Proving MY FAVORITE IDENTITY using DISCRETE DERIVATIVE
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In continues version of identity integration should start from 0. If you start from 1 it will become (A^2/2-1/2)^2 != (A^4/4-1/4).
To tie the discrete to the continuous one should not start with one, but with zero for both. (0+1+...+n)^2 = 0^3 +1^3 +...+ n^3
* continuous
@@terryendicott2939 Are you familiar with Euler's doubly infinite geometric series aka doubly infinite identity?
0 =...n^3 + n^2 + n^1 + n^0 + 1/(n^1) + 1/(n^2) 1/(n^3)...= 0
There's some very deep connection with mediants of 1/0 : 1/1 : 1/0.
The last rest in the video about integration should be from 0 to A , not from 1 to A. Otherwise it will be wrong. The video is very nice, I like it. It is the first time I see discrete derivatives.
I wasn't really understand why there was 1
There is a really in depth video on that topic which Ive watched a while ago and would really recommend.
or he missed a term -1/4
@@Ligatmarping could you please tell me where it is?
I did the proof by induction for practice.
We want to prove (1+...+n)^2 = 1^3 +...+ n^3. We already know some trivial base cases.
So let's set our induction hypothesis: (1+...+k)^2 = 1^3 +...+ k^3; for shorthand, S_k = (1+...+k).
(1+...+k+(k+1))^2 = (S_k +(k+1))^2 = S_k^2 + 2*S_k*(k+1) + (k+1)^2
But recall another way to calculate S_k = k(k+1)/2, thus we have:
= S_k^2 + k(k+1)^2 + (k+1)^2 = S_k^2 + (k+1)(k+1)^2 = S_k^2 + (k+1)^3
= 1^3 +...+ k^3 + (k+1)^3
Proof complete.
To me proof by induction is kinda useless cos I've always felt that unless you could already guess the identity itself or you know the identity is true but do not have a proof, it becomes very difficult to actually derive.
For example how do you guess that (1+2+....+N)^2=1^3+2^3+....+N^3 so that you could prove it by induction?
i think the integrals at the end should be from 0?
Yes, I think so.
He could have done it from A to B. As it turns out, it doesn’t matter.
the end result would not be the same as the squaring on the left hand side comes after the evaluating of the integral. what’s on the board is not correct. It clearly refers to the starting point 0.
If we include 0 and 0^3 in our sums, then we have a direct comparison with the integral starting from 0.
If u see sums as a lebesgue integral with the Dirac measure over the natural numbers, u have to start at 0 in order to make the extrapolation to the real numbers true, cuz if u don’t, the formulas aren’t correct (the sum can start at n=1 but the truth is it starts at n=0 but 0 doesn’t change for this particular formula the result, to get general formulas right u always need A_n to be able to start with A_0 with n=0 cuz the measure u use is over the whole natural numbers including 0)
Pretty elegant, and no explicit induction needed. Love it!
Proof for natural numbers without induction. Use:
1) 0^0=1
2) a^d is infinity(interior) of a-ary d-digit positional(or 0-nested) number system:
1^3 -"- unary 3-digit n/s (point, exterior and interior are not counted)
2^3 -"- binary 3-digit n/s (triangle, 3-vertex simplex, exterior and interior are counted)
3^3 -"- trenary 3-digit n/s (
3D cube: 8 vertices 12 edges 6 square 1 interior ,exterior is not counted.
2D cube : 4 vertices 4 edges 1 square(interior);
1D cube : 2 vertices 1 edge (interior) ; all it will be used in example in "4)" )
4^3 -"- 4-ary 3-digit n/s =2^6 (6-vertex simplex)
5^3 -"- 5-ary 3-digit n/s (3D 5_cube, consisting of 2x2x2 joined together cubes with 125 faces
(4*2=8 interiors 9*3=27 vertices (12*2+4*3=36) squares (12*3+9*2=54) edges )
6^3 -"- 6-ary 3-digit n/s (1D 3-ring 3 times shifted and first and last 3-rings glued (2D 3_donut: 9 vertices 9 squares 18 edges)
or 1D 3-chain pressed into 1 vertex)(2D 6_mebius: (12-3) vertices 6 triangles 3 squares 18 edges)
7^3 -"- 7-ary 3-digit n/s (3D 7_cube, consisting of 3x3x3 joined together cubes with 7^3 faces
And so on.
