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# Tag: times

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## Visualization of a set of data with a shape $ m times n times 3 $ with the last dimension interpreted as a color scale

## java – Programming challenge – report fraudulent banking activity times out

## reference request – The correlation of hitting times of a process with correlated increments

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## visas – Apostille two times of the same document with a translation added

## I Need an Idea For a Site That Can Be Sold Multiple Times

## parametric functions – Integrating Superposition of several sinusoidal waves squared times its derivative

## google sheets – Array Formula to calculate times

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Actually, if the data is shaped as $ m times n times 2 $, functions like `ListLinePlot`

just do the job. But now I want to add colors to the curves according to each 3rd data. But it has stuck me. Any ideas?

How can I optimise this code

there is time out error

Program Challenge:

HackerLand National Bank has a simple policy for warning clients about possible fraudulent account activity. If the amount spent by a client on a particular day is greater than or equal to the client’s median spending for a trailing number of days, they send the client a notification about potential fraud. The bank doesn’t send the client any notifications until they have at least that trailing number of prior days’ transaction data.

Given the number of trailing days and a client’s total daily expenditures for a period of days, find and print the number of times the client will receive a notification over all days.For example,

and . On the first three days, they just collect spending data. At day , we have trailing expenditures of . The median is and the day’s expenditure is . Because , there will be a notice. The next day, our trailing expenditures are and the expenditures are . This is less than so no notice will be sent. Over the period, there was one notice sent.Note: The median of a list of numbers can be found by arranging all the numbers from smallest to greatest. If there is an odd number of numbers, the middle one is picked. If there is an even number of numbers, median is then defined to be the average of the two middle values. (Wikipedia)

Function Description

Complete the function activityNotifications in the editor below. It must return an integer representing the number of client notifications.

activityNotifications has the following parameter(s):

expenditure: an array of integers representing daily expenditures

d: an integer, the lookback days for median spendingInput Format

The first line contains two space-separated integers and, the number of days of transaction data, and the number of trailing days’ data used to calculate median spending.

The second line contains space-separated non-negative integers where each integer denotes.Constraints

Output Format

Print an integer denoting the total number of times the client receives a notification over a period of days.Sample Input 0

9 5

2 3 4 2 3 6 8 4 5Sample Output 0

2

HackerRank fraudulent activity.

```
public static int() CountSort(int ()arr,int si,int ei) {
int ()count=new int(201);
int result()=new int(ei-si);
for(int i=si;i<ei;i++) {
count(arr(i))++;
}
for(int i=1;i<count.length;i++) {
count(i)+=count(i-1);
}
for(int i=ei-1;i>=si;i--) {
result(--count(arr(i)))=arr(i);
}
return result;
}
// Complete the activityNotifications function below.
static int activityNotifications(int() expenditure, int d) {
int notice=0;
for(int i=0;i<=expenditure.length-d;i++) {
int arr()=new int(d);
double median=0;
arr=CountSort(expenditure,i,i+d);
if(d%2==0) {
int mid1=(d)/2;
int mid2=(d)/2 +1;
median=(arr(mid1)+arr(mid2))/2;
}else {
int mid=(d)/2;
median=arr(mid);
}
if(i+d<expenditure.length&&expenditure(i+d)>=2*median) {
notice++;
}
}
return notice;
}
```

The general question

Q1: Suppose a process $X_{s}$ has dependent increments i.e. $Cor(X_{t_{1}},X_{t_{3}}-X_{t_{2}}))neq 0$ (for $t_{1}<t_{2}<t_{3}$), how can one go about studying the correlation of its stopping times increments $T_{a}$ and $T_{c}-T_{b}$? Has anyone seen this question before in some context? Any references will be much appreciated so that I can try to extract any techniques from there.

In the case of a process with independent increments, the strong Markov property immediately implies that the stopping time increments are also independent.

