the easy way to expand (a+b+c)^n, the trinomial theorem
HTML-код
- Опубликовано: 25 ноя 2024
- Sign up for a free account at brilliant.org/... and try their daily problems now. You can also get a 20% off discount for their annual premium subscription so you can get access to ALL of their awesome designed courses!
Here's the easy way to expand (a+b+c)^n, the trinomial theorem. You would see this in a discrete math class.
Binomial and trinomial theorem:
Part 1: • How to expand (a+b)^n ...
part 2 (ft. best friend): • Can we really do "-2 c...
part 3: • the easy way to expand...
#combinatorics #discretemath
a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7+7a^6c+42a^5bc+105a^4b^2c+140a^3b^3c+105a^2b^4c+42ab^5c+7b^6c+21a^5c^2+105a^4bc^2+210a^3b^2c^2+210a^2b^3c^2+105ab^4c^2+21b^5c^2+35a^4c^3+140a^3bc^3+210a^2b^2c^3+140ab^3c^3+35b^4c^3+35a^3c^4+105a^2bc^4+105ab^2c^4+35b^3c^4+21a^2c^5+42abc^5+21b^2c^5+7ac^6+7bc^6+c^7
Im fucking done
36 terms
Kove 21 #ExtremeAlgebra
Oh my lord. My man
use stars and bars principle to get 9 choose 2 = 36 terms
but honestly that is commitment goals i stan
how much time it took you to do this.....
sa fssfs or just choose one of the three variables from each parenthesis, so it's 3 choose 7 with replacement = (3+7-1, 7) = (9, 7) (which is the same as 9 choose 2)
The greatest thing about you is u teach maths in a gentle way that is not painfull you do hard problem with such an ease and with having smile on your face.
Black pen red pen yeah😊
oh man, you just made discrete math soooo much more easier and fun with your previous 3 videos. How knew it wasn't so complicated after all?
I can't thank you enough!
Nikil Kumar you are very welcome!!!! My pleasure to help!!
You just proved the question I had for 5 years: Proof of binomial and multinomial theorem in the starting 3.5 minutes. I love you for that.
Vighnesh Raut thank you!!!!
I got 36 terms... (Correct me if I'm wrong!)
(a+b+c)^7 = [a+(b+c)]^7, so just treat (b+c) as a single value to make things easier.
Expand this out using the binomial theorem (I'm gonna ignore the co-efficients because they don't really matter):
a^7 + a^6 (b+c) + a^5 (b+c)^2 + a^4 (b+c)^3 + a^3 (b+c)^4 + a^2 (b+c)^5 + a (b+c)^6 + (b+c)^7
By observation we know the number of terms from (p + q)^n is gonna be n+1 terms.
So, looking at each term here, we know there's gonna be 1 term in a^7, 2 terms in a^6 (b+c), 3 terms in a^5 (b+c)^2, etc.
Thankfully, there will be no 'merging' or simplifying of the terms because they're multiplied by different powers of a, so adding these together gives:
1+2+3+4+5+6+7+8 = 36 terms.
Hmm, good
I saw i j k, was getting excited for quaternions.
Same lmao 😂
Can you also make a video on the multinomial theorem? 3 IS NOT ENOUGH !😂
I've done some research on it. It's pretty nuts.
Holy moly. You just clarified combinatorics for me. The actual formula is n choose a = n!/(a!(n-a)!). It never occurred to me that this simplified to a terms of n! from n down. Only a numbers are multiplied. That's a big help. Thank you!
You’re very welcome!!
Math is all about thinking, not answers, not numbers
I did it by having (a+b+c)^7 = (a+y)^7 where y=(b+c) to get the coefficient of a^3 y^4 as 35, then finding the coefficient of b^2 c^2 in y^4 to ultimately get 210 too. I imagine this method breaks down somewhat if you had a variable appearing twice inside the bracket such as (a+ab+c)^7 though.
For how many terms, I started with a^7, one term. For a^6, you can have b^1c^0 and b^0c^1, 2 more terms. For a^5 you have b and c ranging from 0 to 2, so 3 more terms. And so on until you have 8 different terms with a^0, with b and c ranging from 0 to 7. Thus 1+2+3+4+5+6+7+8 = 36.
Sir
you teach us Permutations from Combinations.
