The Math Thread

[Kharakov]

What function for a sum = 2 for first 4 terms, 4 for first 9, 6 for the first 16... without using if statements or floor functions.

\(\sum_{n=1}^4 f(n) =2 \) \(\sum_{n=1}^9 f(n) =4 \) \(\sum_{n=1}^{16} f(n) =6 \)


No floor functions or conditionals!

 

[lpetrich]

How might one generalize it?
\(\sum_{n=1}^{k^2} f(n) = 2(k-1)\)

Kharakov, I think that you are asking for some continuous function.

 

[lpetrich]

I'll do the asymptotic case: k -> infinity. We can then do sum -> integral, and the problem becomes much easier:
\( \int_1^{k^2} f(n) \, dn = 2(k-1) \)

Take the derivative with respect to k, and one finds
\( 2k \cdot f(k^2) = 2 \)

giving solution
\( f(n) = \frac{1}{\sqrt{n}} \)


Returning to the sums, I find this solution for Kharakov's three:
\( f(n) = \frac{ 3n^2 - 353 n + 5650 }{10080} \)


Adding the extrapolation to summing up to 1 gives f(1) = 0, and formula
\( f(n) = \frac{ (n - 1) (15 n^2 - 333 n + 1775) }{ 2520 } \)

 

[Kharakov]

Starting at n=2, n^2-(n-1)^2= 2n-1 terms of \(\sum_{k=(n-1)^2}^{n^2}\,\,\, \frac{1}{\sqrt{k}}\) ~ 2. 2 is the limit as n--> infinity

So \(\sum_{n=2}^\infty \,\, -2 + \sum_{k=(n-1)^2}^{n^2}\,\,\, \frac{1}{\sqrt{k}}\) approaches some value that I want to know.


n^3-(n-1)^3 terms of \(\sum_{k=(n-1)^3}^{n^3} \,\,\, \frac{1}{\sqrt[3]{k}}\) ~ 3n-3/2. 3n-3/2 is the limit as n--> infinity for the addition of terms

So \(\sum_{n=2}^\infty \, \, \frac{3}{2} -3n + \sum_{k=(n-1)^3}^{n^3} \,\,\, \frac{1}{\sqrt[3]{k}}\) approaches some value that I'm curious about.

n^4-(n-1)^4-1 terms of \(\sum_{k=(n-1)^4}^{n^4}\,\,\, \frac{1}{\sqrt[4]{k}}\)
~ (2n-1)^2-1/3 is the limit as n--> infinity for that number of terms

So \(\sum_{n=2}^\infty\,\, \frac{1}{3} - (2n-1)^2 + \sum_{k=(n-1)^4}^{n^4}\,\,\, \frac{1}{\sqrt[4]{k}}\) approaches something that I am curious about.

I want a way to input it into Wolfram Alpha, so I can use it to calculate the exact value for the "residuals" when I add up the whole series minus those values. It takes too long on my computer. Probably will time out on Alpha, unless they already have this stuff precalculated.

 

[lpetrich]

Kharakov, what you need is the  Euler–Maclaurin formula:

\( \sum_{k=m}^n f(k) = \int_m^n f(k) \, dk + \frac12 (f(m) + f(n)) + \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!} ( f^{(2k-1)}(n) - f^{(2k-1)}(m) ) \)

where the B_2k are Bernoulli numbers.

 

[Kharakov]

Is there an easy way to implement it in Alpha? I'm assuming much quicker convergence, but if I can't implement it?

It would be easier to find some single sum form (yeah, I know I've asked for a similar thing before), instead of double sum forms like:

\(\alpha_5 = \sum_{n=2}^\infty\,\, -\[\frac{5 \left( 4 {{n}^{3}}-6 {{n}^{2}}+4 n-1\right)}{4}\] +\,\,\,\,\,\,\, \sum_{k=(n-1)^5}^{n^5}\,\,\, \frac{1}{\sqrt[5]{k}}\)

 

[beero1000]

Viewing the sum as estimated by the integral, you'd expect the the subtracting term to be k/(k-1)(n^(k-1) - (n-1)^(k-1)), and the error to be about the Riemann sum error.

