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Mute

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I was reading a paper the other day that made the following claim, and provided no reference for the assertion. I would like to find a reference or figure out how to derive the asymptotic behavior myself.

The situation is as follows:

Suppose we have a function ##f(z)##, defined as a power series about ##z=0##:

$$f(z) = \sum_{k=0}^\infty a_k z^k.$$

We assume ##f(z)## has a radius of convergence ##|z| \leq 1##, and we can take ##f(1) = 1##. ##f(z)## can have a finite number of well-defined derivatives at ##z=1##, but at some order the derivatives do not exist at that point. Hence, ##f(z)## has a branch point at ##z = 1##. It can be analytically continued for ##|z| > 1##, with a branch cut running along the real axis from 1 to ##\infty##. For context, ##f(z)## is a probability-generating function in the paper I was reading, but I don't think that's relevant other than it sets ##f(1) = 1## (which really isn't all that important anyways).

The terms ##a_k## decay as ##k^{-\gamma}## at large k, for positive ##\gamma##.

The paper claims that ##f(z)## has an asymptotic series near ##z=1##,

$$f(z) \simeq 1 - b(1-z)^{\phi(\gamma)},$$

where ##b## is a positive constant and ##\phi(\gamma)## is an exponent that depends on the exponent of ##a_k \sim k^{-\gamma}##.

The actual claim in the paper goes the other direction: if a series ##\sum_k a_k z^k## with radius of convergence 1 has asymptotic behavior ##(1-z)^\phi## near ##z = 1##, then ##a_k \sim k^{-\phi - 1}##. In their claim, the exponent of the decay of ##a_k## is the same as the exponent of ##(1-z)^\phi##, but I think they may be considering a special case. As such, I'm interested in deriving the asymptotic behavior from the series.

Does anyone know any references on how to do this, or which derive the rresult? I know that the series for the Polylogarithm, ##\mbox{Li}_\gamma(z) = \sum_{k=1}^\infty z^k/k^\gamma##, or at least its analytic continuation, has a power series about ##\mu = 0## for ##z = \exp(\mu)##, but I'm wondering if the result claimed by the paper is actually a general result for all series having coefficients that decay as ##k^{-\gamma}##.

Thanks!

The situation is as follows:

Suppose we have a function ##f(z)##, defined as a power series about ##z=0##:

$$f(z) = \sum_{k=0}^\infty a_k z^k.$$

We assume ##f(z)## has a radius of convergence ##|z| \leq 1##, and we can take ##f(1) = 1##. ##f(z)## can have a finite number of well-defined derivatives at ##z=1##, but at some order the derivatives do not exist at that point. Hence, ##f(z)## has a branch point at ##z = 1##. It can be analytically continued for ##|z| > 1##, with a branch cut running along the real axis from 1 to ##\infty##. For context, ##f(z)## is a probability-generating function in the paper I was reading, but I don't think that's relevant other than it sets ##f(1) = 1## (which really isn't all that important anyways).

The terms ##a_k## decay as ##k^{-\gamma}## at large k, for positive ##\gamma##.

The paper claims that ##f(z)## has an asymptotic series near ##z=1##,

$$f(z) \simeq 1 - b(1-z)^{\phi(\gamma)},$$

where ##b## is a positive constant and ##\phi(\gamma)## is an exponent that depends on the exponent of ##a_k \sim k^{-\gamma}##.

The actual claim in the paper goes the other direction: if a series ##\sum_k a_k z^k## with radius of convergence 1 has asymptotic behavior ##(1-z)^\phi## near ##z = 1##, then ##a_k \sim k^{-\phi - 1}##. In their claim, the exponent of the decay of ##a_k## is the same as the exponent of ##(1-z)^\phi##, but I think they may be considering a special case. As such, I'm interested in deriving the asymptotic behavior from the series.

Does anyone know any references on how to do this, or which derive the rresult? I know that the series for the Polylogarithm, ##\mbox{Li}_\gamma(z) = \sum_{k=1}^\infty z^k/k^\gamma##, or at least its analytic continuation, has a power series about ##\mu = 0## for ##z = \exp(\mu)##, but I'm wondering if the result claimed by the paper is actually a general result for all series having coefficients that decay as ##k^{-\gamma}##.

Thanks!

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