Our particular setting is we have a stationary Gaussian process $V(s):(0,1)to mathbb{R}$ with short-range linear correlation

$$E(V(s)V(t))=left{begin{matrix}

a-|t-s|b &, |t-s|< epsilon\

0, & |t-s|geq epsilon

end{matrix}right.$$

for $a,b,epsilon>0$. Since V is a Gaussian process, this means that if $|t-s|geq epsilon$, then $V_{s},V_{t}$ are independent of each other. Then we consider its integrated exponential:

$$mu((0,t)):=int_{0}^{t}e^{V(s)}ds$$

and its inverse measure $T_{a}:=Q((0,a)):=mu^{-1}(0,a)$. The project is trying to understand the correction of consecutive stopping times increments: for $a<b<c$ the correlation of

$$T_{a}text{ and } T_{c}-T_{b}.$$

The measure $mu(t)$ doesn’t have independent increments i.e. $mu(t_{1})$ and $mu((t_{2},t_{3}))$ (for $t_{1}<t_{2}<t_{3}$) are correlated when

$$t_{2}<t_{1}+epsilon<t_{3}.$$

Q2: So inspired by this inequality we try to estimate the mixing condition by that: Can we obtain an estimate of the form

$$ |P(T_{c}-T_{b}geq t| T_{a}geq s)-P(T_{c}-T_{b}geq t)|leq C P(T_{b}leq epsilon+T_{a}<T_{c}),$$

for some constant uniform constant C that doesn’t depent on t,s.

**Approaches**

1)We can write the difference as

$$T_{c}-T_{b}=inf{tgeq 0: X_{b}(t):=mu(T_{b},T_{b}+t)geq c-b }=:T_{c-b}^{X_{b}}.$$

So then the question is what is the correlation of $mu(T_{b},T_{b}+t)$ with ${T_{a}geq s}$. The correlation comes from

$$T_{b}=T_{a}+T_{b-a}^{X_{a}}.$$

In particular by splitting

$$X_{a}(t)=mu(T_{a},T_{a}+epsilon)+mu(T_{a}+epsilon,T_{a}+t)=:g_{a}(epsilon)+widetilde{X}_{a}(t),$$

we understand that $mu(T_{b},T_{b}+t)$ and ${T_{a}geq s}$ are correlated via the pieces $g_{a}(epsilon)$ and $T_{a}$ in

$$T_{b}=T_{a}+T_{b-a}^{widetilde{X}_{a}+g_{a}(epsilon)}.$$

So here the question shifts to getting sharp deterministic bounds $b(epsilon) leq g_{a}(epsilon)leq B(epsilon)$ that are true with high probability. That will then allow to replace by independent random variables:

$$T_{b-a}^{widetilde{X}_{a}+B(epsilon)}leq T_{b-a}^{widetilde{X}_{a}+g_{a}(epsilon)}leq T_{b-a}^{widetilde{X}_{a}+b(epsilon)}$$

with high probability. (There are some details here, such intersecting by the event ${T_{b-a}^{widetilde{X}_{a}+g_{a}(epsilon)}geq epsilon}$).

Q3:I feel this approach must have been done already in some context. So it would be great to study how they structured it.

2)A toy model for this project is in Correlation of stopping times for integral of Brownian motion increment.

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Discussion in ‘Growing and Managing a Business’ started by Kristina Wilson, Jun 22, 2020 at 5:52 AM.

I don’t know what is exactly meant by the these document requirements:

1- certified copy of the high school diploma with apostille;

2- translation of the high school diploma into Italian performed by an official translator (sworn at the Court) and equipped with apostille;

Now, am I required to do two apostilles, this is impossible, because Apostille is issued for a single unique document, either as original or legalized, not translation.

What is meant?

I want to sell starter sites on Flippa but I don’t know where to start. I listed a site recently but no one was interested, no bids at all.

Please help me come up with ideas by sharing your ideas.

I am not looking for a blog or a forum. I want a web app or a similar idea.

Any suggestions would be much appreciated,

Farris

I’m currently finishing some research from a internship I did last summer and part of the project involves calculating the centroids of coronal holes from synoptic maps. If you aren’t aware of what these are, they are basically holes or regions of the corona, an upper layer of the suns atmosphere, where it is less dense. The purpose of this is so I can, track how the motion of these coronal holes change over time. The way this is done is by taking a certain latitude band, say 10 degrees south to 10 degrees north, finding the centroids for coronal holes after each carrington rotation, solar month, and finding the slope with all of these points. In general, calculating these centroids is not an easy task because in general the shapes of these regions, on solar maps, are non uniform and can take on any flat 2D shape. However last summer I did find a solution that should work, at least theoretically. Essentially my idea was to generate parametric plots that look identical to the images of each coronal hole and then use green’s theorem, with the parametric equations, to calculate the centroids through line integration. Calculus 3 was still fresh in my mind at that point and I thought, if we can use Green’s Theorem to calculate area, why not centroids as well? After doing some research I found that these line integrals do in fact exist and I went through the derivations to prove to myself that the line integrals were equivalent to the double integrals used to calculate centroids. If you are wondering what these line integrals are, here they are.