To verify, just look at the series expansion, which will give you the remaining terms after your subtraction.

So, in WolframAlpha, try: Series[HurwitzZeta(1/2, (n-1)^2) - HurwitzZeta(1/2, n^2+1), {n, Infinity, 5}]

which returns 2 + 1/n + 1/(2n^2) + ..., so after subtracting the 2, your first sum grows like the Harmonic series.

For other roots, change the 2's appropriately, e.g. Series[HurwitzZeta(1/3, (n-1)^3) - HurwitzZeta(1/3, n^3+1), {n, Infinity, 5}], returns 3n - 3/2 + 1/n + 1/2n^2 + ... so your subtracted sum grows like the harmonic again, etc.

 

[Kharakov]

So you're saying that the residuals \(\alpha_a\) of:

\(\lim_{n\to\infty} \,\,\,\,\,\, \sum_{k=(n-1)^a}^{n^a} \,\,\,\, \frac{1}{\sqrt[a]{k}} \,\,=\,\, \frac{{{\left( -1\right) }^{a+1}}\, \left( a+1\right) \, \left( {{\left( 1-n\right) }^{a}}+{{\left( -1\right) }^{a+1}}\, {{n}^{a}}\right) }{a}\)


\(\alpha_a = \sum_{n=2}^\infty\,\,\,\, \frac{{{\left( -1\right) }^{a}}\, \left( a+1\right) \, \left( {{\left( 1-n\right) }^{a}}+{{\left( -1\right) }^{a+1}}\, {{n}^{a}}\right) }{a} +\,\,\,\,\,\,\, \sum_{k=(n-1)^a}^{n^a}\,\,\, \frac{1}{\sqrt[a]{k}}\)


Grow like the harmonic series?

How about:

\(\alpha_a = \sum_{n=2}^\infty\,\,\,\, \frac{ \frac{{{\left( -1\right) }^{a}}\, \left( a+1\right) \, \left( {{\left( 1-n\right) }^{a}}+{{\left( -1\right) }^{a+1}}\, {{n}^{a}}\right) }{a} +\,\,\,\,\,\,\, \sum_{k=(n-1)^a}^{n^a}\,\,\, \frac{1}{\sqrt[a]{k}}}{n-1}\)

 

[beero1000]

So you're saying that the residuals \(\alpha_a\) of:

\(\lim_{n\to\infty} \,\,\,\,\,\, \sum_{k=(n-1)^a}^{n^a} \,\,\,\, \frac{1}{\sqrt[a]{k}} \,\,=\,\, \frac{{{\left( -1\right) }^{a+1}}\, \left( a+1\right) \, \left( {{\left( 1-n\right) }^{a}}+{{\left( -1\right) }^{a+1}}\, {{n}^{a}}\right) }{a}\)

No. That equation doesn't make sense, the limit on the left should mean that there shouldn't be any n terms on the right.

I'm saying that:

\(\sum_{k=(n-1)^a}^{n^a} \,\,\,\, \frac{1}{\sqrt[a]{k}} \,\,\approx \,\, \int_{(n-1)^a}^{n^a} \frac{1}{\sqrt[a]{k}} dk = \frac{a}{a-1}\left (n^{a-1} - (n-1)^{a-1}\right)\)

with an error term that's asympotatically \(1/n + O(1/n^2)\). See lpetrich's comment about Euler-Maclaurin.

\(\alpha_a = \sum_{n=2}^\infty\,\,\,\, \frac{{{\left( -1\right) }^{a}}\, \left( a+1\right) \, \left( {{\left( 1-n\right) }^{a}}+{{\left( -1\right) }^{a+1}}\, {{n}^{a}}\right) }{a} +\,\,\,\,\,\,\, \sum_{k=(n-1)^a}^{n^a}\,\,\, \frac{1}{\sqrt[a]{k}}\)


Grow like the harmonic series?

You've introduced a power of -1 and an off-by-1 error, but other than that, yes.

