How Can Dirichlet Series be Inverted to Obtain Coefficients?

[eljose79]

given the Dirichlet series Sum(0,infinite)a(n)n-^s=f(s) would be an analityc formula to invert the Dirichlet series to obtain a(n)?..i have searched it at google but found no results.

 

[shmoe]

First of all your infinite sum should start at 1, not 0, otherwise you have a 1/0 term.

There are many different inversion formulae, try doing a search for Perron's formula, it's one of the simpler versions.

 

[eljose79]

If perron,s formula for inverting a series is known..why we can not compute mobius function,divisor function and so on with that?...using Perron,s formula what would be the integral expresion for moebius function mu(x)?..where could i find.

 

[shmoe]

Take a look at

http://www.maths.ex.ac.uk/~mwatkins/zeta/pnt5.htm

Perron's formula is used here to express the sum von-Mangoldt function from n=0 to x in terms of an integral. You could then use this to get the von-Mangoldt function for any value of n provided you can work out this integral accurately enough.

This is the problem. The resulting integral can be represented very nicely except for an infinite sum over the zeros of the zeta function (follow the "back to the proof outline" in the link above). This infinite sum is not so easy to work with and is what prevents us from getting anything better than an asymptotic result for this summatory function, thus stopping us from evaluating the von-Mangoldt function explicitly.

The more that's known about where the zeros of zeta live, the smaller we can make our error term. However, even in the best possible case (the Riemann Hypothesis), our error term is still too large for explicitly determining von-Mangoldt.

Similar prblems will arise with other arithmetic functions. We can convert to integrals and use analytic techniques to estimate. These estimates unfortunately are usually only good enough to provide averages of our arithmetic function.

 

[eljose79]

Another question ..how do you obtain the convergence radius for a Dirichlet series? in fact is this the criterion? if the series is given by a(n)n^-s then s must satisfy taking logarithms s>nlog[a(n)/a(n+1)] with [x] the modulus of x,is that true?...(i have taken the criterion that [b(n+1)/b(n)]<1 with b(n)=a(n)n-^s it gives the result [a(n+1)/a(n)].[1/1+1/n]^s taking logarithm and knowing [Log(1/1+1/n)]=1/n

Thanks for your explanation fo Perron,s formula..accoridng to that it can not invert the Fourier series but only to give a finite sum of the coefficients a(n) but not any a(n) itself i was looking for a formula that given g(s) sum of idirichlet series would give us an expresion for a(n)

 

[shmoe]

Another question ..how do you obtain the convergence radius for a Dirichlet series? in fact is this the criterion? if the series is given by a(n)n^-s then s must satisfy taking logarithms s>nlog[a(n)/a(n+1)] with [x] the modulus of x,is that true?...(i have taken the criterion that [b(n+1)/b(n)]<1 with b(n)=a(n)n-^s it gives the result [a(n+1)/a(n)].[1/1+1/n]^s taking logarithm and knowing [Log(1/1+1/n)]=1/n

Log(1+1/n) is only approximately 1/n, not equality. The problem is this goes to 0 as n goes to infinity so the ratio test gives no information.

Also, there's no such thing as a "radius of convergence" for a Diriclet series, they converge in half planes, not circles. We have what's called an abcissa of convergence. There will be a number real number b such that if Re(s)>b then your series converges and if Re(s)<b your series diverges. Moreover, for any delta>0 your series converges uniformly on Re(s)>=b+delta. Many of the interesting arithmetic functions satisfy $$a_n=O(n^{\epsilon})$$ for any $$\epsilon>0$$. If this is the case, our associated Dirichlet series has an abscissa of convergence of 1 or less (comparison test).

Thanks for your explanation fo Perron,s formula..accoridng to that it can not invert the Fourier series but only to give a finite sum of the coefficients a(n) but not any a(n) itself i was looking for a formula that given g(s) sum of idirichlet series would give us an expresion for a(n)

Actually knowing the sum up to [x] for all x can get you any term you like. Let $$A(x)=\sum\limits_{1\leq n\leq[x]}a_{n}$$ then $$a_i=A(i)-A(i-1)$$. Perron's gives you A(x), but the known techniques for evaluating the integrals aren't sharp enough to pick off the individual $$a_{i}$$. This is the best technology to date, we have to be content with asymptotics for the averages of our coefficients.