And here are the double integrals that the line integrals are equivalent to.

My mentors had another method that they had planned for me to use on IDL but encouraged me to try applying my method instead. I ended up working on Mathematica instead, because of useful code I found on other forums. From there I added some of my own code and did a bunch of tinkering over the course of several weeks. At the time, I did not know if this was the most efficient method computationally and I still don’t but I would still like to get this to be fully functional. I got the method to work for the most part, but at times I’m still having issues getting an accurate answer. Sometimes it gives an answer that looks reasonable but at other times it gives an answer that is obviously completely off just by eyeballing it. I believe that the issue solely lies in convergence issues with the integrals but I will show all of my code, just in case some of the issues lie elsewhere.

```
param(x_, m_, t_) :=
Module({f, n = Length(x), nf},
f = Chop(Fourier(x))((;; Ceiling(Length(x)/2)));
nf = Length(f);
Total(Rationalize(
2 Abs(f)/Sqrt(n) Sin(
Pi/2 - Arg(f) + 2. Pi Range(0, nf - 1) t), .01)((;;
Min(m, nf)))))
tocurve(Line(data_), m_, t_) := param(#, m, t) & /@ Transpose(data)
img = !(*
GraphicsBox(
TagBox(RasterBox(CompressedData("
1:eJzt229sE3Ucx/HP/f/dXa/XrmUmiFl8hCgzMWbBTDfnnH+GmyYVE+IziJMY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"), {{0, 26}, {367, 0}}, {0, 255},
ColorFunction->RGBColor),
BoxForm`ImageTag("Byte", ColorSpace -> "RGB", Interleaving -> True),
Selectable->False),
DefaultBaseStyle->"ImageGraphics",
ImageSizeRaw->{367, 26},
PlotRange->{{0, 367}, {0, 26}}))
img = Binarize(img~ColorConvert~"Grayscale");
lines = Cases(
Normal@ListContourPlot(Reverse@ImageData(img),
Contours -> {0.5}), _Line, -1);
ParametricPlot(
Evaluate(tocurve(#, 100000000, t) & /@ lines), {t, 0, 1},
Frame -> True, Axes -> False, ImageSize -> Large)
ParametricEquations = Evaluate(tocurve(#, 100000000, t) & /@ lines)
n = Length(ParametricEquations)
img2 = Binarize(img~ColorConvert~"Grayscale");
bsFs = Cases(
Normal@ListContourPlot(Reverse@ImageData(img2), Contours -> {0.5}),
Line(x_) :> BSplineFunction(x), All);
ParametricPlot(Evaluate(Through@bsFs@t), {t, 0, 1}, Frame -> True,
Axes -> False, ImageSize -> Large, Axes -> false, Frame -> True)
stepsize = 0.0001
list = Table(
N(Evaluate(Through@bsFs@t) /. t -> i), {t, 0, 1, stepsize})
xmatrix =
Table(Table(list((i, j, 1)), {i, 1, (1/stepsize) + 1}), {j, 1, n})
ymatrix =
Table(Table(list((i, j, 2)), {i, 1, (1/stepsize) + 1}), {j, 1, n})
Minx = Table(xmatrix((i, Ordering(xmatrix((i)), 1))), {i, 1, n})
Miny = Table(ymatrix((i, Ordering(ymatrix((i)), 1))), {i, 1, n})
x = Table(ParametricEquations((i, 1)), {i, 1, n})
y = Table(ParametricEquations((i, 2)), {i, 1, n})
Mx = Table(NMinValue({x((i)), 0 <= t <= 1}, t), {i, 1, n})
My = Table(NMinValue({y((i)), 0 <= t <= 1}, t), {i, 1, n})
NewParametricEquations =
Table({x((i)) - (Mx((i)) - Minx((i))),
y((i)) - (My((i)) - Miny((i)))}, {i, 1, n})
xCentroids =
Table(NIntegrate(
NewParametricEquations((i, 1))^2*
D(NewParametricEquations((i, 2)), t), {t, 0, 1})/
NIntegrate(
2*NewParametricEquations((i, 1))*
D(NewParametricEquations((i, 2)), t), {t, 0, 1}), {i, 1, n})
yCentroids =
Table(NIntegrate(-NewParametricEquations((i, 2))^2*
D(NewParametricEquations((i, 1)), t), {t, 0, 1})/
NIntegrate(-2*NewParametricEquations((i, 2))*
D(NewParametricEquations((i, 1)), t), {t, 0, 1}), {i, 1, n})
Centroids = Table({xCentroids((i)), yCentroids((i))}, {i, 1, n})
plot = ParametricPlot(NewParametricEquations, {t, 0, 1},
ImageSize -> Large, Axes -> False, Frame -> True, PlotRange -> All);
centroidpoints =
ListPlot(Table({First(xCentroids((i))), First(yCentroids((i)))}, {i,
1, n}) -> Range(n), LabelingFunction -> Bottom,
PlotStyle -> {Red, PointSize(Medium)});
coronalholepoints =
ListPlot(Table({First(N(NewParametricEquations((i, 1)) /. t -> 0)),
First(N(NewParametricEquations((i, 2)) /. t -> 0))}, {i, 1,
n}) -> Range(n), LabelingFunction -> Left,
PlotStyle -> {Blue, PointSize(Medium)});
Show(plot, centroidpoints, coronalholepoints)
```