Thank tou sir🙏🙏
8:34 "don't get too excited" :)
Neil Gerace hahahh
Maybe j is the imaginary unit instead, as in electronic engineering (because current = i already)
They are all imaginary units of a quaternion
Look like unit vectors
Thank you so much blackpenredpen my teacher gives these annoying trinomial expansions which took a lot of time to solve using only binomial theorem but now i can quickly solve these questions
A general formula for the number of terms in (x+y+z)^n is (n+1)(n+2)/2
Same answer with this eq. but simplified :- [ { n( n + 3) } / 2 ] + 1
To calculate how many terms in a trinomial expansion we can use the formula (n+1)(n+2)/2 where n is the power and in a quadnomial expansion it would be (n+1)(n+2)(n+3)/6.
and to calulate the coeficient of a trinomial expansion with any or all of the terms being greater than 1 is:-
Given ((pa)+(qb)+(rc))^n, where (p, q and r) are the coefficients of (a, b and c) respectively. Then the coefficient of each term is calculated as
[n!/(i!*j!*k!)](p^i*q^j*r^k) where (i, j, k) are the powers of a, b and c respectively, for example in the expansions of (2a + b + 3c)^7 and we want to coefficient of the (a^2 b^2 c^3) term we have:-
[7!/(2!*2!*3!)] 2^2*1^2*3^3 = 22680
The full expansion of the above would be:-
128 a^7+448 a^6 b+672 a^5 b^2+560 a^4 b^3+280 a^3 b^4+84 a^2 b^5+14 a b^6+b^7+1344 a^6 c+4032 a^5 b c+5040 a^4 b^2 c+3360 a^3 b^3 c+1260 a^2 b^4 c+252 a b^5 c+21 b^6 c+6048 a^5 c^2+15120 a^4 b c^2+15120 a^3 b^2 c^2+7560 a^2 b^3 c^2+1890 a b^4 c^2+189 b^5 c^2+15120 a^4 c^3+30240 a^3 b c^3+22680 a^2 b^2 c^3+7560 a b^3 c^3+945 b^4 c^3+22680 a^3 c^4+34020 a^2 b c^4+17010 a b^2 c^4+2835 b^3 c^4+20412 a^2 c^5+20412 a b c^5+5103 b^2 c^5+10206 a c^6+5103 b c^6+2187 c^7
Thank you so much for these videos. I love your channel but sometimes I think that all you do is integrals and I love integrals but some change on the subject is always good. Keep up this good work, man
Guilherme Nery Rocha yea I know. Thank you!
this explanation with places is realy awsome, now it is simple to generalize that to any amount of summants
Oh my goooood ! The lengendary one... the PURPLE PEN :o
Let's hope I'm not too late, *there is an extremely easy way to do this* , even IGCSE(O Level), or High School Students in America can do this. Combinatorics/Trinomial Theorem in my opinion is too complicated.
We can exploit the fact that two of the three constants in the expansion have the same degree of power, so we can vary the power of a, and then keep that constant to find the coefficient of b^2 c^2
(a+b+c)^7 "let u=b+c"
(a+u)^7 "use standard binomial theorem to expand, find the coefficient of a^3"
u^7 + 7(u^6)(a) + 21(u^5)(a^2) + 35(u^4)(a^3) + .... "we have reached a^3, expanding further not necessary"
Undoing substitution u=b+c
35(a^3)(b+c)^4 "use standard binomial theorem to expand, find the coefficient of (b^2)(c^2)"
35(a^3) [b^4 + 4(b^3)(c) + 6(b^2)(c^2) + 4(b)(c^3) + c^4] "Cross multiply required term"
35(a^3)(6(b^2)(c^2))
210(a^3)(b^2)(c^2)
Surprisingly, in the end you've done the same calculation. 7!/(4!3!) = 35, and then 4!/(2!2!) = 6, multiplying you get 7!/(3!2!2!) = 210
Nice solution
You? Genius.
@@memeticTrojanCalc thanks man!
this is a way. I did it this way before watching the video when he told to pause. But this is *not an extremely easy way*
what he taught in the video is the extremely easy way. Because you can expand the same ideas to any number of polynomials and find out coefficients of them instantly. For example, the coefficient of a^7.b^2.c^5.d^3.e^2.f^3.g^3 in the expansion of (a+b+c+d+e+f+g+h)^25 is simply 25!/(7!.2!.5!.3!.2!.3!.3!) = 2.968384491x10^16
I don't think you'll ever find out that with your method.
Just use multinomial theorem it is
n!/a!b!c!
Which gives 7!/3!2!2!