3) (ker+1)-ary img-digit number system is equivalent to (img-d)-nested 2-ary d-digit number system:
(ker+1)^img = Sum F[i,img-d]* Sum F[j,d]*ker^(j+i) ,for i:=0 to img-d, for j:=0 to d, F[j,d] - amount of j-faces in d-vertex simplex;
if d=3 then:
(0+1)^3=1*(1*0^0+3*0^1+3*0^2+1*0^3)=1
(1+1)^3=1*(1*1^0+3*1^1+3*1^2+1*1^3)=8
(2+1)^3=1*(1*2^0+3*2^1+3*2^2+1*2^3)=27
(3+1)^3=1*(1*3^0+3*3^1+3*3^2+1*3^3)=64
(4+1)^3=1*(1*4^0+3*4^1+3*4^2+1*4^3)=125
(5+1)^3=1*(1*5^0+3*5^1+3*5^2+1*5^3)=216
(6+1)^3=1*(1*6^0+3*6^1+3*6^2+1*6^3)=343
if d=2 then:
(0+1)^3=1*(1*0^0+2*0^1+1*0^2)+1*(1*0^1+2*0^2+1*0^3)= 1^2+0+2*0^2+0^3
(1+1)^3=1*(1*1^0+2*1^1+1*1^2)+1*(1*1^1+2*1^2+1*1^3)= 2^2+1+2*1^2+1^3
(2+1)^3=1*(1*2^0+2*2^1+1*2^2)+1*(1*2^1+2*2^2+1*2^3)= 3^2+2+2*2^2+2^3
(3+1)^3=1*(1*3^0+2*3^1+1*3^2)+1*(1*3^1+2*3^2+1*3^3)= 4^2+3+2*3^2+3^3
(4+1)^3=1*(1*4^0+2*4^1+1*4^2)+1*(1*4^1+2*4^2+1*4^3)= 5^2+4+2*4^2+4^3
(5+1)^3=1*(1*5^0+2*5^1+1*5^2)+1*(1*5^1+2*5^2+1*5^3)= 6^2+5+2*5^2+5^3
(6+1)^3=1*(1*6^0+2*6^1+1*6^2)+1*(1*6^1+2*6^2+1*6^3)= 7^2+6+2*6^2+6^3
if to notice that last number is a previose sum ,for example 1^3 is a 1^2+0+2*0^2+0^3,
after summarising we receive 1^3+2^3+3^3+4^3+5^3+6^+7^3=(1+2+3+4+5+6+7)^2.
This is just another way of doing an inductive proof. For example say we want to prove a_n = b_n. We first show a_1 = b_1 (you do this part last). Then we need to show that a_n = b_n => a_{n+1} = b_n{+1}. This is equivalent to a_{n+1} - a_n = b_{n+1} - b_n, i.e. we need to show Δ_n = Δ_{n-1}.
I always marvel at the way it can often work out that the concepts that are harder to wrap your head around are often easier to work with once you have. For example:
The real numbers are initially less intuitive than the natural numbers, but once you have the tools of calculus, they make this sort of problem trivially easy.
The complex numbers are even further challenging when first encountered, but they make all kinds of calculations that are more complicated in the reals incredibly easy.
Infinite sums take more work to understand that finite sums, but they're often (though not always) easier to work with.
14:35
This seems like a good opportunity to look at a more general analogies between discrete and continuous derivatives. For instance, say you have sequences aₙ, bₙ , and cₙ
Scalar Operators
Let aₙ = cbₙ for all n. Then ∇aₙ = cbₙ₊₁ - cbₙ = c(bₙ₊₁ - bₙ) = c∇bₙ . (Similar to how (cf(x))' = cf'(x)
Linearity
Let aₙ = sbₙ + tcₙ . Then ∇aₙ = (sbₙ₊₁ + tcₙ₊₁) - (sbₙ + tcₙ) = (sbₙ₊₁ - sbₙ) + (tcₙ₊₁ - tcₙ) = s∇bₙ + t∇cₙ
You can similarly get analogies for the product rule and quotient rule. And, something to maybe fiddle with from the video, there is also an analogy for taking derivatives of powers. It uses a "falling power" which is a bit like a partial factorial. For example, the 3rd "falling power" of n is n(n-1)(n-2). When you work out the discrete derivative of falling powers they behave exactly like continuous derivatives of real powers. Which you can then experiment using these combined rules on the problem in the video, for instance, to maybe prove it in a slightly different way. 🙂
Another way to look at this is to look at the discrete integral Σ, the sequence of partial sums, and then to look at the fundamental theorem of discrete calculus.