Here is an example of an image with some coronal holes I attempted to calculate the centroids of. All other colors have been filtered out in order to generate accurate parametric equations for each coronal hole (the blue regions in this case, which indicates that the coronal holes have a positive polarity).

And here is an image of the parametric plot of all the coronal holes, along with red dots which represent the calculated centroids.

As you can see, the first two centroids look reasonable but the last one is way off and isn’t even inside the coronal hole. As a reference, here is a parametric equations that Mathematica generated for the coronal holes.

```
{{{24.7056 - 1/45 Sin(1/11 - (3349 t)/13) -
1/77 Sin(3/8 - (754 t)/3) - 1/23 Sin(8/9 - (3217 t)/16) -
1/76 Sin(1/18 - (377 t)/2) - 1/28 Sin(1/4 - (1640 t)/9) -
1/30 Sin(7/13 - (1797 t)/11) - 1/53 Sin(2/9 - (1885 t)/12) -
1/23 Sin(22/15 - (4191 t)/29) - 1/28 Sin(5/9 - (2111 t)/16) -
1/22 Sin(7/5 - (2199 t)/25) - 1/16 Sin(5/4 - (1552 t)/19) -
1/9 Sin(1/4 - (553 t)/8) - 1/11 Sin(8/9 - (377 t)/6) -
1/8 Sin(13/10 - (553 t)/11) - 59/15 Sin(13/12 - (88 t)/7) +
1/5 Sin((1891 t)/43) + 19/5 Sin(17/5 + (44 t)/7) +
5/4 Sin(103/26 + (132 t)/7) + 2/3 Sin(25/14 + (201 t)/8) +
6/13 Sin(47/14 + (377 t)/12) + 1/7 Sin(13/4 + (377 t)/10) +
1/9 Sin(25/6 + (509 t)/9) + 1/7 Sin(5/14 + (377 t)/5) +
1/24 Sin(23/14 + (377 t)/4) + 1/24 Sin(45/13 + (1307 t)/13) +
1/16 Sin(8/5 + (1175 t)/11) + 1/11 Sin(94/21 + (1131 t)/10) +
1/71 Sin(31/7 + (955 t)/8) + 1/25 Sin(13/5 + (377 t)/3) +
1/27 Sin(3/14 + (1244 t)/9) + 1/19 Sin(5/4 + (754 t)/5) +
1/33 Sin(65/22 + (2287 t)/13) + 1/23 Sin(58/19 + (1753 t)/9) +
1/27 Sin(3/2 + (3525 t)/17) + 1/55 Sin(20/7 + (1709 t)/8) +
1/65 Sin(14/3 + (2419 t)/11) + 1/28 Sin(10/19 + (1131 t)/5) +
1/85 Sin(49/11 + (3952 t)/17) + 1/43 Sin(10/3 + (3104 t)/13) +
1/33 Sin(5/11 + (5391 t)/22) + 1/56 Sin(7/2 + (2375 t)/9) +
1/43 Sin(14/15 + (3041 t)/11) + 1/92 Sin(41/11 + (1131 t)/4) +
1/80 Sin(32/9 + (2953 t)/10)}, {14.8678 -
1/86 Sin(5/4 - (1885 t)/6) - 1/83 Sin(16/15 - (3349 t)/13) -
1/21 Sin(6/5 - (3217 t)/16) - 1/21 Sin(3/5 - (1640 t)/9) -
1/46 Sin(7/11 - (1797 t)/11) - 1/36 Sin(7/8 - (1244 t)/9) -
1/11 Sin(4/7 - (2111 t)/16) - 1/44 Sin(17/18 - (955 t)/8) -
1/34 Sin(7/6 - (1307 t)/13) - 1/15 Sin(1 - (2199 t)/25) -
1/7 Sin(23/15 - (553 t)/11) - 3/13 Sin(23/15 - (377 t)/10) -
2/5 Sin(7/10 - (377 t)/12) - 8/11 Sin(4/3 - (132 t)/7) -
10/7 Sin(5/6 - (88 t)/7) - 145/18 Sin(13/9 - (44 t)/7) +
4/11 Sin(12/11 + (201 t)/8) + 2/9 Sin(1/3 + (1627 t)/37) +
1/8 Sin(14/5 + (509 t)/9) + 1/24 Sin(137/46 + (377 t)/6) +
1/10 Sin(35/12 + (553 t)/8) + 1/66 Sin(5/8 + (377 t)/5) +
1/9 Sin(57/14 + (1307 t)/16) + 1/30 Sin(5/6 + (1175 t)/11) +
1/22 Sin(50/11 + (1131 t)/10) + 1/21 Sin(8/3 + (377 t)/3) +
1/30 Sin(22/5 + (3324 t)/23) + 1/22 Sin(26/11 + (754 t)/5) +
1/27 Sin(34/15 + (1885 t)/12) + 1/24 Sin(19/7 + (1866 t)/11) +
1/69 Sin(23/8 + (2287 t)/13) + 1/23 Sin(1/5 + (377 t)/2) +
1/30 Sin(11/3 + (1753 t)/9) + 1/21 Sin(37/11 + (1709 t)/8) +
1/64 Sin(29/7 + (2419 t)/11) + 1/38 Sin(10/19 + (1131 t)/5) +
1/31 Sin(4/5 + (5391 t)/22) + 1/36 Sin(29/8 + (2375 t)/9) +
1/86 Sin(5/7 + (3041 t)/11) +
1/64 Sin(48/11 + (1131 t)/4)}}, {{217.089 -
1/41 Sin(36/35 - (2375 t)/9) - 1/65 Sin(2/3 - (3952 t)/17) -
1/17 Sin(1/30 - (3217 t)/16) - 1/18 Sin(31/21 - (2287 t)/13) -
1/19 Sin(1 - (1866 t)/11) - 1/22 Sin(3/13 - (1885 t)/12) -
