that answers the question being asked.
The OP's is a legitimate question. He wants to know how much YU 55 was deflected by Earth's passage. consider: if you were stationary wrt the asteroid and looked in the direction of its travel, you'd see a point it's heading toward. If Earth had not crossed its path that point would move steadily as the asteroid proceeded in its orbit. After Earth fly-by the point would be completely different.
Does that make it clearer to you what the OP is asking?
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Perhaps you should make distinction between the words course/trajectory/orbit?
First I never said orbit. I used the words course and trajectory.
course n. - The route or path taken by something, such as a stream, that moves.
trajectory n. - The path of a projectile or other moving body through space.
I used orbital characteristics to clear any confusion that you had.
orbital characteristics n. - a table of values that gives the positions of astronomical objects in the sky at a given time (see ephemeris).
synonym n. -
1. A word having the same or nearly the same meaning as another word or other words in a language.
2. A word or an expression that serves as a figurative or symbolic substitute for another.
Griz if you can't see how course/trajectory relates to my question, you are thick. Everyone else seems to understand me.
What -you- don't see is the fact that orbits are never the same perfect ellipses. Doesn't happen.
I never said that orbits are consistent perfect ellipses.
Drakkith,
I can't help it if some people are so easily confused. Perhaps they should study a bit?
I also can't help it if all they want are some pretty pictures which they (for some mysterious reason) insist on posting over and over again.
At least you agree that I am "technically" correct.
Thanks.
Now you are acting arrogant and being sarcastic (humor for the small mind). Please quit posting, you are removing value from this topic.
Or to put it another way...
Look at it as an object traveling in a perfectly straight line tnrough some very uneven surface (gravity wells of Earth, Sun, Moon...). The object still rolls/flies in a straght line... But the line is warped.
Fine. How much did Earth warp the straight line?
Perhaps I'm reading too much into the (nicely done) diagram showing the before-and-after orbits of 2005 YU55, but: does it look like the shifted orbit will now make a collision with Mars more likely?
Perhaps I'm reading too much into the (nicely done) diagram showing the before-and-after orbits of 2005 YU55, but: does it look like the shifted orbit will now make a collision with Mars more likely?
You're definitely reading too much into it.
The very fact that the orbit as shown is smooth and curvilinear is a strong indication that YU55 is pretty much never near Mars in its orbit. Space is very big and orbiting bodies very small.
Furthermore, as it stands, YU55 is no more in Mars' gravity well than it was before.
With the "before" orbit of YU55, there would be zero chance of the asteroid colliding with Mars since it never crosses Mars' orbit. With the "after" orbit of YU55, there are two times that the asteroid crosses Mars' orbit. This has to increase the chance of a collision between the two bodies with all things remaining the same. I assume that YU55 and Mars are on the same orbital plane.
With the "before" orbit of YU55, there would be zero chance of the asteroid colliding with Mars since it never crosses Mars' orbit. With the "after" orbit of YU55, there are two times that the asteroid crosses Mars' orbit. This has to increase the chance of a collision between the two bodies with all things remaining the same. I assume that YU55 and Mars are on the same orbital plane.
But Mars is not a billiard ball floating in space. It is a gravity well, much larger than its physical size. Before Earth fly-by, if Mars had been in just the right spot in its orbit, YU55 would have passed through its gravity well. After Earth fly-by, if Mars were in just the right spot, YU55 would pass through it gravity well.
i.e. Not a lot has changed.
And since its interacting with the gravity well, not the body, it would not collide with Mars, it would have a close encounter, altering its orbit, just as with Earth, sending it off on a another trajectory, perhaps never to bother Mars again.
To continue: Even if you fall back on the classical/Newtonian physics the question still doesn't make sense. "changed" compared to what? To what it would be in the Earth's absence? But... hold on a sec, in Earth's absence that asteroid wouldn't even arrive here in the first place - Earth's gravitational pull is part of the reason that asteroid got where it was anyway.
Generally speaking, such a question does make sense. Outside of a certain distance from the Earth, the Sun's gravity dominates the object's trajectory and the Earth's effect can pretty much be ignored for all practical purposes.
This radius defines what is called the "gravitational sphere of influence" of the Earth. It was calculated by Laplace to be:
$$rp = \sqrt[5] {\frac{M_p}{M_s}}^2$$
And for the Earth, works out to being about 927,000 km.
So when we talk about how much the Earth deflects something like 2005 YU55, we are talking about how much its path changes from the time it enters the sphere of influence until it leaves.
You can find the angle of deflection (relative to the Earth) with the formula:
$$b = \frac{GM_e}{v^2} \cot q$$
Here b is the impact parameter, or the closest approach to the Earth the object would have had if it had traveled on a straight line through the sphere of influence.
v is the relative velocity between Earth and object.
and
q is the deflection angle.
Thus
The process would go something like this:
From the Earth's and object's motions relative to the Sun, transform to the object's motion relative to the Earth.
Use the formula above to determine the deflection angle and new trajectory relative to the Earth.
Using the new Earth relative motion, transform back to a Sun relative frame, giving you the new trajectory of the object relative to the Sun.
I assume that YU55 and Mars are on the same orbital plane.
They're close, but not exact, and that translates into lots of km of separation along the z-axis. Although the Earth passage raised YU55's SMA by about 2 million km, it also reduced its inclination. At the point where Mars and YU55's orbits appear to intersect, Mars' orbit is 7.7 million km above the ecliptic. Before the Earth passage, YU55 was about 2.2 million km above the ecliptic at that point. Afterwards, it is about 1.5 million km above the ecliptic, distancing itself from Mars along the z-axis.
In either case, their orbits don't actually intersect. As time goes on, the longitude of perihelion precesses for both objects. Such change is usually measured in arcseconds per century. So from time to time, their orbits can intersect in the very distant future. Since Mars is near aphelion at appearant point of intersection in the image, even without an Earth encounter a change in their longitude of perihelions could still cause their orbits to intersect.
Even if they did line up such that their orbits perfectly intersected, chance of a collision is very low. Mars is very small compared to the size of its orbit. Each time YU55 perfectly crossed Mars' orbit, it would stand about a 1 in 150,000 chance that Mars would be there at that moment.
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# Stonksmaster - Predict Stock prices using Python & ML 📈
Nirvik Agarwal ・Updated on ・8 min read
Newbie to Machine Learning?
Need a nice initial project to get going?
You are on the right article!
In this article, we will try to build a very basic stock prediction application using Machine Learning and its concepts. And as the name suggests it is gonna be useful and fun for sure. So let's get started.
We expect you to have a basic exposure to Data Science and Machine Learning.
"The field of study that gives computers the ability to learn without being explicitly programmed"
is what Arthur Samuel described as Machine Learning.
Machine Learning has found its applications in various fields in recent years, some of which include Virtual Personal Assistants, Online Customer Support, Product Recommendations, etc.
We will use libraries like `numpy`, `pandas`, `matplotlib`, `scikit-learn`, and a few others.
If you are not familiar with these libraries, you can refer to the following resources:
## Steps in Machine Learning
While performing any Machine Learning Task, we generally follow the following steps:
1. Collecting the data
This is the most obvious step. If we want to work on an ML Project we first need data. Be it the raw data from excel, access, text files, or data in the form of images, video, etc., this step forms the foundation of future learning.
2. Preparing the data
Bad data always leads to bad insights that lead to problems. Our prediction results depend on the quality of the data used. One needs to spend time determining the quality of data and then taking steps for fixing issues such as missing data etc.
3. Training the model
This step involves choosing the appropriate algorithm and representation of data in the form of the model. In layman terms, model representation is a process to represent our real-life problem statement into a mathematical model for the computer to understand. The cleaned data is split into three parts – Training, Validation, and Testing - proportionately depending on the scenario. The training part is then given to the model to learn the relationship/function.
4. Evaluating the model
Quite often, we don’t train just one model but many. So, to compare the performance of the different models, we evaluate all these models on the validation data. As it has not been seen by any of the models, validation data helps us evaluate the real-world performance of models.
5. Improving the Performance
Often, the performance of the model is not satisfactory at first and hence we need to revisit earlier choices we made in deciding data representations and model parameters. We may choose to use different variables (features) or even collect some more data. We might need to change the whole architecture to get better performance in the worst case.
6. Reporting the Performance
Once we are satisfied by the performance of the model on the validation set, we evaluate our chosen model on the testing data set and this provides us with a fair idea of the performance of our model on real-world data that it has not seen before.
Now coming to our project, as we are dealing with the stock market and trying to predict stock prices the most important thing is being able to Read Stocks
Reading stock charts, or stock quotes is a crucial skill in being able to understand how a stock is performing, what is happening in the broader market, and how that stock is projected to perform.
Stocks have quote pages or charts, which give both basic and more detailed information about the stock, its performance, and the company on the whole. So, the next question that comes up is what makes up a stock chart?
### Stock Charts
A Stock Chart is a set of information on a particular company's stock that generally shows information about price changes, current trading price, historical highs and lows, dividends, trading volume, and other company financial information.
Also we would like to familiarise you some basic terminologies of the stock market
#### Ticker Symbol
The ticker symbol is the symbol that is used on the stock exchange to delineate a given stock. For example, Apple's ticker is (AAPL) while Snapchat's ticker is (SNAP).
All stock ticker symbols
#### Open Price
The open price is simply the price at which the stock opened on any given day
#### Close Price
The close price is perhaps more significant than the open price for most stocks. The close is the price at which the stock stopped trading during normal trading hours (after-hours trading can impact the stock price as well). If a stock closes above the previous close, it is considered an upward movement for the stock. Vice versa, if a stock's close price is below the previous day's close, the stock is showing a downward movement.
Now its time to get your hands dirty and begin setting up the project
## Initializing our project
### Step 1 : Collecting the data
Use the `iexfinance` library to download the dataframe. The dataframe which we get contains daily data about the stock. The downloaded dataframe gives us a lot of information including Opening Price, Closing Price, Volume, etc. But we are interested in the opening prices with their corresponding dates.
``````import pandas as pd
import numpy as np
import iexfinance
from iexfinance.stocks import get_historical_data
from datetime import datetime, date
# start date should be within 5 years of current date according to iex API we have used
# The more data we have, the better results we get!
start = datetime(2016, 1, 1)
end = date.today()
# use your token in place of token which you will get after signing up on IEX cloud
# Head over to https://iexcloud.io/ and sign-up to get your API token
df = get_historical_data("AAPL", start=start, end=end, output_format="pandas", token="your_token")
``````
### Step 2 : Preparing the data
Also, it would convenient to convert the dates to their corresponding time-stamps. And finally, we will be having a dataframe which will contain our opening prices and time-stamps.
We need to know that the model we created is good. We are going to hold back some data that the algorithms will not get to see and we will use this data to get a second and independent idea of how accurate the best model might actually be.
We will split the loaded dataset into two, 80% of which we will use to train, evaluate, and select among our models, and 20% that we will hold back as a validation dataset.
``````from sklearn.model_selection import train_test_split
prices = df[df.columns[0:1]]
prices.reset_index(level=0, inplace=True)
prices["timestamp"] = pd.to_datetime(prices.date).astype(int) // (10**9)
prices = prices.drop(['date'], axis=1)
prices
dataset = prices.values
X = dataset[:,1].reshape(-1,1)
Y = dataset[:,0:1]
validation_size = 0.15
seed = 7
X_train, X_validation, Y_train, Y_validation = train_test_split(X, Y, test_size=validation_size, random_state=seed)
``````
The function `train_test_split()` comes from the `scikit-learn` library.
scikit-learn (also known as sklearn) is a free software machine learning library for Python. Scikit-learn provides a range of supervised and unsupervised learning algorithms via a consistent interface in Python.
The library is focused on modeling data. It is not focused on loading, manipulating, and summarizing data.
### Step 3 : Training the model
We don’t know which algorithms would be good on this project or what configurations to use.
And So, we are testing with 6 different algorithms:
• Linear Regression (LR)
• Lasso (LASSO)
• Elastic Net (EN)
• KNN (K-Nearest Neighbors)
• CART (Classification and Regression Trees)
• SVR (Support Vector Regression)
``````from sklearn.linear_model import LinearRegression
from sklearn.linear_model import Lasso
from sklearn.linear_model import ElasticNet
from sklearn.tree import DecisionTreeRegressor
from sklearn.neighbors import KNeighborsRegressor
from sklearn.svm import SVR
# Test options and evaluation metric
num_folds = 10
seed = 7
scoring = "r2"
# Spot-Check Algorithms
models = []
models.append((' LR ', LinearRegression()))
models.append((' LASSO ', Lasso()))
models.append((' EN ', ElasticNet()))
models.append((' KNN ', KNeighborsRegressor()))
models.append((' CART ', DecisionTreeRegressor()))
models.append((' SVR ', SVR()))
``````
### Step 4 : Evaluating the model
``````from sklearn.model_selection import KFold
from sklearn.model_selection import cross_val_score
# evaluate each model in turn
results = []
names = []
for name, model in models:
kfold = KFold(n_splits=num_folds, random_state=seed, shuffle=True)
cv_results = cross_val_score(model, X_train, Y_train, cv=kfold, scoring=scoring)
# print(cv_results)
results.append(cv_results)
names.append(name)
msg = "%s: %f (%f)" % (name, cv_results.mean(), cv_results.std())
print(msg)
``````
The output of the above code gives us the accuracy estimations for each of our algorithms. We need to compare the models to each other and select the most accurate.
Once we choose which results in the best accuracy, all we have to do is to
• Define the model
• Fit data into our model
• Make predictions
Plot your predictions along with the actual data and the two plots will nearly overlap.
### Step 5 : Reporting the model and making prediction
``````# Future prediction, add dates here for which you want to predict
dates = ["2020-12-23", "2020-12-24", "2020-12-25", "2020-12-26", "2020-12-27",]
#convert to time stamp
for dt in dates:
datetime_object = datetime.strptime(dt, "%Y-%m-%d")
timestamp = datetime.timestamp(datetime_object)
# to array X
np.append(X, int(timestamp))
from matplotlib import pyplot as plt
from sklearn.metrics import mean_squared_error
# Define model
model = DecisionTreeRegressor()
# Fit to model
model.fit(X_train, Y_train)
# predict
predictions = model.predict(Xp)
print(mean_squared_error(Y, predictions))
# %matplotlib inline
fig= plt.figure(figsize=(24,12))
plt.plot(X,Y)
plt.plot(X,predictions)
plt.show()
``````
Hurrah! You finally built a Stock Predictor. We hope this article was of great help to beginners and everyone else alike. For those who are interested in taking this project to the next level, we recommend you to read on LSTMs neural nets and try implementing it.
Though we are predicting the prices, this model is practically not viable because a lot of other factors have to be considered while making predictions!
## Model References
### Update: We have made a new post following this article in which we have used Ensemble Methods to further enhance our models.
We hope you found this insightful.
This was one of the projects in 10 Days of Code organized by GNU/Linux Users' Group, NIT Durgapur
## Discussion (5)
Peter Hoffmann
This approach is technically interesting but one should not try to sell snake oil here. You can't predict the future and all foreseeable developments are already contained in the stock market. What's making prices change is the element of surprise.
Faris Natour
In the context of the equity market, this is true. Furthermore, the approach in this post is mathematically unsound as stock prices are serially correlated!
Jan Küster
How strong would you indicate the results? Did you actually had success with this approach? I am asking because I read several times, that there is no evidence for beating the market over longer periods (years) using technical analysis.
From the ML perspective it would therefore be interesting which other data could be added to train the models that allows a much more detailed prediction?
Michael Burrows
I'm getting back into the stock market and keen to start learning ML/Python so this ticks two boxes, thanks for the write up :)
Yogesh Singh
Will surely try it.
Nice Article thanks
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31 Aug
# What Is A Table Of Values In Math
Back when I first started traveling for business — I’m 45, my first business trip came when I was 20, so you can do the math.
The heat index is not just a made up number, it’s actually carefully calculated using the mathematics formula. the heat index value can be increased by up to 15°F. As shown in this table, heat.
What is pure mathematics. to do abstract mathematics. In any case, this example should illustrate that the pursuit of purely mathematical problems is a worthwhile cause that can be of tremendous.
Math video demonstrating how to graph a line using a table of values given the equation in slope-intercept form (y=mx+b) Problem 1.
A large p-value indicates that it would be pretty normal to get a sample like ours if the null hypothesis is true. So you can see, there is no reason here to change our minds like we did with a small p-value.
In math, the frequency is the number of times a specific value appears in a data set or list. To find the frequency of these values, one constructs a frequency table and inputs all the different values from the set. Keep Learning.
Math Tables. Another type of math table is one constructed for variable values, typically for graphical data, called a T-Chart. This type of math table would have two columns, one for X and one for Y, creating a series of ordered pairs. This table is most often used for solving graphing equations.
When students work within a classroom community in which both teachers and students value hard work. that deserve a seat at the table. Every child is a mathematician, and every child deserves the.
The t-table (for the t-distribution) is different from the Z-table (for the Z-distribution); make sure you understand the values in the first and last rows. Finding probabilities for various t-distributions, using the t-table, is a valuable statistics skill. Use the t-table as necessary to solve the following problems.
Organic Chemistry Concepts And Applications For Medicinal Chemistry Organic Chemistry Concepts and Applications for Medicinal Chemistry provides a valuable refresher for understanding the relationship between chemical bonding and
Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Pre-Algebra. Linear Equations and Inequalities. Use a Table of Values to Graph the Equation. Substitute for and find the result for. Solve the equation for.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Algebra Examples. Step-by-Step Examples. Algebra. Tables. Find the Function Rule. Check if the function rule is linear. Tap for more steps. To find if the table follows a function rule, check to see if the values follow the linear form. Calculate the value of using each value in the table and compare this value to the given value in the.
To find the answer, the teachers broke apart some of the small, colorful 1-centimeter cubes that were sitting on their table, placing them side by side. off an ambitious revamp of the K-8 literacy.
Nov 05, 2008 · Learn how to create a table of values. This feature is not available right now. Please try again later.
Jun 21, 2017 · A truth table is a handy little logical device that shows up not only in mathematics, but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool.
“We fund our values, right? So, if we really are a nation that believes. credentials of those teachers are considered out-of-field and what they bring to the table. And, I also believe that the.
Table of cosines are the counted values of angles cosines noted in the table from 0° to 360°. Using a table of cosines you can make calculations even if not at hand will be the scientific calculator. To find the cosine of the angle is sufficient to find the value in the table.
Accuplacer College Level Math Practice Problems College-Level Math: 20 questions from the following five content areas:. This site provides sample test questions and a free iPhone
Improve your math knowledge with free questions in "Truth values" and thousands of other math skills.
Physiology Of Hangry Study Hunger and satiety are sensations. Hunger represents the physiological need to eat food. Later studies showed that appetite regulation is
For the First Trust Large Cap Value AlphaDEX Fund ETF, we found that the implied analyst. which is 26.53% above the recent price of \$40.81. Below is a summary table of the current analyst target.
we were assigned a math problem that involved filling one glass with water from another glass. This writer attacked the problem and finished first out of 80 people present. But, following the.
Looking through his email during a free moment in math class, he read the words he had been hoping for. “I scrolled down to a bold ‘you are accepted’ and I just banged the table and everybody looked.
Click here to find out more » We’ve seen computers dominate the world’s best chess players and breeze through math problems so complex that they. relevant to finding an investment’s intrinsic value.
So what values of w are permissible. I hope this explanation makes it clear that all this is backed by pretty solid, rigorous mathematics; yet at the same time, we have not the faintest clue as to.
Math video demonstrating how to graph a line using a table of values given the equation in slope-intercept form (y=mx+b) Problem 1.
We measured peer norms by administering a behavioural challenge-seeking task (the ‘make-a-math-worksheet’ task) at the end of the second intervention session (Fig. 1) and aggregating the values of the.
So the club hosted its July 1 weekly lunch meeting, usually held at Maggiano’s Little Italy, at Horizons Atlanta’s campus at Atlanta International, where at each table three or four. But their time.
Lesson 2 Graphs of inequalities and sign tables. We know that (x + 1)(x − 1) = x2 − 1. We also know that −∙− = + and +∙+ = +. This means that the expression (x + 1)(x − 1) is positive when both brackets have the same sign and negative when they have opposite signs. Now we can complete the table.
Table of cosines are the counted values of angles cosines noted in the table from 0° to 360°. Using a table of cosines you can make calculations even if not at hand will be the scientific calculator. To find the cosine of the angle is sufficient to find the value in the table.
The t-table (for the t-distribution) is different from the Z-table (for the Z-distribution); make sure you understand the values in the first and last rows. Finding probabilities for various t-distributions, using the t-table, is a valuable statistics skill. Use the t-table as necessary to solve the following problems.
How to Graph Lines from a Table of Values Learn, math, science, English, SAT & ACT from expert teachers at Brightstorm. See more graphing lines using a table of values videos at Brightstorm.
4 Grade Science Questions The gap at grade C/4 has narrowed for the second year in a row. 71.7% of entries by girls got
Microeconomics can be, but is not necessarily, math-intensive. Fundamental microeconomic assumptions. For example, students should be able to derive the value of the slope of a line using the.
Solving a tough math problem, listening to music. Your brain gets involved in pattern recognition ("Have I tasted something like this before?"), memory, value judgment, emotion, and pleasure. And.
Fecl2 4h2o Molecular Weight Contact Us · Help. -. Products · Suppliers · Promotions · Quote Request · About · Midland Team · Customer
How to Graph Lines from a Table of Values Learn, math, science, English, SAT & ACT from expert teachers at Brightstorm. See more graphing lines using a table of values videos at Brightstorm.
The value of a function at a certain point refers to the value of a function if you plug in the values given for the variables in the function. Value can also refer to worth of an object in math. In this case, value usually refers to money. To unlock this lesson you must be a Study.com Member.
Mar 30, 2016 · Lesson Summary. A function is a rule that assigns a set of inputs to a set of outputs in such a way that each input has a unique output. A function table in math is a table that describes a function by displaying inputs and corresponding outputs in tabular form.
The same thing happens when you look at a mathematical equation, but don’t see what the x value is (unless you are good at mathematics). Walter Lewin is the author of “For the Love of Physics”. This.
Let’s go through the process of calculating the SD for the SPY. The table below shows a sample of monthly returns for the SPY and TLT from July 2016 to July 2017. We will use these values as sample.
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# numpy division with RuntimeWarning: invalid value encountered in double_scalars
I wrote the following script:
``````import numpy
d = numpy.array([[1089, 1093]])
e = numpy.array([[1000, 4443]])
``````
But I got this result and with the error occurred:
``````nan
C:\Users\Desktop\test.py:16: RuntimeWarning: invalid value encountered in double_scalars
``````
It seems to be that the input element were too small that python turned them to be zeros, but indeed the division has its result.
How to solve this kind of problem?
You can't solve it. Simply `answer1.sum()==0`, and you can't perform a division by zero.
This happens because `answer1` is the exponential of 2 very large, negative numbers, so that the result is rounded to zero.
`nan` is returned in this case because of the division by zero.
Now to solve your problem you could:
• go for a library for high-precision mathematics, like mpmath. But that's less fun.
• as an alternative to a bigger weapon, do some math manipulation, as detailed below.
• go for a tailored `scipy/numpy` function that does exactly what you want! Check out @Warren Weckesser answer.
