what function satisfies these conditions??
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- Опубліковано 11 гру 2024
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The first condition is wrong in the thumbnail. It's x->0 in the video and x->infinty in the thumbnail. Very different conditions there.
Intuitively, the second condition seems designed for g(x) = e^(-x) to satisfy it (since you get equality). But with that g(x) you would get the limit e^x g(x) = 1, not 0. And if pick a g(x) that decreases any faster than e^(-x) then we won't satisfy the integral inequality, because it will decrease too quickly. Essentially, the first inequality says the function must decrease *faster* than an exponential, while the second inequality says it must decrease *no faster* than an exponential. These are incompatible so we get no solution other than the trivial 0 constant function.
Ugh, 15 minutes just to confirm that the only solution is the trivial solution, the zero function.
Not a good place to stop 😔
The fact that the trivial result is the only one isn't necessarily bad news. What's interesting about this outcome is that any CONTINUOUS function other than the trivial one, defined on a BOUNDED and CLOSED interval (compact, i.e., [0,1]), cannot decrease faster than e^(-1/x) as x->+0. This highlights a fundamental characteristic of continuous functions on compact domains. This somewhat restricts the structure of the function set of continuous functions on compact domains.
However, if we relax the continuity condition or remove the boundedness of the domain, a richer structure emerges.
what about e^(-1/x^2 )
I had a go at this problem in the far more specific case where f is analytic and got (obviously) that x=0. I then wanted to look at the case where f is right-continuous and still haven't approached that yet.
(right continuous so the limit to zero from above is defined)
Since f is continuous it can be approximated in the uniform norm by a polynomial, p(x), as a consequence of the Stone Weierstraß theorem. If we replace e^(1/x) by its power series and multiply them as per the limit assumption, we can see that if p(x) is nonzero, its highest degree term will be dominated by infinitely many terms of the form 1/k!x^k, making a limit of 0 as x->0+ impossible. Hence p, and therefore f, must be the zero function.
What you said after stone-weierstrass is wrong.
Here's a counterexample: f(x)=e^(-1/x^2) (with f(0)=0).
It's fine to approximate f with p, but in the limit what happens is that if |f-p|
@ 12:42 Should be -e^x*G(x), not -G(x).
Once you change variables to g(x) instead of f(x), you're basically just reproving the integral version of Gronwall's inequality.
yet again a typo in the thumbnail made me click on the video. is this intentional? lol
All that work for identically nothing.
😡
Yes: 😡!!!
Instructions unclear, I kept looping the video
No effort November is over!
Why is the thumbnail diferent?
He has been doing this for some time. This makes him have more comments/interactions pointing out the mistake. Dirty way, though...
When a function is simply zero, it can’t be a function of x. It’s an inconsistent way to describe what is factually a non-function.
Why? What about f(x)=0•x?
blud does not know what a function is.
Why tho? Function is an injective mapping from one set to another. In this case, it's a mapping from R to {0}
You do not know what a function is. Here's the definition (from Wikipedia) : "In mathematics, a function from a set X to a set Y assigns to each element of X exactly one element of Y.".
Then if you understand the definition you just have to set Y to { 0 } to know that your statement is wrong.
Please stop trying to sound smart in front of people you do not even know when you do not understand the basics.
We are all here to learn because Maths are beautiful ! Cheers.
@self8ting that's the definition of a bijective function, not a general function. f(x)=0 is a function indeed, the zero function. by that definition, f(x)=cos(x) isn't a function either, please review that you copied the correct definition from wiki