4 to 5 digits per month, passive income with your own TV / adult / movie series | NewProxyLists

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A flawless stay

Modeling: Linear or nonlinear time series?

Suppose my box contains 5 balls numbered from 1 to 5 and the bullets are drawn randomly with replacement. Every day, 10 drawings are made, forming a chronological sequence such as

1, 5, 1, 4, 3, 2, 4, 2, 5, 5

I have been given 2000 historical drawings for the last 200 days, and I am free to use them to predict the next drawing with the help of mcmc methods.

My current references are:

  1. Nonlinear Bayesian MCMC Time Series Prediction, by Y. Nakada, T. Kurihara, and T. Matsumoto

  2. Markov Chain Monte Carlo Estimation of autoregressive models with application to the concentration of metal pollutants in sludge, by G. Barnett, R. Kohn and S. Sheather

  3. A Python program at
    https://github.com/pymc-devs/pymc/wiki/Surplus

Should I model it as a linear or nonlinear time series?

Prove this basic inequality related to the Fourier series:

I'm stuck with proving this equality:

$ displaystyle sum_ {n = 1} ^ { infty} dfrac { pi (-1) ^ {n + 1}} {i n} e ^ {in theta} = 2
pi displaystyle sum_ {n = 1} ^ { infty} dfrac { sin n theta} {n} (-1) ^ {n + 1} $

Can any one solve this problem or please give me a vital clue?

sweet question – Bertrand series, unnamed / standard in English?

As a French high school student / student, I discovered the Bertrand series (Bertrand series), things of the form
$$
sum_ {n = 2} ^ infty frac {1} {n ^ alpha log ^ beta n}
$$

in the program. They are considered standard and a Wikipedia page is dedicated to them (with the results and the proofs of their convergence). Since they appear quite often in (examples), I would say that's for a good reason.

However, as far as I know, they are a lot less common and known in the United States (and the United Kingdom?). Even on this website, they seem to be largely unknown (among a large majority of newcomers), and they are not mentioned on Wikipedia (let alone a page entry for them).

Is there a reason? Are not they considered useful, or are they known but "Internet" does not reflect that?

series and series – Convergence radius of $ sum_ {n = 0} ^ infty a_n x ^ n $, with $ a_ {n + 2} = frac {n (n + 1) a_ {n + 1} – a_n} {(n + 2) (n + 1)}, a_2 = -a_0 / 2 $

Problem

Find the convergence radius of the power series

$$
sum_ {n = 0} ^ infty a_n x ^ n
$$

or $ a_n $& # 39; s are defined by the following recurrence relation

$$
begin {aligned}
a_ {n + 2} & = frac {n (n + 1) a_ {n + 1} – a_n} {(n + 2) (n + 1)}, n ge 1 \[8pt]
a_2 & = -a_0 / 2
end {aligned}
$$

with arbitrary $ a_0, a_1 $.


Try1

I've tried to directly apply the ratio test, by dividing the recurrence relationship above by $ a_ {n + 1} $,

$$
r_ {n + 1} = frac {n} {n + 2} – frac {1} {(n + 2) (n + 1)} frac {1} {r_n}
$$

or $ r_n: = a_ {n + 1} / a_n $. We can observe that

$$
r_n r_ {n + 1} = frac {n} {n + 2} r_n – frac {1} {(n + 2) (n + 1)}
$$

but I can not continue to find the expression for

$$
lim_ {n to infty} green r_n green
$$


Try2

Since the recurrence relation depends on the arbitrary choice pf $ a_0, a_1 $, let us proceed leaving $ a_0: = 0 $. We have

$$
begin {aligned}
a_2 & = a_0 = 0 \[7pt]
a_3 & = – frac {1} {6} a_1 \[7pt]
a_4 & = frac {2} {4} a_3 = -a_1 / 12 \[7pt]
a_5 & = frac {3} {5} a_4 – frac {1} {5 cdot 4} a_3 = – frac {1} {24} a_1 \[7pt]
end {aligned}
$$

where I can not find simple rules. Similarly, leaving $ a_1: = 0 $, we have

$$
begin {aligned}
a_2 & = -a_0 / 2 \[7pt]
a_3 & = frac {1} {3} a_2 = – frac {1} {6} a_0 \[7pt]
a_4 & = frac {2} {4} a_3 – frac {1} {4 cdot 3} a_2 = – frac {1} {24} a_0 \[7pt]
a_5 & = frac {3} {5} a_4 – frac {1} {5 cdot 4} a_3 = – frac {1} {60} a_0 \[7pt]
end {aligned}
$$

where again I have not found any rules. So, I can not find the validity interval of the next composition.

$$
sum_ {n = 0} ^ infty a_n x ^ n = a_0 left[ 1 – x^2/2 – x^3/6 – x^4/24 – x^5/60 + cdots right] + a_1 left[ x – x^3/6 – x^4/12 – x^5/24 + cdotsright]
$$

Any help would be appreciated.