 

[eljose79]

But for evaluating integrals in the form Int(c-i8,c+i8) involving Riemann,s function R(s) and its derivatives (8 here means infinity) will be any numerical method won,t it?.. in fact the Perron,s formula with the change of variable is=u wouldn,t be Fourier inverse transform? i think the Mellin inverse transform and the Laplace inverse transform are the Fourier inverse transorm with w=iu)

In fact i developed a formula for getting the a(n) knowing the sum of the dirichlet series f(s) is the way

a(n)= M^-1[f(4-s)]
------------ where R(s) is the Riemann function
M^-1[R(4-s)]

Another question, how do you obtain the half plane of convergence for the Dirichlet series?...

 

[shmoe]

But for evaluating integrals in the form Int(c-i8,c+i8) involving Riemann,s function R(s) and its derivatives (8 here means infinity) will be any numerical method won,t it?

Sure, but your numerical methods will never be accurate enough. If you wanted to get a formula for the mobius function it would be enough to be able to compute this integral to within something like 1/2. This will actually be impossible. You can estimate these integrals, but the absolute error term will be enormous. Think how x^2 and x^2+x are asymptotic to one another as x goes to infinity. The relative difference between the two functions is small, as their ratio goes to 1, but their absolute difference grows without bound.

Another question, how do you obtain the half plane of convergence for the Dirichlet series?...

Depends on the Dirichlet series. The zeta functions half plane of convergence can be found easily with the integral test. I should have mentioned if $$\sum\limits_{n=1}^{\infty}\frac{a_{n}}{n^{\sigma}}$$ converges then so does $$\sum\limits_{n=1}^{\infty}\frac{a_{n}}{n^{\sigma+it}}$$ for all values of t.

Like I said, most interesting arithmetic functions will have their Dirichlet series converge to the right of 1, by the comparison test. They may converge further to the left as well (such as non-trivial Dirichlet Characters), but that's in general not an easy thing to determine. Think about the Mobius function, it's Dirichlet series converges to the right of 1. If you could push this half plane to the left to Real part 1/2 for the Mobius function (whose Dirichlet series is 1/zeta) you'd have proved the Riemann Hypothesis. As you can imagine, this isn't an easy thing to do.

 

[eljose79]

But given a n ntegra of the form Int(c-i8,c+i8)F(s)exp(sx with the change of variable s=c+iu the Integral becomes Int(-8,8)F(c+iu)exp(icx)exp(iux)idu (8 here means infinite) so our integral is almost a Fourier series F^-1[F(c+iu)]exp(cx) can not you compute a Fourier transform accurate enough using Gaussian integration or another numerical method?..

So with the integral criterion if Int(1,8)mu(x)/x^s converges for 1/2<Re(s)<1 The Riemann hypothesis would be true so could not use Perron,s formula to apply the integral criterion to mu(x)?...another question if we proved that Sum(1,n)mu(n)/n^s converges for 1/2+e being e>0 for example e=0.000000000000000000000000000000000...Riemann,s hypothesis is true
integral is convergent for Re(z) (1/2,1) the Riemann hypothesis would be true?,so the problem is to determine mu(x) isn,t it?..

Note: where i could find the prove that for mu(n)=O(n^e) e>0 or that the series with general term mu(n)/n^-s converges for s>1?..

 

[shmoe]

Note: where i could find the prove that for mu(n)=O(n^e) e>0 or that the series with general term mu(n)/n^-s converges for s>1?..

Hi, I'm short on time so I'll get to your other questions in a day or two.

the mobius function is bounded, it hops around between -1, 0, and 1 so it's trivially O(1), which is better than O(n^e) for an e>0. This is what happens with Dirichlet characters as well. You should be able to find proofs that things like the divisor function is O(n^e) in any intro analytic number theory book. Look for something called "Dirichlet's Hyperbola method/trick".

 

[eljose79]

Thanks again Shmoe..in fact i am not mathematician by myself but physicist although i am interested in analytic number theory method.

I found a formula to get the Mobius Mangoldt and other function as expressed above to invert a Dirichlet series similar to Perron,s formula with that mobius function is
M^-1[1/R(4-s)]
mu(x)=---------------- with R(s) Riemann,s function so you can get mu(x)
M^-1[R(4-s)]

but the problem i have is that perhaps you can not compute Mellin transform with numerical method with a great accuracy making the change of variable s=c+iu the mellin transform becomes

f(exp(x))=Int(-8,8)exp(cx)exp(iux)f(c+iu)=F[exp(cx)f(c+iu)] F is the Fourier inverse transform of exp(cx)f(c+iu)

but i do not know if we can compute this integral with great accuracy,this is a big problem i see in number theory