CompassRed

Not sure I understood all the metaphors

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Just say you recently came into some inheritance and that you are looking into investment opportunities. Then they will expect you to be out of your element, so you won't need to try to pretend you're someone you're not. If they ask about the inheritance, say your grandfather made a fortune selling lumber or something boring like that.

A vector space is a collection of vectors in which you can scale vectors and add vectors together such that the scaling and addition operations satisfy some nice relationships. The 2D and 3D vectors that we are used to are common examples. A less common example is polynomials. It's hard to think of a polynomial as having a direction and a magnitude, but it's easy to think of polynomials as elements of the vector space of polynomials.

It's only controversial because the way the therapy has been implemented in the past (and unfortunately in some places still today) is similar to what you describe. However, modern practices don't try to train their clients to act like "normal people" and any serious technician or analyst will only use punishment (or threats of punishments) as a last resort in programs written to target the most imperative behaviors (like running into traffic). Instead, they focus on the use of reinforcement to teach their clients skills that help them to become self sufficient. Following your metaphor, it would be like offering a depressed person $10 for every chore they complete that day rather than holding a gun to their head. The goal is to establish a foundation for life outside of therapy, not to reduce the presentation of autism.

I don't know the reason. I think not having the extra blank lines would be better, but it works just fine as is - even the post admits this much. That's why it's an enhancement. It's possible for software to be functional and consistent and still have room for improvement - that doesn't mean there is a bug.

My point is that someone made the decision for it to do that and that the software works just fine as is. It's not a bug, it's just a weird quirk. The fact that they made the enhancement you requested doesn't make the old behavior buggy. Your post title said "it's not a bug, it's a feature!", but the behavior you reported is not accurately classified as a bug.

It's not a bug just because the software doesn't conform to your personal preferences. You're asking for what would be considered an enhancement - not a bug fix.

It depends. If the variable names are arbitrary, then a map is best. If the variable names are just x_1, x_2, x_3, ..., x_n, then a list or dynamic array would be more natural. If n is constant, then a vector or static array is even better.

I don't recall any socialized courier or food delivery services.

This is just a continuous extension of the discrete case, which is usually proven in an advanced calculus course. It says that given any finite sequence of non-negative real numbers x,

lim_n(Sum_i(x_i^n ))^(1/n)=max_i(x_i).

The proof in this case is simple. Indeed, we know that the limit is always greater than or equal to the max since each term in the sequence is greater or equal to the max. Thus, we only need an upper bound for each term in the sequence that converges to the max as well, and the proof will be completed via the squeeze theorem (sandwich theorem).

Set M=max_i(x_i) and k=dim(x). Since we know that each x_i is less than M, we have that the term in the limit is always less than (kM^n )^(1/n). The limit of this upper bound is easy to compute since if it exists (which it does by bounded monotonicity), then the limit must be equal to the limit of k^(1/n)M. This new limit is clearly M, since the limit of k^(1/n) is equal to 1. Since we have found an upper bound that converges to max_i(x_i), we have completed the proof.

Can you extend this proof to the continuous case?

For fun, prove the related theorem:

lim_n(Sum_i(x_i^(-n) ))^(-1/n)=min_i(x_i).

2 may be the only even prime - that is it's the only prime divisible by 2 - but 3 is the only prime divisible by 3 and 5 is the only prime divisible by 5, so I fail to see how this is unique.