Here I explain how to do some math manipulation that helps on this problem. We have that for the numerator:
``````exp(-x)+exp(-y) = exp(log(exp(-x)+exp(-y)))
= exp(log(exp(-x)*[1+exp(-y+x)]))
= exp(log(exp(-x) + log(1+exp(-y+x)))
= exp(-x + log(1+exp(-y+x)))
``````
where above `x=3* 1089` and `y=3* 1093`. Now, the argument of this exponential is
`-x + log(1+exp(-y+x)) = -x + 6.1441934777474324e-06`
For the denominator you could proceed similarly but obtain that `log(1+exp(-z+k))` is already rounded to `0`, so that the argument of the exponential function at the denominator is simply rounded to `-z=-3000`. You then have that your result is
``````exp(-x + log(1+exp(-y+x)))/exp(-z) = exp(-x+z+log(1+exp(-y+x))
= exp(-266.99999385580668)
``````
which is already extremely close to the result that you would get if you were to keep only the 2 leading terms (i.e. the first number `1089` in the numerator and the first number `1000` at the denominator):
``````exp(3*(1089-1000))=exp(-267)
``````
For the sake of it, let's see how close we are from the solution of Wolfram alpha (link):
``````Log[(exp[-3*1089]+exp[-3*1093])/([exp[-3*1000]+exp[-3*4443])] -> -266.999993855806522267194565420933791813296828742310997510523
``````
The difference between this number and the exponent above is `+1.7053025658242404e-13`, so the approximation we made at the denominator was fine.
The final result is
``````'exp(-266.99999385580668) = 1.1050349147204485e-116
``````
``````1.105034914720621496.. × 10^-116 # Wolfram alpha.
``````
and again, it is safe to use numpy here too.
• But in this case I need to get values of the division by 2 very small values. Jan 5, 2015 at 17:24
• @Heinz I think you meant the case in which a small number is divided by a small number. In that case change your algorithm to scale both numbers up is much better than finding mechanical twists. For example, take logarithm of the analytical equations that your code is trying to simulate. There are many issues with stability of computation when small numbers are involved. It's nicer to avoid having any of them if possible. Nov 8, 2015 at 22:49
You can use `np.logaddexp` (which implements the idea in @gg349's answer):
``````In [33]: d = np.array([[1089, 1093]])
In [34]: e = np.array([[1000, 4443]])
In [36]: log_res
Out[36]: -266.99999385580668
In [37]: res = exp(log_res)
In [38]: res
Out[38]: 1.1050349147204485e-116
``````
Or you can use `scipy.special.logsumexp`:
``````In [52]: from scipy.special import logsumexp
In [53]: res = np.exp(logsumexp(-3*d) - logsumexp(-3*e))
In [54]: res
Out[54]: 1.1050349147204485e-116
``````
This warning also occurs if a zero-length (in some dimension) array is passed to a numpy method that uses division to derive its output (e.g. `np.mean()`, `np.median()`, `np.var`, `np.cov`, etc.).
A common way to get such array is to filter an array using some condition which returns a zero-length array. So if this warning is raised even though you're not working with very small or very large numbers, check the shape of the array.
An example:
``````arr = np.array([1, 2, 3])
arr_sliced = arr[arr < 0] # array([], dtype=int32)
avg = np.mean(arr_sliced) # RuntimeWarning: invalid value encountered in double_scalars
``````
Note the order of z,y,x rather than x,y,z: see the code
``````# one point and w,h ROI creating
x, y = 400, 300
w, h = 50, 80
roi_mask = np.zeros_like(stacked_volume) # create same size all-zero tensor
roi_mask[:, y:y+h, x:x+w] = 1 # note: the order is Z, Y, X
# roi_mask[x:x+w, y:y+h, :] = 1 # this will make the error
``````
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+0
# HELP!!!!!
0
121
1
How do I figure out compound interest in an equation? Do I still do I=PRT and multiply the numbers together like simple interest?
Guest May 10, 2017
Sort:
#1
+90559
+1
No the formula for compound interest is
\(FV=PV(1+r)^n\)
where
FV=future value
PV = present value (at the start)
r is the rate per compouning period expressed as a decimal
n is the number of compouning perods.
so
If you inbest \$250 for 3 years at 6% per annum compounded monthly then
n = 3*12 = 36 (months)
r= 0.06/12 = 0.005 ( this is the rate per month)
so
FV(in three years) = \(250(1+0.005)^{36} = 250*1.005^{36} \)
250*1.005^36 = \$299.1701
Melody May 10, 2017
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基于经验分布的区间数据分析方法
文章快速检索 高级检索
Interval data analysis based on empirical distribution function
WANG Huiwen, WANG Shengshuai, HUANG Lele, WANG Cheng
School of Economics and Management, Beijing University of Aeronautics and Astronautics, Beijing 100191, China
Abstract:Uniform distribution in some closed or tight interval is a basic assumption in the literature about interval data analysis, which is difficult to satisfy in real data processing. To solve this problem, the empirical cumulative distribution function (ECDF) and kernel estimation of cumulative distribution were studied, on the assumption that the date were from some continuous distribution. Based on ECDF and kernel estimation, a transformation to obtain new data was designed, which was uniformly distributed in theory. Then whether the distribution of transformed data was uniform distribution was tested. If the null hypothesis was not rejected, traditional methods in the field of interval data analysis could be utilized based on transformed data. The transform and the test were both for guaranteeing the transformed data were from some uniform distribution. Both simulation and real data example show that, the results based on ECDF and kernel estimation transformed data are more reasonable and with strong explanatory ability.
Key words: interval data uniform distribution kernel estimation empirical distribution hypothesis test
1 基于经验分布函数的变换
X为服从某一连续分布的随机变量,(x1,x2,…,xn)是已得到的一组样本数据,将其转化为区间数据的方法是取其最大值和最小值作为区间的两个端点,假定其他样本在这个区间服从均匀分布[5].这一假定明显过于严格,如果样本服从其他分布,会导致这一假定及其后续分析的结果失效.
X的分布函数为F(t),经验分布函数Fn(t)定义为
2 变换后的假设检验
3 基于变换数据的区间数据分析
4 数据模拟 4.1 数据模拟1
样本量 N(0,1) Exp(2) Cauchy U(0,1) U(5,10) 5 0.115 0.086 0.052 0.865 0.878 10 0.035 0.012 0.004 0.934 0.925 20 0.006 0.002 0 0.965 0.951 40 0.002 0 0 0.949 0.948 50 0 0 0 0.957 0.955 100 0 0 0 0.972 0.952 200 0 0 0 0.956 0.947
4.2 数据模拟2
图 1 对不同分布的分布函数分别采用经验分布函数和核方法进行估计的结果Fig. 1 Simulation results for estimating the cumulative distribution function by empirical distribution and kernel method
分布类型 样本量20 样本量50 样本量100 样本量200 经验分布 核估计 经验分布 核估计 经验分布 核估计 经验分布 核估计 N(0,1) 0.247 6 0.197 8 0.180 6 0.164 9 0.171 4 0.126 4 0.096 4 0.081 9 Exp(2) 0.180 3 0.168 0 0.152 5 0.136 8 0.092 1 0.109 4 0.026 4 0.029 5 Cauchy 0.626 1 0.552 0 0.526 1 0.546 5 0.472 6 0.585 9 0.226 7 0.407 0 U(2,3) 0.086 0 0.079 3 0.057 6 0.040 1 0.028 6 0.020 5 0.011 9 0.016 1 U(5,10) 0.274 5 0.222 0 0.204 4 0.195 8 0.190 7 0.140 9 0.159 8 0.153 8
5 结 论
[1] Sankararaman S, Mahadevan S.Likelihood-based representation of epistemic uncertainty due to sparse point data and/or interval data[J].Reliability Engineering & System Safety,2011,96(7):814-824. Click to display the text [2] Diday E, Noirhomme-Fraiture M.Symbolic data analysis and the SODAS software[M].London:Wiley Online Library,2008:81-92. [3] Billard L. Symbolic data analysis:what is it?[M].New York:Springer,2006:261-268. [4] Diday E, Esposito F.An introduction to symbollic data analysis and the SODAS software[J].Intelligent Data Analysis,2003,7(6): 583-601. Click to display the text [5] Wang H W, Guan R,Wu J J.CIPCA:complete-information-based principal component analysis for interval-valued data[J].Neurocomputing,2012,86:158-169. Click to display the text [6] Wang H W, Guan R,Wu J J.Linear regression of interval-valued data based on complete information in hypercubes[J].Journal of Systems Science and Systems Engineering,2012,21(4):422-442. Click to display the text [7] Yue Z L. A group decision making approach based on aggregating interval data into interval-valued intuitionistic fuzzy information[J].Applied Mathematical Modelling,2014,38(2):683-698. Click to display the text [8] Cerný M, Hladík M.The complexity of computation and approximation of the t-ratio over one-dimensional interval data[J].Computational Statistics and Data Analysis,2014,80:26-43. Click to display the text [9] Yang X J, Yan L L,Peng H,et al.Encoding words into cloud models from interval-valued data via fuzzy statistics and membership function fitting[J].Knowledge-Based Systems,2014,55:114-124. Click to display the text [10] 郭均鹏,陈颖, 李汶华.一般分布区间型符号数据的K均值聚类方法[J].管理科学学报,2013,16(3):21-28. Guo J P,Chen Y,Li W H.K-means clustering of generally distributed interval symbolic data[J].Journal of Management Sciences in China,2013,16(3):21-28(in Chinese). Cited By in Cnki (5) [11] 高飒. 一般分布区间型符号数据的聚类分析方法研究[D].天津:天津大学,2009. Gao S.The clustering analysis of generally distributed interval symbolic data[D].Tianjin:Tianjin University,2009(in Chinese). Cited By in Cnki (6) [12] Silverman B W. Density estimation for statistics and data analysis[M].London:Chapman and Hall,1986:34-48. [13] Fan J Q, Yao Q W.Nonlinear time series: nonparametric and parametric methods[M].New York:Springer Verlag,2003:193-212. [14] Marhuenda Y, Morales D,Pardo M C.Power results of tests for the uniform distribution,I-2005-09[R].Spain:Miguel Hernandez University of Elche,2005. [15] Kolmogorov A N. Sulla determinazione empirica di una legge di distribuzione[J].G Inst Ital Att,1933,4:83-91. [16] Sinclair C D, Spurr B D.Approximations to the distribution function of the anderson:darling test statistic[J].Journal of the American Statistical Association,1988,83(404):1190-1191. Click to display the text [17] Conover W J. Practical nonparametric statistics[M].New York:Wiley,1999:63-70. [18] Zhang J. Powerful goodness-of-fit tests based on the likelihood ratio[J].Journal of the Royal Statistical Society,Series B(Statistical Methodology),2002,64(2):281-294. Click to display the text
#### 文章信息
WANG Huiwen, WANG Shengshuai, HUANG Lele, WANG Cheng
Interval data analysis based on empirical distribution function
Journal of Beijing University of Aeronautics and Astronsutics, 2015, 41(2): 193-197.
http://dx.doi.org/10.13700/j.bh.1001-5965.2014.0435
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# How to calculate the probabilities of observing quantum states using the "expectation_from_wavefunction" in cirq (and why)
I am currently simulating some quantum circuits, and want to calculate the probabilities of observing each individual state. I am able to use Cirq for this, and calculate it using $$P_{00} = |\alpha|^2$$. Code:
import cirq
import sympy
x0, x1 = sympy.symbols('x0 x1')
q = cirq.GridQubit.rect(1, 2)
circuit = cirq.Circuit(
cirq.rx(x0).on(q[0]), cirq.rx(x1).on(q[1]),
cirq.ry(3.14/4).on(q[0]), cirq.ry(3.14/4).on(q[1]))
resolver = cirq.ParamResolver({x0: 0.2, x1: 0.3})
simulator = cirq.Simulator()
results = simulator.simulate(program=circuit, param_resolver=resolver,
qubit_order=q).final_state
print("Internal quantum state:", results)
print("Probabilities of observing each state:", [abs(x)**2 for x in results])
Output:
internal quantum state: [0.8377083+0.08743566j 0.3529709-0.11249491j 0.35297093-0.06251574j 0.13120411-0.08743566j]
probabilities of observing each state: [0.7094002059173512, 0.13724355108492148, 0.12849669756020887, 0.024859516013751914]
However, in multiple tutorials (for instance from TFQ) I see the use of "expectation_from_wavefunction":
z0 = cirq.Z(q[0])
qubit_map={q[0]: 1, q[1]: 1}
z0.expectation_from_wavefunction(results, qubit_map).real
output:
0.6757938265800476
My question: How can I use expectation_from_wavefunction to obtain the probabilities of observing the individual states ($$P_{00}, P_{01}, P_{10}, P_{11}$$)?
Bonus question: why would I favor this approach?
• Why would want to use "expectation_from_wavefunction()" to calculate the probabilities when you can just do what you did; square the values of your wavefunction? The former takes more work than the latter. Mar 22, 2020 at 14:24
• @VictoryOmole Because in the tutorial of TFQ that I refer to, they only use the "expectation_from_wavefunction". All following examples build upon it. I think it helps with batching, but I am not sure. Mar 22, 2020 at 21:06
• sorry, forgot to mention: the top approach works in Cirq. I am trying to run Tensorflow Quantum, and all tutorials here rely on this method. Mar 22, 2020 at 21:23
• Tensorflow Quantum combines Tensorflow with Cirq. If you can "import tfq" you can "import cirq" and thus use all the functionality in Cirq. Mar 23, 2020 at 16:14
• expectation_from_wavefunction is used when you don't want to write the logic for yourself. This is more useful in cases with multi-qubit observables involving the X and Y axies. Apr 21, 2020 at 22:56
why would I favor this approach?
The Expectation value is defined as $$\langle A \rangle= \langle\psi|A|\psi\rangle$$ where $$\psi$$ is the wavefunction and $$A$$ is the operator. Use "expectation_from_wavefunction()" if you don't want to write code that calculates $$\langle\psi|A|\psi\rangle$$.
• Thanks. I see now that the bottom approach calculates the expectation value, which is defined as ⟨𝐴⟩=⟨𝜓|𝐴|𝜓⟩. But how does this relate to the answer from the top approach? Because I see no relation between the numeric outcomes. For a 1-qubit system, would the expectation value not give me the probability of observing the qubit in the state 1? Mar 22, 2020 at 21:49
• If you don't see the relation, what calculation did you perform? What do you get? Mar 23, 2020 at 17:00
• The code and its output should be shown in my question. I embed using Rx followed by an Ry. The top approach provides: internal quantum state: [0.8377083+0.08743566j 0.3529709-0.11249491j 0.35297093-0.06251574j 0.13120411-0.08743566j] probabilities of observing each state: [0.7094002059173512, 0.13724355108492148, 0.12849669756020887, 0.024859516013751914] The bottom approach (that seems to have an extra Rz, which I have also experimented with in the top approach to no affect) gives as output for the real component 0.6757938265800476. Mar 23, 2020 at 21:20
• I meant: did you take the wavefunction and do a pen-and-paper calculation of that expectation value and get the same answer that z0.expectation_from_wavefunction(results, qubit_map).real gives you? Mar 23, 2020 at 23:17
The expectation value that the system provided me with was 0.6757938265800476. This was in a range of [-1,1]. Mapping this to a range of [0,1]: (0.6757938265800476+1)/2=0.837896913290024. The expectation value was the expectation value for qubit 0. This implies that it observed qubit 0 in state 0 83.7896913290024% of the time, and 1-0.837896913290024 in state 1.
The system also printed the following probabilities:
$$P_{00} = 0.7094002059173512$$
$$P_{01} = 0.13724355108492148$$
$$P_{10} = 0.12849669756020887$$
$$P_{11} = 0.024859516013751914$$
The probability to observe qubit 0 (indexing right to left) in state 0:
$$E_{q0} = 0.8378969 = 0.7094002 + 0.1284966 = P_{00} + P_{10}$$
To answer the question "How can I use expectation_from_wavefunction to obtain the probabilities of observing the individual states": you can't. You can only relate this particular expectation value (with Pauli Z) to a set of state probabilities, but not to individual ones without extra information (such as observables in different computational basis)
"Why would I favor this approach": access to the internal quantum state is not realistic with quantum system, and would require more computation.
P.S. this is my current interpretation which I expect to be right on a high level. I think I still have details wrong, such as parts of the explanation. Very open to opinions.
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## What is the solution to the system of equations shown below. 1.(-3,1) 2.(-1,3) 3.(1,-3) 4.(-3,1)
Question
What is the solution to the system of equations shown below.
1.(-3,1)
2.(-1,3)
3.(1,-3)
4.(-3,1)
in progress 0
6 months 2021-07-28T03:25:52+00:00 1 Answers 6 views 0
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Anonymous
Not applicable
## 12 Month Rolling Average (Fiscal Calendar) Calculation
Hi,
I'm trying to recreate the following formula from Excel in DAX:
The 12 month rolling avergae formula is:
=SUM(E81:E133)/SUM(D81:D133)*100000
=SUM(E82:E134)/SUM(D82:D134)*100000
=SUM(E83:E135)/SUM(D83:D135)*100000
Column E is Total X
Column D is Total Y
So it's basically summing the last 53 rows of the 'Total X' column divided by last 53 rows of 'Total Y' column * 100000.
I'm using the company calendar which is fiscal periods so cannot use any of the inbuilt date/time intelligence functions or the quick measures. My calendar looks like this:
Please could someone explain how I can create the same formula in dax...It's driving me mad!
3 REPLIES 3
Anonymous
Not applicable
@amitchandak thank you for this, my figure is still not quite matching up with the excel spreadsheet. I think the fact the column is called '12 month AFR rolling average' might be a bit confusing because the calculation is actually using the last 53 weeks.
So for example the 12 month rolling average for W/C 17/01/2021 is 16.3 which is calculated by adding the data starting from W/C 19/01/2020 to W/C 17/01/2021 (53 weeks)
When using your first measure the figure I get is 17.0 for W/C 17/01/2021 and the second measure gives 17.4
Thank you for your help with this!
Super User
@Anonymous , you do have week option in period, check or you can use 364 days etc
or you can have column like
new columns
Week Start date = 'Date'[Date]+-1*WEEKDAY('Date'[Date],2)+1
Week End date = 'Date'[Date]+ 7-1*WEEKDAY('Date'[Date],2)
Week Rank = RANKX(all('Date'),'Date'[Week Start date],,ASC,Dense)
OR
Week Rank = RANKX(all('Date'),'Date'[Year Week],,ASC,Dense) //YYYYWW format
use WEEKDAY('Date'[Date],1) for sunday week
Last 53 weeks = CALCULATE(sum('Table'[Qty]), FILTER(ALL('Date'),'Date'[Week Rank]>=max('Date'[Week Rank])-53 && 'Date'[Week Rank]<=max('Date'[Week Rank])))
Super User
@Anonymous , Assuming month are not standard, Create a Rank column in date table on period start or yyyypp
Period Rank = RANKX(all('Period'),'Period'[year period],,ASC,Dense)
Example rolling measure
rolling = CALCULATE(sum('order'[Qty]),filter(ALL('Date'),'Date'[Period Rank]>=max('Date'[Period Rank])-12 && 'Date'[Period Rank]<=max('Date'[Period Rank])) )
or
rolling = CALCULATE(Averagex(Values('Date'[Year period]) ,calculate(sum('order'[Qty]))),filter(ALL('Date'),'Date'[Period Rank]>=max('Date'[Period Rank])-12 && 'Date'[Period Rank]<=max('Date'[Period Rank])) )
This Period = CALCULATE(sum('order'[Qty]), FILTER(ALL('Date'),'Date'[Period Rank]=max('Date'[Period Rank])))
Last Period = CALCULATE(sum('order'[Qty]), FILTER(ALL('Date'),'Date'[Period Rank]=max('Date'[Period Rank])-1))
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Improvements on a Numerical Model of Borehole Heat Exchangers
KTH, School of Industrial Engineering and Management (ITM), Energy Technology, Applied Thermodynamics and Refrigeration.
EnergyLab, Spain.
KTH, School of Industrial Engineering and Management (ITM), Energy Technology, Applied Thermodynamics and Refrigeration.ORCID iD: 0000-0002-5093-9070
2016 (English)In: Proceedins of the European Geothermal Congress 2016, 2016Conference paper, Poster (with or without abstract) (Other academic)
##### Abstract [en]
In the mathematical simulation of bore fields for ground coupled heat pump (GCHP) systems, it has become a common practice to consider the boreholes as having a uniform temperature. As a rule the boreholes are hydraulically connected in parallel and the small temperature difference between incoming and outgoing heat carrier fluid justifies the assumption that all boreholes have the same uniform temperature in operation. Two simultaneous boundary conditions usually apply: All borehole walls should have a uniform temperature and the heat flow from the bore field should equal the energy needed by the heat pump. This paper describes improvements applied to a previous numerical approach that employs the concept of a highly conductive material (HCM) embedded in the boreholes and connected to a HCM bar above the ground surface to impose a uniform temperature boundary condition at the borehole wall. The original boundary condition with the uniform fluid temperature comes in conflict with the concept of the uniform borehole wall temperature. Between the fluid and the borehole wall there is a thermal borehole resistance. The heat flux increases at the borehole ends and thus also the temperature changes between borehole wall and the fluid. The borehole wall temperature deviates from the uniform assumption and will cause an error in the simulations. This paper presents a correction to that error. Firstly, the improvements to the HCM model are validated for g-function generation, which presents a good agreement with reference solutions. Secondly, the improvements to the HCM model are illustrated to predict fluid temperatures for measured variable daily loads of a monitored GCHP installation. The predicted fluid temperatures are compared with monitored data for about four years. The predicted fluid temperatures deviate from the measured data by less than 1 K during the last monitored year.
2016.
##### Keywords [en]
Monitored installation, borehole heat exchanger, fluid temperature prediction
##### National Category
Energy Engineering
##### Identifiers
OAI: oai:DiVA.org:kth-192897DiVA, id: diva2:974357
##### Conference
European Geothermal Congress
##### Note
QC 20160928
Available from: 2016-09-26 Created: 2016-09-22 Last updated: 2016-09-28Bibliographically approved
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I taught second grade for 7 years in a small, rural school. After a short break to raise my children, I was offered a kindergarten position and am still there many years later and loving every minute!I can honestly say that I still jump out of bed each school morning eager to begin the day with my little ones!
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해당 항목을 미리보기
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# The complete idiot's guide to statistics
저자: Robert A Donnelly, Jr. New York, NY : Alpha, ©2007. Print book : 영어 : 2nd ed모든 판과 형식 보기 WorldCat From the Publisher: Not a numbers person? No problem! This new edition is aimed at high school and college students who need to take statistics to fulfill a degree requirement and follows a standard statistics curriculum. Readers will find information on frequency distributions; mean, median, and mode; range, variance, and standard deviation; probability; and more. Emphasizes Microsoft Excel for number-crunching and computations. 더 읽기… (아무런 평가가 없습니다.) 0 리뷰와 함께 - 첫번째로 올려주세요.
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## 상세정보
장르/형태: Statistics 책 Robert A Donnelly, Jr. 다음에 대한 추가 정보 찾기: Robert A Donnelly, Jr. 9781592576340 1592576346 123509784 Includes index. xx, 395 pages : illustrations ; 24 cm Part 1: Basics -- 1: Let's get started -- 2: Data, data everywhere and not a drop to drink -- 3: Displaying descriptive statistics -- 4: Calculating descriptive statistics: measures of central tendency (mean, median, and mode) -- 5: Calculating descriptive statistics: measures of dispersion -- Part 2: Probability Topics -- 6: Introduction to probability -- 7: More probability stuff -- 8: Counting principles and probability distributions -- 9: Binomial probability distribution -- 10: Poisson probability distribution -- 11: Normal probability distribution -- Part 3: Inferential Statistics -- 12: Sampling -- 13: Sampling distributions -- 14: Confidence intervals -- 15: Introduction to hypothesis testing -- 16: Hypothesis testing with one sample -- 17: Hypothesis testing with two samples -- Part 4: Advanced Inferential Statistics -- 18: Chi-square probability distribution -- 19: Analysis of variance -- 20: Correlation and simple regression -- Appendixes -- A: Solutions to "your turn" -- B: Statistical tables -- C: Glossary -- Index. Complete idiot's guideStatistics by Robert A. Donnelly.