1/12 Sin(3/7 - (4191 t)/29) - 1/25 Sin(11/12 - (377 t)/3) -
1/7 Sin(11/9 - (1552 t)/19) - 1/4 Sin(5/9 - (377 t)/5) -
4/15 Sin(1/9 - (509 t)/9) - 2/11 Sin(3/10 - (377 t)/12) -
9/7 Sin(7/6 - (201 t)/8) - 24/25 Sin(1/8 - (88 t)/7) +
82/15 Sin(51/13 + (44 t)/7) + 7/9 Sin(4 + (132 t)/7) +
3/11 Sin(34/9 + (377 t)/10) + 1/9 Sin(75/19 + (1627 t)/37) +
2/7 Sin(2/9 + (553 t)/11) + 1/8 Sin(29/7 + (377 t)/6) +
3/11 Sin(25/7 + (553 t)/8) + 1/14 Sin(13/3 + (2023 t)/23) +
1/11 Sin(27/11 + (377 t)/4) + 1/33 Sin(9/7 + (1307 t)/13) +
1/10 Sin(55/12 + (1175 t)/11) + 1/42 Sin(47/11 + (1131 t)/10) +
1/28 Sin(71/18 + (955 t)/8) + 1/13 Sin(11/5 + (2111 t)/16) +
1/22 Sin(40/11 + (1244 t)/9) + 1/17 Sin(35/18 + (754 t)/5) +
1/18 Sin(49/16 + (1797 t)/11) + 1/13 Sin(21/8 + (1640 t)/9) +
1/17 Sin(7/4 + (377 t)/2) + 1/37 Sin(5/8 + (1753 t)/9) +
1/50 Sin(42/11 + (1709 t)/8) + 1/28 Sin(23/5 + (2419 t)/11) +
1/60 Sin(5/4 + (3104 t)/13) + 1/70 Sin(49/15 + (5391 t)/22) +
1/36 Sin(25/6 + (754 t)/3) + 1/24 Sin(8/5 + (3349 t)/13) +
1/99 Sin(7/3 + (2972 t)/11) + 1/63 Sin(159/79 + (1885 t)/6) +
1/64 Sin(16/15 + (2375 t)/7) +
1/75 Sin(25/11 + (2463 t)/7)}, {13.1114 -
1/97 Sin(10/9 - (3041 t)/11) - 1/46 Sin(9/8 - (2375 t)/9) -
1/28 Sin(1/20 - (3952 t)/17) - 1/27 Sin(2/5 - (3217 t)/16) -
1/51 Sin(8/7 - (1753 t)/9) - 1/21 Sin(11/9 - (2287 t)/13) -
1/19 Sin(3/7 - (1866 t)/11) - 1/21 Sin(3/5 - (1885 t)/12) -
1/15 Sin(7/10 - (4191 t)/29) - 1/16 Sin(4/5 - (377 t)/3) -
1/12 Sin(16/13 - (1175 t)/11) - 1/34 Sin(4/13 - (1307 t)/13) -
1/11 Sin(4/13 - (377 t)/6) - 1/6 Sin(1/11 - (509 t)/9) -
11/23 Sin(5/16 - (201 t)/8) - 57/8 Sin(22/23 - (44 t)/7) +
27/13 Sin(47/10 + (88 t)/7) + 3/4 Sin(2/5 + (132 t)/7) +
6/11 Sin(25/7 + (377 t)/12) + 1/76 Sin(14/5 + (377 t)/10) +
1/8 Sin(70/23 + (1627 t)/37) + 2/9 Sin(13/3 + (553 t)/11) +
1/6 Sin(25/7 + (553 t)/8) + 1/8 Sin(1/4 + (377 t)/5) +
1/25 Sin(49/12 + (1307 t)/16) + 1/20 Sin(68/15 + (2023 t)/23) +
1/11 Sin(8/3 + (377 t)/4) + 1/46 Sin(10/7 + (1131 t)/10) +
1/48 Sin(41/9 + (955 t)/8) + 1/15 Sin(23/11 + (2111 t)/16) +
1/20 Sin(11/3 + (1244 t)/9) + 1/15 Sin(37/16 + (754 t)/5) +
1/26 Sin(19/6 + (1797 t)/11) + 1/16 Sin(34/13 + (1640 t)/9) +
1/29 Sin(9/5 + (377 t)/2) + 1/43 Sin(17/4 + (1709 t)/8) +
1/35 Sin(19/5 + (2419 t)/11) + 1/36 Sin(9/5 + (1131 t)/5) +
1/65 Sin(7/6 + (3104 t)/13) + 1/43 Sin(27/7 + (5391 t)/22) +
1/27 Sin(37/8 + (754 t)/3) + 1/32 Sin(18/11 + (3349 t)/13) +
1/68 Sin(31/12 + (2972 t)/11) + 1/79 Sin(1/10 + (1131 t)/4) +
1/72 Sin(3/8 + (3594 t)/11)}}, {{287.833 -
1/39 Sin(8/15 - (1552 t)/19) - 1/36 Sin(9/11 - (377 t)/5) -
1/22 Sin(11/8 - (553 t)/8) - 1/32 Sin(5/9 - (377 t)/6) -
1/5 Sin(7/11 - (1891 t)/43) - 1/16 Sin(5/9 - (377 t)/12) -
1/10 Sin(11/10 - (201 t)/8) + 19/7 Sin(19/6 + (44 t)/7) +
2/5 Sin(1/21 + (88 t)/7) + 1/8 Sin(2/5 + (132 t)/7) +
1/9 Sin(1/72 + (377 t)/10) + 1/37 Sin(27/7 + (553 t)/11) +
1/9 Sin(24/11 + (509 t)/9) + 1/31 Sin(16/7 + (2023 t)/23) +
1/60 Sin(11/5 + (377 t)/4) +
1/43 Sin(9/5 + (1175 t)/11)}, {19.1795 - 1/39 Cos((1307 t)/16) -
1/21 Sin(9/13 - (553 t)/8) - 1/37 Sin(1/11 - (377 t)/6) -
1/11 Sin(14/9 - (1891 t)/43) - 1/21 Sin(8/15 - (377 t)/10) +
23/9 Sin(51/11 + (44 t)/7) + 1/3 Sin(1/5 + (88 t)/7) +
2/7 Sin(2/5 + (132 t)/7) + 1/12 Sin(79/17 + (201 t)/8) +
1/7 Sin(3/4 + (377 t)/12) + 1/15 Sin(9/2 + (553 t)/11) +
1/14 Sin(31/9 + (509 t)/9) + 1/27 Sin(55/12 + (377 t)/5) +
1/39 Sin(13/5 + (2023 t)/23) + 1/47 Sin(5/2 + (1175 t)/11) +
1/72 Sin(51/11 + (1131 t)/10)}}}
```

