## E.J. Janse van Rensburg

Print publication date: 2015

Print ISBN-13: 9780199666577

Published to Oxford Scholarship Online: August 2015

DOI: 10.1093/acprof:oso/9780199666577.001.0001

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# (p.558) Appendix D Asymptotic approximations

Source:
The Statistical Mechanics of Interacting Walks, Polygons, Animals and Vesicles
Publisher:
Oxford University Press

The Stirling approximation for the factorial [187] is given by

(D.1)
$Display mathematics$

Lower and upper bounds on the factorial are

(D.2)
$Display mathematics$

(see references [187, 494]), while the approximation

(D.3)
$Display mathematics$

is from a numerical point of view very accurate.

Stirling’s approximation can also be stated as

(D.4)
$Display mathematics$

from which it follows, for example,

(D.5)
$Display mathematics$

More on Stirling’s approximation may be found in reference [592].

Catalan numbers are defined by

(D.6)
$Display mathematics$

By putting both $n→2n$ and $ℓ=n$ in equation (D.5) and keeping more terms in the Stirling approximation, one finds that

(D.7)
$Display mathematics$

for some fixed $M>0$; see lemma 4.4 in reference [488].

# (p.559) D.1 Approximation of the binomial coefficient

The binomial coefficient can be approximated using approximations for the factorial. A more interesting approach is to consider random walks in $Z$.

Consider a random walk from 0 in $Z$ giving left or right steps. Assign a generating variable t with each right step so that $tˉ≡1t$ represents a left step.

The generating function of walks of length n is $(t+tˉ)n$, and the coefficient of $tk$ is the number of walks ending at $k∈Z$ from the origin. Thus, by the binomial theorem, the number of walks ending at k is $n(n+k)/2$, since the walk ends at k if it gives $12(n+k)$ right steps (and $12(n−k)$ left steps).

Assume that $t∈(0,1)$ and suppose that the walk steps to the right with probability $P+=tt+tˉ$ and to the left with probability $P−=tˉt+tˉ$. Let Xk be the final position of the walk after k steps. Then $X0=0$.

Note that Xk is a random variable and that $Xn=k$ for $k∈[−n,n]$ with probability

(D.8)
$Display mathematics$

The value of $Pr(Xn=k)$ is 0 if n and k do not have the same parity (are not both even, or not both odd).

Asymptotically, n and k have the same parity with probability $12$, and the distribution of Xn is given by

(D.9)
$Display mathematics$

By the central limit theorem, the random variable Xn is asymptotically normally distributed with mean $μn$ and variance $σ2n$, where

$Display mathematics$

Thus, it follows that

(D.10)
$Display mathematics$

This is maximal when $k=μn$, with μ‎ given above. In particular, if $k=an$, then one may solve for t to obtain

(D.11)
$Display mathematics$

Since $t∈(0,1)$, select the positive root $ta+$. In this event, it follows that, if $t=ta+$, then $μn=an+O(n)$. For this choice of k and with $t=ta+$, one may solve for the binomial coefficient in equation (D.9):

(p.560)

(D.12)
$Display mathematics$

Substitute $t=ta+$ into equation (D.12) to obtain the following approximation for the binomial:

Theorem D.1

The binomial coefficient is approximated by

$Display mathematics$

Next, replace $an=2bn−n$ for $b∈[0,1]$ and simplify to obtain

(D.13)
$Display mathematics$

For large values of n, the above can be approximated by

(D.14)
$Display mathematics$

Taking $n→∞$ gives Corollary D.2.

Corollary D.2

The rate of growth of binomial coefficients is given by

$Display mathematics$

for $b∈[0,1]$. The function $B(b)$ has infinite right- and left-derivatives at $b=0$, and $b=1$, respectively. Moreover, $B(0)=B(1)=1$, and $B(12)=2≥B(b)$.

One may replace $an$ or $bn$ by $an$ or $bn$ with few, if any changes, in Corollary D.2.

More generally, for $a,X>0$, and $b∈[0,a]$, one may consider the function

(D.15)
$Display mathematics$

which has a maximum for fixed $a>0$ given by

(D.16)
$Display mathematics$

Similar to $B(b)$ in Corollary D.2, $B(a,b)$ has infinite right- and left-derivatives at $b=0$, and $b=a$, for fixed $a,x>0$.