### 초록:
From the Publisher: Not a numbers person? No problem! This new edition is aimed at high school and college students who need to take statistics to fulfill a degree requirement and follows a standard statistics curriculum. Readers will find information on frequency distributions; mean, median, and mode; range, variance, and standard deviation; probability; and more. Emphasizes Microsoft Excel for number-crunching and computations.
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Analyze the function $f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$
The problem
The task is to find the intervals for which the function $f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$ is constant and for which it takes the form of:
$f \left( x \right) =a \cdot \arcsin \left( x \right) +b, \\a,b=const, a \neq 0$
Constants a,b are also to be calculated.
What I have done so far
I've defined the functions domain, calculated its derivative and the derivatives domain. $D_{f} x \in \left\langle -1,1\right\rangle\\ f' \left( x \right) =\frac{\sqrt{x^{2}}+x}{\sqrt{x^{2}-x^{4}}}\\ D_{f'} x \in \left(-1,0\right) \cup \left(0,1\right)\\ f' \left( x \right) =0 \Leftrightarrow \sqrt{x^{2}}+x=0\\ \left|x\right|=-x \Leftrightarrow x \le 0$
Which indicates that our function is constant for x'es from the interval (-1,0). That's where I currently am. I don't know how to go about the second part of the task, that is finding the intervals for which the function behaves as: $f \left( x \right) =a \cdot \arcsin \left( x \right) +b$ and calculating a,b.
Any help, greatly appreciated.
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As $f$ is defined on $[-1,\,1]$, you may, for any $x\in[-1,\,1]$ take a unique $\alpha\in[-\pi/2,\,\pi/2]$ such that $\sin\alpha=x$. This number is by definition called $\alpha=\arcsin x$. Now you know that $\sin^2\alpha+\cos^2\alpha=1$ and $\cos\alpha\geq0$. This gives $\cos^2\alpha=1-\sin^2\alpha=1-x^2$ and therefore $\cos\alpha=\sqrt{1-x^2}$. Since $\sin(\pi/2-\alpha)=\cos\alpha$, you can conclude that $$\arcsin(\cos\alpha)=\pi/2-\alpha \quad\text{if}\quad\pi/2-\alpha\in[-\pi/2,\,\pi/2],$$ but if $\pi/2-\alpha\notin[-\pi/2,\,\pi/2]$ this means that $\alpha<0$, and thus that $\alpha\in[-\pi/2,\,0)$ you should then use $\sin(\pi/2+\alpha)=\cos\alpha$ and get $$\arcsin(\cos\alpha)=\pi/2+\alpha \quad\text{if}\quad\alpha\in[-\pi/2,\,0].$$ Altogether, you have, for $x\geq0$, $f(x)=\alpha-\pi/2+\alpha=2\alpha-\pi/2$ and for $x\leq0$ $f(x)=\alpha-\pi/2-\alpha=-\pi/2$.
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I really like your approach. One question though, why do we "know" that $\cos\alpha\ \geq0$? – Tom Feb 27 '14 at 14:20
We have $\cos\alpha\geq0$ because $\alpha\in[-\pi/2,\,\pi/2]$. – Tom-Tom Feb 27 '14 at 14:32
I do not know what you are allowed to use. From Abramowitz/Stegun 4.3.45 and 4.4.2 (or http://dlmf.nist.gov/4.16#T3 and http://dlmf.nist.gov/4.23.E11) we have $$\arcsin(\sqrt{1-x^2}) = \arccos x, \quad 0 \le x \le 1$$ $$\arccos x = \frac{\pi}{2} - \arcsin x$$ Combining these results you get for $0 \le x \le 1$ $$\arcsin x-\arcsin(\sqrt{1-x^2}) = \arcsin x -\left(\frac{\pi}{2} - \arcsin x\right) = 2\arcsin x -\frac{\pi}{2}$$ and therefore $a=2$ and $b=-\frac{\pi}{2}.$
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the second equality found in Abramowitz is valid only for positive $x$. For negative $x$, it is $\arccos x=\pi/2+\arcsin x$. – Tom-Tom Feb 27 '14 at 12:30
@V. Rossetto: No, the formula I gave is correct. Look for example at $x=-\frac{1}{2}:$ $$\arccos(-\frac{1}{2}) = \frac{2\pi}{3},\quad \arcsin(-\frac{1}{2}) = -\frac{\pi}{6}.$$ Your formula gives the wrong result $$\frac{\pi}{2}+\arcsin(-\frac{1}{2})=\frac{\pi}{2}-\frac{\pi}{6} = \frac{\pi}{3},$$ while the other formula is correct $$\frac{\pi}{2}-\arcsin(-\frac{1}{2})= \frac{\pi}{2}+\frac{\pi}{6}= \frac{2\pi}{3}$$ – gammatester Feb 27 '14 at 12:48
You are right, I mixed up the propagation of the minus sign. The correct derivation for $x<0$ is $\arcsin\sqrt{1-x^2}=\arcsin\sqrt{1-(-x)^2}=\arccos(-x)=\pi-\arccos x=\pi/2+\arcsin x$. – Tom-Tom Feb 27 '14 at 14:45
Hint
Let us consider $$f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$$ and $$g \left( x \right) =a \cdot \arcsin \left( x \right) +b$$ In order they be identical (and this can only happen in the range [$0<x<1$]), a minimum requirement is that there will be two points where their values should be equal. Let us take the two limiting points $x=0$ and $x=1$. So, $f(0)=-\frac{\pi }{2}$ and $g(0)=b$, $f(1)=\frac{\pi }{2}$ and $g(1)=a \frac{\pi }{2}+b$; you then have two very simple linear equations to solve for $a$ and $b$.
I am sure that you can take from here.
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In your answer you implicitely use that one of the intervals is $[0,\,1]$. – Tom-Tom Feb 27 '14 at 11:09
@V.Rossetto. Yes, indeed. I do not see how $f$ and $g$ could be identical in [$-1,0$] since $f$ is constant over that range and that the OP precised that $a \neq 0$. Please tell me if I am wrong. Thanks. – Claude Leibovici Feb 27 '14 at 11:14
In his question, the OP asked for a constant solution or a solution of the form you use. – Tom-Tom Feb 27 '14 at 12:28
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<<
Introductory Course No. 100 351 Fall 2005 Second Part: Michael Stoll
Contents
1. What Is Algebraic Geometry? 2 2. Affine Spaces and Algebraic Sets 3 3. Projective Spaces and Algebraic Sets 6 4. Projective Closure and Affine Patches 9 5. and Rational Maps 11 6. — Local Properties 14 7. B´ezout’sTheorem 18 2
1. What Is Algebraic Geometry?
Linear can be seen (in parts at least) as the study of systems of linear . In geometric terms, this can be interpreted as the study of linear (or affine) subspaces of Cn (say). Algebraic Geometry generalizes this in a natural way be looking at systems of equations. Their geometric realizations (their solution sets in Cn, say) are called algebraic varieties.
Many questions one can study in various parts of lead in a natural way to (systems of) polynomial equations, to which the methods of Algebraic Geometry can be applied. Algebraic Geometry provides a translation between algebra (solutions of equations) and geometry (points on algebraic varieties). The methods are mostly algebraic, but the geometry provides the intuition.
Compared to Differential Geometry, in Algebraic Geometry we consider a rather restricted class of “” — those given by polynomial equations (we can allow “singularities”, however). For example, y = cos x defines a perfectly nice differentiable in the , but not an . In return, we can get stronger results, for example a criterion for the existence of solutions (in the complex numbers), or statements on the number of solutions (for example when intersecting two curves), or classification results. In some cases, there are close links between both worlds. For example, a compact Riemann (i.e., a one-dimensional complex ) is “the same” as a (smooth projective) algebraic curve over C. As we do not have much time in this course, we will mostly look at the simplest nontrivial (but already very interesting case), which is to consider one in two variables. Such an equation describes a plane algebraic curve.
1.1. Examples. We will use x and y as the variables. The simplest examples are provided by the equations y = 0 and x = 0; they describe the x-axis and y-axis, respectively. More generally, a is given by an equation ax + by = c with a and b not both zero. The equation x2 + y2 = 1 describes the unit . Note that the of its real points (x, y) ∈ R2 is compact, but its set of complex points is not — there are two “branches” extending to infinity, with x/y tending to i and to −i respectively. It turns out that we can compactify the set of complex points by throwing in two additional points “at infinity” corresponding to these two directions.
More formally, we introduce the P2 as the set of points (x : y : z) with (x, y, z) ∈ C3 \{0}, where we identify (x : y : z) and (λx : λy : λz) for λ ∈ C \{0}. We find the usual affine plane A2 = C2 within P2 as the subset of points (x : y : 1); the points (x : y : 0) form the “line at infinity”, and there is one for each direction in the affine plane. The unit circle acquires the two new points (1 : i : 0) and (1 : −i : 0). A projective is now given by a in the three variables x, y, z. To obtain it from the original affine equation, replace x and y by x/z and y/z, respectively and multiply by a suitable power of z to cancel the 3 denominators. For the unit circle we obtain x2 + y2 = z2; a general line in P2 is given by ax+by +cz = 0 with a, b, c not all zero. (The line at infinity has equation z = 0, for example.) One of the great advantages of P2 over A2 is that in P2 any pair of distinct lines has exactly one common point — there is no need to separate the case of lines; every pair of lines stands on the same footing. The fact that two lines always intersect in exactly one point has a far-reaching generalization, known as B´ezout’sTheorem. It says that two projective plane curvers of degrees m and n intersect in exactly mn points (counting multiplicities correctly).
The first question towards a classification of algebraic curves one could ask is to order them in some way according to their complexity. Roughly, one would expect that the curve is more complicated when the degree of its defining polynomial is large. However, this is not true in general, for example, a curve y = f(x) can be transformed to the line y = 0 by a simple substitution, no matter how large the degree of f is. But it is certainly true that a curve given by an equation of low degree cannot be very complicated. It turns out that there is a unique discrete of an algebraic curve: its g. The genus is a nonnegative , and for a plane curve of degree d, we have g ≤ (d − 1)(d − 2)/2. So lines (d = 1) and conic sections (d = 2) are of genus zero, whereas a general cubic curve (d = 3) will have genus one. Some cubic curves will have genus zero, however; it turns out that these are the curves having a point, where the curve is not smooth (not a manifold in the Differential Geometry sense). In general, there is a formula relating the degree d of a projective plane curve, its genus g and contributions δP associated to its singular points P : (d − 1)(d − 2) X g = − δ . 2 P P
1.2. Example. [Iteration z 7→ z2 + c; to be added]
2. Affine Spaces and Algebraic Sets
In the following, we will do everything over the field C of complex numbers. The reason for this choice is that C is algebraically closed, i.e., it satisfies the “Fundamental Theorem of Algebra”:
2.1. Theorem. Let f ∈ C[x] be a non-constant polynomial. Then f has a root in C. By induction, it follows that every non-constant polynomial f ∈ C[x] splits into linear factors: n Y f(x) = c (x − αj) j=1 × where n = deg f is the degree, c ∈ C and the αj ∈ C. Essentially everything we do would work as well over any other algebraically closed field (of zero). 4
The first thing we have to do is to provide the stage for our objects. They will be the solution sets of systems of polynomial equations, so we need the of points that are potential solutions.
2.2. Definition. Let n ≥ 0. Affine n-space, An, is the set Cn of all n-tuples of complex numbers. Note that A0 is just one point (the empty tuple). A1 is also called the affine line, A2 the affine plane.
2.3. Definition.
(1) Let S ⊂ C[x1, . . . , xn] be a subset. The (affine) algebraic set defined by S is n V (S) = {(ξ1, . . . , ξn) ∈ A : f(ξ1, . . . , ξn) = 0 for all f ∈ S} . If I = hSi is the generated by S, then V (S) = V (I). Note that n V (∅) = V (0) = A and V ({1}) = V (C[x1, . . . , xn]) = ∅. An non-empty algebraic set is called an if it is not the union of two proper algebraic subsets. (2) Let V ⊂ An be a subset. The ideal of V is the set
I(V ) = {f ∈ C[x1, . . . , xn]: f(ξ1, . . . , ξn) = 0 for all (ξ1, . . . , ξn) ∈ V } .
It is clear that I(V ) is indeed an ideal of C[x1, . . . , xn].
2.4. Remark. Note that the finite union and arbitrary of algebraic sets is again an algebraic set — we have \ [ V (Sj) = V Sj j∈J j∈J
V (S1) ∪ V (S2) = V (S1S2) where S1S2 = {fg : f ∈ S1, g ∈ S2}. Since the full An and the empty set are also algebraic sets, one can define a on An in which the algebraic sets are exactly the closed sets. This is called the . Since algebraic sets are closed in the usual topology (the solution set of f = 0 is closed as a polynomial f defines a continuous ), this new topology is coarser than the usual toplogy.
2.5. Remark. We obviously have
S1 ⊂ S2 =⇒ V (S1) ⊃ V (S2) and V1 ⊂ V2 =⇒ I(V1) ⊃ I(V2) . By definition, we have S ⊂ I(V (S)) and V ⊂ V (I(V )) . Together, these imply V (I(V (S))) = V (S) and I(V (I(V ))) = I(V ) . This means that we get an inclusion-reversing between algebraic sets and those ideals that are of the form I(V ). Hilbert’s Nullstellensatz tells us what these ideals are. Theorem. Let I be an ideal, V = V (I). If f ∈ I(V ), then f n ∈ I for some n ≥ 1. We can deduce that n I(V (I)) = rad(I) = {f ∈ C[x1, . . . , xn]: f ∈ I for some n ≥ 1} 5 is the radical of I. Note that rad(I) is an ideal (Exercise). Hence I = I(V (I)) if and only if I is a radical ideal, which means that I = rad(I); equivalently, f n ∈ I for some n ≥ 1 implies f ∈ I. Note that rad(rad(I)) = rad(I) (Exercise). So we see that I 7→ V (I), V 7→ I(V ) provide an inclusion-reversing bijection between algebraic sets and radical ideals of C[x1, . . . , xn]. Restricting this to alge- braic varieties, we obtain a bijection between algebraic varieties and prime ideals of C[x1, . . . , xn] (i.e., ideals I such that fg ∈ I implies f ∈ I or g ∈ I).
Note that C[x1, . . . , xn] is a noetherian ; therefore every ideal is finitely gener- ated. In particular, taking S0 to be a finite generating set of the ideal hSi, we see that V (S) = V (S0) — every algebraic set is defined by a finite set of equations.
2.6. Example. Let us consider the algebraic sets and varieties in the affine line A1. An algebraic set is given by an ideal of C[x]. Now C[x] is a principal ideal domain, hence every ideal I is generated by one element: I = hfi. If f = 0, then the algebraic set is all of A1. So we assume now f 6= 0. Then the ideal is radical if and only if f has no multiple roots, and the algebraic set defined by it is just the finite set of points corresponding to the roots of f; these are n points, where n is the degree of f. (This set is empty when n = 0, i.e., f is constant.) So the algebraic sets in A1 are exactly the finite subsets and the whole line. It is then clear that the algebraic varieties in A1 are the whole line and single points (and indeed, the prime ideals of C[x] are the zero ideal and the ideals generated by a linear polynomial x − α).
2.7. Example. Now consider the affine plane A2. The plane A2 itself is an al- gebraic set — A2 = V (∅). Any single point of A2 is an algebraic set (even an algebraic variety) — {(ξ, η)} = V (x − ξ, y − η). Therefore all finite subsets of A2 are algebraic sets. Is there something in between finite sets and the whole plane? Yes: we can consider something like V (x) or V (x2 + y2 − 1). We get an algebraic set that is intuitively “one-dimensional”. Here we look at ideals hF i generated by a single non-constant polynomial F ∈ C[x, y]. Such an ideal is radical iff F has no repeated factors in its prime factorization (recall that C[x, y] is a unique factorization domain), and it is prime iff F is irreducible. We call V (F ) an affine plane algebraic curve; the curve is called irreducible if F is irreducible. Simple examples of affine plane algebraic curves are the lines V (ax + by − c) (with (a, b) 6= (0, 0)) or the “unit circle” V (x2 + y2 − 1), which is a special case of a or — a curve V (F ), where F has (total) degree 2. Note that the “real picture” in R2 of the unit circle is misleading: it does not show the two branches tending to infiniy with of i and −i!
One can show that a general proper algebraic subset of A2 is a finite union of irreducible curves and points.
Finally, we need to introduce two more notions that deal with the functions we want to consider on our algebraic sets. As this is algebra, the only functions we have at our disposal are and quotients of polynomials. If V is an algebraic set, I = I(V ) its radical ideal, then two polynomial functions will agree on V if and only if their difference is in I. This prompts the following definitions. 6
2.8. Definition. Let V ⊂ An be an algebraic set with ideal I = I(V ). The C[V ] := C[x1, . . . , xn]/I is called the affine coordinate ring of V . If V is an algebraic variety (hence I is prime, hence C[V ] is an domain), then the field of fractions C(V ) := Frac(C[V ]) of the affine coordinate ring is called the function field of V . The affine coordinate ring and function field are closely analogous with the ring of holomorphic functions and the field of meromorphic functions on a .
2.9. Definition. Let V ⊂ An be an algebraic set. The elements of the coordinate ring C[V ] are called regular functions on V . If f ∈ C[V ] is a regular function and P ∈ V is a point on V , then f(P ) ∈ C makes sense: take a representative F ∈ C[x1, . . . , xn] of f, then f(P ) := F (P ) is well-defined — if F and G both represent f, then their difference is in the ideal of V , hence vanishes on P . Let V ⊂ An be an algebraic variety. The elements of the function field C(V ) are called rational functions on V . If f ∈ C(V ) is a and P ∈ V is a point on V , then f is regular at P if f can be written f = g/h with g, h ∈ C[V ] such that h(P ) 6= 0. In this case, we can define f(P ) = g(P )/h(P ) ∈ C. The regular functions on V are exactly the rational functions that are regular at all points of V .
3. Projective Spaces and Algebraic Sets
The affine space An has certain shortcomings. For example, its point set Cn is not compact (in the usual topology), and the same is true for any algebraic set that does not just consist of finitely many points. Or, looking at the affine plane, two lines in A2 may intersect in one point or not intersect at all. In order to get nicer objects and a nicer theory, we introduce a larger space. The price we have to pay is that the definition is more involved.
3.1. Definition. Let n ≥ 0. Projective n-space, Pn, is the quotient (Cn+1\{0})/ ∼ of the set of non-zero points in Cn+1 modulo the × (ξ0, . . . , ξn) ∼ (η0, . . . , ηn) ⇐⇒ ∃λ ∈ C : η0 = λξ0, . . . , ηn = λξn .
We write (ξ0 : ... : ξn) for the point represented by a tuple (ξ0, . . . , ξn). Note that P0 is again just one point. Again, P1 is called the , and P2 is called the projective plane.
n n n We can find A inside P in a number of ways. Let Uj be the subset of P of points (ξ0 : ... : ξj : ... : ξn) such that ξj 6= 0. Then there is a bijection between n A and Uj given by ιj :(ξ1, . . . , ξn) 7→ (ξ1 : ... : ξj−1 : 1 : ξj : ... : ξn) and ξ0 ξj−1 ξj+1 ξn (ξ0 : ... : ξn) 7→ ( ,..., , ,..., ). ξj ξj ξj ξj n−1 The complement of Uj is in a natural way a P (dropping the zero coordinate ξj), so we can write Pn = An ∪ Pn−1. In particular, the projective line P1 is A1 with one point “at infinity” added, and the projective plane P2 is A2 with a (projective) line “at infinity” added. If we identify t ∈ A1 with the point (t : 1) ∈ P1, so that (t : u) ∈ P1 corresponds t to u (when u 6= 0), then approaching the “point at infinity” corresponds to letting the denominator tend to zero, keeping the numerator fixed (at 1, say). If we look u at another “chart”, that given by the coordinate t , then in this chart, we approach 7 zero. In this way, we can consider the projective line as being “glued together” from two affine lines with coordinates t and u, identified on the complements of the origins according to tu = 1. In terms of complex points, this is exactly the construction of the Riemann Cˆ = C ∪ {∞}. This shows that P1 is compact (in its complex topology). More generally, Pn is compact (with the quotient topology induced by the quotient map Cn+1\{0} → Pn, or equivalently, with the topology coming from the affine “charts” Uj).
3.2. Definition. A polynomial in C[x0, . . . , xn] is called homogeneous (of de- gree d) if all its non-vanishing terms have the same total degree d. We will write n o X k0 kn C[x0, . . . , xn]d = ak0,...,kn x0 ··· xn : ak0,...,kn ∈ C k0+···+kn=d for the (C-vector) space of homogeneous polynomials of degree d. (The zero poly- nomial is considered to be homogeneous of any degree d.) Note that M C[x0, . . . , xn] = C[x0, . . . , xn]d g≥0 as C-vector spaces: every polynomial is a (finite) sum of homogeneous ones; we write f = f0 + f1 + ··· + fn if f has (total) degree ≤ n, where fd is homogeneous of degree d. Regarding the multiplicative structure, we have that the product of two homogeneous polynomi- als of degrees d and d0, respectively, is homogeneous of degree d + d0. (In algebraic terms, C[x0, . . . , xn] is a . A graded ring is a ring R whose additive L 0 is a direct sum R = d≥0 Rd such that Rd · Rd0 ⊂ Rd+d0 for all d, d ≥ 0.)
3.3. Definition.
(1) Let S ⊂ C[x0, . . . , xn] be a set of homogeneous polynomials. The projective algebraic set defined by S is n V (S) = {(ξ0 : ... : ξn) ∈ P : F (ξ0, . . . , ξn) = 0 for all F ∈ S} . V (S) 6= ∅ is a projective algebraic variety if it is not the union of two proper projective algebraic subsets. (2) Let V ⊂ Pn be a subset. The (homogeneous) ideal of V is M I(V ) = {F ∈ C[x0, . . . , xn]d : F (ξ0, . . . , ξn) = 0 for all (ξ0 : ... : ξn) ∈ V } . d≥0
n 3.4. Remarks. If F is homogeneous, then for (ξ0 : ... : ξn) ∈ P it makes sense to ask whether F (ξ0, . . . , ξn) = 0, as this does not depend on the representative d — F (λξ0, . . . , λξn) = λ F (ξ0, . . . , ξn) if F is homogeneous of degree d. An ideal is called homogeneous, if it is generated by homogeneous polynomials. I(V ) is the homogeneous ideal generated by all the homogeneous polynomials vanishing on all points of V . We get again an inclusion-reversing bijection between projective algebraic sets and homogeneous radical ideals contained in the I0 = hx0, . . . , xni (note that V (I0) = V ({1}) = ∅), and between projective algebraic varieties and homogeneous prime ideals ( I0. 8
Note that a projective algebraic set is compact (in the usual topology), since it is a closed subset of the compact set Pn. In fact, a famous theorem due to Chow states: Every connected compact complex manifold is complex-analytically isomorphic to a projective algebraic variety. Even more is true: via this , the field of meromorphic functions is identified with the function field of the algebraic variety (see below).