Which we get by 7C3×4C2×2C2. ; )
I grouped it as (a+(b+c))^7. From there I know nCr(7,3)*a^(3)*(b+c)^(4) gives me all the a cubed terms. Because I specifically only want the coefficient of the a^(3)*b^(2)*c^(2) term, I know I need nCr(7,3)*a^(3)*nCr(4,2)*b^(2)*(c)^2 = 210.
Also n+2 choose 2 gives the number of terms for the n power of the trinomial and for n= 7
9 choose 2
Stars and Bars theorem says 9c2 terms (36).
Love this channel!!
Thanks!
Thanks : )
Time for making Pascal's hyperpyramid for the quadronomial theorem, be right back.
Do it
Hey buddy try this IIT JEE question: Integration from 1 to 5 of (x-1)(x-2)(x-3)(x-4)(x-5) with respect to dx. BUT don' t solve so long(i.e. don' t expand it) try to get the answer in just 4-5 lines 😊
i did in my head using multinomial theorem, but watching this also fun.
Aman kashyap thanks!!
210. By use of the multinomial theorem, the coefficient is 7!/(3!2!2!) Which simplifying the expression gives (7×6×5×4/(2×2)) which is 210. What's more, I learned this on Brilliant.org. couldn't reccomend it enough
hands down best video yet
Thank you for the video it has helped me a lot
You can just simply use binomial theorem even for this just write the general term by considering any two of those a,b,c as one single unit and then, write the general term again for the unit
This just came up while reviewing and didn't see it on the reviewer given to us :P thanks
Very easy sir :we can use multinomial theorem :Coefficient of a^3 b^2 c^ 2 will be 7!/(3! 2!2!)
kaushik Mahanta good!
The best part is that this fully generalizes.
This is what makes me smile.
14:07 This... is brilliant
I can see how you can use the same logic to find the probability of certain hands of cards in poker
LOVE FOR MATHS....
please do a proof for binomials having real coefficient
You can set i bracket b+c then do the binomial theorem twice
Not twice technically 7 times due to internal expansion
And I would rather leave the question than doing that
No you dont need to expand it just look the coefficient [a^3b^2c^2](a+(b+c))^7=(7over3)*a^3*(b+c)^4=(7over3)*a^3[(4over2)b^2c^2)]=210*a^3b^2c^2
Very nice explanation
In the brilliant.org problem, is phi the golden ratio? if so, for ease I'll call it x. EDIT: the triangle structure forces it to be the golden ratio.
small square = 1
medium square = x²
large square = (1+x)² = 1 + 2x + x²
we know that x/1 = (1+x)/x because golden ratio. Thus x²=x+1
rearrange to x=x²-1. double everything to get 2x=2x²-2
We are now ready. The big square's area was 1+2x+x². Substitute 2x for 2x²-2, so you get 1+2x²-2+x²
which equals 3x²-1. Therefore the solution is adding 3 medium squares and subtracting 1 small square.
brilliant is taking over!
Wearing my best friend today!
Nole Cuber great!!!
More combinatorics!!!
You make this look like kindergarden math, thanks a lotttttt. The outro music is Jay Chou 明明就
Great video!
Why It? Thanks!!
8:36 Steve got me!
OK, let's focus on the nth term of a binomial raised to a power, The rth term of (a + b)^n = nCr * a^(n-r) * b^r
Let's take the trinomial and turn it into a binomial
(a + b + c)^7 = (a + (b + c)) ^ 7, let d = b + c,
clearly we're only interested in the term of this binomial expansion with a^3, that means r = 3, and as n = 7, the first coefficient for the partial expansion of the trinomial is 7C3.
For the second part, we need to take a look at the term with a^3 in it, as r = 3, the part with d in it is raised to power of 7 - 3 = 4. Now as we're looking for b^2c^2, and the powers used are both equal to 2, we can go for the middle term of the expansion of the 2nd binomial, which will have a coefficient of 4C2.
To get the actual answer, we multiply these two coefficients together. The answer is 7C3 * 4C2. = 7!/(3!4!) * 4!/(2!2!)
At this point, you can either use Pascal's triangle to generate the values of nCr up to n = 7, or you can do some number crunching by using nCr = n!/(r!(n-r)!)
7C3 * 4C2 = 7!/(3!4!) * 4!/(2!2!) = 7!/3! * 1/(2!2!) = 7!/(3!2!2!) = 7 * 6 * 5 * 4 * 3! / (3! * 2 * 2), as 2! = 2 * 1 = 2
This simplifies to 7*6*5*4 / (2*2) = 7*6*5*4/4 = 7*6*5 = 7 * 30 = 210
I apologise for the messy one-liner fractions used above, but it'd take way too long to write it properly as it would be done in a book or on a board.