How about:

\(\alpha_a = \sum_{n=2}^\infty\,\,\,\, \frac{ \frac{{{\left( -1\right) }^{a}}\, \left( a+1\right) \, \left( {{\left( 1-n\right) }^{a}}+{{\left( -1\right) }^{a+1}}\, {{n}^{a}}\right) }{a} +\,\,\,\,\,\,\, \sum_{k=(n-1)^a}^{n^a}\,\,\, \frac{1}{\sqrt[a]{k}}}{n-1}\)

Dividing something that asymptotically grows like 1/n by n-1 gives you something that grows like 1/n^2.

 

[Kharakov]

No. That equation doesn't make sense, the limit on the left should mean that there shouldn't be any n terms on the right.

So the following doesn't make sense, replacing an n--> infinity with n=10^4:

Show

(-1)^(a)*(a)*((1-n)^(a-1)+(-1)^(a)*n^(a-1))/(a-1)

\( \,\,\,\,\,\, \sum_{k=(10^4-1)^a}^{10^4^a} \,\,\,\, \frac{1}{\sqrt[a]{k}} \,\,=\,\, \frac{{{\left( -1\right) }^{a}}\, \left( a\right) \, \left( {{\left( 1-10^4\right) }^{a-1}}+{{\left( -1\right) }^{a}}\, {{10^4}^{a}}\right) }{a-1}\)

with an error term that's asympotatically \(1/n + O(1/n^2)\). See lpetrich's comment about Euler-Maclaurin.

Does that mean that the error term approaches the harmonic, although it doesn't do so at the same rate?

There should be some type of manipulation that allows data (some constant) to be extracted if one determines the compared rates.

You've introduced a power of -1 and an off-by-1 error, but other than that, yes.

Figured that out while you put up your post when I saw the binomial expansion terms popping up.

Dividing something that asymptotically grows like 1/n by n-1 gives you something that grows like 1/n^2.

Lol...


Does asymptotic growth like 1/n imply that if I multiply it by some exact constant, the series will have the exact same value as the harmonic series at specific points determined by the relative growth rates of the (2 different) series, or will it only approach the same limit?

 

[beero1000]

So the following doesn't make sense, replacing an n--> infinity with n=10^4:

Show

(-1)^(a)*(a)*((1-n)^(a-1)+(-1)^(a)*n^(a-1))/(a-1)

\( \,\,\,\,\,\, \sum_{k=(10^4-1)^a}^{10^4^a} \,\,\,\, \frac{1}{\sqrt[a]{k}} \,\,=\,\, \frac{{{\left( -1\right) }^{a}}\, \left( a\right) \, \left( {{\left( 1-10^4\right) }^{a-1}}+{{\left( -1\right) }^{a}}\, {{10^4}^{a}}\right) }{a-1}\)

No, that's not right. Checking values, for a = 2, I get 2.000100005... (which is about 2 + 1/10^4 + 1/(2*(10^4)^2)...) for the left hand side, but something on the order of 200 million for the right hand side. If you use my equation of a/(a-1)*(n^(a-1) - (n-1)^(a-1)), with a = 2 and n = 10^4, you get the estimate of 2, with the expected error of around 1/10^4, plus something like 1/(10^4)^2, which is pretty close.

Does that mean that the error term approaches the harmonic, although it doesn't do so at the same rate?

Yes.

There should be some type of manipulation that allows data (some constant) to be extracted if one determines the compared rates.

You've introduced a power of -1 and an off-by-1 error, but other than that, yes.

Figured that out while you put up your post when I saw the binomial expansion terms popping up.

Dividing something that asymptotically grows like 1/n by n-1 gives you something that grows like 1/n^2.

Lol...


Does asymptotic growth like 1/n imply that if I multiply it by some exact constant, the series will have the exact same value as the harmonic series at specific points determined by the relative growth rates of the (2 different) series, or will it only approach the same limit?

You can get it to have any specific value you want, for one value of n at a time. I'd think it's unlikely for you to get more than that though.

 

[Kharakov]

I didn't want to recode the other stuff, so I switched the variables in this equation: a is n, and n is the nth term. Sorry, it works, it's just written backwards for convenience with code, which is a bad habit when asking others about the code you write.

\(\[\frac{\, {{\left( -1\right) }^{a}} a\, \left( {{\left( 1-n\right) }^{a-1}}+{{\left( -1\right) }^{a}}\, {{n}^{a-1}}\right) }{a-1}\]\)



I was hoping to find specific residuals with connections to something else. Silly.