It would be nice to hear more about discrete calculus. But in this video, if you know the sum of numbers up to n, you could've used it right at the begining on the left hand side... would've made things a bit easier. Also would be nice to the a derivation for the continuous case more analogous, like perhaps taking the derivative in A from the integral.
Looks like the lower bound of the integrals should be 0.
Now that you mention discrete derivative, it seems so natural, since the integral version is also satisfied
∫_0^x u du = x²/2
∫_0^x u³du = x⁴/4
= (∫_0^x u du)²
Now it got me wondering if this applies more widely...
(∫_0^x u^(b-1) du)²
= (u^b / b)²
= u^2b / b²
= (2/b)*∫_0^x u^(2b-1) du
The dominant terms of
∑_0^n k^(b-1) and ∑_0^n k^(2b-1) are just the integrals above, so these match the identity, but it doesn't mean the rest of the terms do
About the change from from discrete to continuous in the end: why is it still valid? Can a generalization of calculus be used to explain this?
Anyway seeing how the "same" result holds on both discrete and continuous cases reminded me of Umbral calculus. Umbral calculus is about something else but, just imagine that if you had the right theorems, you could prove the harder discrete case by transporting the problem to the easier continuous case!
(Also, it would actually be cool if you made a video about Umbral calculus)
Love this video. Cool identity and even cooler approach. Love the use of this 'finite difference'.
That is really beautiful. I had not seen this identity before. Honestly 😃 Thanks for sharing.
You could have also summed both sides and showed the sum is the same.
The lhs would be n^2(n+1)^2/4
The rhs is the sum of cubes. Which you can find by converting to factorial powers first and then taking the sum.
For proving the lemma I like to prove the stronger a_n - a_1 = Σ Δ_k (k goes from 1 to n-1) which is easy just by splitting into two sums and reindexing (or just view it as a 'telescoping series'). Thus Δ_n = 0 implies a_n - a_1 = 0 => a_n = a_1 for all n. No induction needed.
I did a problem like this in a math olympiad one time, but the question asked to generalize this to the exponent n and n-1, and then find all the values of n such that the equation holds. Quite similar, but a bit different.
HW:
let a_n = 1+2+...+n, b_n=n(n+1)/2, for all positive integers n
Da_n = a_n+1 - a_n = n+1
Db_n + b_n+1 - b_n = [(n+1)(n+2)-(n+1)n]/2=(n+1)2/2= n+1 = Da_n
By the lemma, Da_n = Db_n implies the functions differ only by a constant, which is a_1-b_1=1-1(2)/2=0. //
Funny how close this winds up to being inductive. In the end we are talking about starting at the same point and incrementing change by the same amount.
My favorite way to think of this identity is to see how each side of the equation arises when you build it up from adding up entries in a multiplication table.
That's great! Thanks a lot!
Pretty clever!
Nobody steals my hackers! Also, very cool proof! Also also, the lower bounds of both integrals should be 0.
A easy relation would be by building upon this one i.e.
n x (Integral of x^(n-1) from 0 to A)=
(Integral of 2x from 0 to A) ^ n/2
Or
Sum over 0 to A of X^n-1 =
(2^ n/2 x (Sum over 0 to A of X)^n/2)/n
=> 1^(n-1) + 2^(n-1) + ... + A^(n-1)
= [{(A)(A+1)}^n/2] /n
Even n only.
Nice one. Now I want you to solve for n the equation 1ⁿ+2ⁿ+...+mⁿ= (1+2+...+m)^{n-1}, for all natural numbers m.
