Beueghwh

[quiet tide]

Bruegey

[wooden delta]

1. Do not ping the Moderators, unless someone is breaking the rules.
2. Do not ping the Helper Moderators, unless there is a conflict between helpers.
3. Do not ping other members randomly for help.
4. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
5. Wait patiently for a helper to come along.
6. If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[quiet tide]

@opaque adder @terse tangle

Type something

[opaque adder]

something

[quiet tide]

@terse tangle

[opaque adder]

i would like to conjecture that it’s because many messages were sent after exiled’s last message in the thread

however, i know nothing of the bot or code itself

[devout moss]

hullo

[terse tangle]

hello

hi people

Prove that for |z|>1 that $\lim_{x\to\infty} P(x)z^{-x}=0$.\
To prove this, we must prove that $z^x$ grows faster than $x^n$. This is proven when the derivative of $z^x$ is greater than $x^n$ as x approaches infinity. Let’s take a base case $n=0$. Then we have $\lim_{x\to\infty} z^{-x}$, which is obviously zero. Multiplied by any constant, we have $\lim_{x\to\infty} c\cdot z^{-x}=0$. We can write a polynomial as $\sum_{k=0}^{n} a_kx^k$, where the $a_k$ are coefficients of the polynomial. Thus we have $\lim_{x\to\infty} a_0z^{-x}+\lim_{x\to\infty} a_1xz^{-x}...\lim_{x\to\infty} a_nx^nz^{-x}$. We know already that the first term of this limit is equal to zero. Now, we have to prove that $(z^x)’>a_1(x)’$ for the next limit. Thus, the inequality is $\ln(z)z^x>a_1$, which is obviously true as x tends to infinity. Thus, $\lim_{x\to\infty} a_1x\cdot z^{-x}=0$. Now we must prove that $\ln(z)z^x>2x$ as x tends to infinity but we have already proved this when we said that $\lim_{x\to\infty} a_1xz^{-x}=0$! Thus, we can repeat this process $n$ times, because the derivative of $x^n=nx^{n-1}$, and thus we have $0+0+0...+0=0$ as our limit, Q.E.D.

HAHAHAHAHAHAHAHAHHAHAHAHA

[sullen lichen]

flying green people eater

[terse tangle]

[devout moss]

?tag get-help

[livid creek]

🔹 To receive math help, navigate to #1015578016606343218 or #1020426321261756536. In there, the pinned entry will contain instructions on how to create your post. Once created, please wait patiently for a helper to come along.

Please do not ping moderators or random users unsolicited, this will not make help arrive any faster. Once done, thank your helper, close your post with +close, and optionally nominate them as helper of the week in #helper-nominations.

[terse tangle]

you need to get help

[quiet tide]

@opaque adder I see the issues, it only fetches 50 Message to ensure performance, the thread probably had more than that which resuled in Exiled not being recognized

[opaque adder]

interesting

that’s fair, i suppose

[quiet tide]

I will be increasing the limit to 200 using pagination

[dim briar]

you could consider keeping track of messages as they come

which would let you consider the entire thread (for new threads) no matter how large they are, without requesting either spamming the API or ignoring old messages

[opaque adder]

would that not cause every new message to be checked to see if it belongs to a previous user?

[quiet tide]

The limit is 100 message for the discord API

i will just make it load if it fetches more than 100 message, since i will be making it fetch in batches

[dim briar]

In these threads, I suppose? rn the bot waits until the end of the thread, the checks the past 50 messages all at once. I was pointing out that the bot could check the messages in help threads as they come, instead of waiting for the end

[quiet tide]

i will be making it fetch in batches if the messages is more than 100

+close

[elder finch]

Please thank your Helpers before closing!

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