I also tried to do the fractions in a way to avoid using a calculator, and to use a method that allows you to rely purely on easy mental arithmetic to get the job done. Believe in factorials, not calculators.
This channel helps me a lot thank you!
could you show us how to use pascals pyramid and how to get the exponent of said terms
YOU ARE THE BEST!
Multinomial Therom
4D calculus. This is shit I need
1+4+9+16+25+... = 0 next?
i wonder if it would be 1/144?
No it just diverges.
The sum doesn't approach anything; it's just a sum of squares of every positive integer.
Unless this is some Zeta-function level stuff, in which case I'm not even going to help you there
@@dog_owner the joke is in reference to 1+2+3...=-1/12
Yes of course I'm not stupid I know what the Zeta function is I literally just mentioned it in my last comment
Love you man.
I have became your fan from yesterday.
Seriously
.......🙏😘
Anubhav Taneja thank you!!!
thanks to this I discovered soemthing more powerful than pascal's triangle
Very helpful video
Please sir can you show us how to get pyramid from Pascal triangle
I got 210 by multiplying ( 7 3 ) and ( 4 2 ). I guess the point is to do it without a calculator (you could construct Pascal's Triangle). But then again, I guess your method would be more suitable if you don't have a prior understanding of binomial expansions.
brilliant !
Thank you sir..
I was always confused about this, because when you do combinations, the order in which the elements are chosen doesn't matter, that's why it's a combination. But when something like this happens, the problem turns into a permutation with repetition, in which order does matter, theoretically at least. Because in this case, the order of the elements does not matter at all, it's a multiplication, and yet it's a permutation. Why does this happen?
Are you still confused?
Please Do a Proof of the multinominaltheorem
Was waiting for this!
Love you man!
Great video!!!
Purplepen !!!
And for the stars and sticks argument, even. It's x-mas!
I love ure problems bcz i find a lot of fun
Great video
Also called multinomial theorem
That was a great video i liked the question and solution with the answer 36.The idea was very creative.
There is just one small criticism that I have with the beginning. You did not explain why 9 choose 3 corresponds to picking 3 things that are the same from 9 Thomas.
[pre-apology for diverging] ... when you wrote abbacca, I caught myself hoping you'd work in some r's and 'd' so it could permute to 'abracadabra'...but I suppose a problem with circles would be needed for those vars. [ Thanks for the vid :) ]
0:42 pause the video?
*Coming back after 2 hrs
woah.
my whole life has led up to this moment.
i don't know what to say.
but
woah
Hailfire08 : )
There's 36 terms in the expansion of (a+b+c)⁷
Hold on a second, doesn't this mean that the coefficient of a^3b^2c^2 is the same as that for a^2b^3c^2 and a^2b^2c^3. If so, that's a massive shortcut! The problem is reduced to a matter of how many ways you can partition "n" (for example, a^3b^2c^2 vs a^7 vs b^4c^3, et cetera) and calculating the coefficient for each. Really goes to show the symmetry of the whole thing.
On another note, combinatorics (as a whole) is so simple and intuitive, it feels like cheating xD
That's one of those stars and bars problems.
Actually you can compute this directly from 7!/(3!*2!*2!), there is no need to compute this using 7 choose 3 ect method
Kamil Bizoń
I explained both ways
@@blackpenredpen I skipped forward a bit to check my answer and didn't notice, sorry
How many terms in the expention of a trinomial raised to the nth power : answer is (n+1)(n+2)/2 of course the triangular numbers .
What if you are looking for the coefficient when a, b and c are actual integers? How would our answer change if we use combinatorics to find it?
E.g (x^2-1/x+7)^8 find the coef of x^6 or smth.
And is there a way, if specifically using this method, to find the independent of x?
Hi blackpenredpen, can you show us the Leibenz integral (differentiation under the integral sign)? Thanks
Me: I am finally done with binomial theorem
Goes to trinomial theorom next: I give up
Answer must be exactly
3 medium squares
- 1 small square
Yup i solved it using fi^2=fi+1 where fi is the golden ratio
The Gaming Nut *phi
@@NYsummertimeCHI u got the idea right? So no need for correction
The Gaming Nut Just because I understood in this instance doesn't mean it will always be clear what you mean to everyone who reads it. When you spell things correctly it makes it easier for people to understand. No need to be defensive I just assumed you hadn't seen it written down before so was letting you know the correct spelling.