3.5. Example. Consider the (projective) algebraic subsets of P1. Any nonzero homogeneous polynomial F (x, y) in two variables is a product of linear factors βx − αy (with α and β not both zero). Such a linear polynomial has the single point (α : β) as its algebraic set. The algebraic set defined by a general (squarefree) homogeneous polynomial is therefore again a finite set of points, and the same is true for a homogeneous ideal, since its algebraic set is the intersection of the algebraic sets of its generators. (In fact, the homogeneous ideals of C[x, y] are all generated by one element.)
3.6. Example. In the projective plane P2, we again have finite sets of points as algebraic sets (which are varieties when they consist of just one point). The whole plane P2 is a projective algebraic variety. As before, there are also “one- dimensional” algebraic sets and varieties; they are again defined by single equa- tions, which are now given by (non-constant) homogeneous polynomials F (x, y, z) in three variables. They are (surprisingly) called projective plane algebraic curves.
As before, we can define coordinate rings and function fields.
3.7. Definition. Let V ⊂ Pn be an algebraic set with homogeneous ideal I = I(V ). The quotient ring C[V ] := C[x0, x1, . . . , xn]/I is called the homogeneous coordinate ring of V . If V is a projective algebraic variety, then the function field of V is defined as
nf f and g both have representatives o C(V ) := : f, g ∈ C[V ], g 6= 0, . g in C[x0, . . . , xn]d for some d ≥ 0
It is something like the “degree zero part” of the field of fractions of C[V ].
3.8. Definition. Let V ⊂ Pn be a projective algebraic variety. The elements of the function field C(V ) are called rational functions on V . If f ∈ C(V ) is a rational function and P ∈ V is a point on V , then f is regular at P if f can be written f = g/h with g, h ∈ C[V ] such that h(P ) 6= 0. In this case, we can define f(P ) = g(P )/h(P ) ∈ C. Note that this is well-defined, since g and h are represented by homogeneous polynomials of the same degree d:
d g(λξ0, . . . , λξn) λ g(ξ0, . . . , ξn) f (λξ0 : ... : λξn) = = d = f (ξ0 : ... : ξn) h(λξ0, . . . , λξn) λ h(ξ0, . . . , ξn) 9
3.9. Example. The concept of a rational function on a projective algebraic vari- ety is at first sight a bit involved. An example will help to clarify it. Consider the projective version of the unit circle, given by the equation x2 + y2 − z2 = 0. Then f = (y−z)/x defines a rational function (since numerator y−z and denominator x have the same degree, and the denominator is not in the homogeneous ideal of the curve). Let us find out at which points of the curve f is regular. This is certainly the case for all points with x 6= 0. Let us look at the points where x vanishes. These are (0 : 1 : 1) and (0 : 1 : −1) (recall that projective coordinates are only determined up to scaling). At (0 : 1 : −1), the numerator does not vanish, which implies that f is not regular (Exercise!). At (0 : 1 : 1), the numerator and the denominator both vanish, so we have to find an alternative representation. Note that we have (in the function field) y − z (y − z)(y + z) y2 − z2 −x2 x = = = = − . x x(y + z) x(y + z) x(y + z) y + z In this last representation, the denominator does not vanish at (0 : 1 : 1), so f is regular there (and in fact takes the value zero).
There is an important result (which has some analogy to Liouville’s Theorem in , which can be formulated to say that any holomorphic function on a compact is constant).
3.10. Theorem. If V ⊂ Pn is a projective algebraic variety and f ∈ C(V ) is regular everywhere on V , then f is constant.
4. Projective Closure and Affine Patches
We now are faced with an obvious question: how do we go between affine and projective algebraic sets or varieties? There should be some correspondence related to the idea that going from affine to projective means to add some points in order to “close up” the algebraic set.
4.1. Definition. For a polynomial f ∈ C[x1, . . . , xn] of (total) degree d, we define ˜ d x1 xn f = x0 f ,..., ∈ C[x0, x1, . . . , xn]d . x0 x0 This operation corresponds to multiplying every term in f with a suitable power of x0 in order to make the total degree equal to d. This process is sometimes called homogenization.
4.2. Definition. Let V ⊂ An be an affine algebraic set, with ideal I = I(V ) ⊂ ˜ C[x1, . . . , xn]. The projective closure V of V (with respect to the ι0 : An → Pn) is the projective algebraic set given by the equations f˜ = 0 for f ∈ I. It can be shown that V˜ really is the topological closure (in both the usual and the Zariski on Pn) in Pn of V ⊂ An ⊂ Pn, thus justifying the name.
4.3. Definition. Let V ⊂ Pn be a projective algebraic set, given by equations f = S 0 for f ∈ S ⊂ d C[x0, x1, . . . , xn]d. Let 0 ≤ j ≤ n. The jth affine patch of V is the affine algebraic set Vj given by the equations f(x0, . . . , xj−1, 1, xj+1, . . . , xn) = 0 n for f ∈ S. (Here, we use C[x0, . . . , xj−1, xj+1, . . . , xn] as the coordinate ring of A .) The following is quite immediate. 10
n ˜ 4.4. Proposition. If V ⊂ A is an affine algebraic set, then (V )0 = V .
Proof. Exercise.
The converse needs more care.
4.5. Proposition. Let V ⊂ Pn be a projective algebraic variety such that V is n not contained in the “hyperplane at infinity”, i.e., the complement P \ U0. Then (gV0) = V .
Proof. Exercise.
Note that V0 = ∅ when V is contained in the hyperplane at infinity.
4.6. Examples. If we consider a line L in A2, given by the equation ax + by = c, say (with (a, b) 6= (0, 0)), then L˜ is given by ax + by − cz = 0 (writing z for the additional coordinate on P2). There is exactly one “point at infinity” in L˜ \ L; it has coordinates (b : −a : 0). This is the point common to all lines parallel to L.
Conversely, if we have a (projective) line Λ ⊂ P2, given by ax + by + cz = 0 (with 2 (a, b, c) 6= (0, 0, 0)), then Λ0 ⊂ A is given by ax + by = −c. If (a, b) 6= (0, 0), this is an affine line, otherwise it is the empty set (since then c 6= 0).
Now consider the “unit circle” C : x2 + y2 = 1 in the affine plane. Its projective closure is C˜ : x2 + y2 − z2 = 0. The zeroth affine patch of this is of course again C. ˜ ˜ 2 2 The first affine patch of C is (set x = 1) C1 : z = 1 + y . So in this sense, the circle is “the same” as a . In the projective closure, the unit circle acquires the two new points “at infinity” with coordinates (1 : i : 0) and (1 : −i : 0) — they come from the factorization of the leading term x2 + y2 into a product of linear forms. More generally, if we have any circle C in the affine plane, given by (x − a)2 + (y − b)2 = r2, then C˜ still has the two points (1 : i : 0) and (1 : −i : 0): they are common to all !
This explains, by the way, why two circles intersect at most in two points (in A2), even though one would generically expect four points of intersection (since we are intersecting two curves of degree 2) — two of the intersection points are out at infinity (and not defined over R in addition to that).
4.7. Proposition. Let V ⊂ An be an affine algebraic variety, V˜ its projective closure. Then the function fields C(V ) and C(V˜ ) are canonically isomorphic, and the affine coordinate ring C[V ] can be identified with the set of rational functions on V˜ (or on V ) that are regular everywhere on V .
Proof. The proof of the first statement is an exercise. The second statement follows from the fact that C[V ] is the subset of C(V ) consisting of functions that are regular on all of V . 11
5. Morphisms and Rational Maps
So far we have defined what our objects are (affine or porjective algebraic sets or varieties), and we have defined functions on them. But we also want to relate these objects with each other; in particular, we want to say when two such objects should be considered “the same”. So we need to define a suitable class of maps between the objects. In the affine case, this is quite straight-forward.
5.1. Definition. Let V ⊂ An and W ⊂ Am be two affine algebraic sets. A mor- phism V → W is a map φ : V → W that is given by polynomials: if C[x1, . . . , xn] n m is the coordinate ring of A and C[y1, . . . , ym] is the coordinate ring of A , then there are polynomials F1,...,Fm ∈ C[x1, . . . , xn] such that
φ(ξ1, . . . , ξn) = (F1(ξ1, . . . , ξn),...,Fm(ξ1, . . . , ξn)) .
In order to make sure that the is contained in W , it is necessary and sufficient that for every G ∈ I(W ), we have G(F1,...,Fm) ∈ I(V ).
Note that the Fj are only determined modulo I(V ); therefore they can also be considered as elements of the coordinate ring C[V ]. The condition above then amounts to saying that φ corresponds to a ring φ∗ : C[W ] → C[V ] that sends G to G(F1,...,Fm). (It is really a C-algebra homomorphism, as it has to preserve the constants from C.) Conversely, every such leads to a V → W : the polynomials Fj are obtained (modulo I(V )) as the images of y1, . . . , ym. As usual, a morphism is called an isomorphism if there is an inverse morphism; in this case V and W are called isomorphic, V ∼= W . Note that it is not sufficient to require φ to be a bijective map between the points of V and W . This is analogous to the situation in topology, where a bijective continuous map is not necessarily a . ∼ ∼ Note also that V = W if and only if C[V ] = C[W ] as C-. It is clear how to compose morphisms; the composition of two morphisms is again a morphism.
5.2. Example. Let us show that what we consider a line in A2 really is isomorphic to the affine line A1. Let the line L be given by y = ax + b (this excludes the case x = c, which can be dealt with in a similar way). Let C[t] be the coordinate ring of A1. We set up two morphims:
1 1 φ : A −→ L , t 7−→ (t, at + b); ψ : L → A , (x, y) 7−→ x Both maps are given by polynomials, as required, and the image of the first is contained in L — it satisfies the equation. (There is nothing to check in this respect for the second map, as there is no equation to be satisfied.) It is then ∼ 1 obvious that the two morphisms are inverses of each other, hence L = A . In terms of coordinate rings, we have an isomorphism between C[x, y]/hy − ax − bi and C[t], which comes essentially down to the fact that we can eliminate y. 12
5.3. Example. Now consider the cuspidal cubic curve C : y2 = x3 in A2. There is a morphism 1 2 3 φ : A −→ C , t 7−→ (t , t ) . φ is even bijective on points — if (x, y) ∈ C is not (0, 0), then the unique t that maps to it is y/x, and (0, 0) has the unique preimage 0 — but it is not an isomorphism. One way of checking this is to notice that the image of the corresponding C-algebra homomorphism φ∗ : C[x, y]/hy2 − x3i → C[t] does not contain t. In fact, there is no isomorphism between A1 and C, and this is a good thing, since the point (0, 0) on C is bad (it is “singular”; we will come to that), but there are no bad points on A1.
The projective case is a little bit more involved.
5.4. Definition. Let V ⊂ Pn and W ⊂ Pm be two projective algebraic sets. A morphism V → W is a map φ : V → W that is “locally given by homogeneous polynomials of the same degree”. What this means is that there are (m + 1)- (j) (j) (j) tuples (F0 ,...,Fm ) of homogeneous polynomials Fk ∈ C[x0, . . . , xn]dj of the same degree dj, for j in some index set J, such that
(j) (j) (1) for all homogeneous G ∈ I(W ) and all j ∈ J, G(F0 ,...,Fm ) ∈ I(V ); 0 0 (j) (j0) (j) (j0) (2) for all j, j ∈ J, 0 ≤ k < k ≤ m, we have Fk Fk0 − Fk0 Fk ∈ I(V ); (j) (3) for all P ∈ V , there is some j ∈ J such that not all of Fk (P ) vanish, 0 ≤ k ≤ n, and in this case
(j) (j) φ(P ) = (F0 (ξ0, . . . , ξn): ... : Fn (ξ0, . . . , ξn)) ,
if P = (ξ0 : ... : ξn). The first condition ensures that the image is contained in W . The second condition ensures that φ is well-defined at points where more than one tuple of polynomials would give a result. The third condition ensures that φ is defined everywhere. We can again compose morphisms between projective algebraic sets (by plugging in; note that we may have to use all possible combinations of defining polynomials for both morphisms); the composition is again a morphism.
5.5. Example. In a similar way as for the affine case, one can show that a line L : ax + by + cz = 0 in P2 is isomorphic to the projective line P1. Let us do something more interesting here: we show that P1 is isomorphic to the unit circle C : x2 + y2 − z2 = 0. The geometric intuition behind this is the idea that one can “rationally parametrize” the points on the unit circle by fixing one point P on it and considering all the lines through P . The lines are parametrized by their slope, and each line intersects C in a unique second point. In this way, there is a correspondence between and points on C. In formulas, this leads to the morphism 1 2 2 2 2 φ : P −→ C, (t : u) 7−→ (t − u : 2 tu : t + u ) . The polynomials on the right hand side are homogeneous of the same degree 2. It is easy to see that the image is contained in C. Also, it suffices to give just this one set of defining polynomials, since they never vanish all three at the same time (recall that t and u do not both vanish for a point in P1). 13
What about the inverse? In the affine picture (where z = 1), the point P is (−1, 0), and u/t is the slope of the line. So we should have u/t = y/(x + 1). Writing this in homogeneous terms leads to 1 ψ : C −→ P , (x : y : z) 7−→ (x + z : y) . But here we have a problem: at P = (−1 : 0 : 1), both polynomials x + z and y vanish, and the map is not defined. So we need to find an alternative representation that is defined at P . We can proceed as follows. (x + z : y) = ((x + z)(z − x): y(z − x)) = (z2 − x2 : y(z − x)) = (y2 : y(z − x)) = (y : z − x) This version is defined at P (and gives the image (0 : 1)), however, it is not defined at (1 : 0 : 1). So we see that we really need two different sets of defining polynomials in this case.
5.6. Remark. There are no interesting morphisms from projective to affine alge- braic sets. One can consider morphisms from affine to projective algebraic sets, however; they are defined in a similar way as above in the projective case, with the modification that the defining polynomials need not be homogeneous. For example, we can define a morphism from the affine unit circle C0 : x2 + y2 = 1 to the projective line: 0 0 1 ψ : C −→ P , (x, y) 7−→ (x + 1 : y) or (y : 1 − x) In this sense, the canonical inclusion of an affine algebraic set V into its projective closure V˜ is a morphism.
Sometimes one wants to be less demanding and does not require the maps to be defined everywhere, as long as they are defined on a sufficiently large subset. This leads to the notion of rational map.
5.7. Definition. Let V ⊂ Pn and W ⊂ Pm be two projective algebraic varieties. A rational map φ : V → W is defined in the same way as a morphism, with the exception that we do not require the last condition for all points P ∈ V , but just for at least one point. If P is a point that satisfies this condition, then we say that φ is defined at P . It is not hard to see that φ is defined on the complement of a proper algebraic subset of V . Now it is not true that we can always compose rational maps: the image of the first map may be outside the (maximal) domain of definition of the second one. Therefore, we single out a subclass of rational maps that are better behaved in this respect. Consider some affine patch W 0 of W such that W 0 meets the image of φ. Then we can pull back regular functions on W 0 to rational functions on V via φ: we get a C-algebra homomorphism C[W 0] → C(V ). We call φ dominant if this homomorphism is injective. (It can be shown that this does not depend on the affine patch chosen and that it is equivalent to saying that the image of φ is dense in the Zariski topology of W .) In this case, the homomorphism extends to a C-algebra homomorphism φ∗ : C(W ) = Frac(C[W 0]) → C(V ). Explicitly, if φ is given by homogeneous polynomials F0,...,Fm of the same degree and f = g/h is a quotient of homogeneous elements of C[W ] of the same degree, ∗ then φ (f) = g(F0,...,Fm)/h(F0,...,Fm). Conversely, every such homomorphism C(W ) → C(V ) comes from a dominant rational map. 14
We can compose dominant rational maps; the composition is again a dominant rational map. More generally, if φ : V → W is a dominant rational map and ψ : W → X is any rational map, then ψ ◦ φ is defined and a rational map. The (dominant) rational map φ is called birational or a birational isomorphism if there is an inverse (dominant) rational map. In view of the preceding discussion, this is equivalent to saying that the function fields of V and W are isomorphic (as C-algebras). Intuitively speaking, a birational isomorphism between two algebraic varieties is an isomorphims “modulo proper algebraic subsets.” Since the function fields of an affine algebraic variety V and of its projective closure V˜ can be identified, we can extend the notion of rational maps to affine algebraic varieties.
5.8. Example. Let us look at the cuspidal cubic curve again and show that it is birationally isomorphic to the line. This time, we are looking at the projective situation. Then the cuspidal cubic is given by C : x3 − y2z = 0 and the line is P1. We still have the morphism 1 2 3 3 φ : P −→ C, (t : u) 7−→ (t u : t : u ) , but now we claim that it is invertible as a rational map. The inverse is 1 ψ : C −→ P , (x : y : z) 7−→ (y : x) . This is not a morphism, since it is not defined at (0 : 0 : 1), but it is a rational map. It is easy to check that φ and ψ are inverses: (t : u) 7−→ (t2u : t3 : u3) 7−→ (t3 : t2u) = (t : u) (x : y : z) 7−→ (y : x) 7−→ (xy2 : y3 : x3) = (xy2 : y3 : y2z) = (x : y : z) In terms of function fields, the isomorphism is given by identifying t/u with y/x.
6. Curves — Local Properties
Starting with this section, we will restrict ourselves to the consideration of plane algebraic curves. In this section, we will consider the behavior of a curve near one of its points.
6.1. Definition. Let C : f(x, y) = 0 be an affine plane algebraic curve, and let P = (ξ, η) ∈ C be a point on C. Let fx = ∂f/∂x and fy = ∂f/∂y be the two partial of the polynomial f. We say that P is a regular point of C, and that C is regular or smooth or nonsingular at P , if fx(ξ, η) and fy(ξ, η) do not both vanish. In this case, the equation fx(ξ, η)(x − ξ) + fy(ξ, η)(y − η) = 0 describes a line through the point P ; it is called the line to C at P . If fx(ξ, η) = fy(ξ, η) = 0, we call P a singular point or a singularity of C, and we say that C is singular at P . We call the affine curve C smooth if it has no singular points. Writing f(x + ξ, y + η) = f0(x, y) + f1(x, y) + ··· + fd(x, y) with fj homogeneous of degree j, we have f0 = 0 (since P is on C) and f1 = 0 if and only if P is a singularity of C. We call the smallest j such that fj 6= 0 the multiplicity of P . So a singular point is one with multiplicity at least 2. If the multiplicity is n, we speak of an n-uple point of C (e.g., double, triple point). If the point is regular, then f1(x − ξ, y − η) = 0 gives the tangent line. 15
Intuitively, the local behavior of C near P is determined by the lowest order non- vanishing term fn (where n is the multiplicity of P ). If n = 1, then this lowest order term is linear, and the curve looks “like a line” near P ; in particular, there is a well-defined tangent line. If n > 1, then we can factor fn as a product of n linear forms, which correspond to the tangent directions of the various “branches” of the curve at P ; these directions may occur with multiplicities. If fn is a product of n pairwise non-proportional linear forms (so that there are n distinct tangent directions), we call the singularity ordinary. So an ordinary double point (also called a node) is a point of multiplicity 2 with two distinct tangent directions.
6.2. Definition. Let C : F (x, y, z) = 0 be a projective plane algebraic curve, P ∈ C a point on C. Let Ci be an affine patch of C that contains P . Then we define the notions of regular/singular point etc. for P ∈ C by those for P ∈ Ci. It is easy to check that this does not depend on the choice of affine patch when several are possible (Exercise). We say that C is smooth if it has no singular points.
It can be shown that P ∈ C is singular if and only if Fx(P ) = Fy(P ) = Fz(P ) = 0 (Exercise). Furthermore, if P = (ξ : η : ζ) is smooth on C, then the tangent line to C at P has equation
Fx(ξ, η, ζ)x + Fy(ξ, η, ζ)y + Fz(ξ, η, ζ)z = 0 (Exercise).
6.3. Example. An affine line ax + by = c ((a, b) 6= (0, 0)) is smooth: the partial derivatives are a and b, and at least one of them is nonzero. Similarly, a projective line ax + by + cz = 0 ((a, b, c) 6= (0, 0, 0)) is smooth.
6.4. Example. The affine and projective unit circles are smooth. Consider the affine circle x2 + y2 − 1 = 0. The partial derivatives are 2x and 2y, so they vanish together only at the origin, but the origin is not on the curve. The same kind of argument works for the other affine patches of the projective unit circle.
6.5. Example. The reducible affine curve xy = 0 has an ordinary double point at the origin. Indeed, both partial derivatives y and x vanish there, and the first nonvanishing term in the local expansion is xy of degree 2, which factors into the two non-proportional linear forms x and y.
6.6. Example. Let us consider the nodal cubic curve N : y2 = x2(x + 1). Writing 2 3 2 2 f(x, y) = y − x − x , the partial derivatives are fx = −3x − 2x and fy = 2y. If they both vanish, we must have y = 0 and then x2(x + 1) = x(3x + 2) = 0, which 2 2 implies x = 0. So the origin is singular. There we have f2 = y −x = (y−x)(y+x) 3 (and f3 = −x ), so we have an ordinary double point, with tangent directions of slopes 1 and −1.
6.7. Example. Now look at the cuspidal cubic curve C : y2 = x3. The partial derivatives are −3x2 and 2y, so the origin is again the only singularity. This time, 2 f2 = y is a , so there is only one tangent direction (in this case, along the x-axis). Such a singularity (where f2 is a nonzero square and f2 and f3 have no common divisors) is called a . 16
6.8. Example. Consider the affine curve C : y = x3. It is smooth, since the y- is constant 1. However, if we look at the projective closure yz2 −x3 = 0, we find that there is a singularity at (0 : 1 : 0). Therefore it is not true that the projective closure of a smooth affine plane curve is again smooth.
Our next topic is the local behavior of rational functions on a curve.
6.9. Definition. Let C be an irreducible affine plane curve, P ∈ C a smooth point, and consider a regular function 0 6= φ ∈ C[C]. We want to define the order of vanishing of φ at P . For this, let the equation of the curve be f(x, y) = 0 and assume (we can make a translation if necessary) that P = (0, 0). Then
f(x, y) = ax + by + f2(x, y) + ··· + fd(x, y) with (a, b) 6= (0, 0). Assume that b 6= 0 (otherwise interchange x and y). We define the order of vanishing or just order of φ at P to be n φ o v (φ) = max j : is regular at P . P xj j Here φ/x is considered as a rational function on C. We set vP (0) = +∞. The idea here is that in the situation described, x = 0 defines a line that meets the curve transversally at P (i.e., not in the tangent direction) and therefore x has a simple zero at P . So we should have vP (x) = 1. Also, if φ is regular and non-zero at P , we want to have vP (φ) = 0. That the definition makes sense follows from the following result.
6.10. Lemma. Consider the situation described in the definition above.
(1) If φ is regular at P and vanishes there, then vP (φ) > 0. n (2) Let n = vP (φ). Then φ/x is regular at P and has a non-zero value there. This property determines n uniquely.
(3) vP (φ1φ2) = vP (φ1) + vP (φ2).
(4) vP (φ1 + φ2) ≥ min{vP (φ1), vP (φ2)}, with equality if vP (φ1) 6= vP (φ2).