(p.561) The function for $x>0$, and $a∈(0,1)$, defined by

(D.17)
$Display mathematics$

similarly has the maximum

(D.18)
$Display mathematics$

for fixed $a>0$. The function $C(a)$ has infinite left- and right-derivatives at $a=0$, and $a=1$, respectively.

Lemma D.3

Let $0 and suppose $γ∈[0,qq+1]$. Then

$Display mathematics$

Proof

Let $p=qq+1$ so that $0≤γ≤p<1$. Then

$Display mathematics$

Consider p to be the probability that a biased tossed coin will land heads. Let η‎ be the number of heads in n tosses of the coin. Then the probability that there are at most $γn$ heads in n tosses is

$Display mathematics$

If $ζ$ is a random variable equal to 1 if $η≤γn$, and 0 otherwise, then the expectation of $ζ$ is $Eζ=Pr(η≤γn$. By Jensen’s inequality,

$Display mathematics$

The right-hand side is a minimum when $γn(1−p)=p(n−γn)e−t$. Thus, substituting for t gives

(D.19)
$Display mathematics$

Substitute $p=qq+1$, take the power $1n$ and take $n→∞$.

This completes the proof.

# (p.562) D.2 Approximation of trinomial coefficients

Trinomial coefficients are defined by

(D.20)
$Display mathematics$

The term ‘trinomial coefficient’ is used as in reference [9]; that is, ‘trinomial coefficient’ is used synonymously with ‘central trinomial coefficient’.

Trinomial coefficients are given in terms of binomial coefficients by

(D.21)
$Display mathematics$

The properties of a random walker in $Z$ can be used to find approximations to trinomial coefficients. Let the position of the walker be xn and suppose it gives right or left unit length steps in $Z$, or a neutral step where it stays put at xn, so that $xn+1=xn±1$, or $xn+1=xn$.

Let $x0=0$ and suppose the random walker gives right, neutral and left steps with probabilities $P+$, P0 and $P−$, respectively. Suppose $t∈(0,1)$, let $tˉ=1t$ and let

$Display mathematics$

The probability that $xn=k$ for $k∈[−n,n]$ is given by

(D.22)
$Display mathematics$

since $xn=k$ if the walk gives $k+p$ right steps and p left steps, and since the number of walks ending at k is exactly given by the trinomial coefficient.

The central limit theorem states that the random variable xn is asymptotically normal with mean $μn$ and variance $σ2n$, where

(D.23)
$Display mathematics$

and

(D.24)
$Display mathematics$

Thus, it follows that

(D.25)
$Display mathematics$

Substitute $k=αn$, substitute $μ=t−tˉt+1+tˉ$ and solve for t:

(p.563)

(D.26)
$Display mathematics$

Since $t∈(0,1)$, select $tα+$ so that $μn=αn+O(n)$ and solve for the trinomial coefficient in equation (D.22). If $k=αn$, and $t=tα+$, then this gives

(D.27)
$Display mathematics$

Simplification gives theorem D.4.

Theorem D.4

The trinomial coefficient is approximated by

$Display mathematics$

where

$Display mathematics$

For large values of n, one may use Corollary D.5.

Corollary D.5

The trinomial coefficient

$Display mathematics$

where $β=4−3α2$.

The rate of exponential growth of the trinomial coefficient is given by Corollary D.6.

Corollary D.6

The limit

$Display mathematics$

The rate of growth of the generating function of trinomial coefficients is defined by

(D.28)
$Display mathematics$

Given an n, there is a value of j, say $j=jn$, which maximises the summand.

(p.564) Assuming that $jn=ϵmn+o(n)$, where $ϵm∈(0,1)$ is a function of z, it follows that

(D.29)
$Display mathematics$

By replacing jn in the above by $ϵn+o(n)$ and then using equation (D.6) when taking $n→∞$ in equation (D.29), it follows that

(D.30)
$Display mathematics$

By differentiating and solving for $ϵ$, the supremum is realised when

(D.31)
$Display mathematics$

It follows that $ϵm∈(0,1)$ if $z>1$, and so the supremum is achieved when $ϵm=0$ when $z≤1$ at $ϵm$. These results show that, after simplification,

Theorem D.7

The function $F(z)$ is given by

(D.32)
$Display mathematics$

where $zˉ=1z$.

