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.
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
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 first condition is wrong in the thumbnail. It's x->0 in the video and x->infinty in the thumbnail. Very different conditions there.
@ 12:42 Should be -e^x*G(x), not -G(x).
yet again a typo in the thumbnail made me click on the video. is this intentional? lol
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...
Once you change variables to g(x) instead of f(x), you're basically just reproving the integral version of Gronwall's inequality.
All that work for identically nothing.
😡
Yes: 😡!!!
Instructions unclear, I kept looping the video
No effort November is over!
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