Buying – bot/macro/Automate/etc upload videos/filehost, to series page | Proxies-free

I have an series page, I am an uploader, I upload episodes manually, and I know that it could be automated.

Obviously, details are required for you to think about how to automate it, but it is “advanced” because it will require many things (connect with an app that can upload a file to multiple different filehosts, without using a browser/change pages> once all downloads are finished> copy links> and put the links on my page in a specific order/place etc etc

(I can pay ~210 usd or less, it DEPENDS on if the app/script has gui, parameterized to work in any file, etc)

Obviously, details are required for you to think about how to automate it

Customized for my page, although you can also change part of the page to make it work.It puts the embed link in itself, meaning that in some uploads / player as mega I must first put “embed” to the link.


sequences and series – Yet another “what’s the next number” question…

2, 6, 14, 29, 51, 97, 180, 250, 278, ?, ?

These are the answers to a topological enumeration problem for $n=1$ to $9.$ The sequence is increasing and bounded; it’s very possible that 278 is the final value but my intuition says it isn’t.

I am currently working on $n=10$ and $11$ ($n$ represents the cardinality of an otherwise arbitrary topological space) and expect to have those numbers in a reasonably short time (less than a month). If they are not equal, then I am out of luck for the time being because it would probably take my current laptop at least a year to calculate the value for $n=12$. My intuition says they will be equal.

The three evens followed by three odds followed by three evens pattern is surely coincidental, but just in case it isn’t, my stab is that the next two values will both be 281.

Guesses are welcome. I will describe the problem once I have the next two values (assuming I am able to compute them).

convergence divergence – Calculating Error for Taylor Series (Calculus II)


Estimate the error if P3(X) = x – x^3/6 is used to estimate the value of sin(x) at x = 0.5

Can someone explain how I’m getting this wrong? I’m using the formula (M * |x-1|^(n+1)) / (n+1).

So far, what I understand are my variables are M = 1, n = 3, and x = 0.5. (Or at least I think so)

When I plug those in I get 0.0026042, rounded to the 7 decimals place. Any info about where I went wrong would be greatly appreciated!

sequences and series – Changing the order of summation – check for correctness

Assume $|r|<1$, I’m working with

begin{align*}A &=sum_{i=1}^{infty} sum_{j=i+1}^{infty} r^{j-i}
sum_{u=1}^{i}frac{r^{i-u}}{u} sum_{v=1}^{j}frac{r^{j-v}}{v} \
&=sum_{u=1}^{infty} frac{1}{u} sum_{v=1}^{infty} frac{1}{v} sum_{i=u}^{infty} sum_{j: j geq i+1 & j geq v }^{infty} r^{j-i}
r^{i-u}r^{j-v} \
&=sum_{u=1}^{infty} frac{1}{u} sum_{v=1}^{infty} frac{1}{v} sum_{i=u}^{infty} sum_{j = min{i, v} }^{infty} r^{2j-u-v} \
&=sum_{u=1}^{infty} frac{1}{u} sum_{v=1}^{infty} frac{1}{v}left( sum_{i: i geq u & i leq v}^{infty} sum_{j = i }^{infty} r^{2j-u-v} + sum_{i: i geq u & i > v}^{infty} sum_{j = v }^{infty} r^{2j-u-v} right) \
end{align*} $$

The initial sum seems to converge as per numerical simulations, however, further expanding by change of the order of the summations, I keep arriving at diverging or complex values.

Am I missing something during the change of the summation order?

gm.general mathematics – I need help with this geometric series problem

An ancient chessboard puzzle has one grain of rice on one square, the next square has double the number of grains, and so on. E.g. 5th square, 2^4 = 16.

(c) According to one website there are about 15,000 grains of rice in a kilogram. And according to an- other website, 1 kg of rice has a volume of 1.24 Litres.
If the surface of the Earth was covered in the rice from the chessboard construction, estimate the depth of rice we would obtain. Assume the Earth is a sphere of radius 6,371 km. Assume also, that the rice can be laid over the (smooth) ocean.

sequences and series – I am looking to make recursive in this expression

I would like to know given the following definition of the function X(n) if it is possible to express two consecutive values of X(n) (example: X(1) and X(2) ) to obtain a recursive expression.

