how to install azure functions on apple M1 pro, facing issues when installed with Rosetta

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Onki

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Onki Asks: how to install azure functions on apple M1 pro, facing issues when installed with Rosetta
I just had my first Mac book but that too with M1 processor and since then I am struggling with software compatibility issues.

Considering that I dont have much experience with Mac, I really need some help in resolving this issue.

I want to run azure functions on VS Code locally and for that, I need python, VS Code and azure tools but I kept on getting errors saying that Architecture ARM is not supported for language python.

I tried with python 3.8, 3.9 and 3.10 (3.10 was not accepted by VS Code.)

So, now what I understand is that using Rosetta, we can use the same softwares which we use on intel processor, so I followed below protocol for that.

  1. installed Rosetta : /usr/sbin/softwareupdate --install-rosetta
  2. /bin/bash -c "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/HEAD/install.sh)"
  3. cd ~/Downloads
  4. mkdir homebrew
  5. curl -L https://github.com/Homebrew/brew/tarball/master | tar xz --strip 1 -C homebrew
  6. sudo mv homebrew /usr/local/homebrew
  7. export PATH=$HOME/bin:/usr/local/bin:$PATH
  8. Let’s add an alias and path in the ~/.zshrc file:

-If you come from bash you might have to change your $PATH.

-need this for x86_64 brew

export PATH=$HOME/bin:/usr/local/bin:$PATH alias axbrew='arch -x86_64 /usr/local/homebrew/bin/brew'

LINK I referred for this: How to Install x86_64 Homebrew Packages on Apple M1 MacBook

  1. Now you can install apps for Intel processors: axbrew install package-name

so, now I thought I am finally done.

  1. I installed python -> axbrew install python@3.9
  2. I downloaded VS Code Universal.
  3. ran the function, it threw error for "you must have download azure function core tools". Used below commands to download it : [IMG alt="-[![enter image description here][1]][1]
    [![enter image description here][2]][2]"]https://i.stack.imgur.com/u7V98.png[/IMG]

but still I keep getting this error from VS Code. VS code doesnt recognize the azure tools functions I installed.

I tried with VSCode universal, for intel etc, but still the same.

Any leads on why I am facing this issue and how to come out of it.

VS Code error

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Suppose $G=\left\langle x, y, t|x^7=y^7=t^3=1, txt^{-1}=x^2, tyt^{-1}=y\right\rangle$. Show that $y\in Z(G)$.

Dick Grayson Asks: Suppose $G=\left\langle x, y, t|x^7=y^7=t^3=1, txt^{-1}=x^2, tyt^{-1}=y\right\rangle$. Show that $y\in Z(G)$.
The Problem: Suppose $G=\left\langle x, y, t|x^7=y^7=t^3=1, txt^{-1}=x^2, tyt^{-1}=y\right\rangle$. Show that $y\in Z(G)$.

My Attempt: Clearly $y$ commutes with $t$, so $y$ commutes with $t^2$ as well. Thus to show $y\in Z(G)$, it suffices to show that $y$ commutes with $x$. But I struggle to prove that $xy=yx$. It seems a rather silly place to get stuck-any HINT would be greatly appreciated.

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Inner and Outer Radius of a Set in Isotropic Position

user6952886 Asks: Inner and Outer Radius of a Set in Isotropic Position
For a compact convex set $C\in \mathbb{R}^d$ in isotropic position, a classical result in convex geometry says that there exists $r,R>0$ such that $B_2^d(r)\subseteq C \subseteq B_2^d(R)$ and $R/r \leq O(d)$.

Does the same remain true when the $\ell_2$ balls are replaced by $\ell_{\infty}$ balls? That is, does there exist $r,R>0$ such that $B_\infty^d(r)\subseteq C \subseteq B_\infty^d(R)$ and $R/r \leq O(d)$?

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Prove $e$ is irrational.

Julian Neira Asks: Prove $e$ is irrational.
I'm stucked in this problem: For every $n\in\mathbb{N}$, show that

$$0<e-\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}\right)<\frac{1}{n!n}$$ From this, prove $e$ is irrational.

I don't know how to prove $e-\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}\right)<\frac{1}{n!n}$. I already used induction, but it didn't work. Thanks for your help.

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What are the total number of state combinations for this dice game?