I think it might be 3, but I dunno...
Of course m=0 or m=1 is a solution for any value of n. But, given a value of n, answering this for any natural number m is a bit more complicated, but we can easily say something in the case where n is fixed and m→∞. In that case, the left hand side ~ c * m^(n+1) for some constant c. The right hand side ~ d * m^(2(n-1)) for some constant d. If these sequences were the same, the exponents should be equal so: n+1 = 2(n-1). This has only one solution, namely n=3. This shows that n = 3 is the only case where your equality is true for all natural numbers m.
Better start the integral at 0 and not at 1, which is natural, but then the results are also correct.
This method is interesting and leads to the continuous case, but the proof by induction is much more straightforward.
a good mnemonic for getting students to remember this for life is "sum 'n' square is cube 'n' sum"
Faulhaber's identities for odd powers fall out quite naturally from the integral calculus approach.
cool
in the last board, lower bound of integrals must be 0 not 1
WAIT !!! Integral of x.dx between 1 & A is (1/2 . x^2) for X = A minus X=1 so A^2 - 1 and, squared gives 1/4 (A^4 - 2A^2 + 1) different from 1/4(A^4-1) the integral of x^3 dx taken between 1 & A........
Indeed! To make it work, we should take the integral from 0 to A.
9:32 can you do it without using the closed form ?
S = 1 + 2 + 3 + ... + n
S = n + (n-1) + (n-2) + ... + 1
2S = n*(n+1)
S = n*(n+1)/2
Proof by Gauss :)
should the integrals be integrated from 0 instead?
Would it be possible to develop all the main content of calculus from discrete first principles instead of limits, secant lines, areas under curves, etc?
While there's a great deal you can do, I'm not sure you can get (as you say) "all the main content," because the entire epsilon-delta proof depends on being able to choose _arbitrarily_ small gaps. You'd need to develop a new definition of discrete continuity which...I'm not sure if that's even possible. A thing being discrete is generally understood to be inherently discontinuous.
2*(1 + 2 + ... + n)^4 == (1^5 + 2^5 + ... + n^5) + (1^7 + 2^7 + ... + n^7)
(Int X dx)^2=Int x^3 dx
very nice video.
Does this mean that the analytic continuation of the Riemann-Zeta-Function evaluated at -3 is 1/144?
It's 1/120
Is this true for any other exponents?
Pretty funny
4:37... it's 0;
But we all understand what you mean... the closest *without being 0*. It's just me being pedantic.
At 14:01 the integral from 1 to A of x dx = 0.5 A^2 - .5. So the square of this is not equal to ((A^2)/2)^2. Also the integral from 1 to A of x^3 dx
= 0.25 A^4 - 0.25, not A^4/4. The lower bounds of both integrals should be 0, then they are the same.
cool
Quick, confuse the physicists! Take n to infinity so that (-1/12)² = ζ(-3)
Does this identity have a name ? Who found it ? What is its history ?
Nicomachus' Identity. Source: ruclips.net/user/shortsSZsMVGqUiic
Probably it was known since ancient times
Nicomachus's theorem
Your continuous analogue is false, but the version that starts the integrals at 0 is true; correspondingly, if the empty sum is taken to be 0, then the discrete form of that identity still holds true, because 0^2=0; the identity could be further extended if you take the sum from a greater number to a smaller to be the negative of that sum from the smaller to the greater.
If it's true if the integral starts at 0 it must be true if the integral starts at 1, because integrating from 0 to A is the same as integrating from 0 to 1 plus integrating from 1 to A, then you can cancel out the 0 to 1 part
@@ere4t4t4rrrrr4, the integral of x from 1 to A is ½A^2−½; the square of that integral is ¼A^4−½A^2+¼. Meanwhile, the integral of x^3 from 1 to A is ¼A^4−¼; the problem with your line of argument is that you didn't consider the cross term, or to make it shorter, that (a+b)^2≠a^2+b^2 unless a=0 or b=0.
What does it mean a_n = b_n? Like how we get it judging by only n=1, what if we take n=2, isn’t a_2 = 2 and b_2 = 8 or we take the base only without powers?
a and b aren't the individual terms, they are sums. so a_2 is (1+2)^2 = 9, and b_2 is 1^3+2^3= 1+8=9
@@silver6054 thanks!