Show

I don't know if you remember a few years ago when I was exploring infinite nested inverse functions (log vs. exp, root vs. power, cos vs. acos, etc.) .

Stuff like:

\(x= \lim_{roots\to\infty} \,\,\, \sqrt[n]{x^n-x+\sqrt[n]{x^n-x+\sqrt[n]{x^n-x +\dots}}} \)

The successive terms (nth root/ n+1th root = nx^(n-1)) of that series tended to the derivative (or 1/derivative) of x^n, so I'd subtract say, x-x (1000 nestings) and multiply it by its derivative^1000 power. First one for 2^2 is ??pi^2??, then a lot of things close to that that... I haven't checked on the inverse symbolic calculator. Just curious about that kind of stuff- stuff at the edge (limits), where something approaches a specific value following specific rules. I suppose I think I'll find something new, and that's not true. Sometimes I think I'll learn something new, and do, but.. still. It's not a career path for me, obviously, so... mehh. anyway. Time for a break

 

[Kharakov]

I was trying to find the 0s of this function earlier:

\( f(a)= \lim_{k\to\infty} \,\,\, \frac{\left( a-1\right) ! {{\left( -1\right) }^{a}}\, {{k}^{a}}}{{{e}^{\frac{1.0b0}{k}}}}\,\,- \frac {1}{a} \,\,\,-\sum_{n=0}^{a-1}{\left. {{\left( -k\right) }^{n+1}}\, {{\left( a-n\right) }_n}\right.} \)

\(f(a)= \lim_{k\to\infty} \frac{(a-1)! (-k)^a}{\sqrt[k]{e}} - \frac {1}{a} -\sum_{n=0}^{a}{\left. {{\left( -k\right) }^{n+1}}\, {{\left( a-n\right) }_n}\right.}\)

\(a_n\) denotes the falling factorial (Pochhammer?) of a.


I'm feeling slightly annoyed by it (probably need some exercise, which I will get to in a minute). Since we can cancel out the last term of the sum with the first part of function, we end up with 1/a = a bunch of really large numbers added and subtracted to one another.

It's annoying as the numberphile video, but looks fricken more justified. A bit. Not really. I know I'm supposed to accept limits as limits, not as actual values. But if I do the limit of the one (dividing by e^(1/infty))... and add it to the limit of the other (the sum as k-->infty) the limits obviously differ by the specific amount. But it looks like the (a-1)!k^a /lim (e^(1/infty) should simply subtract out the last of the sum terms.


What is that type of mathematical rule breaking called?

 

[lpetrich]

I'll now try to derive the semisimple Lie algebras. I find the theory of them to be most elegant, so that's why I'm posting on them. If you've ever taken a course in quantum mechanics at the advanced-undergraduate or graduate level, you will have run into the ladder-operator formalism of angular momentum. The QM angular-momentum operators are the smallest SSLA, and QM angular-momentum states are that algebra's representations, ways of constructing that algebra. One then extends the QM AM techniques to what is appropriate for larger SSLA's.

Lie algebras are defined with commutation: if A and B are in algebra L, then [A,B] = A.B - B.A is also in L. Not surprisingly, commutation is antisymmetric, with [B,A] = - [A,B]. It also satisfies the Jacobi identity: [[A,B],C] + [[B,C],A] + [[C,A],B] = 0

We divide the algebra into a Cartan subalgebra with generators Hi and additional elements E(a) for some a:
[Hi,E(a)] = ai*E(a)
for a components ai. The a's are called roots.

Let us impose the constraint that H is Hermitian: Hermitian conjugate or HC of H = H. Since the a's are associated with eigenvalues of H on some state, those eigenvalues being real implies that the a's are also real. This also means HC(E(a)) = E(-a).

Let's look at the other commutators.

