Відео

@@nickmcstuffins1836 Thank you! Really glad you enjoyed.

@@akshatsharma8151 Thank you so much!

@@maths_505 Thank you!!!

@@alphalunamare Thank you so much!! Really appreciate it.

@@Silver-bq6td Oh good to know thanks!

@@jaysn1683 Thank you!! It was very hard...

@@Kàrtik21140 London! You?

@@OscgrMaths India

@@OscgrMaths I m from India 🇮🇳 so you have been graduated 🎓

Is it mean yes bro

Not yet but soon!

@@Kàrtik21140 Thank you!

@@user-cw1ow1cv2f Okay thank you!

@@MohdAbuolwan Thank you!

@@OscgrMaths I only recommend that you take it slowly on the viewer because not all people math geniuses like yourself😅

@@MohdAbuolwan Sorry!! I'll keep that in mind next time. Thanks again for the comment.

Thank you!

@@Aman-b5k Okay thank you! 🇮🇳

@@blew319 I'll give it a shot! Thanks!

You could try it! I'm not sure how it would work because the derivative of 1/(1+x^2)^n is messy but let me know if you get somewhere!

@@paulpinecone2464 😂😂😂 Great response!

Thanks so much!!

Thanks for the comment!

😮😮😮😮bom dia

@@tiktik9413 bom dia! 😄

Hello! Bom dia! 🇧🇷

Hey, no I've never come across this before as it seems to be mostly a big thing in the US - but it looks really interesting! Definitely love the idea of having a focus on problem solving.

@@OscgrMaths Right! Thanks for the videos. Most all AoPS students are math competition participants. Although AoPS advertises for a younger crowd, the biggest community is around the Olympiad training. You have to score in the top 1 percent (literally), on two tests (you have to pass the first test in order to take the second), before you can take a year-round maths training to make it onto the American Olympiad team. There are only 5 people on each team, and there's only 1 team per country. There are over 100 countries that participate, yearly! It really gets intense as test results near, or when the end of a class nears. Usually AoPS classes are 6 or 7 months long, and there isn't a single test for the whole of the 6 months! You usually get a proof that you need to write each week, and then 7 or so problems that you need to solve each week (they only count the total amount of problems solved over the 6 months, and it doesn't matter what your weekly score is). There are usually no instructions for how to solve these problems, but instead there is a book of similar problems with loads of proofs and solutions. It takes nearly 2 or 3 hours to solve a single problem (1 hour if it's a really good day). As one student of mathematics to another, seriously ignore small-fry stuff that most people are comfortable with (non of your videos are like that, but people will probably ask for stuff like that in the future). It's special to do something special, and not something watered-down or just okay. I appreciate your willingness to stick it out. Problem solving is most intuition and persistence, and that is something special which most people aren't very comfortable with. Your video topics are excellent! It takes some imagination to do proper math (most proofs -- if not all -- are based on intuition, and then the justification of it); so don't get bogged-down if people say otherwise. I really appreciate the fact that you are keeping the footage and representation basic, and not cluttering it with complex lighting or video edits. Thanks again!👍😄

@@booshkoosh7994 Thank you so much for this comment! Having my videos get bigger has been amazing but it has led to more negative comments too and so your kind words really mean a lot. Glad to know there's people out there who understand what I'm going for. I also completely agree with your view of maths on problem solving and I've found the equivalent program in the UK - the first exam is around October, so I'll definitely go for it then. Thanks!

@@OscgrMaths Absolutely! Happy to help! Good luck on the exam! 😄

Thanks so much! Very kind of you.

Ah another molchat doma fan!! I could try and put together a playlist for you - what other kind of music do you like?

Thank you so much! The usual way is the weak approximation but introducing the gamma function is what helps provide the sqrt2pin part! Glad you enjoyed.

@@OscgrMaths I shold have thought of it. This approximation for exponential integrals is a standard in asymptotic analysis

These numbers are 1000 consecutive non-primes!! All of them can be factored in a way other than themselves and one so therefore they are non-primes!

@@OscgrMaths but it would be interesting to see where the first sequence of 1000 consecutive non-primes numbers occur amoung the integers.

@@NXT_LVL_DVL Oh that's a really interesting question! Could probably do that by coding a prime sieve in python or excel.

I did spend some time in Huxley! Then they put me in Skempton as there was a free room with a whiteboard and it's just down the road.

