Відео

This is just beta a+1,b+1, can derive it using ibp

How can you assume all f(x) can be expressed as x(x-1)(x-2)... ? For instance, f(x) = ax^2 + bx + c = (x-r1)(x-r2), where r1 and r2 are roots of f(x), but not always consecutive natural numbers.

Nice! So, same can said for 13, 53, 73, etc.

Another way to sort of skip to the answer here, You can realize the integral is equivalent to: The probability that, given a balls of color A, b balls of color B and 1 ball of color C, the balls are arranged in the order (all A) (c) (all b). Assuming the balls are indistinguishable, there is only one arrangement that satisfies this. But there are (a + b choose b) * (a + b + 1) possible arrangements of the balls. Thus the integral is 1/((a + b + 1)( a + b choose b)). For further clarification, the integral is treating the balls as points in the interval [0,1] and conditioning the locaiton of the C ball at t (with prob dt) and the probability that that all a of the A balls are less than t is t^a and all B balls are greater than t is (1 - t)^b, hence the equivalence. Thanks for the problem!

condolences to the one student who didn't get an offer o7

Beta function!

Taking thêta as an integer variable is wild

Hi jpi, this is unrelated (and a slightly strange question) but will the MAT format change ( it being online) make a difference to the type of questions they give in the long form questions at the end. Im just wondering because, your expected to type your answers, so does this completely rule out stuff like integration and geometry ( things you cant type) ? Looking at this years paper there problems were algebra/logic questions. So yeah , do you think theyll get rid questions about integration and geometry etc (in the long form section) for the sake of not being able to type solutions out?

Thanks Sir.

Where was the simple hypothesis used?

nice proof for part (a), I proved it by using the result that a bounded monotonic series must be convergent, since the sum of a_n's converges, it is bounded by some number A, say, and since all the a_n's are positive, the sum of a_n's squared should be bounded above by A squared for example. The a_n squared are all positive so the sequence of partial sums is monotonically increasing and bounded above so it must converge.

This years BMO2 paper came out recently, itd be nice ig you could go through some of the problems, i definelty prefer you over the ukmt channel : ). Problems 1 and 4 in particular look interesting!

Really cool that you use descending Pochhammer polynomials or, as I like to call them, base-decreasing powers polynomials!

For D: I'd multiply the top and bottom by (√2 +1)³ instead. Since (√2 -1)(√2 +1) = 2 - 1 = 1, we eliminate the square root at the denominator. We have (√2 +1)³/4 Since √2 > 1, (√2 +1)³/4 > 2³/4 = 2, the next step is the same as you.

It’d be appreciated if you did some more individual mat and tmua problems!

There’s a typo in your ‘exercise’ solution: it’s (t,2,3t).

Very intelligently solved 🔥🔥

Thank you !

watching a BMO problem instead of doing last minute recap before my calculator mock exam tomorrow.

I know it's not to do with the video but, thank you for all of your videos on interview questions. They were extremely useful as well as the demo interviews and helped me get an offer for maths at Oxford🎉 Once again, thank you❤

Just got in for CS, your vids were so helpful for interview prep! Thanks

How can you claim that l is rational

Thank you mucho

how to think for the need to use this identity , what extra knowledge helps here?

2023=7*(17^2) . Not a prime !!

And, each has 1011 lattice points in it.

Where is your fixed point video,it was uploaded i thought i willnwatch later but now coundnt find

It’s been a while since you’ve done a BMO problem. Do you plan on going through some soon? Thanks!

Thank you for this! Offer day is on Tuesday and I'm terrified

In your thumbnail it looks like you're going to tackle the root x integral which is a bit deceiving.

This video is extremely clear, I understood it way better than I expected Thank you

Thank you. Although I am applying to Cambridge, I just need motiviaton right now to finish my mock preparation (thankfully I still have 2 weeks), and your videos help.

I'm stupid at math but can't believe i actually solved it...i guess if I change the ceiling function to the floor function, the answer would change to 2^0 + 2^1 + ... 2^98 ? not sure though

Hi, thanks for the awesome video! pretty specific question but do english IGCSE matter for Oxford math or is the MAT and then interview that matter most ?

The integral starts at 0, and floor(0) is 0, so you have to add an extra 1. Another way of thinking about its final value is that it would just be 99 1s in binary, which is 2^100 - 1.

0 to 1 = 0 so start form 2^0+2^1+...

Getting nervous for the January sitting, I've heard so many people saying how hard the last one was!

Can you do some more tmua problems !

King's property! :D

Hi, why isn’t that right: 16^16^16^16 is equal to (2^4)^(2^4)^(2^4)^(2^4) which is equal to (2^2^2)^(2^2^2)^(2^2^2)^(2^2^2) And that are 12 twoes

This is so cool, its kind of scary to think that that series is forced that way, hope he is not claustrofobic. Love the thumbnail, keep going!!

assume f(a)!=a, i.e. f(a)>a define x_0 := sup{x|f(x)>x} and eps := f(x_0)-x_0 if eps>0 then f(x_0+eps/2)>=f(x_0)-k*|eps/2|>x_0+eps/2 -> contradiction to definition of x_0 if eps<0 then f(x_0+eps/2)<=f(x_0)+k*|eps/2|<x_0+eps/2 > contradiction to definition of x_0 therefore we have eps = 0 and thus x_0 is our fixpoint

Ana2 immer spaßig

👀

Fucking Legend dude😂😂😂😂😂(the mathematics is also cool😅)

I have No-one to share this with

I appreciate you using the old imperial logo in the thumbnail, because the new one is just atrocious

Jajaaj

Happy new year ⭐❤

Theres no way you did this lmao