It goes without saying but these equations are very long. And on top of that for the integrand you have to square these equations and multiply by the derivative. That’s a pretty complicated integral, which on top of the the large number of terms, I imagine, is also highly oscillating. Since the integrals are so complicated, I imagine that’s what is causing the centroids to be inaccurate at times. Whenever I run my code, I’ve also been getting an error message, like this, which seems to support this.

```
NIntegrate::levmaxord: (287.833 -1/39 Sin(8/15-(1552 t)/19)-1/36 Sin(9/11-(377 t)/5)-1/22 Sin(11/8-(553 t)/8)-1/32 Sin(5/9-(377 t)/6)-1/5 Sin(7/11-(1891 t)/43)-1/16 Sin(5/9-(377 t)/12)-1/10 Sin(11/10-(201 t)/8)+19/7 Sin(19/6+(44 t)/7)+2/5 Sin(1/21+(88 t)/7)+1/8 Sin(2/5+(132 t)/7)+1/9 Sin(1/72+(377 t)/10)+1/37 Sin(27/7+(553 t)/11)+1/9 Sin(24/11+(509 t)/9)+1/31 Sin(16/7+(2023 t)/23)+1/60 Sin(11/5+(377 t)/4)+1/43 Sin(9/5+(1175 t)/11))^2 is a Levin function of differential order 1088 which exceeds value of option "MaxOrder" -> 50. Treating (287.833 -1/39 Sin(8/15-(1552 t)/19)-1/36 Sin(9/11-(377 t)/5)-1/22 Sin(11/8-(553 t)/8)-1/32 Sin(5/9-(377 t)/6)-1/5 Sin(7/11-(1891 t)/43)-1/16 Sin(5/9-(377 t)/12)-1/10 Sin(11/10-(201 t)/8)+19/7 Sin(19/6+(44 t)/7)+2/5 Sin(1/21+(88 t)/7)+1/8 Sin(2/5+(132 t)/7)+1/9 Sin(1/72+(377 t)/10)+1/37 Sin(27/7+(553 t)/11)+1/9 Sin(24/11+(509 t)/9)+1/31 Sin(16/7+(2023 t)/23)+1/60 Sin(11/5+(377 t)/4)+1/43 Sin(9/5+(1175 t)/11))^2 as a non-Levin function.
```

