And now for the same question but with a_k instead of a_n in the ln (which was my initial read of the problem)... If I'm not mistaken, it should be the same result: same proof, but bounding the sum only from N onward, the N first terms being finite (a little work needed here to deal with the remaining /n) they will disappear when multiplied by 1/n. A much more counter-intuitive result!
Omg! I finally know why I could always omit the added terms that go to 0 in the limit when I transformed riemann sums to integrals. I knew it worked but I didnt know **why** it worked. Thanks
6:20 wait I don’t understand. How can we have that natural log inequality if the epsilon bound only applies for n>N? I don’t think we can guarantee it for 1
I originally thought the same, but the index in the sum is over k, not n. So with respect to the sum, the a_n is constant and changes with the limit and then so does the upper bound of the sum.
Seems like something that can be calculated using Stirling's approximation, even though there's a pesky square root factor that seems to invalidate it.
i misread the question but its honestly more interesting so heres an interesting varient: whats the limit as n-> infty of 1/n*\sum_{k=1}^{n} ln(k/n + a_k) (HINT: split up the sum)
I now realize why a lot of content creators use a virtual blackboard instead of a physical one. Visual challenged people like me have a hard time seeing the contrast with a recorded physical blackboard and can't just walk closer to it if it is on a screen.
The int(ln x dx, x = 0..1) could be done better by realizing it's equivalent to int(exp(y) dy, y = -inf..0) because they're inverse functions. At that point, the integral becomes almost trivial. Only need to incorporate the - sign.
A bounded set D ⊆ R^n is called Jordan measurable if the characteristic function 1_D is integrable on some n-hyperrectangle R contains D. The Jordan measure of D is defined as |D| = ∫1_D on R. Prop. D is Jordan measurable iff ∂D is Jordan measutable and |∂D| = 0. A set C ⊆ R^n is called of content zero if ∀ε>0, ∃ n-hyperrectangles R1, ..., Rm such that |R1| + ... + |Rm| < ε and C ⊆ (R1∪...∪Rm) Let E ⊆ R^n be bounded. Show that E is of content zero if and only if E is Jordan measurable of zero measure. Can anyone help me with this problem ? Thank
the goal limit clearly depends on the a_n series - it can't be just some fixed value. If a_n is 1/n, the value is clearly less than if a_1 is a googol and a_n (n > 1) is 1/(n-1)....
I might be missing something, but we only know that a_n < € for n > N, so i dont understand how we can jump to the sum which sum for k from 0 to n
And now for the same question but with a_k instead of a_n in the ln (which was my initial read of the problem)...
If I'm not mistaken, it should be the same result: same proof, but bounding the sum only from N onward, the N first terms being finite (a little work needed here to deal with the remaining /n) they will disappear when multiplied by 1/n.
A much more counter-intuitive result!
Yes that part is sketchy as hell
Omg! I finally know why I could always omit the added terms that go to 0 in the limit when I transformed riemann sums to integrals.
I knew it worked but I didnt know **why** it worked. Thanks
6:20 wait I don’t understand. How can we have that natural log inequality if the epsilon bound only applies for n>N? I don’t think we can guarantee it for 1
I originally thought the same, but the index in the sum is over k, not n. So with respect to the sum, the a_n is constant and changes with the limit and then so does the upper bound of the sum.
@@Happy_Abeoh!😮 Ok, got it. Thanks.
@@GreenMeansGOF you’re welcome
11:28 Homework
11:54 Good Place To Stop
Seems like something that can be calculated using Stirling's approximation, even though there's a pesky square root factor that seems to invalidate it.
i misread the question but its honestly more interesting so heres an interesting varient: whats the limit as n-> infty of 1/n*\sum_{k=1}^{n} ln(k/n + a_k)
(HINT: split up the sum)
@ 10:56 Should be dx, not dt.
Masterfully resolved. 👍👏
That really is sweet! Nice mathematics!!!
I now realize why a lot of content creators use a virtual blackboard instead of a physical one. Visual challenged people like me have a hard time seeing the contrast with a recorded physical blackboard and can't just walk closer to it if it is on a screen.
The int(ln x dx, x = 0..1) could be done better by realizing it's equivalent to int(exp(y) dy, y = -inf..0) because they're inverse functions. At that point, the integral becomes almost trivial. Only need to incorporate the - sign.
What would be a good book on Rieman sums?
Is this the standard analytical way of defining this type of improper integral?
A bounded set D ⊆ R^n is called Jordan measurable if the characteristic function 1_D is integrable on some n-hyperrectangle R contains D. The Jordan measure of D is defined as |D| = ∫1_D on R.
Prop. D is Jordan measurable iff ∂D is Jordan measutable and |∂D| = 0.
A set C ⊆ R^n is called of content zero if ∀ε>0, ∃ n-hyperrectangles R1, ..., Rm such that |R1| + ... + |Rm| < ε and C ⊆ (R1∪...∪Rm)
Let E ⊆ R^n be bounded. Show that E is of content zero if and only if E is Jordan measurable of zero measure.
Can anyone help me with this problem ? Thank
Indeed :)
I watched so many Maths-Videos, where can I do my exam? 😀
Noted
We've had an unfortunate return to the meaningless video titles.
just look at the thumbnail
@@GrifGreylol
What happened to the cool video titles and thumbnails? Who used to make these?
Editor on summer break I assume.
@@stephenbeck7222 Editor: can i get paid?
Michael: And thats a good place to stop.
i-th
the goal limit clearly depends on the a_n series - it can't be just some fixed value. If a_n is 1/n, the value is clearly less than if a_1 is a googol and a_n (n > 1) is 1/(n-1)....
Ooops. I see it - the 1/n in front of the sum does win out eventually....
I love your content, but please get yourself an editor that will properly remove your gaffes. It's annoying and kind of embarrassing.
no.
This has the same energy as “I think you’re a good person, but please go f**k yourself”
please get yourself someone in your life that will properly advise against these kinds of immature comments. It's annoying and kind of embarrassing
grow up
He has usually had an editor (a student I think), but it's summer vacation now.