@@NYsummertimeCHI k thankyou for edification on a matter that is literally bullcrap...
I think it would be more comfortable using clipped mic rather than what you have. Anyway, good instruction.
I have this type of questions in class +1 in chapter binomail theorem
In total there are 36 terms.
Yea!
Why not use Sir Pinsky’s triangular pyramid?
bprp can you do a video on this
sqrt(2)^(sqrt(2)^(sqrt(2)^(sqrt(2)^(sqrt(2)... infinite times
the limit is 2.
sqrt(x)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)^sqrt(2)...infinite times -> limit is x
bprp i'm 13
Can you generalize this even further taking the nth power of k-terms?
Nacho Yes, this is called the multinomial theorem, it is well known in combinatorics and probability.
it's not much more different than this, just a little more math involved.
for example, the coefficient of (w^a)*(x^b)*(y^c)*(z^d)*(u^e) in (w+x+y+z+u)^(n) is just
(n!)/(a!)(b!)(c!)(d!)(e!).
To find the total number of terms in that expansion, it will just be the number of ways you can form 'n' out of 5 numbers such that
a+b+c+d+e =5. Using the trick from the video, it'll end up being (5+4) choose 4.
@@angelmendez-rivera351 la voy a buscar,muchas gracias por responder
Tl;dr: I derive a formula for the number of distinct terms in the fully-simplified/'canonical form' expression of (a + b + c)ⁿ; this formula can be found in the antepenultimate paragraph with a very simplified form in the final paragraph. When n = 7, the answer is 36.
Take n ∈ ℤ: n > 1. This will be required throughout this analysis, even if not explicitly specified infra.
There are p(n, m ≤ 3) ≜ round((n + 3)² / 12) distinct partions of n into at most 3 parts. Each such partition corresponds to an ordered triple of numbers, with the entries of the triple being arranged so that the first entry is w.l.o.g. the largest part and each subsequent entry is a part which is not greater than the one immediately prior to it (id est: the entries are monotonically non-increasing); if we run out of parts, then we can pad the triple with entries which are 0. We can then order the triples in, say, reverse lexicographic order. For example, the distinct partitions of n = 6 into at most 3 parts are:
(6, 0, 0)
(5, 1, 0)
(4, 2, 0)
(4, 1, 1)
(3, 3, 0)
(3, 2, 1)
(2, 2, 2)
Notice that there are seven entries in this ordered list, which aligns with our formula for p(n, m ≤ 3).
These triples are going to be helpful to us. We can now assign the exponents of a, b, and c to the entries in an orderes triple. For example, (3, 2, 1) represents a³b²c¹. But, obviously, there are assignments missing from our prior list; for example, (3, 1, 2) is not distinguished from (3, 2, 1) in the list supra, but would represent a different multivariable polynomial (a³b¹c² vs. a³b²c¹, resp.).
So, we need to permute the entries of the ordered triples. That is nice and easy: we multiply by 3! = 6.
But, wait, some triples have repeated entries. (6, 0₁, 0₂) is not actually different from (6, 0₂, 0₁). We cannot multiply these by 3!. Luckily, there are only two options for what these triples can be like: either their entries belong to a set of cardinality 2 or to singleton (set of cardinality 1). For example, (6, 0, 0) has entries in {0, 6}, which has cardinality 2, whereas (2, 2, 2) has entries in {2}, which is a singleton. In the former case (cardinality 2), we can just specify the position of the non-repeated entry (exempli gratia: 6 in (6, 0, 0)) and see that it has three positions; these ones get multiplied by 3, rather than 3! when we are counting options. In the latter case, when we have the entries belonging to a singleton, there is only one distinct way of writing the triple ((2, 2, 2) is the same as (2, 2, 2), after all), so we multiply by 1 for these.
Notice that the entries belong to a singleton ⇔ the triple is (x, x, x) ⇔ n = x + x + x = 3x ⇔ 3|n ⇔ n is rhedde. (Meaning: n has to be a multiple of 3). That is why n = 6 has (2, 2, 2) as a partition. For n = 7, we do not need to worry about this because, well, 7 is grean, which means that 7 % 3 = 1; in other words, 7 ≡ 1 (mod 3).
So, we just need to count our partitions of n into at most 3 parts such that exactly one part is repeated with multiplicity exactly 2. The options for this are either that the repeat part is 0, or that the repeat part is nonzero.