Proof. (1) We have to show that φ/x is regular at P . By assumption, φ is represented by a polynomial without constant term. It is therefore sufficient to show that y/x is regular at P , since φ/x is represented by a polynomial (regular at P ) plus a polynomial times y/x. Now we can write f(x, y) = x(a + F (x, y)) + y(b + G(x, y)) where F (0, 0) = G(0, 0) = 0. Then y(b + G(x, y)) ≡ −x(a + F (x, y)) mod f , hence y/x = −(a + F (x, y))/(b + G(x, y)) in C(C), where the denominator does not vanish at P = (0, 0) (and y/x takes the value −a/b). (2) φ/xn is regular at P by definition. If φ/xn vanishes at P , then by part (1), φ/xn+1 is also regular, contradicting the definition of n. [The proof of part (1) also works for rational functions.] So (φ/xn)(P ) 6= 0. It is clear that there can be at most one n such that φ/xn is regular and non-vanishing at P .
n1+n2 (3) Let n1 = vP (φ1) and n2 = vP (φ2), then φ1φ2/x is regular and non- vanishing at P , hence vP (φ1φ2) = n1 + n2. 17
n (4) Keep the notations from (3) and let n = min{n1, n2}. Then (φ1 + φ2)/x n is regular at P , so vP (φ1 + φ2) ≥ n. If n1 < n2 (say), then φ1/x is non-zero n n at P , whereas φ2/x vanishes, so (φ1 + φ2)/x is non-zero at P , and we have vP (φ1 + φ2) = n.
6.11. Definition. If C is an irreducible affine curve, P ∈ C is a smooth point, and φ ∈ C(C) is a rational function represented by the quotient f/g of regular functions, then we define the order of φ at P to be vP (φ) = vP (f) − vP (g). If C is an irreducible projective curve, P ∈ C is a smooth point, and φ ∈ C(C) is a rational function, then we define the order of φ at P to be the order of φ at P for any affine patch of C containing P . It can be checked that the definitions do not depend on the choices made (of numerator and denominator or affine patch).
6.12. Definition. If P ∈ C is a smooth point on an irreducible curve, then any rational function t ∈ C(C) such that vP (t) = 1 is called a uniformizing or uniformizer at P . We then have again for all 0 6= φ ∈ C(C) that φ v (φ) = n ⇐⇒ is regular and non-vanishing at P . P tn
6.13. Lemma. Let C be an irreducible curve, P ∈ C a smooth point and φ ∈ C(C) a rational function. Then φ is regular at P if and only if vP (φ) ≥ 0.
Proof. We work in an affine patch containing P . Write φ = f/g as a quotient of regular functions, and let m = vP (f), n = vP (g). Let t be a uniformizer m n at P . Then f = t f0 and g = t g0 with f0 and g0 regular and non-vanishing m−n at P . Hence φ = t f0/g0, where f0/g0 is regular and non-vanishing at P . If m−n m − n = vP (φ) ≥ 0, then t and therefore φ is regular at P . If m − n < 0, n−m then φ = f0/(t g0) is a quotient of regular functions such that the denominator vanishes at P , but the numerator does not, hence φ is not regular at P .
6.14. Example. Let us see what these notions mean for the line (which, to satisfy the assumptions of the definitions, we can consider as being embedded in the plane, for example as a coordinate axis). Let t be the coordinate on A1, then a rational function is a quotient of polynomials φ = f(t)/g(t), which we can assume is in lowest terms. At a point P = τ, a uniformizer is given by t − τ, and we see that n vP (f) = max{n :(t − τ) divides f} is the multiplicity of the zero τ of f. In particular, X X vP (f) = deg f and vP (φ) = deg f − deg g . 1 1 P ∈A P ∈A Now let us consider A1 ⊂ P1 and find the order of φ at ∞. There, 1/t is a uniformizer, and a quotient f(t)/g(t) is regular if and only if deg g ≥ deg f. These facts together imply that v∞(φ) = deg g − deg f and therefore X vP (φ) = 0 . 1 P ∈P This is a very important result: a rational function has as many zeros as it has poles (counted with multiplicity). 18
6.15. Example. Consider again the (affine) unit circle C : x2 + y2 = 1. At the point P = (0, 1) ∈ C, x is a uniformizer (since the tangent direction is horizontal, x = 0 intersects C transversally at P ). So vP (x) = 1. On the other hand, 2 2 = vP (x ) = vP (y − 1) + vP (y + 1) = vP (y − 1). This shows again that (y − 1)/x is regular at P and vanishes there.
6.16. Remark. Let V be an algebraic variety (affine or projective) and let W ⊂ Pm be a projective algebraic variety. Then every collection of rational functions F0,...,Fm ∈ C(V ), not all the zero function, determines a rational map V → W by evaluating (F0 : ... : Fm) (at points where F0,...,Fm all are regular and at least one of them does not vanish). Conversely, every rational map can be written in this way (there is nothing to do when V is affine; in the projective case, the map is given by homogeneous polynomials of the same degree d, and we can just divide every polynomial by a fixed homogeneous polynomial of degree d that does not vanish on all of V .)
6.17. Proposition. Let C be an irreducible plane algebraic curve and φ : C → Pm a rational map. Let P ∈ C be a smooth point. Then φ is (or can be) defined at P .
Proof. By the remark above, we can write φ = (F0 : ... : Fm) with ratio- nal functions F0,...,Fm ∈ C(C). Let t be a uniformizer at P , and let n = −n −n min{vP (Fj): j = 0, . . . , m}. Then φ = (t F0 : ... : t Fm) as well, and for all j, −n −n vP (t Fj) = vP (Fj) − n ≥ 0, with equality for at least one j = j0. Hence t Fj is −n regular at P for all j, and (t Fj0 )(P ) 6= 0. Therefore, we get a well-defined point m φ(P ) ∈ P . 6.18. Corollary. Let C be an irreducible smooth plane curve. Then every rational map C → Pm is already a morphism.
Proof. By the proposition, φ is defined on all of C. 6.19. Corollary. Let C be an irreducible smooth projective plane curve. Then there is a natural bijection between non-constant rational functions on C and dom- inant morphisms C → P1.
Proof. The map is given by associating to a rational function φ ∈ C(C) the rational map (φ : 1). By the previous corollary, this rational map is already a morphism; it is dominant if φ is non-constant. Conversely, a dominant morphism gives rise to an injective C-algebra homomorphism C[t] = C[A1] → C(C); the corresponding rational function on C is the image of t under this homomorphism. 6.20. Example. Consider again the nodal cubic curve N : y2 = x2(x+1) and the rational function y/x ∈ C(N). It is not possible to extend it to the point (0, 0) (which is a singular point of N): y/x is the slope of the line connecting (0, 0) to (y, x), and so it will tend to 1 if you approach (0, 0) within N along the branch where y ≈ x, and it will tend to −1 along the branch where y ≈ −x.
7. Bezout’s´ Theorem
B´ezout’s Theorem is a very important statement on the intersection of two pro- jective plane curves. It generalizes the statement that two distinct lines always intersect in exactly one point. We begin with a simple case. 19
7.1. Definition. Let C : F (x, y) = 0 be a plane affine curve and L : ax + by = c an affine line such that L is not contained in C. Let P = (ξ, η) ∈ C ∩ L. If b 6= 0, then we set (C · L)P equal to the multiplicity of the zero ξ of the polynomial F (x, (c − ax)/b). If a 6= 0, we set (C · L)P equal to the multiplicity of the zero η of the polynomial F ((c − by)/a, y). If P/∈ C ∩ L, then we set (C · L)P = 0. The number (C · L)P is called the intersection multiplicity of C and L in P .
7.2. Lemma. The intersection multiplicity is well-defined: if both a and b are nonzero in the above, then both definitions produce the same number.
c−ax c−by Proof. Let f(x) = F (x, b ) and g(y) = F ( a , y). Then g(y) = f((c − by)/a), f(x) = g((c−ax)/b), and the two maps x 7→ (c−by)/a, y 7→ (c−ax)/b are inverse between the polynomial rings C[x] and C[y], mapping x − ξ to y − η (up to scaling) and conversely. Therefore the multiplicities agree. (Also note that f and g are nonzero, since the defining polynomial of C does not vanish everywhere on L.)
7.3. Definition. Let C be a plane projective curve, L 6⊂ C a projective line, and 2 0 0 P ∈ P a point. We define (C · L)P to be the intersection multiplicity of (C · L )P for a suitable affine patch of P2 that contains P (and the corresponding affine patches C0 and L0 of C and L). (If L0 is empty, the number is zero.) It can be checked that the number does not depend on the affine patch chosen if there are several possibilities. Also, by homogenizing the affine situation, we see that (C · L)P is the multiplicity of the linear factor ηx − ξy in the homogeneous ax+by polynomial F (x, y, − c ), if C : F (x, y, z) = 0, L : ax + by + cz = 0 with c 6= 0, and P = (ξ : η : ζ). Similar statements are true when a 6= 0 or b 6= 0.
7.4. Theorem. Let C be a projective plane curve of degree d and L ⊂ P2 a projective line that is not contained in C. Then X C · L := (C · L)P = d . 2 P ∈P In words, C and L intersect in exactly d points, counting multiplicities.
Proof. Let C : F (x, y, z) = 0 and L : ax + by + cz = 0. Without loss of generality, assume that c 6= 0. Consider the homogeneous polynomial ax + by f(x, y) = F x, y, − c of degree d. Note that f 6= 0 since L 6⊂ C. For a point P = (ξ : η : ζ) ∈ L, we have that (C ·L)P is the multiplicity of ηx−ξy in f (and P is uniquely determined by (ξ : η)). The sum of these multiplicities is d, since f is a product of d linear factors. For all other P , the multiplicity is zero. This proves the claim.
Let us look more closely at the intersection multiplicities. 20
7.5. Proposition. Let C be a plane affine curve and P ∈ C.
(1) If L is a line through P , then (C · L)P ≥ 1.
(2) If P is smooth, then for all lines L through P , we have (C ·L)P = 1, except for the tangent line to C at P , which has intersection multiplicity at least 2 with C at P .
(3) If P is not smooth, then for all lines L through P , (C · L)P ≥ m, where m is the multiplicity of P as a point of C. We have equality except for the finitely many lines whose slopes are the tangent directions to C at P .
Proof. It suffices to prove the last part. Without loss of generality, assume that P = (0, 0), and write
f(x, y) = f1(x, y) + f2(x, y) + ··· + fd(x, y) , where f(x, y) = 0 is the equation of C, and fj is homogeneous of degree j. The multiplicity m is the smallest j such that fj 6= 0. If we plug in λx for y (to find the intersection multiplicity with the line y = λx), then we get a polynomial in x m that is divisible by x , hence (C · L)P ≥ m. Similarly if we plug in 0 for x (when the line is the vertical x = 0). On the other hand, the polynomial in x will start m x fm(1, λ) (plus higher order terms); therefore the intersection multiplicity will be exactly m unless λ is a slope corresponding to one of the tangent directions (and this extends to the vertical case).
7.6. Definition. If C is a plane curve, P ∈ C a smooth point, then P is called an inflection point or flex point or flex of C, if (C · L)P ≥ 3, where L is the tangent line to C at P .
7.7. Remark. The property of a point P ∈ C to be a flex point is not an “in- trinsic” property — it depends on the embedding of the curve in the plane. For example, if C is a smooth projective cubic curve and P is any point on C, then there is an isomorphism of C with another smooth plane cubic curve C0 such that the image of P on C0 is a flex point.
7.8. Theorem. Let C : F (x, y, z) = 0 be a projective plane curve. Define the Hessian of F to be 2 2 2 ∂ F ∂ F ∂ F
∂x2 ∂x ∂y ∂x ∂z
2 2 2 ∂ F ∂ F ∂ F HF = . ∂y ∂x ∂y2 ∂y ∂z
2 2 2 ∂ F ∂ F ∂ F
∂z ∂x ∂z ∂y ∂z2
Then a smooth point P ∈ C is an inflection point if and only if HF (P ) = 0.
Proof. Exercise.
7.9. Example. If C is a smooth conic section (so of degree 2), then C does not have flex points. Indeed, the Hessian is constant and nonzero in this case. 21
7.10. Example. Consider C : x4 + z4 − y2z2 = 0. The partial derivatives of the defining polynomial are (4 x3, −2 yz2, 4 z3 − 2 y2z); they all vanish at P0 = (0 : 1 : 0), which is therefore a singularity. The Hessian is
12 x2 0 0 2 2 2 2 2 0 −2 z −4 yz = −144 x z (y + 2 z ) . 2 2 0 −4 yz 12 z − 2 y
If x = 0, then z = 0 (this gives the singularity P0) or y = ±z; this gives two flex points (0 : ±1 : 1). (In these two points, the tangent line meets the curve even with multiplicity 4.) If z = 0, then x = 0, and we find P0 again. Finally, if 2 2 4 4 y = −2z√, then we√ obtain the equation x = −3z , and we find eight more flex points (ik 4 −3 : ± −2 : 1) (with k = 0, 1, 2, 3).
When looking at the intersection of a curve and a line, we could solve the equation of the line for one of the variables and then simply plug this into the equation of the curve. So we have eliminated one variable and reduced the problem to a one-dimensional one, which was easy. If we intersect with a curve of higher degree, elimination is not so straight-forward. However, it is possible. The tool that comes in handy here is the .
7.11. Definition. Let f1, . . . , fn be polynomials of degree less than n with coeffi- n−1 n−2 cients in a . Write fi = ai1x + ai2x + ··· + ain. We set for the following
det(f1, . . . , fn) = det(aij) .
n m Let f = anx + ··· + a1x + a0 and g = bmx + ··· + b1x + b0 be two polynomials with coefficients in a commutative ring. The resultant of f and g (with respect to the variable x) is the (n + m) × (n + m) determinant
m−1 n−1 Res(f, g) = Resx(n, m; f, g) = det(x f,...,xf,f,x g, . . . xg, g) .
7.12. Lemma. Keeping the notations of the definition above, assume that f is monic of degree n and that R is a field. Then Res(n, m; f, g) is the determinant of the endomorphism φ of the R- V = R[x]/hfi, that is given by v 7→ g · v.
Proof. If h is a polynomial of degree < n + m, then by performing row operations on the matrix whose rows are the coefficient vectors of xm−1f,...,xf,f,h, we can change the last row into r, where h = qf + r and deg r < deg f. Applying this to xn−1g, . . . , xg, g, and denoting by h the remainder of h mod f (r in the above), we see that Res(n, m; f, g) = det(xm−1f,...,xf,f, xn−1g, . . . , xg, g) = det(xn−1g, . . . , xg, g) . (For the second equality, note that the matrix on the left is a block matrix whose upper left block is an upper triangular matrix with 1s on the and whose lower left block is a zero matrix.) The last matrix represents φ in the xn−1,..., x, 1, whence the claim. 22
7.13. Corollary. Let f be a monic polynomial of degree n over a field, let g and h be polynomials of degree ≤ m and ≤ `, respectively. (1) Res(n, m + k; f, g) = Res(n, m; f, g) for all k ≥ 0. So we can just write Res(n, ∗; f, g) to denote Res(n, k; f, g) for any k ≥ deg g. (2) Res(n, ∗; f, g + hf) = Res(n, ∗; f, g). (3) Res(n, ∗; f, gh) = Res(n, ∗; f, g) Res(n, ∗; f, h).
Proof. (1) By Lemma 7.12, both are equal to the determinant of the same linear map. (2) Since g + hf = g, both sides are again equal to the determinant of the same linear map. (3) This follows from Lemma 7.12 and the multiplicativity of determinants. 7.14. Lemma. Let R be a general commutative ring, f and g as in the definition. (1) For c ∈ R, Res(n, m; cf, g) = cm Res(n, m; f, g). (2) Res(n, m; f, g) = (−1)mn Res(m, n; g, f).
Proof. (1) This is clear from the definition: we multiply the upper m rows of the matrix by c. (2) The two determinants differ by the nth power of a cyclic permutation of the (n + m) rows. The sign of this permutation is (−1)n(n+m+1 = (−1)mn(−1)n(n+1) = mn (−1) (since n(n + 1) is always even). 7.15. Proposition. Let R, f, g as before, and let h be a polynomial over R of degree ≤ `. (1) Res(n, m + `; f, gh) = Res(n, m; f, g) Res(n, `; f, h). (2) If f is monic of degree n, then Res(n, m + k; f, g) = Res(n, m; f, g) =: Res(n, ∗; f, g) =: Res(f, g) for all k ≥ 0. (3) If f is monic of degree n, then Res(n, ∗; f, g + hf) = Res(n, ∗; f, g).
Proof. All identities are indentities between polynomials with integral coefficients in the coefficients of f, g, h. Consider all these coefficients as independent vari- ables, and let K be the field of fractions of the P in all these variables over Z. Let F, G, H be the polynomials over P whose coefficients are the corresponding variables. Then parts (2) and (3) are valid for F, G, H in place of f, g, h and over K, by Cor. 7.13. Part (1) is true over any field if f is monic of degree n. By Lemma 7.14, the statement continues to hold when f has nonzero leading coefficient (scale f to be monic and observe that both sides scale in the same way). Since the leading coefficient of our “generic polynomial” F is nonzero (it is a variable), part (1) holds for F, G, H over K. Now since both sides of all the equalities are in P , they are also true over P . But this means that they hold in general. (For any ring R and polynomials f, g, h over R, there is a (unique) ring homomorphism P [x] → R[x] that sends F to f, G to g and H to h, hence it maps the identities that hold over P to the ones we want over R.) The most important property of the resultant is the following. 23
7.16. Theorem. Let R be a Unique Factorization Domain and let f, g be poly- nomials over R of degrees ≤ n and ≤ m, respectively. Then Res(n, m; f, g) = 0 if and only if either deg f < n and deg g < m, or f and g have a nonconstant common divisor in R[x].
Proof. If deg f < n and deg g < m, then the resultant obviously vanishes, since the first column of the matrix is zero. So we can assume that (say) deg f = n. Let F be the field of fractions of R; then we can write f = cp1 ··· pk ∈ F [x] with a constant c 6= 0 and irreducible monic polynomials pj. By Prop. 7.15, we have m Res(n, m; f, g) = c Res(p1, g) ··· Res(pk, g) . Now if p ∈ F [x] is irreducible, then g is either invertible mod p or divisible by p. In the first case, multiplication by g on F [x]/hpi is an invertible map, hence has nonzero determinant. In the second case, we have the zero map. Therefore:
Res(pj, g) = 0 ⇐⇒ pj | g So Res(n, m; f, g) = 0 if and only if f and g have a nonconstant common divisor in F [x]. But by Gauss’ Lemma, this is equivalent to the existence of a nonconstant common divisor in R[x]. In other words, the vanishing of the resultant indicates that f and g have a common root (in a suitable extension field). Now we want to use the resultant to get information on the intersecion of two curves.
7.17. Lemma. Let f, g ∈ C[x, y, z] be homogeneous of degree n and m, respec- tively. Then Resz(n, m; f, g) is homogeneous of degree nm in the remaining two variables x and y.
Proof. Consider the relevant matrix. The entry in the ith row and jth column is homogeneous in x and y of degree j − i if 1 ≤ i ≤ m and of degree m + j − i if m < i ≤ m + n. Every term in the expansion of the determinant as a polynomial in the entries therefore is homogeneous of degree n+m m n+m X X X (n + m)(n + m + 1) m(m + 1) n(n + 1) j − i− (i−m) = − − = nm . 2 2 2 j=1 i=1 i=m+1 7.18. Corollary. Let C and D be two projective plane curves of degrees n ≥ 1 and m ≥ 1, respectively. Then C ∩ D is nonempty. If C and D do not have a component in common, then #(C ∩ D) ≤ nm.
Proof. Let F (x, y, z) = 0 and G(x, y, z) = 0 be the equations of C and D. After a linear change of variables, we can assume that (0 : 0 : 1) is not in C ∩ D. Let R(x, y) = Resz(F,G); then R is homogeneous of degree nm by Lemma 7.17. If R = 0, then F and G have a nonconstant common factor by Thm. 7.16, i.e., C and D have a component in common. (Note that the coefficient of zn in F or the coefficient of zm in G is nonzero since (0 : 0 : 1) ∈/ C ∩D.) In this case, we certainly have C ∩D 6= ∅, hence we can assume that R 6= 0. Then R is a product of nm ≥ 1 linear forms. Therefore, there is some (ξ : η) ∈ P1 such that R(ξ, η) = 0. But then by Thm. 7.16 again, the polynomials F (ξ, η, z) and G(ξ, η, z) in C[z] have a common root ζ. But this means that (ξ : η : ζ) ∈ C ∩ D. 24
Now assume that there are nm+1 distinct intersection points P1,...,Pnm+1. After perhaps another linear change of variables, we can assume that Pj = (ξj : ηj : ζj) 6= (0 : 0 : 1) such that all Qj = (ξj : ηj) are distinct. Then R(ξj, ηj) = 0 for all j (note that ζj is a common root of F (ξj, ηj, z) and G(ξj, ηj, z)). But R is homogeneous of degree nm, so this implies that R = 0, contradicting the assumption that C and D do not have a component in common.
Now this prompts the following definition.
7.19. Definition. Let C and D be two projective plane curves without common component. Let P ∈ C ∩ D. If necessary, make a linear change of variables as in the second part of the proof above; then we define the intersection multiplicity of C and D at P to be the multiplicity of the factor ηx − ξy in R(x, y), where P = (ξ : η : ζ). As before, we write (C · D)P for that number. If P/∈ C ∩ D, we set (C · D)P = 0.
With this definition, the following theorem is immediate.
7.20. B´ezout’s Theorem. Let C and D be projective plane curves of degrees n and m, without common component. Then X C · D := (C · D)P = nm . 2 P ∈P In words, C and D intersect in exactly nm points, if we count the points according to intersection multiplicity.
7.21. Example. Let us look at the points of intersection of the two circles with 2 2 2 2 affine equations C1 : x +y = 1 and C2 :(x−2) +y = 4. The projective closures are given by the polynomials 2 2 2 2 2 F1 = x + y − z and F2 = x − 4xz + y . We compute the resultant (with respect to z)
−1 0 x2 + y2 0 2 2 0 −1 0 x + y 2 2 2 2 R = 2 2 = −(x + y )(15x − y ) . 0 −4x x + y 0 2 2 0 0 −4x x + y √ √ Its zeros correspond to the points (i : 1), (−i : 1), (1 : 15), (1 : − 15) ∈ P1. We find the intersection points of intersection multiplicity 1 √ √ (i : 1 : 0) , (−i : 1 : 0) , (1 : 15 : 4) , (1 : − 15 : 4) .
2 2 2 7.22. Example. Now consider two concentric circles C1 : x + y = 1, C2 : x + y2 = 4. This time,
−1 0 x2 + y2 0 2 2 0 −1 0 x + y 2 2 2 R = 2 2 = 9(x + y ) . −4 0 x + y 0 2 2 0 −4 0 x + y R now has double zeros at (±i : 1), and we get two intersection points (±i : 1 : 0) of multiplicity 2: concentric circles touch at infinity! 25
7.23. Example. Let us look at our old friends, the nodal and cuspidal cubics. The projective equations are N : x3 + x2z − y2z = 0 ,C : x3 − y2z = 0 . The point (0 : 0 : 1) is on both curves, as is (0 : 1 : 0), so we take the resultant with respect to x to avoid problems. 2 1 z 0 −y z 0 0 2 0 1 z 0 −y z 0 2 0 0 1 z 0 −y z 4 5 R = 2 = −y z . 1 0 0 −y z 0 0 0 1 0 0 −y2z 0
0 0 1 0 0 −y2z The roots are at (η : ζ) = (0 : 1) and (1 : 0), of multiplicities 4 and 5, respectively. So the intersection points are (0 : 0 : 1) of intersection multiplicity 4 and (0 : 1 : 0) of intersection multiplicity 5.