# D.3 The Euler-Maclaurin formula

The Euler-Maclaurin formula (see for example [273]) is a useful method for approximating series by integrals. The formula is useful because it gives precise remainders, allowing estimates of the error in the approximation.

Theorem D.8 (Euler-Maclaurin)

If $f∈C2m[0,N]$ for some $m,N∈N$, and both $N>0$ and $m≥1$, then

$Display mathematics$

where the remainder term is given by

$Display mathematics$

The function $Bn(x)$ is the n-th Bernoulli polynomial; $Bn=Bn(0)$, and $Bn(x−x)$ is the n-th periodic Bernoulli polynomial.

(p.565) In some applications the remainder term Rm approaches 0 with $m→∞$ (derivatives of f exist to all orders), and one may able to show that the series is convergent.

The remainder term can be bounded by

$Display mathematics$

The Bernoulli functions are defined by the recurrence $ddxBn(x)=nBn−1(x)$, with $B0(x)=1$, and $∫01Bn(x)dx=0$, for $n≥1$. The first few are given by $B1(x)=x−12$; $B2(x)=x2−x+16$; $B3(x)=x3−32x2+12x$; and $B4(x)=x4−2x3+x2−130$.

The Bernoulli numbers $Bn$ are defined by the exponential generating function $tet−1=∑m=0∞Bmm!tm$. Notably, $B2n+1=0$ if $n≥1$ (that is $B3=B5=B7=⋯=0$), and the first few non-zero Bernoullui numbers are $B1=−12$; $B2=16$; $B4=−130$; and $B6=142$.

The Stirling approximation for $n!$ is obtained as follows. Consider the integral

$Display mathematics$

Applying the Euler-Maclaurin formula to the integral on the left-hand side gives

$Display mathematics$

The infinite series in the above is a polygamma function and

$Display mathematics$

Take the anti-derivative of this twice to obtain

$Display mathematics$

The constant C is computed from the Legendre duplication formula

$Display mathematics$

This gives $C=12log(2π)$. Finally,

(D.33)
$Display mathematics$

This is an asymptotic series approximating $logz!$. By truncating the infinite series, the error in the approximation is less than the magnitude of the first omitted term. If the series is discarded in its entirety, then the Stirling approximation in equation (D.1) is found.

# (p.566) D.4 Saddle point approximations of the integral

Suppose that f is an analytic complex function of the complex variable $z=x+iy$. Then $f=u+iv$, and ${u,v}$ satisfies the Cauchy-Riemann equations. Consequently, it may be shown that u satisfies the Laplace equation:

(D.34)
$Display mathematics$

At any stationary point of f, if $∂2∂x2u>0$, then $∂2∂y2u<0$. That is, if f is real valued, then its stationary points are necessarily saddle points.

The function f may be approximated in the vicinity of a saddle point z0 by

(D.35)
$Display mathematics$

This approximation of f near its saddle points allows the approximation of integrals of the form

(D.36)
$Display mathematics$

where N is large, and $g(z)$ is slowly varying.

Insert the approximation of f in the above and replace $g(z)≈g(z0)$. Deform the contour of integration from a to b into a contour C in the complex plane passing through z0. This gives

(D.37)
$Display mathematics$

Put $z−z0=reiϕ$ and let $f′′(z0)=f′′(z0)eiθ$. The integration will be along r, with ϕ‎ fixed at a convenient value, namely, $θ+2ϕ=π$. The integration is approximated by integrating from $r=−∞$ to $r=∞$. These substitutions simplify the integral above into a Gaussian integral:

(D.38)
$Display mathematics$

Evaluating the integral gives the saddle point approximation

(D.39)
$Display mathematics$

The phase ϕ‎ is determined by solving for $ϕ=π−θ2$.

It is possible to refine the above approximations by keeping more terms in the expansion (D.35). This approximation also has the implicit assumption that the only significant contribution to the integral along the contour C is from the vicinity of the saddle point (this means that $u(z0)≫u(z)$ if $|z0−z|$ is ‘large’. This may not be necessarily the case. For more on this method, which is related to steepest descent methods, see, for example, reference [13].

