The best teacher on yt rn Thanks a lot for these lectures ❤️🙏🏻
@ajaxbjax1029
6 жыл бұрын
Your last formula on the # of partitions with no summands divisible by 3 is also true if you rephrase it to allow the summand to be divisible by 3 but not the quantity (multiplier). For example: one 2 and two 2's are allowed, but not three 2's. Likewise, one 3 and two 3's are allowed, but not three 3's or six 3's etc. The generating function becomes (1+x+x^2 + 0 + x^4 +x^5 +0 +x^7+ ...) * (1+x^2 x^4 +0 +x^8 +x^10 +0 +x^14+ ...) *(1+x^3 x^6 +0 +x^12+x^15 +0+x^21+ ...) which is the same as your (1-x^3)/(1-x) * (1-x^6)/(1-x^2) * (1-x^9)/(1-x^3)...
@parnadbhattacharjee6368
3 жыл бұрын
How is it working ? Please explain. I can't understand how you got the generating function in the closed form (precisely the last line of your comment) .
@envalemdor
9 жыл бұрын
In your last example, could you explain why the generating function for 2s is (1 - x^6) / (1 - x^2) ? I thought the formula is (1- x^n+1) / (1 - x) since n = 4, shouldn't it be (1 - x^5) / (1 - x^2) ? Thank you for the great tutorials!
@SreeramHareesh
8 жыл бұрын
Consider t=x^2, now the function for 2 would be (1+t+t^2) which is equal to (1-t^3)/(1-t) according to the formula(1+x+x^2+..x^n)=(1-x^(n+1))/(1-x). Now replacing t by x^2, it would be (1-x^6)/(1-x^2)
@GiovannaIwishyou
2 жыл бұрын
Maybe I'm too late but hope this helps someone. In that case, yes you still use the formula (1-x^n+1)/(1-x) but this time, instead of x, you are calculating with x^2. So the series goes 1+x^2+x^4 = 1+(x^2)^1+(x^2)^2 which is, when we apply the formula (1-(x^2)^(2+1))/(1-x^2). The same goes for x^3 in the next parethesis.
@WOLF91
4 жыл бұрын
Good morning, Do you have a video on set partitions?
@rlira0908
4 жыл бұрын
bruh, where's the composition video? this is great btw.
@loftkey
7 жыл бұрын
Thank you so much!!! 1,000 x better than my teacher!!!
@MarvinOGarza
6 жыл бұрын
Hello, great tutorial!! Thanks for sharing. Could you please tell us what's the name of the book you were using for this, the one you took out the problem from. Thanks!
All the possible division of a positive integer n is called partition number of that integer. But we could find only p division, how could we find? As 3 division of 5 is(1+1+3) and (1+ 2+2) is 2,but partition of 5 is 7.
@alimahdavi4282
4 жыл бұрын
Thanks a lot for sharing
@harpuneetkalsi5960
4 жыл бұрын
can we use bell numbers for partitioning??
@peterbeach4194
6 жыл бұрын
Thank you. Thank you. Thank you.
@muralidharrao5831
4 жыл бұрын
For the last problem, while considering the summand not divisible by 3, by considering x^3 in the 1s, arent you considering a summand divisible by 3?
@parnadbhattacharjee6368
3 жыл бұрын
no. Summand, here refers to the integers. So, since 1 is not divisible by 3, we can take it as many times as we want as Trev has shown. You may refer to the comment by @AjaxBjax
@syedmdabid7191
3 жыл бұрын
But p(5) =7.
@obzen12
2 жыл бұрын
in the last question was it has to go to infinity 7 is not divisible by 3, 8 is not divisible 3, 9 is divisible so it is not included and so on?
@obzen12
2 жыл бұрын
why he stopped at 5
@pemudahijrah2454
7 ай бұрын
why there is no x^6 in 9:44
@flight8594
4 жыл бұрын
7:40 can't we have x2 as b > and = 2
@yusufhrj
4 жыл бұрын
I'm still a kid, and i think i have to learn this
@alex.k4371
2 жыл бұрын
how do i get the coefficient
@AltuğBeyhan
6 жыл бұрын
what does [x^n] mean? x^n 's coefficients?
@jonathannissim2774
5 жыл бұрын
yes.
@unremove
4 жыл бұрын
Is there a mistake at 11m:13sec? Why isn't the equation x^n = (1/1-x)( 1/1-x^4)(1/1-x^12) - - similiar to what was written at 7:45 where x^n = x^n = (1/1-x)( 1/1-x^2)(1/1-x^4) ?
@GiovannaIwishyou
2 жыл бұрын
Because we had starting summand x^4 so we take it out. What's left is 1+x^2+... and we know how to sum that series but we still have that x^4 that we took out so it goes into numerator.
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