7.24. Example. It is true (Exercise) that the curve H given by the Hessian of a polynomial F defining the curve C meets C with multiplicity 1 (i.e., transversally) in a simple flex point (i.e., such that the intersection multiplicity of the tangent line and the curve is just 3 and not larger). If C is a smooth cubic curve, then any line meets C in at most three points (counting multiplicities), so there are only simple flex points. Therefore, H meets C only transversally, hence there are exactly 9 intersection points (H is also a curve of degree 3): A smooth plane cubic curve has exactly nine inflection points. If the coefficients of the polynomial defining C are real numbers, then exactly three of the inflection points are real (the other six come in three pairs of complex conjugates).
7.25. Corollary. If C is a smooth plane projective curve, then C is irreducible.
Proof. Otherwise, C would have at least two components C1 and C2, which would have to meet in some point P . But then P is a point of multiplicity at least 2 on C, hence a singularity. Now let us look at bitangents of quartic curves. A bitangent to a curve C is a line that meets C in two points with multiplicity at least 2. (As a boundary case, we include that the line meets C in some point with multiplicity at least 4; in this case, the two points of tangency can be thought of as coinciding.)
7.26. Example. As an example, consider the Fermat Quartic F : x4 +y4 +z4 = 0. Let P = (ξ : η : ζ) ∈ F be a point; then the tangent line to K at P has equation L : ξ3x + η3y + ζ3z = 0 . For now, assume that ζ 6= 0; then we can set ζ = 1. We eliminate z = −ξ3x − η3y from the equation for F ; we get a homogeneous polynomial in x and y of degree 4 that is divisible by (ηx − ξy)2. Dividing off this factor (one has to use that ξ4 + η4 + 1 = 0 in this computation), a quadratic polynomial remains whose is a constant times ξ2η2(ξ4 + η4 + ξ4η4). Taking into account points at infinity leads to G = x2y2z2(x4y4 + y4z4 + z4x4) 26 such that the points of tangency of the bitangents are exactly the points of inter- section of F with the curve defined by G. By B´ezout, there are exactly 14 · 4 = 56 intersection points (counting multiplicity), and since every bitangent accounts for two points, there are 28 bitangents. There are twelve lines that intersect F four- fold in points like (0 : ζ1+2k : 1), where ζ = eπi/4 is a primitive eighth root of unity. They correspond to the factors x2y2z2 in G, and the lines are given by y − ζ1+2kz = 0. The other factors of G lead to points (τ 1+3k : τ 2+3` : 1) and (τ 2+3` : τ 1+3k : 1) with τ = eπi/6. They are the points of tangency of 16 other bitangents iax + iby + z = 0. These considerations can be extended to arbitrary smooth plane quartic curves: A smooth plane quartic curve has exactly 28 bitangents.
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# Analysis of variance (Anova)
16/06/2019
In this post, I would like to look into Anova hypothesis testing. The dataset I am going to use is published in https://smartcities.data.gov.in which is a Government of India project under the National Data Sharing and Accessibility Policy.
I want to analyse if the unemployment across Bangalore is similar or are there pockets with high unemployment. The Unemployment_Rate_Bengaluru.csv dataset has total employed and unemployed people in Bangalore at a Ward level. For simplicity sake, I want to concentrate on three zones, Bangalore-east, Mahadevpura and Rajeshwari Nagar.
As I want to test that there is significant difference between unemployment rate in different zones in Bangalore, the null and alternate hypothesis will be as follows:
H0: μBangalore-east = μMahadevpura = μRajeshwari-Nagar
H1: Not all μ are equal
Sample unemployment data set:
``````## # A tibble: 5 x 8
## cityName zoneName wardName wardNo unemployed.no employed.no total.labour
## <chr> <chr> <chr> <int> <int> <int> <int>
## 1 Bengalu~ East Jayacha~ 46 19148 12301 31449
## 2 Bengalu~ Rajaraj~ HMT Ward 105 16096 12259 28355
## 3 Bengalu~ East SanJaya~ 19 19564 12927 32491
## 4 Bengalu~ East Kadugon~ 97 20156 15565 35721
## 5 Bengalu~ East Maruthi~ 126 16381 12938 29319
## # ... with 1 more variable: rate <dbl>``````
Visualizing the difference of unemployment in each zone.
``````ggplot(unemployment, aes(x=zoneName, y= rate, group = zoneName)) +
geom_boxplot() +
labs(x='Zone', y='Unemployment %') +
theme_minimal()``````
The summary statistics of unemployment across different zones are as follows:
``````data.summary <- unemployment %>%
rename(group = zoneName) %>%
group_by(group) %>%
summarise(count = n(),
mean = mean(rate),
sd = sd(rate),
skewness = skewness(rate),
kurtosis = kurtosis(rate))
data.summary``````
``````## # A tibble: 3 x 6
## group count mean sd skewness kurtosis
## <chr> <int> <dbl> <dbl> <dbl> <dbl>
## 1 East 81 0.574 0.0361 -0.228 0.747
## 2 Mahadevapura 24 0.567 0.0449 -0.812 1.69
## 3 Rajarajeswari Nagar 18 0.548 0.0499 0.00421 -1.24``````
From the above table I observe that the unemployment rate is between 54% and 57% in all the three zones.
One of the conditions to perform anova is that the population response variable follows a normal distribution in each group. The distribution of unemployment rate in diffeent zones are:
``````ggplot() +
geom_density(data=unemployment, aes(x=rate, group=zoneName, color=zoneName), adjust=2) +
labs(x = 'Unemployment Rate', y='Density', title = 'Testing normality among response variables') +
theme_minimal()+theme(legend.position="bottom")``````
From the above graph, it looks like the three groups are normally distributed. To test for normality, we could use Shapiro-Wilk test.
The test for each group would have the following hypothesis:
H0: The unemployment rate in the group is normally distributed
H1: The unemployment rate in the group is not normally distributed
``````for(i in 1:nrow(data.summary)){
test.dist <- (unemployment %>% dplyr::filter(zoneName == data.summary\$group[i]))\$rate
cat('Testing for group ', data.summary\$group[i], '\n')
print(shapiro.test(test.dist))
norm.plot <- ggplot() +
geom_qq(aes(sample = test.dist)) +
stat_qq_line(aes(sample = test.dist)) +
ggtitle(paste0("Normal distribution Q-Q plot for group ",data.summary\$group[i])) +
theme_minimal()
plot(norm.plot)
}``````
``````## Testing for group East
##
## Shapiro-Wilk normality test
##
## data: test.dist
## W = 0.97147, p-value = 0.06734``````
``````## Testing for group Mahadevapura
##
## Shapiro-Wilk normality test
##
## data: test.dist
## W = 0.90815, p-value = 0.03216``````
``````## Testing for group Rajarajeswari Nagar
##
## Shapiro-Wilk normality test
##
## data: test.dist
## W = 0.95852, p-value = 0.5734``````
As p > α, where α = 0.01, retaining the Null hypothesis in all the three groups. Therefore we can assume that unemployment rate is normally distributed among all groups.
Another condition for anova is that the population variances are assumed to be same. This is an assumption I am willing to take at this point.
Taking these assumptions, the ideal distributions of the sample are as follows. These distributions can be compared to a simulation that I created where change in F value (and significance of anova) can be visualised by increasing the between variance (increasing the distance between group means)
``````plot.normal.groups <- function(data.summary, mean, sd, label, title){
common.group.sd <- mean(data.summary\$sd)
range <- seq(mean-3*sd, mean+3*sd, by = sd*0.001)
norm.dist <- data.frame(range = range, dist = dnorm(x = range, mean = mean, sd = sd))
# Plotting sampling distribution and x_bar value with cutoff
norm.aov.plot <- ggplot(data = norm.dist, aes(x = range,y = dist))
for (i in 1:nrow(data.summary)) {
norm.aov.plot <- norm.aov.plot +
stat_function(fun = dnorm, color=colors()[sample(50:100, 1)], size = 1,
args = list(mean = data.summary\$mean[i], sd = common.group.sd))
}
norm.aov.plot + labs(x = label, y='Density', title = title) +
theme_minimal()+theme(legend.position="bottom")
}
set.seed(9)
mean <- mean(unemployment\$rate)
sd <- sd(unemployment\$rate)
plot.normal.groups(data.summary, mean, sd, 'Travel time (sec)', 'Assuming normality among response variables')``````
Performing Anova with
H0: μBangalore-east = μMahadevpura = μRajeshwari-Nagar
H1: Not all μ are equal
``````# Functions used in anova-test
f.plot <- function(pop.mean=0, alpha = 0.05, f, df1, df2,
label = 'F distribution',title = 'Anova test'){
# Creating a sample F distribution
range <- seq(qf(0.0001, df1, df2), qf(0.9999, df1, df2), by = (qf(0.9999, df1, df2)-qf(0.0001, df1, df2))*0.001)
f.dist <- data.frame(range = range, dist = df(x = range, ncp = pop.mean, df1 = df1, df2 = df2)) %>%
dplyr::mutate(H0 = if_else(range <= qf(p = 1-alpha, ncp = pop.mean, df1 = df1,df2 = df2),'Retain', 'Reject'))
# Plotting sampling distribution and F value with cutoff
plot.test <- ggplot(data = f.dist, aes(x = range,y = dist)) +
geom_area(aes(fill = H0)) +
scale_color_manual(drop = TRUE, values = c('Retain' = "#00BFC4", 'Reject' = "#F8766D"), aesthetics = 'fill') +
geom_vline(xintercept = f, size = 2) +
geom_text(aes(x = f, label = paste0('F = ', round(f,3)), y = mean(dist)), colour="blue", vjust = 1.2) +
labs(x = label, y='Density', title = title) +
theme_minimal()+theme(legend.position="bottom")
plot(plot.test)
}``````
``````anva <- aov(rate ~ zoneName, unemployment)
anova.summary <- summary(anva)
print(anova.summary)``````
``````## Df Sum Sq Mean Sq F value Pr(>F)
## zoneName 2 0.01065 0.005323 3.315 0.0397 *
## Residuals 120 0.19269 0.001606
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1``````
``f.plot(f = anova.summary[[1]]\$F[1], df1 = anova.summary[[1]]\$Df[1], df2 = anova.summary[[1]]\$Df[2])``
Here p < α, where α = 0.05 Hence rejecting the Null Hypothesis. From this test we can observe that not all groups have same mean. But to find out which groups are similar and which are different, I am conducting a TukeyHSD test.
``````tukey.test <- TukeyHSD(anva)
print(tukey.test)``````
``````## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = rate ~ zoneName, data = unemployment)
##
## \$zoneName
## diff lwr upr
## Mahadevapura-East -0.007071899 -0.02917267 0.015028872
## Rajarajeswari Nagar-East -0.026768634 -0.05154855 -0.001988723
## Rajarajeswari Nagar-Mahadevapura -0.019696735 -0.04934803 0.009954561
## Rajarajeswari Nagar-East 0.0309193
## Rajarajeswari Nagar-Mahadevapura 0.2597730``````
``````# Plot pairwise TukeyHSD comparisons and color by significance level
tukey.df <- as.data.frame(tukey.test\$zoneName)
tukey.df\$pair = rownames(tukey.df)
ggplot(tukey.df, aes(colour=cut(`p adj`, c(0, 0.01, 0.05, 1),
label=c("p<0.01","p<0.05","Non-Sig")))) +
geom_hline(yintercept=0, lty="11", colour="grey30") +
geom_errorbar(aes(pair, ymin=lwr, ymax=upr), width=0.2) +
geom_point(aes(pair, diff)) +
labs(x = 'Groups', y='Density', colour="", title = 'Tukey HSD Test') +
theme_minimal()+theme(legend.position="bottom")``````
From this test we can see that there is significant difference (with α = 0.05 confidence) between Rajeshwari Nagar and Bangalore East.
Created on 17th June 2019, Achyuthuni Sri Harsha
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# Question: How Do You Find The Value Of Sin Pi?
## Is the value of sin 90?
Therefore, sin 90 degree equals to the fractional value of 1/ 1..
## Which value of sin is?
Answer) In Mathematics, the value of sin 0 degree is always equal to 0….Sin 0⁰ Value = 0.NameAbbreviationRelationSineSinSin (θ)= Opposite/HypotenuseCosineCosCos (θ)= Adjacent/HypotenuseTangentTanTan (θ)= Opposite/Adjacent
## What is the sine value of π?
0The sine value of π, denoted sin(π), is 0.
## Why do we use sine?
The sine function is defined as the ratio of the side of the triangle opposite the angle divided by the hypotenuse. This ratio can be used to solve problems involving distance or height, or if you need to know an angle measure. Example: … To find the length of the side opposite the angle, d, we use the sine function.
## How is sine calculated?
In a right triangle, the sine of an angle is the length of the opposite side divided by the length of the hypotenuse. … In any right triangle, the sine of an angle x is the length of the opposite side (O) divided by the length of the hypotenuse (H).
## What sine means?
1 : the trigonometric function that for an acute angle is the ratio between the leg opposite the angle when it is considered part of a right triangle and the hypotenuse.
## What is the value of sin 2 pi by 3?
The exact value of sin(π3) sin ( π 3 ) is √32 .
## What is the value of sin 45 degree?
0.7071067812The value of Sin 45 degree in decimal form is 0.7071067812. Sine is considered as one of the most important functions in trigonometry as it is used to find out the unknown values of the angles and length of the sides of a right-angle triangle.
## What is the exact value of sin pi over 4?
The exact value of sin(π4) sin ( π 4 ) is √22 .
## How do you find the value of sin 180 degrees?
We know that the exact value of sin 0 degree is 0. Therefore, the value of sin 180 degrees = 0.
## What is the value of pi by 6?
In this case, π6 refers to the angle in radians, an alternate unit of measurement for angles ( π rad = 180°). The point on the unit circle that is intersected by this line is ( √32 , 12 ). Finally, the function, sin( θ ) returns a value equal to the y-coordinate of the point, giving us an answer of 12 .
## What is value of Cos Pi?
The value of Cos pi= -1 and Sin pi=0. The period of sin is also 2pi or 360° and its value repeats after 2pi or 360°.
## What is the value of cos 4 pi by 3?
Trigonometry Examples The exact value of cos(π3) cos ( π 3 ) is 12 .
## What is the value of sin pi over 6?
12The exact value of sin(π6) sin ( π 6 ) is 12 .
## Why Sine is called sine?
The word “sine” (Latin “sinus”) comes from a Latin mistranslation by Robert of Chester of the Arabic jiba, which is a transliteration of the Sanskrit word for half the chord, jya-ardha.
## What is the exact value of sin 0 degrees?
Sine Definition In Terms of Sin 0Sine Degrees/RadiansValuesSin 000Sin 300 or Sin π/61/2Sin 450 or Sin π/41/\sqrt{2}Sin 600 or Sin π/3\sqrt{3}/24 more rows
## What is the value of Cosec 180 degree?
-1Therefore, the value of cos 180° is -1. The value of trigonometric ratios for different angles are given for the reference.
## What is the value of sin PI by 12?
Check by calculator. Calculator –> sin(π12)=sin15∘=0.26 .
## What is the value of sin Pi by 3?
The exact value of sin(π3) sin ( π 3 ) is √32 .
## Is the value of sin 60?
From the above equations, we get sin 60 degrees exact value as √3/2.
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https://stats.stackexchange.com/questions/552768/forecasting-multi-variate-data-using-arima-errors-with-fourier-terms-and-covaria?noredirect=1
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# Forecasting Multi-variate data using Arima errors with Fourier terms and covariate on a weekly data in R
I'm planning to do a multivariate time series forecasting using arima errors with fourier terms. Data assumptions-moderate seasonality. one independent variable(x) and dependent variable(y) Weekly data collected for 148 weeks. y has some good correlation with x. So planning to forecast y based on the correlation with x along with the fourier values of y.
df.y<-ts(y, frequency=52)
df.x<-ts(x,frequency=52)
zx.f<-fourier(df.x, K=2,h=26) #fourier values of x variable
forecast.x<-auto.arima(df.x,xreg=zx.f,approximation=F,h=26) #forecast values of x
zy.f<-fourier(df.y,K=2,h=26) #fourier values of y variable
forecast.y<-auto.arima(df.y,xreg=cbind(forecast.x,zy.f),approximation=F,h=26) #forecast y using
fourier values of y and forecast values of x
Is it right to use forecasted values of x and Fourier values of Y when forecasting the Y variable? i have just used cbind on the xreg parameter to use both forecasted variable(x) and Fourier values of y
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## What are alternate angles simple definition?
: one of a pair of angles with different vertices and on opposite sides of a transversal at its intersection with two other lines: a : one of a pair of angles inside the two intersected lines. — called also alternate interior angle .
## What is alternate interior and exterior angles?
Angles that are in the area between the parallel lines like angle 2 and 8 above are called interior angles whereas the angles that are on the outside of the two parallel lines like 1 and 6 are called exterior angles . Angles that are on the opposite sides of the transversal are called alternate angles e.g. 1 + 8.
## Do alternate interior angles add up to 180?
Alternate angles form a ‘Z’ shape and are sometimes called ‘Z angles ‘. d and f are interior angles . These add up to 180 degrees (e and c are also interior ). Any two angles that add up to 180 degrees are known as supplementary angles .
## What do alternate interior angles look like?
Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. In this example, these are two pairs of Alternate Interior Angles : c and f.
## What is the difference between alternate interior angles and consecutive interior angles?
If they are on the same side, then the angles are considered consecutive . If they are on opposite sides, then the angles are considered alternate . Are the angles inside or outside of the two intersected lines? If they are inside the two lines, then they will be classified as interior .
You might be interested: How to build a interior wall
## What are alternate angles with diagram?
Alternate angles are angles that are in opposite positions relative to a transversal intersecting two lines. Examples. If the alternate angles are between the two lines intersected by the transversal, they are called alternate interior angles . In each illustration below, LINE 1 is a transversal of LINE 2 and LINE 3.
## Is same side interior angles congruent?
Same side interior angles are on the same side of the transversal. Same side interior angles are congruent when lines are parallel.
## What are the properties of alternate angles?
What Are The Properties of Alternate Interior Angles? Alternate Interior angles are congruent. The sum of the angles formed on the same side of the transversal which are inside the two parallel lines is equal to 180°. Alternate interior angles don’t have any specific properties, in case of non-parallel lines.
## What are the 7 types of angles?
Types of Angles – Acute, Right, Obtuse , Straight and Reflex Acute angle. Right angle. Obtuse angle. Straight angle. Reflex angle.
## What do alternate interior angles equal?
The angles are positioned at the inner corners of the intersections and lie on opposite sides of the transversal. Alternate interior angles are equal if the lines intersected by the transversal are parallel. Alternate interior angles formed when a transversal crosses two non-parallel lines have no geometrical relation.
## What is the sum of alternate interior angles?
These angles are congruent. Sum of the angles formed on the same side of the transversal which are inside the two parallel lines is always equal to 180°. In the case of non – parallel lines, alternate interior angles don’t have any specific properties.
### Releated
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How do you measure for interior window trim? Measure from inside corner to inside corner across the bottom of the window . Add 1/2-inch to the measurement for the bottom piece of trim . The extra 1/2 inch provides a 1/4-inch gap — also known as the reveal — on all sides between the casing […]
#### How to paint a house interior walls
What is the correct order to paint a room? Pros usually follow a certain order when painting a room . They paint the trim first, then the ceiling, then the walls. That’s because it’s easier (and faster) to tape off the trim than to tape off the walls. And you certainly don’t want to tape […]
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What is 5 to the 3rd power?
"What is 5 to the 3rd power? (What is Five to the third power?)" is the math problem we are solving here. In mathematical terms, 5 is the base and 3 is the exponent as follows:
53
The problem above (5 to the 3rd power) means that you multiply 5 by 5 by 5. In other words, 5 × 5 × 5. When we do that, we get the answer as follows:
125
3rd Power Calculator
Here you can submit another base number that we will calculate to the 3rd power.
What is to the 3rd power?
What is 6 to the 3rd power?
Do you want to see the answer to a different problem that we solved related to the third power? Try the next number on our list!
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A034386 Primorial numbers (second definition): n# = product of primes <= n. 247
1, 1, 2, 6, 6, 30, 30, 210, 210, 210, 210, 2310, 2310, 30030, 30030, 30030, 30030, 510510, 510510, 9699690, 9699690, 9699690, 9699690, 223092870, 223092870, 223092870, 223092870, 223092870, 223092870, 6469693230, 6469693230, 200560490130, 200560490130 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Squarefree kernel of both n! and lcm(1, 2, 3, ..., n). a(n) = lcm(core(1), core(2), core(3), ..., core(n)) where core(x) denotes the squarefree part of x, the smallest integer such that x*core(x) is a square. - Benoit Cloitre, May 31 2002 The sequence can also be obtained by taking a(1) = 1 and then multiplying the previous term by n if n is coprime to the previous term a(n-1) and taking a(n) = a(n-1) otherwise. - Amarnath Murthy, Oct 30 2002; corrected by Franklin T. Adams-Watters, Dec 13 2006 REFERENCES Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3, p. 14, "n?". József Sándor, Dragoslav S. Mitrinovic, Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, Section VII.35, p. 268. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..2370 (first 401 terms from T. D. Noe) Jens Askgaard, On the additive period length of the Sprague-Grundy function of certain Nim-like games, arXiv:1902.06299 [math.CO], 2019. Klaus Dohmen and Martin Trinks, An Abstraction of Whitney's Broken Circuit Theorem, arXiv:1404.5480 [math.CO], 2014. Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv:1202.3670 [math.HO], 2012. - From N. J. A. Sloane, Jun 13 2012 J. Barkley Rosser and Lowell Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math., Vol. 6, No. 1 (1962), 64-94. Eric Weisstein's World of Mathematics, Primorial. FORMULA a(n) = n# = A002110(A000720(n)) = A007947(A003418(n)) = A007947(A000142(n)). Asymptotic expression for a(n): exp((1 + o(1)) * n) where o(1) is the "little o" notation. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 08 2001 For n > 0, log(a(n)) < 1.01624*n. [Rosser and Schoenfeld, 1962; Sándor et al., 2005] - N. J. A. Sloane, Apr 04 2017 a(n) <= A179215(n). - Reinhard Zumkeller, Jul 05 2010 a(n) = lcm(A006530(n), a(n-1)). - Jon Maiga, Nov 10 2018 Sum_{n>=0} 1/a(n) = A249270. - Amiram Eldar, Nov 08 2020 EXAMPLE a(5) = a(6) = 2*3*5 = 30; a(7) = 2*3*5*7 = 210. MAPLE A034386 := n -> mul(k, k=select(isprime, [\$1..n])); # Peter Luschny, Jun 19 2009 # second Maple program: a:= proc(n) option remember; `if`(n=0, 1, `if`(isprime(n), n, 1)*a(n-1)) end: seq(a(n), n=0..36); # Alois P. Heinz, Nov 26 2020 MATHEMATICA q[x_]:=Apply[Times, Table[Prime[w], {w, 1, PrimePi[x]}]]; Table[q[w], {w, 1, 30}] With[{pr=FoldList[Times, 1, Prime[Range[20]]]}, Table[pr[[PrimePi[n]+1]], {n, 0, 40}]] (* Harvey P. Dale, Apr 05 2012 *) Table[ResourceFunction["Primorial"][i], {i, 1, 40}] (* Navvye Anand, May 22 2024 *) PROG (PARI) a(n)=my(v=primes(primepi(n))); prod(i=1, #v, v[i]) \\ Charles R Greathouse IV, Jun 15 2011 (PARI) a(n)=lcm(primes([2, n])) \\ Jeppe Stig Nielsen, Mar 10 2019 (SageMath) def sharp_primorial(n): return sloane.A002110(prime_pi(n)) [sharp_primorial(n) for n in (0..30)] # Giuseppe Coppoletta, Jan 26 2015 (Python) from sympy import primorial def A034386(n): return 1 if n == 0 else primorial(n, nth=False) # Chai Wah Wu, Jan 11 2022 (Magma) [n eq 0 select 1 else LCM(PrimesInInterval(1, n)) : n in [0..50]]; // G. C. Greubel, Jul 21 2023 CROSSREFS Cf. A002110, A057872, A249270. Cf. A073838, A034387. - Reinhard Zumkeller, Jul 05 2010 The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358. Sequence in context: A147299 A090549 A080326 * A083907 A084343 A025552 Adjacent sequences: A034383 A034384 A034385 * A034387 A034388 A034389 KEYWORD nonn,easy,nice AUTHOR N. J. A. Sloane EXTENSIONS Offset changed and initial term added by Arkadiusz Wesolowski, Jun 04 2011 STATUS approved
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# Extract area from elipsoidal/cirle trajectory.