What I’m looking for is something like this: $mathbf{X}(n+1) = mathbf{X}(n) + f(mathbf{w},mathbf{x})$ in which x in f(w,x) less than N times.

$mathbf{x}(k)$ is an arbitrary function.


where: $c_1 , c_2$ are constants $inBbb{R}$; $n,Nin Bbb{N}$ ($N$ is constant) ; $kin Bbb{Z}$.

While $mathbf{w}(k)$ is defined as as: $$mathbf{w}(k)={sin(ak)over ak} mathbf{h}(k) $$ where $a$ is constant and $mathbf{h}(k)$ is the Hamming function.

I could write this:
$$mathbf{X}(n+1)=c_1e^{-jc2}Biggl(sum_{k=n+1}^{n+N}mathbf{x}(k)mathbf{w}(k-n-1)e^{-jk}Biggl ) + c_1e^{-jc2}Big(mathbf{x}(n+1+N)mathbf{w}(N)Big)$$
and the same way:
$$mathbf{X}(n)=c_1e^{-jc2}Biggl(sum_{k=n+1}^{n+N}mathbf{x}(k)mathbf{w}(k-n)e^{-jk}Biggl ) + c_1e^{-jc2}Big(mathbf{x}(n)mathbf{w}(0)Big)$$

Now how can I proceed? i want to get rid of this: $$Biggl(sum_{k=n+1}^{n+N}mathbf{x}(k)mathbf{w}(k-n)e^{-jk}Biggl )$$ or at least reduce it as a sum of fewer elements. Thank you!

functional – Power series expansion in terms of a function

I have a two variable function z(x,y) = f(x,y) + g(x,y), such that I know the functional form of f(x,y) but not of g(x,y). I have to do some symbolic calculations with the function z(x,y), but I would like to keep only the first order in g(x,y) (treating g as small). So, for example, I would like Mathematica to approximate (z(x,y))^3 = (f(x,y) + g(x,y))^3 = f(x,y)^3 + 3*f(x,y)^2*g(x,y), or (D(z(x,y), x))^2 = (D(f(x,y), x))^2 + 2*D(f(x,y), x)*D(g(x,y), x). Is there a way to do it? I have tried the most naive way, namely to use Series('exp'(z),{g, 0, 1}), treating g(x,y) as a parameter rather than a function, but (as expected) it doesn’t work. Do you know a way to do it?

Thank you very much in advance for anyone who will reply!

real analysis – Prove difference of infinite series of decreasing function and its integral converges

$f(x)$ is continuous and decreasing on $(0, infty)$, and $f(n) to 0$. Let ${a_n} = f(0) + f(1) + … + f(n-1) – int^{n}_{0}f(x)dx$. Show $a_n$ converges (from Mattuck Analysis).


For any continuous function $f$ and integers $a, b$, with $b > a$, let $L(f, a, b)$ be the Riemann sum of the lower bound of each unit subinterval $(a, a+1), (a+1, a+2)…(b-2, b-1), (b-1, b)$, and $U(f, a, b)$ be the Riemann sum of the upper bound of each such unit subinterval. Then, by definition of integral, $L(f, a, b) leq int_a^b f(x)dx leq U(f,a,b)$. Since $f(x)$ is decreasing, for any unit subinterval $(m, m+1)$, $f$‘s upper bound is $f(m)$ and its lower bound is $f(m+1)$. Therefore, $a_n = U(f,0,n) – int^n_0f(x)dx$, and $a_{n+m} = a_n + U(f, n, n+m) – int^{n+m}_n f(x)dx$.

Furthermore, note that $f$ is always nonnegative, since it is decreasing and has limit zero.

Since $f(n) to 0$, for any $epsilon > 0$, there exists $n$ such that for any $m > 0$, $f(n+m) < epsilon$, so $0 leq L(f, n, n+m) leq int_n^{n+m}f(x) leq U(f, n, n+m) < mepsilon$. Therefore, for any $epsilon > 0$, there exists $n$ such that for any $m >0$, $|a_{n+m} – a_n| < epsilon$, and $a_n$ is Cauchy and converges to a limit. QED.