How can I run a series of tasks in Gulp 4? Does not work

I've looked at gulp docs, tried to make all functions asynchronous, tried a done () callback, tried to add return instructions in each function. I do not understand. Their documents look a lot like the kind of syntax I have in mine

Here is the result when I run distribute for example:

$ gulp distribute
[22:09:26] Using gulpfile ~ / code / projects / sss / gulpfile.js
[22:09:26] From & # 39; distribute & # 39; ...
[22:09:26] Done & distribute after 1.73 ms
Made in 0.56s.

but no remote folder is created, nor the series performing all the other tasks

gulpfile.js

const gulp = require (& gt; gulp & # 39;),
less = require (& gulp-less & # 39;),
uglify = require (& gulp-uglify & # 39;),
babel = require ("gulp-babel"),
shell = require (& gulp-shell & # 39;),
mocha = require ("gulp-mocha"),
browserify = require (gulp-browserify),
rename = require (& gulp-rename & # 39;),
del = require (& # 39;) & # 39;
{series} = require (& gt; gulp & gt;

require (& # 39; dotenv & # 39;). config ();

const config = {
watch: {
src: & # 39; ./ src / *. js & # 39;
}
test: {
way: {
specifications: & # 39; ./ test / specs / * - spec.js & # 39;
}
Mocha: {
reporter: 'spec'
}
}
};

precompile function () {
gulp.src (& # 39; ./ src / client / less / *. less & # 39;)
.pipe (minus ({
compress: true
}))
.pipe (gulp.dest (& # 39; ./ build / client / css & # 39;));
}

function build () {
series (clean, transpile, copyBuild);
}

clean function () {
of the ([build, distribute])
}

transpile function () {
gulp.src (['./src/**/*.js', '!./src/client/less/*.js'])
.pipe (babel ())
.pipe (gulp.dest (build));
}

copyBuild () function
gulp.src (['src/shared/**/*.json']) .pipe (gulp.dest ('build / shared'));
}

distribute () {function
returns series (build,
createDist,
copyBuildAndDist,
copyExt,
package,
precompile,
CopyData,
copyAssets,
copy styles);
}

copyExt function () {
gulp.src (['ext/**/*']) .pipe (gulp.dest (& dist / client / lib));
}

copyData () function
gulp.src (['src/shared/**']) .pipe (gulp.dest ('dist / shared'));
}

copyAssets function () {
gulp.src (['src/client/assets/**']) .pipe (gulp.dest (& dist / client / lib / assets & # 39;));
}

copyStyles () function
gulp.src (['./build/client/css/*.css']) .pipe (gulp.dest (& dist / client / lib / css & nbsp;));
}

copyFeed () function
gulp.src (['feed.xml']) .pipe (gulp.dest ('dist / client'));
}

function copyBuildAndDist () {
gulp.src (['build/server.js', 'build/api.js', 'build/shared', 'src/**/index.html', 'package.json']) .pipe (gulp.dest (& dist;));


bundle of functions () {
back gulp
.src (build / client / index.js & # 39;)
.pipe (browserify ({
debugging: false
}))
.pipe (rename (& # 39; app.bundle.js & # 39;))
.pipe (uglify ())
.pipe (gulp.dest ('dist / client / scripts /'));
}


function createDist () {
gulp.src (['dist/client/scripts'], {}). pipe (gulp.dest (& dist; client / scripts & # 39;));
}


exports.build = build;
exports.clean = clean;
exports.precompile = precompile;
exports.transpile = transpile;
exports.copyBuild = copyBuild;
exports.copyExt = copyExt;
exports.distribute = distribute;
exports.copyData = copyData;
exports.copyAssets = copyAssets;
exports.copyStyles = copyStyles;
exports.copyFeed = copyFeed;
exports.copyBuildAndDist = copyBuildAndDist;
exports.bundle = bundle;
exports.createDist = createDist;

Asymptotic Series: Why does not mathematica calculate it with a power of $ 1 / x $?