# (p.567) D.5 Asymptotic formulae for the q-factorial and related functions

The q-deformations of special functions are obtained from an arithmetic based on q-integers, and defined by

(D.40)
$Display mathematics$

where $q∈C$. These q-integers retain many of the properties of the usual integers. A q-integer is sometimes called a q-bracket or a basic integer.

The q-Pochhammer function is a finite product formally defined by

(D.41)
$Display mathematics$

$(t;q)∞$ is an infinite product.

The q-Pochhammer function $(t;q)∞$ is an analytic function inside the unit disk in the q-plane. Euler’s function is defined by $(q;q)∞$. Observe that $(q)∞=(q;q)∞$ and

(D.42)
$Display mathematics$

Put $n=∞$ in the above to find an infinite product representation of $(q)∞$.

A partition identity (see for example reference [8]) shows that

(D.43)
$Display mathematics$

Note that the usual factorial in the definition of the exponential is replaced by $(q)k$. For this reason, $e(t,q)$ is sometimes called the q-exponential; however, see equation (D.46) below.

(p.568) The natural definition of the q-analogue of the factorial is in terms of q-integers:

(D.44)
$Display mathematics$

and these are also called basic factorials or Gaussian factorials. This gives rise to the q-analogue of the gamma function or the q-gamma function:

(D.45)
$Display mathematics$

which becomes $Γq(k+1)=[k]q!$ if $k∈N$. Observe that, if $q→1−$ in each of the above, then the usual integers, factorials and gamma functions are recovered.

The natural definition of the q-exponential is

(D.46)
$Display mathematics$

or, alternatively,

(D.47)
$Display mathematics$

Observe that $Eqt=e1/qt=1/eq−t$.

These definitions of the exponential give rise to q-analogues of the trigonometric functions sine and cosine. There are also q-analogues of special functions. For example, q-deformed Bessel functions are defined by

(D.48)
$Display mathematics$

More on q-calculus can be found, for example, in reference [345].

## D.5.1 The q-binomial coefficient

The above definitions extend to the q-binomial coefficient, defined by

(D.49)
$Display mathematics$

and, if $n∈N$, and $k∈N$, then this becomes

(D.50)
$Display mathematics$

Observe that $[n]q=n1$ and that $nk$ satisfies the following q-binomial identities:

(D.51)
$Display mathematics$

Lemma D.9 is not obvious.

Lemma D.9

If $n∈N0$, and $k∈N0$, then the Gaussian binomial coefficient is a polynomial of degree $k(n−k)$ with non-negative integer coefficients. In other words,

$Display mathematics$

where $Nnkα∈N0$.

(p.569)

Proof

Observe that $n0=1$ and is a polynomial of degree 0.

The proof proceeds by induction.

Suppose that $mj$ is a polynomial with coefficients in $N0$ and with degree $j(m−j)$, where $m≤n$, and $j≤k$. By equation (D.51),

$Display mathematics$

The largest power of q in the above is $k+k(n−k)=k(n+1−k)$ as required, and the coefficients of $qα$ on the right-hand side are sums over integers in $N0$.

Similarly, again by equation (D.51),

$Display mathematics$

Hence, if $n−1k+1$ is a polynomial in q of degree $(k+1)((n−1)−(k+1))$ and with coefficients in $N0$, then $nk+1$ is a polynomial in q of degree $(k+1)(n−(k+1))$ and with coefficients in $N0$. Notice that $k+1k+1=1$.

The q-binomial theorem is usually stated as

(D.52)
$Display mathematics$

but it is also encountered as

(D.53)
$Display mathematics$

The proof of the last equality proceeds by the counting of partitions.

The q-binomial theorem gives rise to the Jacobi triple product identity

(D.54)
$Display mathematics$

where $tˉ=1t$.

