Something for sci.math's amateur mathematicians? (Part 2)
Dave L. Renfro
http://groups.google.com/group/sci.math/msg/722ec5cae2c3f48a
Ioannis Galidakis showed that the problem I raised there,
which I had obviously not given much thought to, seems
to be either fairly uninteresting or virtually intractable
(depending on how it's interpreted). I think the problem
this time will be more interesting. The problem -- really
two problems -- is at the end of this post.
The n'th triangular number is
T(n) = 1 + 2 + 3 + ... + n = n(n+1)/2.
The n'th tetrahedral (or pyramidal) number is
T(1) + T(2) + T(3) + ... + T(n) = n(n+1)(n+2)/6.
You can get this last result by replacing each
T(k), for k = 1, 2, ..., n, with (1/2)k + (1/2)k^2.
Then regroup terms so that all the (1/2)k terms are
together (and use the triangular number formula
to get a closed form expression for their sum)
and all the (1/2)k^2 terms are together (and use
the sum of the first n squares formula to get
a closed form expression for their sum):
[ (1/2)(1) + (1/2)(1^2) ]
+ [ (1/2)(2) + (1/2)(2^2) ]
+ [ (1/2)(3) + (1/2)(3^2) ]
...
+ [ (1/2)(n) + (1/2)(n^2) ]
= (1/2)*(1 + 2 + 3 + ... + n)
+ (1/2)*(1^2 + 2^2 + 3^2 + ... + n^2)
= (1/2)*[ n(n+1)/2 ] + (1/2)*[ n(n+1)(2n+1) ]
= (1/12)n(n+1)*[ 3 + (2n+1) ]
= (1/12)n(n+1)(2n+4) = (1/6)n(n+1)(n+2)
Let's denote the n'th triangular number by T(n,2)
and the n'th tetrahedral number by T(n,3). Then
the n'th hypertetrahedral number is
T(n,4) = T(1,3) + T(2,3) + T(3,3) + ... + T(n,3),
and expanding things like I did above, you'll get
T(n,4) = n(n+1)(n+2)(n+3)/24.
This can be continued, and one can show that
(follow the URL below)
T(n,k) = C(n+k-1, k), where C(i,j) is the
usual binomial coefficient number "i choose j".
http://www.math.toronto.edu/mathnet/questionCorner/tetnumbers.html
I seem to recall that Fermat proved this identity
for T(n,k).
Instead of the triangular numbers, if we begin with
the sum of the first n p'th powers, there are two natural
ways we could continue:
(1) Take the sum of the first n of these sum-of-p'th-powers
numbers, and continue the process k times like above.
(2) Take the sum of the p'th powers of these sum-of-p'th-powers
numbers, and continue the process k times like above.
I would imagine that various relations analogous to
T(n,k) = C(n+k-1, k) are known in each case, but I'm
not aware of them. (This doesn't mean much, because I
know very little about the field of combinatorics.)
If such relations are not known, it seems to me that
working them out might be something useful to work on
and it ought to be publishable in some expository or
recreational math journal.
Dave L. Renfro
Rupert
One special case of this is discussed in Spivak's "Calculus". There is
an identity involving the Bernoulli numbers.
Gottfried Helms
> For Part 1, see:
>
> http://groups.google.com/group/sci.math/msg/722ec5cae2c3f48a
>
> Ioannis Galidakis showed that the problem I raised there,
> which I had obviously not given much thought to, seems
> to be either fairly uninteresting or virtually intractable
> (depending on how it's interpreted). I think the problem
> this time will be more interesting. The problem -- really
> two problems -- is at the end of this post.
>
> The n'th triangular number is
>
> T(n) = 1 + 2 + 3 + ... + n = n(n+1)/2.
>
> The n'th tetrahedral (or pyramidal) number is
>
> T(1) + T(2) + T(3) + ... + T(n) = n(n+1)(n+2)/6.
>
Yes, you are right. I also found this an intersting
question, and it was one of my first "recreational
math" (as I call it now) findings and treatises,
when I started to learn number theory on my own.
One can find a triangle of numbers, which relate
the (finite) sums of powers and the k-hedral
numbers by solve some polynomial with matrix
inversion, which was already known to Euler.
I don't post the solution here for not to
spoil the fun of finding it by own try.
But you may also see the page
http://go.helms-net.de/math/potenzsummen/potenzsummen_1.htm
where I compiled my findings...
It is in german, but the derivation is simple and
does not need understanding of the language.
For me there was, and still is, is a certain appeal of
beauty with that relations....
Regards -
Gottfried Helms
kunzmilan
A table starts usually at upper left, a graph from bottom left. This is
a graph:
1 4 10 20
1 3 6 10
1 2 3 4
1 1 1 1
It shows the number of ways, how the corresponding points of the square
can be reached by unit steps, a nad b, from the beginning point 1.
The table was obtained by truncation of the full table of binomial
coefficients. This technique can be used for triangular numbers as well
as for any dimensional positive quadrants. Cubes are obtained by
truncation of plane complexes.
Only difficulty is, that we can not see them, and we must rely on
calculations. We start with generating functions and arrange results
into tables of combinatorial functions. Here we reduce too rich space
of strings at first to get equivalent points of the lattice.
We can generate n dimensional cubes directly but then we miss some
functions, this space is not complete.
kunzmilan
Dave L. Renfro
[snip]
The problem I posed at
http://groups.google.com/group/sci.math/msg/ddd62dd5b9befbb7
and which Gottfried Helms cited some work he had done
on it at
http://go.helms-net.de/math/potenzsummen/potenzsummen_1.htm
seems to be essentially solved in a 1950/51 American
Mathematical Monthly problem.
P. A. Piza posed the following problem:
Monthly Problem #4380 -- posed in American Mathematical
Monthly 57 (1950), p. 119; solution by Roger Lessard in
American Mathematical Monthly 58 (1951), pp. 429-430.
"For arbitrary positive integers n and k let
S_1(n,k) = 1^k + 2^k + ... + n^k,
and put
S_{p+1}(n,k) = S_p(1,k) + S_p(2,k) + ... + S_p(n,k),
with p = 1, 2, . . .. Thus S_p(n,k) is the p'th iterated
sum, the sum of the sum of . . . the sum of the first
n perfect k'th powers.
Show that S_p(n,k) is a polynomial in n of degree p + k
and determine the form of the polynomial for k = 3, 4, 5.
If possible, determine the form for general k. (The case
k = 2 is the subject of problem 22, Mathematics Magazine,
vol. XXII, no. 1, p. 51.)"
The solution to the k = 2 case is in Mathematics Magazine
22 #3 (Jan./Feb. 1949), pp. 162-163.
Roger Lessard gives two forms for the general k case,
neither of which is simple enough for me to want to
reproduce here.
Dave L. Renfro