Мy favorite proof of this formula is summation by parts. But i think this proof looks similar.
Це тобі не допоможе.
@@vladalgov1770 Уже допоможло.
@@user-fw9ej9gj1h Дискретне перетворення Абеля не підходить для цієї задачі.
@@vladalgov1770 тебе пишут по английски, а ты отвечаешь по своему. Культуре диалога тебя не учили?
@@danilbutygin238 В такому випадку, тебе теж не вчили, адже ти відповідаєш третьою мовою:)
Nice argument
You tell me to gently touch the like button, so what do I do? I smash it! Like it's a cockroach!!! You're not the boss of me!
No funny story in the description today :(
Your explaination is hard because you don't give all details.
4) (ker+coker)-ary img-digit number system is equivalent to (img-d)-nested (1+coker)-ary d-digit number system:
(ker+coker)^img = Sum F[coker,i,img-d]* Sum F[coker,j,d]*G[ker,coker,d,coker_shape]^(j+i) ,for i:=0 to img-d, for j:=0 to d, F[coker,j,d] - amount of j-faces in d-coker_shape(if coker=1 then it is d-vertex simplex (in this case above formula is equivalent to Newtons binom if (img=d) and (G=ker)) else if coker=even number then it is d-coker_cube shape(exterior isn't counted, face-vectors 8 12 6 1 fo 3D cube ; 4 4 1 for 2D cube; 2 vertices 1 edge for 1D cube; all will be used in example for coker=2;) else if coker=odd number then it is d-coker_donut shape;
Example for coker=2.
if (d=3)and(coker=2) then:(because exterior isn't counted in this shape ,unlike simplex, we start with vertices)
(0+2)^3=1*(8*0^0+12*0^1+6*0^2+1*0^3)=8
(1+2)^3=1*(8*1^0+12*1^1+6*1^2+1*1^3)=27
(2+2)^3=1*(8*2^0+12*2^1+6*2^2+1*2^3)=64
(3+2)^3=1*(8*3^0+12*3^1+6*3^2+1*3^3)=125
if (d=2)and(coker=2) then:
(0+2)^3=2*(4*0^0+4*0^1+1*0^2)+1*(4*0^1+4*0^2+1*0^3)= 8
(1+2)^3=2*(4*1^0+4*1^1+1*1^2)+1*(4*1^1+4*1^2+1*1^3)= 27
(2+2)^3=2*(4*2^0+4*2^1+1*2^2)+1*(4*2^1+4*2^2+1*2^3)= 64
(3+2)^3=2*(4*3^0+4*3^1+1*3^2)+1*(4*3^1+4*3^2+1*3^3)= 125
if (d=1)and(coker=2) then:
(0+2)^3=4*(2*0^0+1*0^1)+4*(2*0^1+1*0^2)+1*(2*0^2+1*0^3)= 8
(1+2)^3=4*(2*1^0+1*1^1)+4*(2*1^1+1*1^2)+1*(2*1^2+1*1^3)= 27
(2+2)^3=4*(2*2^0+1*2^1)+4*(2*2^1+1*2^2)+1*(2*2^2+1*2^3)= 64
(3+2)^3=4*(2*3^0+1*3^1)+4*(2*3^1+1*3^2)+1*(2*3^2+1*3^3)= 125
3) lets coker=3. This case is equivalent to case coker=1,because (1+3)-ary d-digit number system is equivalent to (1+1)-ary 2*d-digit number system and d-coker_3_shape is 2*d-coker_1_shape which is 2*d-vertex simplex.
4) lets (coker=4). This case is similar to case (coker=2) but has 2 types of shape. This means
4.1)if (ker=even) then d-coker_4_shape is equivalent to 1-times nested and only in vertices previouse d-coker_shape wich is d-chain1(d-cube). It means that inside each vertice of visible d-cube nested invisible d-cube. After that make "visible-invisible" inversion and we receive d-coker_4_shape for even ker.
It has : F[coker_4_shape,i,d]=F[coker_2_shape,1,d]*F[coker_2_shape,i,d].