The identity

(D.55)
$Display mathematics$

is also useful [8]. By the q-binomial theorem in equation (D.53), it follows that

(D.56)
$Display mathematics$

## D.5.2 Directed paths and the q-binomial coefficient

There is a close connection between Gaussian binomial coefficients and D.9 (see reference [467]). Consider the square lattice $L2$ and let $L+2$ be the positive half-lattice with the boundary $∂L+2$ as the x-axis. In figure D.1, directed paths from the origin in $L+2$ give only north and east steps. The path oversteps area α‎ in the first quadrant of $L2$ underneath the path and to the left of the endpoint of the path.

Then the number of such paths of length n is related to $Nnki$ and Gaussian binomial coefficients (see lemma D.9) by the following lemma. (p.570)

Fig. D.1 A staircase walk from the origin to the point $(r,n−r)$ overstepping an area α‎.

Lemma D.10

The number of square lattice directed paths in $L2$, of length n, from the origin to the point $(r,n−r)$, giving only north and east steps and which overstep an area of size α‎ in the first quadrant of $L2$ is $Nnrα$.

Proof

This is true for any α‎ if $n=1$, and $0≤r≤1$. Suppose that the lemma is true for any $m≤n$.

Consider $m=n+1$. By equation (D.51),

$Display mathematics$

If coefficients of $qα$ are compared, then

$Display mathematics$

This result is interpreted in figure D.2. By the induction hypothesis, the term $Nnrα$ counts the number of walks of length n, from $(0,0)$ to $(r,n−r)$, and overstepping area α‎.

Appending a vertical step to paths turn these into walks of length $n+1$ which contribute towards $N(n+1)rα$

In a similar fashion, $Nn(r−1)(α+r−n−1)$ counts the number of walks from the origin to $(r−1,n−r+1)$ and which enclose the area $α+r−n−1$. If a horizontal edge is appended from $(r−1,n−r+1)$ to $(r,n−r+1)$, then it encloses area α‎. Thus, $N(n+1)rα$ counts the number of walks from $(0,0)$ to $(r,n−r+1)$ and which overstep area α‎.

(p.571)

Fig. D.2 Any staircase walk of length $n+1$ from the origin to $(r,n+1−r)$ and which oversteps area α‎ has either a north step or an east step as its last step. If the last step was north, then it can be deleted to find $Nnrα$. If it was to the east, then deleting the last step also removes the last column, and this gives $Nn(r−1)(α+r−n−1)$.

Putting $q=1$ gives the corollary

(D.57)
$Display mathematics$

Figure D.1 is also a Ferrer’s diagram of a partition. Thus, let the number of partitions of $α∈N$ into r parts with largest part k be denoted $ρk,r(α)$. Then $ρk,r(α)=N(k+r)rα$.

The generating function of $ρk,r(α)$ for fixed k and r is

(D.58)
$Display mathematics$

See reference [531] for a different proof of this result.

The generating function of partitions follows from this as

(D.59)
$Display mathematics$

where t is the generating variable conjugate to perimeter length.

## D.5.3 Theta functions, and partial theta functions

The Jacobi θ‎-function is defined by

(D.60)
$Display mathematics$

(p.572) $ϕ(q)=θ(1,q)$ is the Ramanujan θ‎-function. The Jacobi triple product identity may be cast as

(D.61)
$Display mathematics$

By expanding the second and third products on the left-hand side, this becomes

(D.62)
$Display mathematics$

where $t=e2πiz$. Putting $t=1$ gives

(D.63)
$Display mathematics$

This last identity may also be obtained from the Ramanujan θ‎-function, which is defined as

(D.64)
$Display mathematics$

The Jacobi triple product becomes

(D.65)
$Display mathematics$

By choosing $a=b=q$ in the above, Ramanujan’s ϕ‎-function $ϕ(q)=f(q,q)$ is recovered:

(D.66)
$Display mathematics$

Other identities may be recovered; for example, if $a=q$, and $b=q3$, then

(D.67)
$Display mathematics$

and, if $a=−q$, and $b=−q2$, then

(D.68)
$Display mathematics$

This identity is the famous pentagonal number theorem [184].