I’ve tried a bunch of different integration methods, but nothing seems to work better than the method Mathematica automatically uses. I wanted to have an alternative way to calculate the centroid, so I could compare my results and see how far off some of the more reasonable answers are. So I came up with an alternative method that turns the parametric plot into a region using DiscretizeGraphics and then calculates the centroid using RegionCentroid and I believe that this gives more accurate results. All three x coordinates for each method are different by at least a couple of pixels. This may not seem significant, but based on the size of the image 1 pixel in the x-direction is approximately 1 degree in longitude, which could make a significant difference when I’m calculating the slope. Do you know any way for these integrals to converge more accurately or perhaps some other unforeseen problems that are causing the numbers to be off? Any help would be greatly appreciated. Sorry in advance if my question is obvious. I would still consider myself a novice when it comes to Mathematica, so there’s a lot I don’t know still. Also, sorry that my post was so long, I may have wrote more than I needed but I wanted to give enough background to make clear the full context of the problem.

I have a sheet I use to update my scheduler about call offs and last minute changes to the schedule. I’m trying to get the system to calculate hours by using `=arrayformula((IF(D5:J42=C45:C87, C5:C42-B5:B42)))`

, which as I understand it should be looking through the cells in the main area, and calcualting the time differences in the C and B columns if their name matches.

It seems to be calculating in a strange way and I’m not really sure why. I originally planned to manually plug in each row, which I got halfway done with in Cell D44. if anyone has any better suggestions I’m open to them 🙂 Example Sheet

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