So face-vector of
1-coker_4_shape is 2*(2 1)= 4 2
2-coker_4_shape is 4*(4 4 1)= 16 16 4
3-coker_4_shape is 8*(8 12 6 1)= 64 96 48 8
4.2)if (ker=odd) then d-coker_4_shape is equivalent d-chain2. It means that
1-chain2 is point shifted 2 times with face-vector 3 2
2-chain2 is 1-chain2 shifted 2 times with face-vector 9 12 4
3-chain2 is 2-chain2 shifted 2 times with face-vector 27 54 36 8
Example for coker= 4.
if (d=img=3) then:
if ker=2*n :
(0+4)^3=1*(64*0^0+96*0^1+48*0^2+8*0^3)=64
(2+4)^3=1*(64*1^0+96*1^1+48*1^2+8*1^3)=216
(4+4)^3=1*(64*2^0+96*2^1+48*2^2+8*2^3)=512
if ker=2*n+1 :
(1+4)^3=1*(27*1^0+54*1^1+36*1^2+8*1^3)=125
(3+4)^3=1*(27*2^0+54*2^1+36*2^2+8*2^3)=343
(5+4)^3=1*(27*3^0+54*3^1+36*3^2+8*3^3)=729
if (d=2)and(img=3) then:
if ker=2*n :
(0+4)^3=4*(16*0^0+16*0^1+4*0^2)+2*(16*0^1+16*0^2+4*0^3)=64
(2+4)^3=4*(16*1^0+16*1^1+4*1^2)+2*(16*1^1+16*1^2+4*1^3)=216
(4+4)^3=4*(16*2^0+16*2^1+4*2^2)+2*(16*2^1+16*2^2+4*2^3)=512
if ker=2*n+1 :
(1+4)^3=3*(9*1^0+12*1^1+4*1^2)+2*(9*1^1+12*1^2+4*1^3)=125
(3+4)^3=3*(9*2^0+12*2^1+4*2^2)+2*(9*2^1+12*2^2+4*2^3)=343
(5+4)^3=3*(9*3^0+12*3^1+4*3^2)+2*(9*3^1+12*3^2+4*3^3)=729
(**********EOF (coker=4)**********************)
5) lets (coker=5). This case has infinitely shapes,
5.1)if (ker=5*n) then d-coker_5_shape is equivalent to d-donut5 ( 5-ring 5 times shifted first and last 5-rings are glued is 2-donut5).
So face-vector of
1-donut5 is 5 5
2-donut5 is 25 50 25
3-donut5 is 125 375 375 125
Example 5.1.3 if (d=img=3):
(0+5)^3 =1*(125*0^0+375*0^1+375*0^2+125*0^3)=125
(5+5)^3 =1*(125*1^0+375*1^1+375*1^2+125*1^3)=1000
(10+5)^3=1*(125*2^0+375*2^1+375*2^2+125*2^3)=3375
Example 5.1.2 if (d=2)and(img=3):
(0+5)^3 =5*(25*0^0+50*0^1+25*0^2)+5*(25*0^1+50*0^2+25*0^3)=125
(5+5)^3 =5*(25*1^0+50*1^1+25*1^2)+5*(25*1^1+50*1^2+25*1^3)=1000
(10+5)^3=5*(25*2^0+50*2^1+25*2^2)+5*(25*2^1+50*2^2+25*2^3)=3375
Example 5.1.1 if (d=1)and(img=3):
(0+5)^3 =25*(5*0^0+5*0^1)+50*(5*0^1+5*0^2)+25*(5*0^2+5*0^3)=125
(5+5)^3 =25*(5*1^0+5*1^1)+50*(5*1^1+5*1^2)+25*(5*1^2+5*1^3)=1000
(10+5)^3=25*(5*2^0+5*2^1)+50*(5*2^1+5*2^2)+25*(5*2^2+5*2^3)=3375
5.2)if (ker =3*n+1) then d-coker_5_shape is equivalent d-donut3. It means that
1-donut3 is 3-ring(point shifted 3 times first and last points are glued) with face-vector 3 3