(p.573) The summation in the series definitions of θ‎-functions given above is over all of $Z$. It may in some instances be necessary to have to have the summation only over $N$. This restriction gives rise to partial θ-functions. An example is encountered in equation (D.63), which may be stated as

(D.69)
$Display mathematics$

This is related to Gauss’s formula (see for example [8]):

(D.70)
$Display mathematics$

More generally, a partial θ‎-function has the form

(D.71)
$Display mathematics$

Some identities for partial θ‎-functions are known. For example, in reference [10] it is shown that

(D.72)
$Display mathematics$

## D.5.4 Asymptotics for $(q;q)∞$

A good starting point is the identity in equation (D.68). The left-hand side converges quickly if $|q|≪1$, but, when q is close to 1, another approach is necessary.

Theorem D.11 (Poisson’s sum formula)

Suppose that f is continuous for all $t∈R$, and $∑k=−∞∞f(t+k)$ is uniformly convergent for $t∈[0,1]$. Then

$Display mathematics$

for all values of t where the right-hand side converges.

Proof

The function

$Display mathematics$

is continuous and periodic with period $T=1$. In a Fourier expansion of h the coefficients would be

$Display mathematics$

Moreover, by Fejér’s theorem, $∑n=−∞∞ane−2πint$ is Cesàro summable to $h(t)$ for all t (since h is continuous and periodic).

(p.574) Hence, by replacing h in the last integral with the uniformly convergent series over f, one obtains

$Display mathematics$

The right-hand side evaluates to an, and therefore the Fourier series for $h(t)$ is as claimed above.

Corollary D.12 (The θ‎-transformation formula)

For all values of $t,s∈C$, such that $Re(s)>0$,

$Display mathematics$

Proof

Put $f(t)=e−πst2$ in theorem D.11. Then $∑n=−∞∞f(t+n)$ converges uniformly for $t∈[0,1]$, and $Re(s)>0$. The Poisson sum formula gives

(D.73)
$Display mathematics$

where the integral above gives the Fourier coefficients. It remains to evaluate these coefficients.

Assume that $s>0$ and substitute $y=sx$:

$Display mathematics$

The contour integral on the right evaluates as

$Display mathematics$

using Cauchy’s theorem. This completes the evaluation of the Fourier coefficients in equation (D.73).

Substitution of the result and replacing $n→−n$ in the summation on the right-hand side of equation (D.73) completes the proof.

Corollary D.12 gives a suitable expansion for $(q;q)∞$; see, for example, reference [359] for an expansive treatment.

Theorem D.13

For Euler’s function,

$Display mathematics$

where $r=e4π2logq$. For $q∈(0,1)$ the right-hand side converges quickly.

(p.575)

Proof

Put $t=16+π3ilogq$, put $q3=e−2πs$ and define r by $(logr)(logq)=4π2$. Substitute these variables in the left-hand side of Corollary D.12 and then simplify. This shows that

$Display mathematics$

Comparison to equation (D.68) gives

$Display mathematics$

It remains to simplify the right-hand side.

Substitute t and s on the right-hand side and simplify:

$Display mathematics$

The sum on the right-hand side is absolutely convergent if $|r|<1$. Collecting terms,

$Display mathematics$

The right-hand side is evaluated by summing along n in residue classes modulo 6: replace $n→6n+ℓ$ and sum over ℓ from $−3$ to 2.

$Display mathematics$

Comparing the terms in the sum over ℓ shows that the terms for $ℓ=−2$ and $ℓ=1$ cancel. The terms for both $ℓ=−3$ and $ℓ=2$ are equal under the exchange of $n→−n$, up to phase factors. This is similarly the case for $ℓ=−1$ and $ℓ=0$. This simplifies the expressions above to

$Display mathematics$

Evaluate the cosines and replace $s=−3logq2π$ to complete the proof.

Taking logarithms of the result in theorem D.13 and keeping the $n=0$ term gives the following asymptotic formula for $(q;q)∞$ as $q→1−$:

Theorem D.14

The function $log(q;q)∞=π26logq+12log2π−logq+O(logq)$.

(p.576) This gives the approximation

(D.74)
$Display mathematics$

For additional results, see, for example, the paper by Moak [425].

## D.5.5 Asymptotics for $(t;q)∞$

First examine the limit $limn→∞(t;q)n$ for $q∈(0,1)$ and assume that $0 for all $k≥1$.

Theorem D.15

If $q<1$, and $0 for all $k≥1$, then $0<(t;q)∞≤1$.