RTT Asks: What are the total number of state combinations for this dice game?
Lets say we have the dice game explained below. What are the total number of game states? In other words, how many numerical combinations of lives, totalScore, and scratchScore exist?

Initial State:

lives = 5, totalScore = 0, scratchScore = 0

The Rules:

Each turn you roll a 6-sided dice. If the roll is [1-5] then you add that score to your scratchScore. If you roll a 6, then your scratchScore is reset to 0 and you lose 1 life.

Before each roll, you may decided to yell "BANK" at which point you add your scratchScore to your totalScore and reset the scratchScore to 0. Everytime you yell "BANK" though you will lose 1 life.

If lives=0 you lose. Otherwise youc an win by achieving a totalScore that is >= 40. The idea is to only yell "BANK" when your scratchScore is somewhat large.

Here is an example game:

  1. lives = 5; total score = 0; scratch score = 0. You roll a 2 and do not yell ”bank.”
  2. lives = 5; total score = 0; scratch score = 2. You roll a 3 and do not yell ”bank.”
  3. lives = 5; total score = 0; scratch score = 5. You roll a 4 and yell ”bank.”
  4. lives = 4; total score = 9; scratch score = 0. You roll a 2 and do not yell ”bank.”
  5. lives = 4; total score = 9; scratch score = 2. You roll a 6.
  6. lives = 3; total score = 9; scratch score = 0
  7. ... repeat untill lives=0 or totalScore >= 40

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Indefinite Integral of ${\tan(x)}^{p/q}$

Mailbox Asks: Indefinite Integral of ${\tan(x)}^{p/q}$
I have seen a lot of videos in which people integrate functions like $\sqrt{\tan(x)}$, $\sqrt[3]{\tan^2(x)}$, etc. I was wondering if there was a closed-form expression for $$\int (\tan{x})^{\frac{p}{q}} dx$$

Where $p,q$ are integers, and $q \neq 0$. Eventually, I would like to generalize the result to an integrand ${\tan(x)}^r$, for $r \in \mathbb{R}$, but for now we can stay with the rationals.

My attempt at solving this:

Let $u = \tan(x) \implies x = \arctan(u), dx = \frac{du}{1+u^2}$. This yields $$\int \frac{u^{\frac{p}{q}}}{1+u^2}du$$

But I'm not sure where to go from here. My first thought was long division, but that can't work here. I also considered treating $\frac{1}{1+u^2}$ like a geometric series, but that has a finite radius of convergence.

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How does a Phantom Rogue's Ghost Walk ability movement work?

  • Rykara
  • Social
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Rykara Asks: How does a Phantom Rogue's Ghost Walk ability movement work?
At level 13, the Phantom Rogue gets the ghost walk ability which grants (in part):

As a bonus action, you assume a spectral form. While in this form, you have a flying speed of 10 feet, you can hover [...]

Does this mean that the phantom rogue has (only) 10 feet of flying movement (and no other movement)?

Or does the phantom rogue gain 10ft of flying movement (in addition to their normal movement)?

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What do the spear and watermelon jokes in "Airplane!" mean?

  • freeling10
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freeling10 Asks: What do the spear and watermelon jokes in "Airplane!" mean?
In the awesome 1980 comedy Airplane!, there are two jokes that make no sense and I've never been able to find an adequate explanation for them. Is anyone able to shed some light?

Near the end of the film, Rex Kramer (Robert Stack) arrives at the control tower to help Ted Striker (Robert Hays) land Trans-American Airlines' endangered flight 209. Kramer confers with Steve McCroskey (Lloyd Bridges) and, while they're discussing the situation, they have the following exchange:

McCroskey: Right now, things aren't so good.

Kramer: Let me tell you something, Steve. Ted Striker was a top notch squadron leader. A long time ago.

At that moment, a spear whizzes across the room and into a nearby bulletin board. Without referencing the projectile, the conversation continues:

McCroskey: I want you to get on the horn and talk that guy down. Now you're going to have to let him get the feel of that airplane on the way and you'll have to talk him onto the approach. So help me, you'll have to talk him right down to the ground.

At this point, a watermelon drops from above and smashes onto the corner of a desk in the background. It is not acknowledged by either character.

Everything else in the movie is crystal clear, but these two gags have always confounded me. Surely somebody must know!

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