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Topic: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Replies: 9   Last Post: Feb 18, 2013 2:34 PM

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Gottfried Helms

Posts: 1,926
Registered: 12/6/04
Re: Bernoulli numbers and sqrt(1)+sqrt(2)+sqrt(3) + ... sqrt(1000)
Posted: Feb 18, 2013 7:17 AM
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I've to correct some obvious typing errors:

Am 18.02.2013 09:38 schrieb David Bernier:
> On 02/18/2013 02:26 AM, Gottfried Helms wrote:

> The summatory polynomial for the k'th powers of 1, 2, ... n,
> P(x), has the property that P(x) - P(x-1) = x^k,
> at least for positive integers x.
> I assume k is a positive integer.
> So, does there exist a continuous f: [a, oo) such that
> f(x) - f(x-1) = sqrt(x) for any x in [a, oo) ?
> Ramanujan wrote a paper on sum of consecutive square roots:
> david bernier

Hmm, my usual tools give only much diverging series for which my
procedures of divergent summation do not work well if I use -say-
only 64 or 128 terms for the power series.
But I've now converted the problem into one which employs the rationale:

f(x) = sqrt(1+x^2)
such that f(1) -> sqrt(2) f(f(1)) = sqrt(3) and so on such that we
can write
S(a,b) = sqrt(a) + f(sqrt(a)) + f(f(sqrt(a))) + ... + f...f(sqrt(a))
= f°0(sqrt(a)) + f°1(sqrt(a)) + f°2(sqrt(a)) + ... + f°d(sqrt(a))

where d=b-a and the circle at f°k(x) indicates the k'th iterate.

After that the problem can be attacked by the concept of Carleman-matrixes
and their powers. Let F be th carleman-matrix for the function f(x),
then let G = I - F then, if we could invert G in the sense that

M = G^-1 = I + F + F^2 + F^3 + F^4 + ... (see Neumann-series, wikipedia)

then we had a solution in terms of a formal power series for

m(x) = x + f(x) + f°2(x) + f°3(x) + ....

and m(sqrt(a)) - m(sqrt(b)) would give the required sum-of-sqrt from
sqrt(a) to sqrt(b) in 1-steps progression from its argument a.

However, G has a zero in the top-left entry(and the whole first column)
and cannot be inverted.
Now there is a proposal which I've seen a couple of times that we invert
simply the upper square submatrix after removing the first (empty) column
in G, let's call this H, and this gives often an -at least- usable
approximate solution, if not arbitrarily exact.

But again - trying this using Pari/GP leads to nonconclusive results; the
inversion process seems to run in divergent series again.

Here we have now the possibility to LDU-factorize H into triangular
factors, which each can be inverted, so we have formally

H = L * D * U
M = H^-1 = U^-1 * (D^-1 * L^-1)

We cannot perform that multiplication due to still strong divergences
when evaluating the row-column-dotproducts (which is only making explicite
the helpless Pari/GP-attempts for inverting H)

But we can use the upper triangular U^-1 in the sense of a "change of base"-
operation. Our goal is to have M such that we can write

(V(sqrt(a)) - V(sqrt(b))) * M[,2] = s(a,b+1) = sqrt(a)+sqrt(a+1)+...+sqrt(b)

where V(x) means a vandermondevector V(x) = [1,x,x^2,x^3,x^4,....]

But now we can proceed from the formal formula

(V(sqrt(a)) - V(sqrt(b))) * M[,2]
= (V(sqrt(a)) - V(sqrt(a))) * U^-1 * ( D^-1 * L^-1)[,2]

and can compute

X(sqrt(a),sqrt(a)) = (V(sqrt(a)) - V(sqrt(a))) * U^-1

exactly to any truncation size because U^-1 is upper triangular and thus

Then we can as well do

Q = ( D^-1 * L^-1)[,2]

which is -besides the truncation to finite size- an exact expression.

Still we have, that the dot-product

s(a,b) = X(sqrt(a),sqrt(a)) * Q

is divergent, but now it seems, that we can apply Euler-summation
for the evaluation of the divergent dot-product.
The results are nice approximations for the first couple of
sums s(1,4), s(2,4) and some more. s(1,9) requires Euler-summation
of some higher order such that I get

s(1,9) ~ 16.3060
(where 16.3060005260 is exact to the first 11 digits)

s(5,10)= sqrt(5)+sqrt(6)+ ... + sqrt(10)
~ 16.32199
where 16.3220138163 is exact to the first 11 digits)

Don't know how to do better at the moment; surely if the composition
of the entries in the matrices were better known to me, one could
make better, possibly even analytical or at least less diverget, expressions.

(But I've not enough time to step into this deeper, I'm afraid)

Gottfried Helms

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