Proof

Surely, $0≤(t;q)n≤1$ for all $n≥0$, and $(t;q)n$ is monotone decreasing, so the limit $limn→∞(t;q)n$ exists and is less than or equal to 1.

It remains to show that the limit is not 0.

For any positive $p<1$, $log(1−p)≥−p1−p$. Moreover, for any small $ϵ>0$, there exists a $N0∈N$ such that $0<1−ϵ<1−tqk$ if $k>N0$, since $q<1$ and since $t≠0$. Hence, $log(1−tqk)≥−tqk1−ϵ$ if $k>N0$.

Thus, assume that $n>N0+1$; then

$Display mathematics$

Take $n→∞$ to obtain

$Display mathematics$

which is greater than 0.

Asymptotic approximations for the q-Pochhammer function $(t;q)n$ can be determined using the Euler-Maclaurin theorem. This can be determined directly or indirectly by first determining an approximation for the q-gamma function (see, for example, references [425] and [468]).

Define $Li2$ to be the dilogarithm:

(D.75)
$Display mathematics$

(p.577) Theorem D.16

If $0, and $0, then

$Display mathematics$

where

$Display mathematics$

For fixed $t<1$, this term vanishes as $q↗1−$.

Proof

The Euler-Maclaurin formula gives

$Display mathematics$

where

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The integral over $log(1−tqx)$ can be done by expanding first the logarithm and then integrating term by term; the result is a sequence of terms giving rise to dilogarithms and logarithms.

It remains only to examine the remainder. Observe that

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Next, consider the derivative in R1 above and compute it to see that

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This completes the proof.

Taking $n→∞$ in theorem D.16 gives the following corollary:

Corollary D.17

If $ϵ>0$, and $t∈[0,1−ϵ]$, and if $q∈(0,1)$, then

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or if $Li2(t)+Li2(1−t)=π26−log(t)log(1−t)$ is used, then

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where R1 approaches 0 uniformly as $q↗1−$ for any $t∈[0,1−ϵ]$.

(p.578) By this corollary,

(D.76)
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where $R1→0$ uniformly as $q→1−$. For example, if $t=qn for $n∈N$, then

(D.77)
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# D.6 Asymptotics from the generating function

If the generating function $g(t)$ of a function qn on $N$, defined by

(D.78)
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is known, then qn may in principle be found by using Darboux’s theorem. Darboux’s theorem has the following formulation (see, for example, reference [443]).

Theorem D.18 (Darboux’s theorem)

Let $ψ(t)$ be an analytic function with Laurent expansion

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in the annulus $0<|t|.

Let $χ(t)$ be a function which is analytic in $0<|t| and with Laurent expansion given by

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in $0<|t|.

Suppose that the difference of the m-th derivatives of $ψ(t)$ and $χ(t)$ has a finite number of singularities at $t=tj$ such that

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for some positive constants $σj$, as $t→tj$. Then

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# (p.579) D.7 Convergence of continued fractions

Finite fractions are denoted by

(D.79)
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(For example, $S1=a1b1$, and $S2=a1b1+a2b2$).

Taking $n→∞$ gives the infinite continued fraction $S∞$. The n-th approximation to $S∞$ is $Sn(s)$, obtained by truncating the fraction at n and putting $bn=c$:

(D.80)
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Useful theorems on the convergence of continued fractions are the following.

Theorem D.19

(Pringsheim [474]) Let $an$ and $bn$ be sequences in $R$ such that $|bn|>|an|+1$. Then the continued fraction $k=1∞akbk$ converges absolutely to $f∈R$ with $0<|f|<1$.

Theorem D.20

(Worpitzky [572]) Let $an$ be a sequence in $C$ and let K be a continued fraction given by

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If $0<|an|≤14$ for all $n≥2$, then K converges absolutely and $0<|K|≤12$. Moreover, if, in addition, $Sn(b)=Kk=1nak1$, where $an=b$ and $|b|≤12$ is the n-th approximant of K, then $0<|Sn(b)|≤12$.

Theorem D.21 (Van Vleck [554])

If $rn>0$, then a condition that is necessary for the continued fraction

(p.580)

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to converge for all values of $θn$ is that the series

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is convergent. This condition is also sufficient if $rn<1$ for all values of n.