2-donut3 is 1-donut3 shifted 3 times first and last 1-donut3 glued with face-vector 9 18 9
3-donut3 is 2-donut3 shifted 3 times first and last 2-donut3 glued with face-vector 27 81 81 27
Example 5.2.3 if (d=img=3):
(1+5)^3 =1*(27*1^0+81*1^1+81*1^2+27*1^3)=216
(4+5)^3 =1*(27*2^0+81*2^1+81*2^2+27*2^3)=729
(7+5)^3 =1*(27*3^0+81*3^1+81*3^2+27*3^3)=1728
Example 5.2.2 if (d=2)and(img=3):
(1+5)^3 =3*(9*1^0+18*1^1+9*1^2)+3*(9*1^1+18*1^2+9*1^3)=216
(4+5)^3 =3*(9*2^0+18*2^1+9*2^2)+3*(9*2^1+18*2^2+9*2^3)=729
(7+5)^3 =3*(9*3^0+18*3^1+9*3^2)+3*(9*3^1+18*3^2+9*3^3)=1728
Example 5.2.1 if (d=1)and(img=3):
(1+5)^3 =9*(3*1^0+3*1^1)+18*(3*1^1+3*1^2)+9*(3*1^2+3*1^3)=216
(4+5)^3 =9*(3*2^0+3*2^1)+18*(3*2^1+3*2^2)+9*(3*2^2+3*2^3)=729
(7+5)^3 =9*(3*3^0+3*3^1)+18*(3*3^1+3*3^2)+9*(3*3^2+3*3^3)=1728
5.3)if (ker =3*n+2) then d-coker_5_shape is equivalent to d-nested d-tube3. It means that
0-tube3 is point face-vector 1
1-tube3 is 3-ring(point shifted 3 times first and last points are glued) with face-vector 3 3
2-tube3 is 1-tube3 shifted 3 times with face-vector 12 21 9 (0*3+3*4 3*3+3*4 3*3+0*4)
3-tube3 is 2-tube3 shifted 3 times with face-vector 48 120 99 27 (0*3+12*4 12*3+21*4 21*3+9*4 9*3 +0*4)
Example 5.3.3 if (d=img=3):
(2+5)^3 =1*(48*1^0+120*1^1+99*1^2+27*1^3)+(2+5)^2=343
(5+5)^3 =1*(48*2^0+120*2^1+99*2^2+27*2^3)+(5+5)^2=1000
(8+5)^3 =1*(48*3^0+120*3^1+99*3^2+27*3^3)+(8+5)^2=2197
Example 5.3.2 if (d=2)and(img=3):
(2+5)^3 =3*(((2+5)^1+12)*1^0+21*1^1+9*1^2)+3*(((2+5)^1+12)*1^1+21*1^2+9*1^3)+(2+5)^2=343
(5+5)^3 =3*(((5+5)^1+12)*2^0+21*2^1+9*2^2)+3*(((5+5)^1+12)*2^1+21*2^2+9*2^3)+(5+5)^2=1000
(8+5)^3 =3*(((8+5)^1+12)*3^0+21*3^1+9*3^2)+3*(((8+5)^1+12)*3^1+21*3^2+9*3^3)+(8+5)^2=2197
Example 5.3.1 if (d=1)and(img=3):
(2+5)^3 =12*(((2+5)^0+3)*1^0+3*1^1)+21*(((2+5)^0+3)*1^1+3*1^2)+9*(((2+5)^0+3)*1^2+3*1^3)+(2+5)^2=343
(5+5)^3 =12*(((5+5)^0+3)*2^0+3*2^1)+21*(((5+5)^0+3)*2^1+3*2^2)+9*(((2+5)^0+3)*2^2+3*2^3)+(5+5)^2=1000
(8+5)^3 =12*(((8+5)^0+3)*3^0+3*3^1)+21*(((8+5)^0+3)*3^1+3*3^2)+9*(((2+5)^0+3)*3^2+3*3^3)+(8+5)^2=2197
5.4)if (ker =4*n+3) then d-coker_5_shape is equivalent to d-donut4 . It means that
1-donut4 is 4-ring(point shifted 3 times first and last points glued) with face-vector 4 4
2-donut4 is 1-donut4 shifted 3 times first and last 1-donut4 are glued with face-vector 16 32 16
