Analytic Number Theory: Introduction to analytic number theory - 4th Year Student Lecture
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- Опубліковано 13 жов 2024
- In this Oxford Mathematics 4th year student lecture, Fields Medallist James Maynard gives an overview of some of the key results of the course, and proves some motivational results about primes.
You can watch many other student lectures via our main Student Lectures playlist (also check out specific student lectures playlists): • Student Lectures - All...
All first and second year lectures are followed by tutorials where students meet their tutor in pairs to go through the lecture and associated problem sheet and to talk and think more about the maths. Third and fourth year lectures are followed by classes.
can't believe james maynard is only a 4th year student
Well done
😂
@@deepdockproletarianarchive4539 its a joke
One of the greats!
😂
Not only is Maynard a great researcher, but from what I can tell from this video, he seems to be a great lecturer as well.
During a chat with a different professor at end of their lecture, I realised that some previous lecturer had forgotten to wipe off some sections of their lectures from the whiteboard at the end. The professor said 'we'll leave it up for the lecturer to remove it tomorrow', then we realised that it was Prof. Maynard's writing and proceeded to clear it up ourselves
it´s an excellent lecture Professor !!!
That's a beautiful lecture. I have been involved studying this area. Is it possible to upload all the videos of this course?
one just to tease and advertise
@@ericb7223 typical oxford
Is the whole course going to be uploaded?
No, just one more lecture to give a flavour of the topic. We do have some full courses among our 100+ student lectures that we have made available and will increase the number of full courses in future.
That is the saddest thing I read today.
please, for the love of god and all that is sacred WE NEED THE COMPLETE SET OF LECTURES 🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
At 20:30 I did not understand what happened with the denominators. We can't just replace them all with log(x) and use a geometric series formula, because that would make the terms smaller, not larger. We can't replace them all with the smallest denominator and roll it into the constant because the denominators are tending to 0. I thought we could switch to a proof by induction, so we would only have to show D n / log(n) + C (n/2) / log(n/2)
I'm confused there too. Which is probably why I'm not studying maths at Oxford!
Remember that Cx/ln(x) is an upper bound, not an exact value of the sum. I believe the logic is as follows:
x/ln(x) is strictly increasing for x > e, so for n >= 3, n/ln(n) < 2n/ln(2n). With this in mind, we can bound all the terms for x/k >=3 above by x/ln(x), and the case of x/k < 3 is taken care of by simply adding 3 to include the prime numbers 2, 3 and 5 (as 2x/k = 3 + #{k >= 2 : x/k >= 3} as x/ln(x) > 1 for x > 1. We can assume x > e as we care about the values of \pi(x) for large x.
@felix.henson I don't understand this part of your comment: "With this in mind, we can bound all the terms for x/k >=3 above by x/ln(x)". What's k? I do agree that n/ln(n) < 2n/ln(2n) but how is it useful here? Sure, we could bound the terms like (x/2)/ln(x/2) by x/ln(x) but the result would be a sum of log_2(x) copies of x/ln(x) so you'd get an upper bound like C*x in the end, instead of C*x/log(x).
@@duncanlevear7413 You're absolutely right, in my method I failed to see the dependence of C on x.
Is this a new upload? I remember him teaching in an older lecture.
Yes, new upload. This is a fourth year lecture, the others were first year or Public Lectures. Another lecture from the same course next week and a full meeting between James and one of his postdocs to come.
@@OxfordMathematics Okay, that's great!
29：00 the third term under sum notation should be n
On my 4th year I was eating my booger. This guy is already a fields medalist
at 42:20 professor wrote 2n/p^m in to the floor function but shouldn't it be 2n! instead of 2n?. Or am I missing a point?
No 2n look at legendre formula
@@yasser_elmoussaed oh thank you, I did not think legendre formula.
@@muhammetboran8782 you're so dumb lol
Matematic is powerfull 🔥
I would like to watch the whole course
Sure pay £9250 per year and you can
thanks a lot ❤❤
No problem
Your welcome
Is there any one who can help solve this problem, even my teacher is having problem with it , “given that h(x)=integral of ((f’(x)x-f(x))/x^2) , where 2h(2)=f(2)+4 and f(-1)=5 , what is the value of h’(-1) ?
There are some steps that i have done, h(x)=f(x)/x+C and i found the value of C which is 2, thus h(x)=f(x)/x+2 and h(-1)=-3 , now how can i find the value of h’(-1) ?
Appreciate your help
❤
Hello,Who can help me on how to enter this university? I visited their official website and did not find anything to help me
View the clip "Maths at Oxford Online Open Day 2024" ...
Thank you
Ok !
First' comment Award 🥇
💯
this lecture rules
what kind of whiteboard is that? why??
Wait I am in x grade and what I am doing here😂
First
Those are paper lecture boards? I can’t tell. Seems a waste.
apparently they use soft flexible whiteboards. at the end of lectures students help lecturers clean the boards
Это не математика, это грëбаное шоу
В смысле?
2=0 cause 0 plus 2 is 2
IT'S NOT INSPIRATIONAL...DOESN'T ACTIVATE THE AUDIENCE...WASTING TIME ONLY........
I can't believe what a mess this lecture is:
1. At the beginning be mentions "analytic number theory" and doesn't define 'analytic'
2. Then he erases 'analytic' and just says that we use 'analysis' as a tool, yet he implies that analysis does not belong to Number Theory.....this is complete nonsense. Number Theory is wholly a part of analysis.
The definition of 'analytic' is that we can find exact differentials and then form them into a conservation equation using Mdy + Ndx = 0. Then we can integrate these for a solution, usually a Hamiltonian
This is the 'starting point' of "analytic number theory".....which he obviously does not understand
Pardon my ignorance but how is this suppose to help me in my daily life?
Pure mathematics is about understanding the universe we live in, not helping our daily lives. However, you're using prime numbers and number theory every time you make a secure transaction online since it's the basis of modern cryptography. If the Riemann hypothesis is solved it could change that. Pure mathematics has a habit of finding practical applications eventually.
Yes, an accurate estimation of the number of primes less than x for an arbitrary x is crucial for your daily life.
Ahh shit. I'm even more confused now,lol. Fuk.
A matemática básica é p=np divisão no sentido anti-horário junto com toda estrutura do diagrama dos números batizados como hindu-arábicos. ➕ a conjectura que é sim um teorema de goldbach. A diferença que ele e o resto dos matemáticos não aprenderam a dividir pelo menos ➖ o número 0 zero no sentido anti-horário com seus divisores e quocientes no infinito. E o grande problema não é o infinito do zero, mas sim o que vêm depois do seu infinito.
É o mínimo que uma pessoa deveria saber para ensinar as crianças nas instituições de ensino.