3 Ansichten (letzte 30 Tage)
Claudio Iturra am 28 Mai 2024
Bearbeitet: Matt J am 28 Mai 2024
Dear all, how are you?
I created the following code to simulate an anticlockwise rotating trajectory forced by an external velocity. Now, I want to calculate the area each time the trajectory makes a closed circle, as described in the attached image. I would appreciate any help in finding the best way to identify the areas inside the circles for each loop/circle in the trajectory.
For circle detection, I used the polyshape function as described in the attached figure. Unfortunately, this requires me to cut the series each time a circular trajectory is identified, and then manually choose the area I need, which takes too much time
clear all,close all;
Fs = 1/0.5;
duration_hours = 25 * 24;
t = 0:1/Fs:duration_hours;
period = 20;
phase = -pi/2;
num_periods = floor(duration_hours / period);
amplitudes = 0.1/2 + 0.1 * rand(1, num_periods); % Random amplitudes between 0.1 and 0.3
y = zeros(size(t));
for i = 1:num_periods
start_idx = (i-1) * period * Fs + 1;
end_idx = i * period * Fs;
uI(start_idx:end_idx) = amplitudes(i) * sin(2*pi*(t(start_idx:end_idx) - t(start_idx))/period);
end
amplitudes = 0.1/2 + 0.1 * rand(1, num_periods); % Random amplitudes between 0.1 and 0.3
for i = 1:num_periods
start_idx = (i-1) * period * Fs + 1;
end_idx = i * period * Fs;
vI(start_idx:end_idx) = amplitudes(i) * sin(2*pi*(t(start_idx:end_idx) - t(start_idx))/period);
end
uI = uI(10:end);
vI = vI(1:length(uI));
cv = complex(uI*100,vI*100);
dt=0.5/24;
dt=dt*3600*24;
cxI=cumsum(cv)*dt;
cxI=cxI/100/1000;
x = real(cxI + complex(cumsum(repmat(0.1,1191,1)),0)' );
y = imag(cxI + complex(cumsum(repmat(0.1,1191,1)),0)' );
clearvars -except x y;
plot(x,y),axis equal;
% for the first circle detected
pgonAll = polyshape(x(1:48),y(1:48),'Simplify',false);
pgonAll = simplify(pgonAll);
pgonEach = regions(pgonAll);
figure
plot(x(1:48),y(1:48))
hold on
plot(pgonEach(3))
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### Akzeptierte Antwort
Matt J am 28 Mai 2024
Bearbeitet: Matt J am 28 Mai 2024
ymin=min(y);
[xx,yy]=deal(x,y);
xx(end+1)=x(end); yy(end+1)=ymin;
p=regions(polyshape(xx,yy));
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
h=arrayfun(@(P)min(P.Vertices(:,2)), p );
p=p(h>0.001+ymin);
Areas=area(p)'
Areas = 1x29
1.8247 0.9686 0.0081 1.0051 0.5726 0.0032 0.1837 0.5910 2.8787 1.0814 0.0241 1.7894 2.5199 0.7346 1.0110 0.0168 0.3967 2.1162 1.7427 2.5049 0.7968 1.8758 1.2651 0.6767 1.3316 1.8829 0.2996 1.1132 0.0252
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
plot(x,y); hold on
plot(p,'FaceColor','r'); hold off
axis([50,100,-5,5]); daspect([1,1,1])
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# What is Position Time Graph?
• Independent variable is always taken on the x-axis.
• We plot the dependent variable on the y-axis.
• The dependent variable is dependent on the independent variable through some function. Know more about Functions in Maths
Let us observe the case of Kinematic equations for uniformly accelerated Motion. Time is our fundamental independent variable and position is a dependent variable. Along with position, displacement, velocity, and acceleration are also dependent variables. There are three basic types of motion graphs
1. Position-Time Graph (x-t graph)
2. Velocity-Time Graph (v-t graph)
3. Acceleration-Time Graph
## What is a position-time graph?
It is a graph where instantaneous position $x$ of a particle is plotted on the y-axis and the time $t$ on the x-axis as shown below in the figure:
## Position time graph for uniform motion
• When an object is in uniform motion it covers equal distance in equal intervals of time.
• x-t for an object in uniform rectilinear motion is a straight line inclined to the time axis.
### Position vs time graph slope for uniform motion
Position vs time graph’s slope gives the velocity of the object. To find the slope of position vs time graph consider the figure given below
From this figure we can see that at time $t=0$ object is at $x_0$, at time $t=t_1$ object is at $x_1$ and at time $t=t_2$ object is at $x_2$. Here $theta$ is the angle which inclined straight line i.e., the x-t plot makes with the x-axis.
Position vs time graph slope = slope of line AB
$= \tan \theta = \frac{QR}{PR} = \frac{x_2 – x_1}{t_2 – t_1}$
$=\frac{Displacement}{time}=velocity\,\,(v)$
Hence Position vs time graph slope gives the velocity of the object.
## x-t graph for uniformly accelerated motion
Position time relation for uniformly accelerated motion along a straight line is
$x=x_0+v_0 t+\frac{1}{2}t^2$
dependence of $x$ on $t^2$ shows that it is a quadratic equation or quadratic function of $t$. So, the position-time graph for uniform accelerated motion is a parabola as shown below in the figure.
### What does the slope of a position-time graph represent for uniform accelerated motion
As established earlier, $x$ vs $t$ graph slope represents the velocity of the object. In the case of uniform accelerated motion slope of the position-time graph gives the instantaneous velocity of the object. Slope of the x-t graph at time $t=0$ represents the initial velocity $v_0$ of the object. Let us now calculate the slope of the $x-t$ for uniform accelerated motion.
#### The slope of the x-t graph
$=\frac{\text{Small change in vertical co-ordinate}}{{Small change in horizontal co-ordinate}}$
$=\frac{dx}{dt}=\text{velocity at instant t}$
### Important points about the slope of the x-t graph
• The slope of the tangent drawn at any point to x-t curve gives the instantaneous velocity.
• Steeper is the slope at any point, more is the speed at that point.
• The numerical value of the slope of the tangent drawn to the x-t curve gives speed.
• We can find at which particular instant the direction of motion of the particle is changing, these points are where the slope of $x-t$ plot is zero.
• We can find from x-t plot, the instant when the particle is moving with the most speed. The instant where the tangent is drawn to $x-t$ plot is steepest the speed would be maximum.
Position time graph at rest
The figure below shows the x-t graph at rest.
When the object under consideration is at rest then its position does not change with the passage of time.
### How to find average velocity on a position-time graph
The slope of the chord joining any two points on the curve gives us the average velocity for a particular time -interval. Now consider the figure given below
The average velocity of the object between point $A$ and $B$ is
the ratio of the displacement $\Delta x=x_2 – x_1$ of the particle in the time interval $\Delta t=t_2-t_1$.
Here $x_1=x_0$ , $x_2=x$ , $t_1=0$ and $t_2=t$ so,
$v_{avg}=\frac{\Delta x}{\Delta t}=\frac{x-x_0}{t}$
## x-t graph vs Displacement time vs distance time graph
We can have displacement of particle and distance traveled by particle is any time interval from plot of x-t graph.
The position-time graph (x-t) is the same as the displacement-time graph. This graph will tell you the exact change in position of a body. So it tells about the shortest distance between initial and final position w.r.t time irrespective of the actual path used.
A distance-time graph will tell you the distance covered, that is the exact length of the actual path used to travel from the initial point to the final point, with respect to time.
A distance-time graph does not care about what direction objects are traveling.
## Representing accelerated motion
We can find out whether a particle is accelerating or decelerating at a particular instant or rather in a time interval.
If the sign of acceleration and velocity are same then the particle is accelerating otherwise decelerating.
We can find the sign of acceleration from $x-t$ plot but its value can be determined only if we have a detailed description of the x-t plot.
x-t graph for negative acceleration
If $x-t$ plot is concave down then
$\frac{d^2x}{dt^2} < 0$
i.e., acceleration is negative.
x-t graph for positive acceleration
If $x-t$ plot is concave up then
$\frac{d^2x}{dt^2}>0$,
i.e., acceleration is positive.
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CC-MAIN-2019-39
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longest
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en
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http://math.stackexchange.com/questions/289993/second-solution-for-search-of-negative-roots
| 1,469,294,557,000,000,000
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text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00273-ip-10-185-27-174.ec2.internal.warc.gz
| 163,548,140
| 18,018
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# Second solution for search of negative roots
How many $\underline{\text{negative roots}}$ does the equation $x^4-5x^3-4x^2-7x+4=0$ have?
My reasoning:
I rewrote the equation like: $$x^4-5x^3-4x^2-7x+4=0 \Rightarrow (x^2-2)^2 = 5x^3+7x$$
For any negative $x$, the outcome is never a negative term in the left member of the equation, and always a negative term in the right member of the equation. That's impossible, so the answer is that there are no negative roots for the original equation.
Does anyone see another way out?
-
You mean $(x^2 - 2)^2 = 5x^3 + 7x$? – Deven Ware Jan 29 '13 at 19:41
O yeah, my bad. Thx. – Sawyier Jan 29 '13 at 19:42
Either way, you are correct, there are no negative roots. – Deven Ware Jan 29 '13 at 19:43
Are yu familiar with Descartes' rule of signs? Regards – Amzoti Jan 29 '13 at 19:45
For any negative $x$ tested? Can't you see immediately that $(x^2 - 2)^2 \ge 0$ for all real $x$, while $5 x^3 + 7 x < 0$ for all negative $x$? – Robert Israel Jan 29 '13 at 19:47
You might want to say explicitly that $\forall x < 0,\; x\neq -\sqrt{2}$, the left-hand side of the equation is positive (whatever the value of $x$ - save for $x = \pm \sqrt{2}$, in which case the LHS $= 0$), while the right hand side of the equation is negative $\forall x<0$, since $5x^3 + 7x = x(x^2 + 7)$, evaluates to the sign of $x$. This is impossible for any equation. Hence, there is no solution (i.e., root) $x$ such that $x \lt 0$.
| 496
| 1,453
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| 3.953125
| 4
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CC-MAIN-2016-30
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latest
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en
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https://tech.paayi.com/use-microsoft-excel-floor-function
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text/html
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crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00072.warc.gz
| 562,313,164
| 19,431
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# Learn How to Use Microsoft Excel FLOOR Function
Written by | 0 Comments | 358 Views
In this article, you will learn how to use the Microsoft Excel FLOOR function and its prime function in Microsoft Excel. You will also get to know the Microsoft Excel FLOOR function return value and syntax with the help of some examples.
Microsoft Excel FLOOR Function
The main function of the Microsoft Excel FLOOR function is to round down the number to the nearest specified multiple. So, with the help of the FLOOR function, you can able to round up the given number, to the nearest specified multiple. It should be noted that the FLOOR function operates similarly as the MROUND function. The only difference is that the FLOOR function always rounds down. So, with the help of the Microsoft Excel FLOOR function, you can easily round down the number to the nearest multiple.
Return Value of FLOOR Function
The return value will be the round-down number.
Syntax of FLOOR Function
=FLOOR(number, significance)
Where the arguments:
• number: This is the number of which you want to round down.
• significance: This is multiple to use while rounding down the number.
How to Use Microsoft Excel FLOOR Function?
So we know that Microsoft Excel FLOOR function you can able to round down the number to the nearest specified multiple. The multiple to use for rounding is provided as the significance argument. If the number is already an exact multiple, no rounding happens and the original number is returned. So, with the help of the FLOOR function, you can able to round up the given number, to the nearest specified multiple.
It should be noted that the FLOOR function operates similarly as the MROUND function. The only difference is that the FLOOR function always rounds down. So, with the help of the Microsoft Excel FLOOR function, you can easily round down the number to the nearest multiple.
| 388
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{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.25
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CC-MAIN-2023-06
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longest
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https://blogmepost.com/348/what-is-the-current-libor-rate
| 1,696,249,005,000,000,000
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text/html
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crawl-data/CC-MAIN-2023-40/segments/1695233510994.61/warc/CC-MAIN-20231002100910-20231002130910-00397.warc.gz
| 147,928,149
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# What is the current LIBOR rate?
in Credit
What is the current LIBOR rate?
Quote the current LIBOR rate and talk about the importance of LIBOR as it relates to spreads and pricing of other credit instruments.
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{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
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http://roseindia.net/answers/viewqa/Design-concepts-design-patterns/27076-Problems-in-Stringtokenizer.html
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Siba mishra
Problems in Stringtokenizer
0 Answer(s) 4 years and 8 months ago
Posted in : Design concepts & design patterns
```hi
here is my code
import java.util.*;
import java.io.*;
import java.util.Scanner;
class Node
{
public boolean marked;
public int node_no;
public String node_type;
//public float value;
//public float control_value;
public String usevar[];
public String defvar[];
public int child[];
public int par[];
public int actIn[];
public int actOut[];
}
public class Input
{
public int n,nn,vn; //vn total no of variables, n total no of node ,nn node no.
public int i,use_var,def_var,p_c,c_c; //
public String c,f,var_nam[];
public Node node[];
StringTokenizer str,str1;
Input() throws Exception
{
Scanner fr=new Scanner(df);
str=new StringTokenizer(fr.nextLine(),"\$");
this.n=Integer.parseInt(str.nextToken()); //total no of nodes
this.vn=Integer.parseInt(str.nextToken()); //toal no of variables
var_nam=new String[vn]; //for total variables
str=new StringTokenizer(fr.nextLine(),"\$");
System.out.println("rtryutrty"+n+vn);
for(int i=0;i<vn;i++)
{
String a=str.nextToken();
var_nam[i]=a;
}
node=new Node[n+1];
for(int i=1;i<=n;i++)
{
node[i]=new Node();
}
i=1;
while(fr.hasNextLine())
{
c=fr.nextLine();
str=new StringTokenizer(c,"\$"); //node no is parsed
nn=Integer.parseInt(str.nextToken());
System.out.println(nn);
node[i].node_no=nn; //node no is assigned
str1=new StringTokenizer(str.nextToken()," ");
use_var=str1.countTokens(); // no of use var
node[i].usevar=new String[use_var];
for(int j=0;j<use_var;j++)
{
node[i].usevar[j]=str1.nextToken();
}
for(int j=0;j<use_var;j++)
{
System.out.println(node[i].usevar[j]);
}
str1=new StringTokenizer(str.nextToken()," "); // for def var
def_var=str1.countTokens();
node[i].defvar=new String[def_var];
for(int j=0;j<def_var;j++)
{
node[i].defvar[j]=str1.nextToken();
}
for(int j=0;j<def_var;j++)
{
System.out.println(node[i].defvar[j]);
}
str1=new StringTokenizer(str.nextToken()," "); // parents node info
p_c=str1.countTokens();
node[i].par=new int[p_c];
for(int j=0;j<p_c;j++)
{
node[i].par[j]=Integer.parseInt(str1.nextToken());
}
for(int j=0;j<p_c;j++)
{
System.out.println(node[i].par[j]);
}
System.out.println("erhiwe");
str1=new StringTokenizer(str.nextToken()," "); // child node info
c_c=str1.countTokens();
node[i].child=new int[c_c];
for(int j=0;j<c_c;j++)
{
node[i].child[j]=Integer.parseInt(str1.nextToken());
}
for(int j=0;j<c_c;j++)
{
System.out.println(node[i].child[j]);
}
node[i].node_type=str.nextToken(); // node type
//System.out.println("sfdsrsr\n\n\n");
int temp=Integer.parseInt(str.nextToken());
System.out.println(temp);
node[i].actIn= new int[temp];
str1=new StringTokenizer(str.nextToken()," ");
for(int temp1=0;str1.hasMoreTokens();temp1++)
node[i].actIn[temp1]=Integer.parseInt(str1.nextToken()); //total actIn are Obtained
temp=Integer.parseInt(str.nextToken());
System.out.println(temp);
node[i].actOut= new int[temp];
str1=new StringTokenizer(str.nextToken()," ");
for(int temp1=0;str1.hasMoreTokens();temp1++)
node[i].actOut[temp1]=Integer.parseInt(str1.nextToken()); //total actOut are Obtained
i++; //next node
}
}
public static void main(String args[])throws Exception
{
Input i=new Input();
System.out.println("\nnew main\n");
for(int k=1;k<=i.n;k++)
{
//System.out.println("\nnew main\n");
System.out.print(i.node[k].marked);
System.out.print(i.node[k].node_no+" a121a ");
for(int j=0;j<i.node[k].usevar.length;j++)
System.out.print(i.node[k].usevar[j]+" ");
for(int j=0;j<i.node[k].defvar.length;j++)
System.out.print(i.node[k].defvar[j]+" ");
for(int j=0;j<i.node[k].child.length;j++)
System.out.print(i.node[k].child[j]+" ");
for(int j=0;j<i.node[k].par.length;j++)
System.out.print(i.node[k].par[j]+" ");
for(int j=0;j<i.node[k].actOut.length;j++)
System.out.print(i.node[k].actOut[j]+" ");
for(int j=0;j<i.node[k].actIn.length;j++)
System.out.print(i.node[k].actIn[j]+" ");
System.out.print(i.node[k].node_type+" ");
System.out.println(" ");
}
}
}
```
When i am trying to compile the code i found the following error
Error: Exception in thread "main" java.util.NoSuchElementException at java.util.StringTokenizer.nextToken(Unknown Source) at Input.(Input.java:84) at Input.main(Input.java:130)
Actually this is an implementation of a slicing algorithm. The class Input.java specifies the Input to the program with the help of a text file. Below I am attaching the text file named Input.txt. the input file is too large in size. Please let me know more about the flaws.
```58\$14
57\$\$\$\$0\$58\$\$\$\$\$
58\$\$\$\$57\$59 60\$\$\$\$\$
59\$\$\$\$58\$6\$\$\$\$\$
60\$\$\$\$58\$33\$\$\$\$\$
6\$7 33 34\$7\$59\$7 8 9 10 12\$\$\$\$\$\$
33\$34 6 7\$34\$60\$67 34 35 36 37 38 39\$\$\$\$\$\$
7\$8 15 34 35\$7 34 \$6\$8\$\$\$\$\$\$
34\$35 7 8\$7 34\$33\$35\$\$\$\$\$\$
35\$36 37 5 7 9 10 14 15\$35\$33 34\$36 37\$\$\$\$\$\$
59\$\$\$\$58\$6\$\$\$\$\$
10\$\$8\$6 8 9\$11 14 17\$\$\$\$\$\$
8\$9 10 14 15 35 36 37\$8\$6 7\$10 15\$\$\$\$\$\$
9\$\$8 35\$6 8\$10 15\$\$\$\$\$\$
36\$\$8 35\$33 35\$37\$\$\$\$\$\$
37\$\$9 35\$6 8\$10 15\$\$\$\$\$\$
38\$\$\$33\$40\$\$\$\$\$\$
39\$\$\$33\$52\$\$\$\$\$\$
40\$\$\$37 38 69\$41 45 46 50\$\$\$\$\$\$
41\$47 48 50 51 70 71 72 74\$41\$40\$47 48\$\$\$\$\$\$
46\$47 49 45\$45\$40\$47 48 49 70\$\$\$\$\$\$
45\$49 45 46\$45\$40\$49\$\$\$\$\$\$
47\$48 45 46 49\$41\$41 45\$48 49\$\$\$\$\$\$
48\$47 49 45 46 70 71 72 74\$41\$41 46 47 50 51\$0\$\$\$\$\$\$
49\$47 49 45 46\$45\$45 46 47\$0\$\$\$\$\$\$
50\$51 41 48\$41\$40\$70 51 48\$\$\$\$\$\$
51\$41 48 50 51 71 72 74 70\$41\$40 50 75\$48\$\$\$\$\$\$
52\$\$\$49\$37 39\$53 54 55 56\$\$\$\$\$
53\$56\$53\$52\$56\$\$\$\$\$\$
54\$56 27 28 29\$54 27\$52 27\$56\$\$\$\$\$\$
55\$56 28 29\$55\$52 28\$56\$\$\$\$\$\$
56\$53 54 55 27 28 29\$53 54 55 27 28\$52 53 54 55\$0\$\$\$\$\$\$
11\$\$\$10\$26\$\$\$\$\$\$
12\$\$\$6 24 66\$13 16\$\$\$\$\$\$
13\$15\$13\$12\$14 15\$\$\$\$\$\$
14\$8 9 10 15 14 35 36 37\$8 35\$10 13\$15\$sleep node\$\$\$\$\$
15\$8 9 10 13 14 15 35 36 37\$8 35 13\$8 9 13 14\$16\$\$\$\$\$\$
16\$\$\$12 15\$0\$\$\$\$\$\$
17\$\$\$\$29 64 10\$19 23 24\$\$\$\$\$
19\$24\$19\$17\$24\$\$\$\$\$\$
23\$23 24 25\$\$17\$24\$\$\$\$\$\$
24\$19 23 24 25\$19\$17 19 23\$12 25\$\$\$\$\$\$
25\$23 24 25\$\$24\$65\$\$\$\$\$\$
26\$\$\$11\$27 28 29\$\$\$\$\$\$
27\$29 54 56\$27 54\$26\$29\$\$\$\$\$\$
28\$29 55 56\$28 55\$26\$29 55\$\$\$\$\$\$
29\$27 28 29 54 55 56\$27 28 54 55\$26 27 28\$17 62\$\$\$\$\$\$
62\$\$\$29\$63 65\$pointcut vertex\$\$\$\$\$
64\$\$\$63\$17\$\$\$\$\$\$
66\$\$\$65\$12\$\$\$\$\$\$
67\$\$\$33\$68 70\$poincut vertex\$\$\$\$\$
69\$\$\$68\$40\$\$\$\$\$\$
70\$71 72 74 41 48 50 70 51\$41\$46 50 67\$71 75 f1_in\$\$1\$f1_in\$\$\$ startingadvice vertex
71\$70 71 72 74 41 48 50 51\$41\$70\$72 74\$\$\$\$\$\$
72\$70 71 72 74 41 48 50 51\$41\$f1_in\$0\$\$\$\$\$\$
74\$70 71 72 74 41 48 50 51\$41\$71 f1_in\$0\$\$\$\$\$\$
75\$\$\$70\$51\$\$\$\$\$\$
```
The algorithm is as follows:
```import java.util.*;
class DynamicSlice
{
String var_name;
int par[];
DynamicSlice(int n)
{
par=new int[n+1]; //no of nodes is=n
}
}
class Algo extends Input
{
int zc=n+1;
DynamicSlice dn[][]=new DynamicSlice[zc][vn];
int rec_def[];
Algo() throws Exception
{}
public void init()
{
rec_def =new int[vn];
for(int i=1;i<n+1;i++)
for(int j=0;j<vn;j++)
dn[i][j]=new DynamicSlice(n);
//set Dynamic slice is creATED
for(int i=1;i<n+1;i++)
for(int j=0;j<vn;j++)
{
dn[i][j].var_name=var_nam[j];
}
}
//variable name is assigned to each position
public int varat(String x)
{
for(int i=0;i<vn;i++)
if(x.equals(var_nam[i]))return(i);
return 0;
} // knowing the position of var in vaiable list
public void a2(int nodeno)
{
for(int l=0; l<node[nodeno].usevar.length;l++)
{
int index = varat(node[nodeno].usevar[l]);
for(int i=1;i<n+1;i++){
int k=dn[nodeno][index].par[i];
if(k!=0)
if(node[k].marked=true)
{
dn[nodeno][index].par[k]=1;
for(int j=1;j<=n;j++)
{
dn[nodeno][index].par[j]=(dn[k][index].par[j]==1)?1:0;
}
}
}
}
}
public void call2a()
{
for(int i=1;i<=n;i++)
a2(node[i].node_no);
}
/* 2 A over*/
/* 2B starts */
public void call2b()
{
for(int i=1;i<=n;i++)
{
for(int j=0;j<node[i].defvar.length;j++)
{
int index = varat(node[i].defvar[j]); //index of def var in varnam list
dn[i][index].par[i]=1; //dynamic slice updated
rec_def[index]=i;
node[rec_def[index]].marked=false;
}
node[i].marked=true;
}
}
/* 2 C and 2 B over */
/* rest start*/
public void other()
{
for(int i=1;i<n+1;i++)
{
if(node[i].node_type.equals("pointcut"))
{
node[i].marked=true;
for(int j=0;j<node[i].actIn.length;j++)
node[node[i].actIn[j]].marked=true;
for(int j=0;j<node[i].actOut.length;j++)
node[node[i].actOut[j]].marked=true;
if(node[i].advice>0) //advice vertex is marked along with the formal in formal out
{
}
}
if(node[i].node_type.equals("sleep"))
{
node[i].marked=true;
}
}
}
public void call3a()
{
System.out.println(" enter the var name and the node no");
int sw,sq;
String arg;
Scanner scan=new Scanner(System.in);
arg=scan.next();
sw=scan.nextInt();
sq=varat(arg); //index no of arg
for(int i=0;i<n;i++)
{
if(dn[sw][sq].par[i]==1)
System.out.print(dn[sw][sq].par[i]+1);
}
}
public static void main(String arg[]) throws Exception
{
//Input input=new Input();
Algo alg=new Algo();
alg.init();
System.out.println("\n\n\n\n\\n\n\n");
alg.call2a();
alg.call2b();
alg.other();
while(true)
{
alg.call3a();
System.out.println(" want to exit then enter 0");
Scanner scan=new Scanner(System.in);
int i=scan.nextInt();
if(i==0)break;
}
}/*end of main*/
}
```
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# 08.100 V2.241 = −
```COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 6.
v 2 − 12 v 0 v 0 − 10
+
+
=0
4
6
2
i1 + i2 + i3 = 0
or v0 = 8.727 V
Chapter 3, Solution 8.