Consider the following:

series[Cos[Pi/2*Sqrt[n/x]]- Cos[Pi/2*Sqrt[(n + 1)/x]], {x, Infinity,
2}]

I'm waiting for mathematica to find me $ 1 / x $ disturbance expansion.
Analytically, the first order is if I have not made any mistakes: $ frac { pi ^ 2} {8} frac {1} {x} $

But mathematica answers me:

Cos[1/2 [Pi] sqrt[n/x]+ O[1/x]^ 5]cos[1/2 [Pi] sqrt[(1+n)/x]+ O[1/x]^ 6]

He does not really answer my question. How can I formulate my question to mathematica more correctly?

Commutative algebra – a ring of generalized power series

Let $ Bbbk $ to be a field; I am interested in the next ring (which, I suppose, is a field). Its elements are formal expressions that resemble

$$ sum_ {n = 0} ^ { infty} a_n x ^ {b_n} $$

or $ a_n in Bbbk $ and $ b_n in mathbb {R} $with $ b_n $ strictly increasing, and $ lim_ {n to infty} b_n = infty $. (Technically, we should quote them by a relationship allowing us to insert and delete terms when $ a_n = $ 0. Or we may need all $ a_n $Let not be zero, but we should also include finite sums. Or we could represent them by functions $ a: mathbb {R} to Bbbk $ assign a coefficient to each exponent, whose support is left-finite.) We can add and multiply these expressions quite obviously; the condition on $ b_n $ ensures multiplication works (that is, the resulting set of exponents can be re-enumerated with the order type $ omega $ and limit $ infty $).

This ring of serial type power objects is closely related to others. More specifically, if $ Bbbk = mathbb {R} $ then it contains the Levi-Civita field as elements for which each $ b_n in mathbb {Q} $, while it is contained in the field of the Hahn series $ Bbbk[[x^{mathbb{R}}]]$. Note that the set of all Hahn series with order type $ omega $ is not closed under multiplication; it's a natural subset of it. I think it's also the whole series of Hahn with order type $ omega $ that come together in the field evaluation topology of Hahn's series, and also that it is the closing of the field $ Bbbk (x ^ { mathbb {R}} $ generalized rational functions within the field of the Hahn series, and the completion of Cauchy's $ Bbbk (x ^ { mathbb {R}} $ in its uniformity of evaluation.

Does this field have a standard name and / or notation? Is it of a more general construction (for example, replace $ mathbb {R} $ by something more general, which would then probably also include the Levi-Civita field as the case of $ mathbb {Q} $)?

E Series Tapered Roller Bearing – Advertising, Offers

Timken has developed a new type of high-efficiency E-Series tapered roller bearing. The company also introduced the new standard for bearing performance.

T series tapered roller bearings, designed by Timken, can operate in extremely difficult conditions. He adds a sealed double design; As a result, particles and other objects will be less polluted than other products in the same category. The newly added coating may well protect the tread surface because the corrosion resistance capability is increased. This type of product also offers better performance because it is equipped with Timken multipurpose grease lubricant, which contains certain additives used to resist corrosion and friction. In addition, there is also a wide range of E-Series tapered roller bearings sizes and structures to meet the requirements of different industries.

As everyone knows, Timken is one of the world's leading suppliers of bearings. There are different types of products manufactured by this company that can meet the different needs of customers. The products are all very good. In addition, the company can also provide good services to customers. The innovation of the company allows it to provide better products to customers. Apart from this, the company also has advanced technologies such as advanced materials technology, advanced process technology, design and analysis technology, and so on. With these technologies, the company can produce more refined products.

Overall, the company always considers customers. He is doing his best to improve technologies and services to respond to market development. Obviously, the new E-Series tapered roller bearings are very innovative and can meet the specific needs of customers.

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fibonacci numbers – How to evaluate the series $ sum_ {n = 1} ^ infty frac {1} {F_ {4n}} $? Where $ F_1 = 1 $, $ F_2 = 1 $ and $ F_n = F_ {n-1} + F_ {n-2} $

How could I evaluate the following series: $$ sum_ {n = 1} ^ infty frac {1} {F_ {4n}} $$

Or $ F_1 = $ 1, $ F_2 = $ 1 and $ F_n = F_ {n-1} + F_ {n-2} $ for $ n $ 3

You do not know how to do it, except maybe using the close-up form for the nth Fibonacci number, but it sounds way too arithmetical. Ideas?