From the Jacobi identity on Hi, E(a), and E(b), [Hi,[E(a),E(b)]] = (ai + bi)*[E(a),E(b)]
Thus, [E(a),E(b)] = N(a,b)*E(a+b) for some N, which we make real. Thus,
N(a,b) = - N(b,a) = - N(-a,-b) = N(-b,-a)

The remaining commutator has form [E(a),E(-a)] = h(a).H where h is a real-valued function of roots a.
This means that h(-a) = - h(a). Using the Jacobi identity on E(a), E(b), and E(-a-b) gives
N(a,b)*h(a+b) + N(-a-b,a)*h(-b) + N(b,-a-b)*h(-a) = 0
For N(-a-b,a) = N(b,-a-b) = N(a,b),
h(a+b) = h(a) + h(b)
meaning that h(a) is a linear function of a: h(a) = a.h for constant real matrix h.
Thus, [E(a),E(-a)] = a.h.H = (a,H)

So we are ready for the next step: constraints on the N's and a's.

 

[lpetrich]

We use the Jacobi identity on E(a), E(b), and E(-b): N(a,b)*N(a+b,-b) + (b,a) + N(-b,a)*N(a-b,b) = 0
Using some of the previously-derived identities,
N(a+b,-b)^2 - N(a,-b)^2 + (b,a) = 0

Let us say that a and b are "simple roots", that is, a - b is not a root, making N(a,-b) = 0. This gives us
N(a+b,-b)^2 = - (b,a)
N(a+2b,-b)^2 = - (b,a) - (b,a+b)
etc. until we come to some a+n*b that is not a root. Then,
0 = - (b,a) - (b,a+b) - (b,a+2b) - ... - (b,a+n*b)
Doing the sum gives us
0 = - (n+1)*(b,a) - n(n+1)/2*(b,b)
giving us
(b,a) = - (n/2) * (b,b)

For simple roots ai, aj, this equation is (ai,aj) = - (nij/2) * (ai,ai) for nonnegative integers nij.

This equation strongly constrains the semisimple Lie algebras.

 

[Kharakov]

'ear is a 3d zeta function fractal, minimal iterations, because it looks more natural this way.

 

[lpetrich]

Nice diagram, Kharakov. It looks like some kind of plant.

-

Let us consider what values the n's may have. This can easily be done with
(a1,a2) = - (n12/2) * (a1,a1) = - (n21/2) * (a2,a2)
Multiplying the two equations together gives
(a1,a2)^2 = (n12*n21/4) * (a1,a1) * (a2,a2)

Since h is positive definite, (a1,a2)^2 <= (a1,a1) * (a2,a2) with equality only for a2 proportional to a1. Since we want linearly independent roots, we get
n12*n21 < 4

That means that each connected pair of roots (n != 0) has a short one and a long one. Let a1 be short and a2 be long. Then, (a2,a2) = n*(a1,a1), n12 = n, and n21 = 1, for n <= 3. For n = 1, both roots are equally long.

One can visualize the connections with a "Dynkin diagram", turning the roots into dots and their connections into lines. For n = 1, 2, and 3, the diagram has 1, 2, and 3 lines connecting two dots. Long roots are represented with open dots, and short ones with closed dots.

A set of roots is for a "simple" Lie algebra if one can go from any root to any other root using root connections. Otherwise, it has at least two disconnected simple parts. Thus, if one can find the simple algebras, one thus has all the semisimple ones.

The linear independence of roots makes an additional constraint on their lengths and connections. Construct a "root metric" matrix Mij = (ai,aj). Find its determinant. If it is zero, then the roots have some linear dependencies. Otherwise, they are linearly independent. One can do this test on subsets of the roots as well as on the whole diagram. For n of the a's, if they are n-component vectors, then det(M) = det(a's)^2 * det(h), meaning that det(M) > 0, a further constraint.

-

Applying the det(root-metric) test to a triple connection, I find that I get zero or negative for any root that can possibly be connected to two triply-connected roots. I checked on connecting to one of the triply-connected roots, and to both of them.

So there is only one Lie algebra with a triple connection, and that is G2.

 

[lpetrich]

Turning to double connections, I find that root-length sequences 1-2-4, 1-2-1, 2-1-2 are invalid along with looped versions of 1-2-1 and 2-1-2.

Likewise, 1-2-2-1 and 2-1-1-2 are invalid, including looped versions of them.