3-donut4 is 2-donut4 shifted first and last 1-donut4 are glued with face-vector 64 192 192 64
4-donut4 has face-vector 256 1024 1536 1024 256 ( 0*3+64*4 64*4+192*4 192*4+192*4 192*4+64*4 64*4+0*4 )
Example 5.4.3 if (d=img=3):
(3+5)^3 =1*(64*1^0+192*1^1+192*1^2+64*1^3)=512
(7+5)^3 =1*(64*2^0+192*2^1+192*2^2+64*2^3)=1728
(11+5)^3 =1*(64*3^0+192*3^1+192*3^2+64*3^3)=4096
Example 5.4.2 if (d=2)and(img=3):
(3+5)^3 =4*(16*1^0+32*1^1+16*1^2)+4*(16*1^1+32*1^2+16*1^3)=512
(7+5)^3 =4*(16*2^0+32*2^1+16*2^2)+4*(16*2^1+32*2^2+16*2^3)=1728
(11+5)^3 =4*(16*3^0+32*3^1+16*3^2)+4*(16*3^1+32*3^2+16*3^3)=4096
Example 5.4.1 if (d=1)and(img=3):
(3+5)^3 =16*(4*1^0+4*1^1)+32*(4*1^1+4*1^2)+16*(4*1^2+4*1^3)=512
(7+5)^3 =16*(4*2^0+4*2^1)+32*(4*2^1+4*2^2)+16*(4*2^2+4*2^3)=1728
(11+5)^3 =16*(4*3^0+4*3^1)+32*(4*3^1+4*3^2)+16*(4*3^2+4*3^3)=4096
Example 5.4.0 if (d=img=4):
(3+5)^4 =1*(256*1^0+1024*1^1+1536*1^2+1024*1^3+256*1^4)=4096
(7+5)^4 =1*(256*2^0+1024*2^1+1536*2^2+1024*2^3+256*2^4)=20736
(11+5)^4 =1*(256*3^0+1024*3^1+1536*3^2+1024*3^3+256*3^4)=65536
5.5)if (ker =5*n+6) then d-coker_5_shape is equivalent to d-cube5 . It means that
1-cube5 is 5-chain(point shifted 5 times with face-vector 6 5
2-cube5 is 1-cube5 shifted 3 times first and last 1-donut4 are glued with face-vector 36 60 25
3-cube5 is 2-cube5 shifted first and last 1-donut4 are glued with face-vector 216 540 450 125 (0*5+36*6 36*5+60*6 60*5+25*6 25*5)
Example 5.4.3 if (d=img=3):
(6+5)^3 =1*(216*1^0+540*1^1+450*1^2+125*1^3)=1331
(11+5)^3 =1*(216*2^0+540*2^1+450*2^2+125*2^3)=4096
(16+5)^3 =1*(216*3^0+540*3^1+450*3^2+125*3^3)=9261
Example 5.4.2 if (d=2)and(img=3):
(6+5)^3 =6*(36*1^0+60*1^1+25*1^2)+5*(36*1^1+60*1^2+25*1^3)=1331
(11+5)^3 =6*(36*2^0+60*2^1+25*2^2)+5*(36*2^1+60*2^2+25*2^3)=4096
(16+5)^3 =6*(36*3^0+60*3^1+25*3^2)+5*(36*3^1+60*3^2+25*3^3)=9261
Example 5.4.1 if (d=1)and(img=3):
(6+5)^3 =36*(6*1^0+5*1^1)+60*(6*1^1+5*1^2)+25*(6*1^2+5*1^3)=1331
(11+5)^3 =36*(6*2^0+5*2^1)+60*(6*2^1+5*2^2)+25*(6*2^2+5*2^3)=4096
(16+5)^3 =36*(6*3^0+5*3^1)+60*(6*3^1+5*3^2)+25*(6*3^2+5*3^3)=9261
and so on infinitely.
6) case coker=even is similar to (coker=4) or (coker=2)
7) case coker=odd is similar to (coker=1) or (coker=3)
So we see (ker+coker)'s unclosure wich is pieces of broken coker_shape (like matryoshka doll unclosure).