I was not able to confirm this bug fact, hard to search a math theorem without a name. If anyone knows what this theorem is called and if it is true id be glad to know
Not a theorem really just homework problems
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).
This kind kf stuff makes my skin crawl. And I'm a programmer
I'm not a programmer and me too
That APOSTROPHE there 🤢🤮
Programming is much less math and much more logic. Unless you're a weirdo using haskell or something.
Cool, didn't think of that one. But it would still work, since you could consider that a constant in front of the f(x) not raised to the nth power (easier to imagine if we have a constant function, then its just (b-a)). The nth root will then normalise it to 1 for any real factor.
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