3Ω
i1
v1
i3
5Ω
i2
3V
+
V0
2Ω
+
–
+ 4V0
–
–
1Ω
v1 v1 − 3 v1 − 4 v 0
+
+
=0
5
1
5
2
8
v 0 = v1 so that v1 + 5v1 - 15 + v1 - v1 = 0
5
5
or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V
i1 + i2 + i3 = 0
But
Chapter 3, Solution 9.
Let V1 be the unknown node voltage to the right of the 250-Ω resistor. Let the ground
reference be placed at the bottom of the 50-Ω resistor. This leads to the following nodal
equation:
V1 − 24 V1 − 0 V1 − 60I b − 0
+
+
=0
250
50
150
simplifying we get
3V1 − 72 + 15V1 + 5V1 − 300I b = 0
But I b =
24 − V1
. Substituting this into the nodal equation leads to
250
24.2V1 − 100.8 = 0 or V1 = 4.165 V.
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Thus,
Ib = (24 – 4.165)/250 = 79.34 mA.
Chapter 3, Solution 13.
At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts
But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts
Chapter 3, Solution 20.
Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence
V1 V2 V3
+
+
=0
→ V1 + 4V2 + V3 = 0
(1)
4
1
4
.
V1
.
4Ω
V2
2Ω
1Ω
Between nodes 1 and 3,
− V1 + 12 + V3 = 0
→ V3 = V1 − 12
Similarly, between nodes 1 and 2,
V1 = V2 + 2i
But i = V3 / 4 . Combining this with (2) and (3) gives
. V2
V3
= 6 + V1 / 2
Solving (1), (2), and (4) leads to
V1 = −3V, V2 = 4.5V, V3 = −15V
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
4Ω
(2)
(3)
(4)
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 23.
We apply nodal analysis to the circuit shown below.
1Ω
30 V
2 Vo
4Ω
Vo
+
V1
–
+
+
_
2Ω
Vo
16 Ω
_
3A
At node o,
Vo − 30 Vo − 0 Vo − (2Vo + V1 )
+
+
= 0 → 1.25Vo − 0.25V1 = 30
1
2
4
(1)
At node 1,
(2Vo + V1 ) − Vo V1 − 0
+
− 3 = 0 → 5V1 + 4Vo = 48
4
16
(2)
From (1), V1 = 5Vo – 120. Substituting this into (2) yields
29Vo = 648 or Vo = 22.34 V.
Chapter 3, Solution 34.
(a)
This is a planar circuit because it can be redrawn as shown below,
7Ω
2Ω
1Ω
3Ω
6Ω
10 V
+
–
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
4Ω
© 2007 The McGraw-Hill Companies.
5Ω
COSMOS: Complete Online Solutions Manual Organization System
(b)
This is a non-planar circuit.
Chapter 3, Solution 36.
10 V
4Ω
+–
i1
12 V
+
i2
I1
–
i3
I2
2Ω
6Ω
Applying mesh analysis gives,
12 = 10I1 – 6I2
-10 = -6I1 + 8I2
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
6 5 − 3 I 1
− 5 = − 3 4 I
2
or
∆=
−3
6 −3
5
6
= 11, ∆1 =
= 9, ∆ 2 =
= −7
−3 4
−5 4
−3 −5
5
I1 =
∆1
9 I = ∆2 = − 7
=
, 2
∆
11
∆
11
i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 – I2 = 10/11 = 1.4545 A.
vo = 6i2 = 6x1.4545 = 8.727 V.
Chapter 3, Solution 38.
Consider the circuit below with the mesh currents.
4Ω
+
_
24 V
3Ω
I3
I4
1Ω
4A
2Ω
2Ω
Io
1Ω
1Ω
I1
I2
4Ω
2A
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
+
_
9V
COSMOS: Complete Online Solutions Manual Organization System
I1 =-2 A
(1)
1(I2–I1) + 2(I2–I4) + 9 + 4I2 = 0
7I2 – I4 = –11
(2)
–24 + 4I3 + 3I4 + 1I4 + 2(I4–I2) + 2(I3 – I1) = 0 (super mesh)
–2I2 + 6 I3 + 6I4 = +24 – 4 = 20
(3)
But, we need one more equation, so we use the constraint equation –I3 + I4 = 4. This now
gives us three equations with three unknowns.
0 − 1 I 2 − 11
7
− 2 6 6 I = 20
3
0 − 1 1 I 4 4
We can now use MATLAB to solve the problem.
>> Z=[7,0,-1;-2,6,6;0,-1,0]
Z=
7 0 -1
-2 6 6
0 -1 0
>> V=[-11,20,4]'
V=
-11
20
4
>> I=inv(Z)*V
I=
-0.5500
-4.0000
7.1500
Io = I1 – I2 = –2 – 4 = –6 A.
Chapter 3, Solution 86.
Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then,
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
[(0.03 – v1)/1k] + 400i = v1/2k
(1)
Assume that i is in mA. But, i = (0.03 – v1)/1
(2)
Combining (1) and (2) yields,
v1 = 29.963 mVolts and i = 37.4 nA, therefore,
v0 = -5000x400x37.4x10-9 = -74.8 mvolts
Chapter 4, Solution 1.
1Ω
1V
8 (5 + 3) = 4Ω , i =
io =
5Ω
i
+
−
io
8Ω
3Ω
1
1
=
1+ 4 5
1
1
i=
= 0.1A
2
10
Since the resistance remains the same we get i = 10/5 = 2A which leads to
io = (1/2)i = (1/2)2 = 1A.
Chapter 4, Solution 3.
R
3R
io
Vs
+
−
3R
3R
+
1V
R
+
−
3R
vo
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
(a)
(b)
1.5R
COSMOS: Complete Online Solutions Manual Organization System
(a) We transform the Y sub-circuit to the equivalent ∆ .
3R 2 3
3
3
3
R 3R =
= R, R + R = R
4R
4
4
4
2
v
v o = s independent of R
2
io = vo/(R)
When vs = 1V, vo = 0.5V, io = 0.5A
(b)
(c)
When vs = 10V, vo = 5V, io = 5A
When vs = 10V and R = 10Ω,
vo = 5V, io = 10/(10) = 500mA
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
```
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# 5.2.5.40 UK invalidity benefit - March 1989 to November 1996
## Summary
This topic contains historical rates for UK invalidity benefit from 16 March 1989 to 18 June 1992. This topic contains the following tables:
Table Number Explanation
1 Rates for a recipient WITHOUT children
2 Rates for a partnered (1.1.P.85) recipient WITH children
3 Exchange rate - 9 December 1976 to 19 February 1987
4 Exchange rate - 4 February 1988 to 28 November 1996
## Table 1: Rates for a recipient WITHOUT children
The following table shows the UK invalidity benefit for a recipient WITHOUT children. Payment rates are shown in dollars per fortnight. Exchange rates show the value of \$A1.00 in Pounds Sterling.
Date Exchange rate Single Partnered
16/03/1989 0.4800 329.20 329.20
12/10/1989 0.5000 211.20 316.00
15/03/1990 0.4500 234.70 351.10
10/05/1990 0.4513 252.16 377.13
05/07/1990 0.4574 248.80 372.10
27/09/1990 0.4286 265.52 397.11
20/12/1990 0.4165 273.23 408.64
28/03/1991 0.4069 310.15 463.75
20/06/1991 0.4477 281.90 421.50
26/09/1991 0.4720 267.40 399.80
19/12/1991 0.4643 271.80 406.40
26/03/1992 0.4159 316.00 472.50
18/06/1992 0.4383 299.80 448.30
## Table 2: Rates for a partnered recipient WITH children
The following table shows the UK invalidity benefit for a partnered recipient WITH children. Rates are shown in dollars per fortnight.
Date Exchange rate Number of children
1 2 3 4 5 6
16/03/1989 0.4800 366.50 403.80 441.00 478.30 515.60 552.90
12/10/1989 0.5000 351.80 387.60 423.40 459.20 495.00 530.80
15/03/1990 0.4500 390.90 430.70 470.40 510.20 550.00 589.80
10/05/1990 0.4513 419.90 462.66 505.43 548.19 590.96 633.72
05/07/1990 0.4574 414.30 456.49 498.69 540.88 583.08 625.27
27/09/1990 0.4286 442.14 487.17 532.20 577.23 622.26 667.29
20/12/1990 0.4165 454.98 501.32 547.66 594.00 640.34 686.67
28/03/1991 0.4069 516.34 568.94 621.53 674.12 726.71 779.31
20/06/1991 0.4477 469.30 517.10 564.90 612.70 660.50 708.30
26/09/1991 0.4720 445.20 490.50 535.90 581.20 626.50 671.90
19/12/1991 0.4643 452.50 498.60 544.70 590.80 636.90 683.00
26/03/1992 0.4159 524.70 576.80 629.00 681.20 733.40 785.50
18/06/1992
Note C
0.4383 497.80 547.30 596.90 646.40 695.90 745.40
### Notes for Table 2
Notes Explanation
A When applying the formula on other dates, the notional exchange rate for that date/s, shown in the following blocks, must be used. Each time the UK notional exchange rate was varied, a National Instruction setting out UK invalidity benefit rates in \$A was also issued.
B A revised Agreement with the UK came into effect on pension payday 2 July 1992. Under the revised agreement reciprocal coverage for DSP was no longer included. However, where an Australian DSP was paid under the previous agreement (with its rate limited to the rate of UK indefinite sickness benefit) that pension can continue to be paid but the rate is assessed in accordance with the usual rating provisions of the Act.
C Payday 18 June 1992 is the last rate update.
## Table 3: Exchange rate - 9 December 1976 to 19 February 1987
The following table shows the exchange rate for one Pound Sterling to \$A from 9 December 1976 to 19 February 1987.
Date of effect Australian dollars
09/12/1976 1.60
31/03/1977 1.49
21/07/1977 1.55
02/02/1978 1.60
30/03/1978 1.68
06/07/1978 1.60
18/01/1979 1.72
10/05/1979 1.85
27/09/1979 1.95
27/03/1980 2.05
12/02/1981 2.00
04/06/1981 1.80
16/07/1981 1.70
27/08/1981 1.60
22/10/1981 1.55
03/12/1981 1.60
25/02/1982 1.65
26/08/1982 1.70
02/12/1982 1.75
30/12/1982 1.65
10/02/1983 1.60
07/04/1983 1.70
21/04/1983 1.65
02/06/1983 1.75
06/10/1983 1.70
15/12/1983 1.60
09/02/1984 1.55
22/03/1984 1.50
23/08/1984 1.55
18/10/1984 1.45
07/02/1985 1.40
18/04/1985 1.55
30/05/1985 1.85
08/08/1985 1.95
28/11/1985 2.00
09/01/1986 2.15
20/02/1986 1.95
03/04/1986 2.00
26/06/1986 2.05
24/07/1986 2.15
07/08/1986 2.30
18/09/1986 2.40
16/10/1986 2.30
27/11/1986 2.15
19/02/1987 2.25
## Table 4: Exchange rate - 4 February 1988 to 28 November 1996
From 4 February, the exchange rate between \$A and Pounds Sterling was expressed differently. The following table shows the exchange rate for \$A1.00 to Pounds Sterling from 4 February 1988 to 28 November 1996.
Date Pound Sterling
04/02/1988 0.4000
26/05/1988 0.4150
07/07/1988 0.4500
21/07/1988 0.4750
16/02/1989 0.4900
02/03/1989 0.5200
16/03/1989 0.4800
12/10/1989 0.5000
15/03/1990 0.4500
10/05/1990 0.4513
05/07/1990 0.4574
27/09/1990 0.4286
20/12/1990 0.4165
28/03/1991 0.4069
20/06/1991 0.4477
26/09/1991 0.4720
19/12/1991 0.4643
26/03/1992 0.4159
18/06/1992 0.4383
24/09/1992 0.3914
19/11/1992 0.4485
25/02/1993 0.4707
08/04/1993 0.4910
06/05/1993 0.4693
17/06/1993 0.4499
26/08/1993 0.4639
07/10/1993 0.4277
04/11/1993 0.4399
18/11/1993 0.4559
10/02/1994 0.4708
10/03/1994 0.4902
28/07/1994 0.4762
17/11/1994 0.4576
01/12/1994 0.4723
15/12/1994 0.4898
09/03/1995 0.4770
20/04/1995 0.4576
07/09/1995 0.4774
05/10/1995 0.4899
04/04/1996 0.5050
02/05/1996 0.5191
05/09/1996 0.5059
28/11/1996 0.4873
Last reviewed: 12 August 2019
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## NY-GEO-G.CO.10: Prove and apply theorems about triangles.
There are 78 resources.
Title Description Thumbnail Image Curriculum Topics
## Math Examples Collection: Solving Equations Using Triangle Properties
This collection aggregates all the math examples around the topic of Solving Equations Using Triangle Properties. There are a total of 10 Math Examples. This collection of resources is made up of downloadable PNG images that you can easily incorporate into your lesson plans. Applications of Equations and Inequalities and Applications of Triangles
## Math Definitions Collection: Geometry Theorems and Postulates
This collection aggregates all the definition image cards around the topic of Theorems and Postulates terms and vocabulary. There are a total of 18 terms. Definition of a Triangle, Definition of an Angle, Right Triangles and Definition of a Line
## Math Clip Art Collection: Angles
OverviewThis collection aggregates all the math clip art around the topic of Angles. There are a total of 22 images. Angles and Applications of Angles and Planes
## Math Examples Collection: Geometric Proofs
OveThis collection aggregates all the math examples around the topic of Geometric Proofs. There are a total of 13 Math Examples. Parallel Lines and Definition of a Triangle
## Math Examples Collection: Equations with Angle Measures
This collection aggregates all the math examples around the topic of Solving Equations with Angle Measures. There are a total of 25 Math Examples. Solving Multistep Equations, Applications of Angles and Planes, Applications of Quadrilaterals, Applications of Equations and Inequalities and Definition of an Angle
## Definition--Theorems and Postulates--AAA Theorem
Definition--Theorems and Postulates--AAA Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of a Triangle
## Definition--Theorems and Postulates--AAS Theorem
Definition--Theorems and Postulates--AAS Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of a Triangle
## Definition--Theorems and Postulates--Alternate Interior Angles Theorem
Definition--Theorems and Postulates--Alternate Interior Angles Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of an Angle
## Definition--Theorems and Postulates--ASA Theorem
Definition--Theorems and Postulates--ASA Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of an Angle
## Definition--Theorems and Postulates--Converse of Same-Side Interior Angles Postulate
Definition--Theorems and Postulates--Converse of Same-Side Interior Angles Postulate
This is part of a collection of definitions of geometric theorems and postulates.
Definition of an Angle
## Definition--Theorems and Postulates--Converse of the Alternate Interior Angles Theorem
Definition--Theorems and Postulates--Converse of the Alternate Interior Angles Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of an Angle
## Definition--Theorems and Postulates--Converse of the Corresponding Angles Theorem
Definition--Theorems and Postulates--Converse of the Corresponding Angles Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of an Angle
## Definition--Theorems and Postulates--Corresponding Angles Theorem
Definition | Theorems and Postulates | Corresponding Angles Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of an Angle
## Definition--Theorems and Postulates--Same-Side Interior Angles Postulate
Definition--Theorems and Postulates--Same-Side Interior Angles Postulate
This is part of a collection of definitions of geometric theorems and postulates.
Definition of a Triangle
## Definition--Theorems and Postulates--SAS Similarity Theorem
Definition--Theorems and Postulates--SAS Similarity Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of a Triangle
## Definition--Theorems and Postulates--SAS Theorem
Definition--Theorems and Postulates--SAS Theorem
This is part of a collection of definitions of geometric theorems and postulates.
Definition of a Triangle
## Definition--Theorems and Postulates--SSS Postulate
Definition--Theorems and Postulates--SSS Postulate
This is part of a collection of definitions of geometric theorems and postulates.
Definition of a Triangle
## Math Clip Art--Angle Illustrations--Supplementary Angles--Labeled
Math Clip Art--Angle Illustrations--Supplementary Angles--Labeled
This is part of a collection of clip art images showing different types of angles. Each type of angle measure includes a labeled and unlabeled version.
Angles and Applications of Angles and Planes
## Math Clip Art--Angle Illustrations--Supplementary Angles--Unlabeled
Math Clip Art--Angle Illustrations--Supplementary Angles--Unlabeled
This is part of a collection of clip art images showing different types of angles. Each type of angle measure includes a labeled and unlabeled version.
Angles and Applications of Angles and Planes
## Math Example--Geometric Proofs--Example 1
Math Example--Geometric Proofs--Example 1
This is part of a collection of math examples that focus on geometric proofs.
Parallel Lines and Definition of a Triangle
## Math Example--Geometric Proofs--Example 10
Math Example--Geometric Proofs--Example 10
This is part of a collection of math examples that focus on geometric proofs.
Parallel Lines and Definition of a Triangle
## Math Example--Geometric Proofs--Example 11
Math Example--Geometric Proofs--Example 11
This is part of a collection of math examples that focus on geometric proofs.
Parallel Lines and Definition of a Triangle
## Math Example--Geometric Proofs--Example 12
Math Example--Geometric Proofs--Example 12
This is part of a collection of math examples that focus on geometric proofs.
Parallel Lines and Definition of a Triangle
## Math Example--Geometric Proofs--Example 13
Math Example--Geometric Proofs--Example 13
This is part of a collection of math examples that focus on geometric proofs.
Parallel Lines and Definition of a Triangle
## Math Example--Geometric Proofs--Example 2
Math Example--Geometric Proofs--Example 2
This is part of a collection of math examples that focus on geometric proofs.
Parallel Lines and Definition of a Triangle
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# dendropy.calculate.statistics: General Statistics¶
Functions to calculate some general statistics.
class dendropy.calculate.statistics.FishersExactTest(table)[source]
Given a 2x2 table:
a b c d
represented by a list of lists:
[[a,b],[c,d]]
this calculates the sum of the probability of this table and all others more extreme under the null hypothesis that there is no association between the categories represented by the vertical and horizontal axes.
left_tail_p()[source]
Returns the sum of probabilities of this table and all others more extreme.
static probability_of_table()[source]
Given a 2x2 table:
a b c d
represented by a list of lists:
[[a,b],[c,d]]
this returns the probability of this table under the null hypothesis of no association between rows and columns, which was shown by Fisher to be a hypergeometric distribution:
p = ( choose(a+b, a) * choose(c+d, c) ) / choose(a+b+c+d, a+c)
right_tail_p()[source]
Returns the sum of probabilities of this table and all others more extreme.
two_tail_p()[source]
Returns the sum of probabilities of this table and all others more extreme.
dendropy.calculate.statistics.empirical_cdf(values, v)[source]
Returns the proportion of values in values <= v.
dendropy.calculate.statistics.empirical_hpd(values, conf=0.05)[source]
Assuming a unimodal distribution, returns the 0.95 highest posterior density (HPD) interval for a set of samples from a posterior distribution. Adapted from emp.hpd in the “TeachingDemos” R package (Copyright Greg Snow; licensed under the Artistic License).
dendropy.calculate.statistics.mean_and_population_variance(values)[source]
Returns the mean and population variance while only passing over the elements in values once.
dendropy.calculate.statistics.mean_and_sample_variance(values)[source]
Returns the mean and sample variance while only passing over the elements in values once.
dendropy.calculate.statistics.median(pool)[source]
Returns median of sample. From: http://wiki.python.org/moin/SimplePrograms
dendropy.calculate.statistics.mode(values, bin_size=0.1)[source]
Returns the mode of a set of values.
dendropy.calculate.statistics.quantile_5_95(values)[source]
Returns 5% and 95% quantiles.
dendropy.calculate.statistics.rank(value_to_be_ranked, value_providing_rank)[source]
Returns the rank of value_to_be_ranked in set of values, values. Works even if values is a non-orderable collection (e.g., a set). A binary search would be an optimized way of doing this if we can constrain values to be an ordered collection.
dendropy.calculate.statistics.summarize(values)[source]
Summarizes a sample of values:
• range : tuple pair representing minimum and maximum values
• mean : mean of sample
• median : median of sample
• var : (sample) variance
• sd : (sample) standard deviation
• hpd95 : tuple pair representing 5% and 95% HPD
• quant_5_95 : tuple pair representing 5% and 95% quantile
dendropy.calculate.statistics.variance_covariance(data, population_variance=False)[source]
Returns the Variance-Covariance matrix for data. From: http://www.python-forum.org/pythonforum/viewtopic.php?f=3&t=17441
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