@@Aman-b5k Thank you! ❤️

@@dakcom-mk6mp Thanks!!

It's at the university Imperial College in London! My other videos are on my own smaller whiteboard. Hoping to make more on the bigger ones at the uni since they're easier to lay working out on. Thanks for the comment!

@@JAzzWoods-ik4vv Oh that's a really helpful suggestion thanks so much!

@@timofeysobolev7498 Thank you so much!!

@@OscgrMaths Thanks to you!)

I've given this problem a go before... if it's bounded you can use kings rule quite nicely but as an indefinite integral it's very long 😅😅!

@@that_one_reply8805 Sure! Which part is it that you haven't seen before?

@@eppssilon School can make maths seem so dull - but it's not always like that! Thanks for the comment!!

Thanks so much!! Very kind of you.

Wow that's so interesting!! Thanks for the comment this is great.

@@OscgrMaths You are very welcome!!Also your video on stirlings approximation is amazing but I think that's obvious by now! By the way if you are interested in seeing the accuracy of it here is an example π!=7.18808272898 (with gamma function of pi+1). With extended stirliing's approximation π!=7.18808287903. You can see how close the 2 evaluations are.Infact, especially after I found out about this, this became my favourite analetic formula, because I think that it captures the behaviour so well that it can be considered an approximate analetic formula for x! for all real x.I have written a paper about it too!😄

​@@OscgrMaths Yeah I know right!!Also you are very welcome, and by the way your video on the original stirling's approximation is exceptional, and so was your smooth derivation with integration!!!Also, if you are interested in seeing the accuracy of the formula I ll write how it works for π!.Here: π!=7.188082729... (Γ(π+1)) π!=7.18808287903...(Extended stirling's approximation).By the way this is my favourite analetic formula in mathematics, as it is so accurate for all x greater than -1 and x less than that with its extension to all negatives with the recursive property, that I think it can be regarded as an approximate analetic expession for x! for all real x rother than an asymptotic expansion!!!

@@OscgrMathsYou are very welcome!!By the way amazing explanation of Stirlings approximation, your derivation was amazing!!!Also this is now my favourite analetic formula because it is so accurate that I like to regard it as an approximate analetic expansion for x! rather than an asymptotic expansion!!

@@Georgeclassified Absolutely, it's quite incredible how accurate it is!

-pi^2/6 is the correct answer... You can check it in an integral calculator if you aren't convinced!

Prettier to write -Z(2).​@@OscgrMaths

Thanks for the comment! If it were a Maclaurin expansion (centred at 0) for a function of n it would be only be valid for small values of n, but since it's a Taylor expansion around the point n for a function of t, whatever value we choose for n (even if it's very large) it just means the approximation will be most accurate roughly around that area. Hope that makes sense! Feel free to ask any more questions you have.

@@liamturman Thanks so much! You might be interested in this document from the university of connecticut which provides a few different and very interesting proofs for this (one of which involves bernoulli numbers!) kconrad.math.uconn.edu/blurbs/analysis/stirling.pdf

@@muskyoxes Glad it makes sense now!! Thanks for the comment.

@Gameboygenius I think it might be a British thing? I always said it as ell-enn because I first learnt most of my calculus online from american channels but my teachers and classmates have recently been calling it lun and I've reluctantly converted... Thanks for the great comment!

@@OscgrMaths lun is great. Hope it catches on !!

@suzum0978 Yeah that part with n in the bounds is a little rough - that's why it's definitely an approximation! But thanks for the comment either way. If you want a slightly more rigorous (but less accessible which is why I didn't include it) derivation, it can actually be done with a contour! By considering 1/n! as a taylor coefficient in the maclaurin series of e^z and then computing it with the cauchy integral formula. You can try and work out the line integral using the saddle point method. The dominant portion near the saddle point is approximated by a real integral and Laplace's method. Actually Laplace's method is useful in evaluating this integral in general so is also worth considering (especially since the saddle point method is kind of a complex extension of laplace's method). Hope this is interesting and thanks for the comment.

@@OscgrMaths this sounds exciting, I hope you make video about it if you have time in the future!

@@suzum0978 Okay great!

@@adw1z Yes! And each will be more and more accurate. Thanks so much for the comment!

@@lucahaines4655 Thanks so much!

@@dakcom-mk6mp Thank you!