(Note on the reasoning here: If more than one distinct part is repeated with a multiplicity of at least 2 (each), then we do not have a triple. If zero parts are repeated (each being of multiplicity 1), then we are in the earlier case that we handled. So, we must have exactly one part with repetition. If it has multiplicity greater than 3, then we do not have a triple. If it has multiplicity 1 (or less), then we have a contradiction. If it has multiplicity 3, then we are in the (x, x, x) case, which we already handled and have dismissed for n = 7. So, we now want to concern ourselves with the part having multiplicity exactly 2, as stated).
In this circumstance, our triple is of form either (x, y, y) or (x, x, y), where x ≠ y, and the set to which the entries belong, in either case, is {x, y} and has cardinality 2. I am going to blend these two cases together; in other words, we will ignore the monotonic ordering requirement within the triple, which we can temporarily do because addition is commutative and associative, and we do not really care about which type of triple of the aforementioned two options we are dealing with. W.l.o.g., let x be the repeated element, and y ≠ x. So, we have n = x + x + y = 2x + y.
∴ x ≤ floor(n/2) ∵ x > floor(n/2) ⇒ 2x > n. This bound on x is our main working criterion for this case.
Note: n ∈ 3(ℤ⁺∪{0}) ⇒ x ≜ n/3 = y ≤ floor(n/2), with strict inequality ∀n ∈ 3ℤ: n ≥ 6, so our rhedde case is included by the condition (of our working criterion) immediately supra. We do not want this, so we will have to manually subtract out such instances.
We also have x ∈ ℤ⁺∪{0}, by assumption. (In particular, x = 0 ⇒ y = n, which is fine).
∀n ∈ ℤ⁺, y is determined by x ∵ y = n - 2x. ∴∀n, we need to count only the possibilities for x.
Thus, the number of partitions of n into at most 3 parts with exactly 1 part repeating with multiplicity exactly 2 is N(n) ≜ |ℤ ∩ [0, floor(n/2)]| - δ(n, 3ℤ). The indicator function at the end ensures that we subtract 1 in the case of n being rhedde again (in order to account for us ignoring the case of (x, x, x)).
Notice that this works for n = 6; N(6) = 4 - 1 = 3; these three options were listed as (6, 0, 0), (4, 1, 1), and (3, 3, 0), with (2, 2, 2) being ignored because that is the multiplicity 3 case.
So, we are now ready to put everything together in order to find the formula for the number of distinct terms in the fully-simplified/'canonical form' expression of
(a + b + c)ⁿ. The answer is:
Ω(n) ≜ 3! (p(n, m ≤ 3) - N(n)) + 3 N(n) + 1 δ(n, 3ℤ),
where: δ(ξ, A) ≜ (1 ⇔ ξ ∈ A; else: 0) is the indicator function ∀A ⊆ ℝ; and, as before, p(n, m ≤ 3) ≜ round((n + 3)² / 12) and N(n) ≜ |ℤ ∩ [0, floor(n/2)]| - δ(n, 3ℤ).
∴The answer to the question is:
Ω(7) = 3! (8 - 4) + 3•4 - 1•0 = 24 + 12 = 36
∵ p(7, m ≤ 3) = round((7 + 3)²/12) = round(100/12) = round(8 + (1/3)) = 8,
& δ(7, 3ℤ) = 0,
& N(7) = |ℤ ∩ [0, floor(7/2)]| - δ(7, 3ℤ) = |ℤ ∩ [0, 3]| - 0 = 4.
I like this formula because it is very clear on where each value is originating. However, with a bit of simplification, we can derive the formula:
Ω(n) = (n + 1)(n + 2) / 2.
There is a nice direct combinatorial proof of this, but I still believe it to be a little less transparent. Additionally, the formula which I just derived generalizes readily.
Do arcsinx/x integral from 0 to 1
Hello 👋, pleas solve integral ((sinx) *sqrt(sinx))
Habeeb Shaalan Can't be done
I am very needy
3^7 terms
Somehow the i j k reminded me of quaternions
Oon Han they reminded me so too!
blackpenredpen you should do a video on quaternions!
Oon Han I don’t know it too well tho.
Oon Han we do a collab soon, yea?
blackpenredpen yea!
I got jebaited by the ijk thinking it could have something to do with quaternions ;-;
I wonder why this vide is unlisted (at least when I'm watching)
The teacher is interested in who is paying attention in class.
Because the video is not completely ready for release, this is a test video so far, although he plans to release a final version soon.
to make whoever found the video special
11:56 what if I can do it by using a function that is made to do this kind of work? nHr(3,7) is your answer.
Three large squares minus one small square
imaginary factorial? quaternion factorial?