For single connections, a 3-root loop is invalid.

A nice bit of simplification for single connections. Consider two roots, a1 and a2, with the same length and connected to each other. Each one is not connected to the roots that the other one is connected to. Merge the two roots: a1+a2. If the original roots were legitimate algebra roots, then the new roots will be. If one runs into an invalid set of roots, then the original one and the intermediate steps must also be invalid.

Thus, one can shrink a singly-connected loop to a simply connected loop of 3 roots. Since that is invalid, then all singly-connected loops are invalid. One can also shrink a loop with double connections to one the 3-root ones. Thus, all loops with double connections in them are invalid. With the invalidity of loops with triple connections, that means that no valid Lie algebra has a root loop in it. So every valid set of roots has either a straight chain or a branched one.

-

If a root set has more than one double connection, then by root merging, two of these connections can be brought together. Likewise, if there is a double connection and a branching, those two features can be brought together by root merging. Both of them are invalid, so the only valid root sets with a double connection are straight-chained with only one double connection.

The double connection being at one end is always possible, but the only root configuration for the connection away from the end is 2-2-1-1. Both 2-2-2-1-1 and 2-2-1-1-1 are invalid.

So we have these algebras:
B - SO(2n+1) - (n-1) long roots, and a short root and the double connection at an end.
C - Sp(2n) - (n-1) short roots, and a long root and the double connection at an end.
F4 - the 2-2-1-1 one, with the double connection in the middle.

-

Now for pure single connections. One can have a straight chain, of course, and also branched ones. A root can be connected to three other roots, but not to four other roots. If there are two branch points, then they can be merged with root merging to have a four-way branch or more. Thus, there is at most one branch point.

The set of branch lengths (n,1,1) is always valid. However, (2,2,2), (1,3,3), and (1,2,5) are invalid.

This, the algebras with a single root length are:
A - SU(n+1) - a straight chain of n roots.
D - SO(2n) - a root with branches from it with lengths 1, 1, and n-3
E6 - likewise, with lengths 1, 2, 2
E7 - likewise, with lengths 1, 2, 3
E8 - likewise, with lengths 1, 2, 4

So that's it.

Four infinite families of simple Lie algebras: A, B, C, D, and five exceptional ones: G2, F4, E6, E7, E8.

 

[lpetrich]

It's easy to see that the smallest (semi)simple Lie algebras have some degeneracies:

• A1 = B1 = C1
• B2 = C2
• D2 = A1 * A1
• A3 = D3

There is also an interesting curiosity about E8. Remove roots from the long branch and keep on going in the removal direction after the long branch becomes short. The results:

• E8
• E7
• E6
• E5 = D5 -- SO(10)
• E4 = A4 -- SU(5)
• E3 = A2 * A1 -- SU(3) * SU(2)

There is a possible physical application of this odd sequence. The Standard Model's gauge symmetry is SU(3) * SU(2) * U(1) or E3 * U(1). The gauge symmetries of the most plausible GUT's are, in order, SU(5) = A4 = E4, SO(10) = D5 = E5, E6, and E8. So could there be some connection?

 

[Kharakov]

The diagram is a 3d rendering of a pseudo-zeta function fractal. I extended the Riemann to 3d (triplex numbers), however it wasn't quite as nice as I'd like. A sort of hacked version produced plant/flower type things, although it's basic form is quite similar to quaternion based "fractals", or some z^2 mbulbs.

Here is a starfish. Sort of simplistic application of mathematics, but they can be quite pretty animated.... but it's slow on my system Gotta get some friends to do the GPU intensive stuff for me... I really like to cycle the colors... psychedelic starfish. Maybe tomorrow. I only have one "ancient" computer though.



lol.. is the following true if a>1 and k>1? time for me to sleep.

(that's limit a->infty || or || lim k->infty )

\(0= \lim_{a\to\infty} \,\,\,|OR|\,\,\, \lim_{k\to\infty} \,\,\,\frac{(a-1)! (-k)^a}{\sqrt[k]{e}} - \frac {1}{a} -\sum_{n=0}^{a}{\left. {{\left( -k\right) }^{n+1}}\, {{\left( a-n\right) }_n}\right.}\)