Theory of Uncertainty Laws

Abstract

This chapter discusses the theory behind the uncertainty laws in Chap. 15. It is intended for researchers who are interested in understanding the mathematics of the theory or learning the techniques for further development.

Appendices

Appendix A Asymptotics of $\varSigma_{k} D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c}$

This appendix derives an asymptotic expression for $\sum\nolimits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} }$, where $a$, $b$ and $c$ are integers and $b \ge 0$; $D_{k} = [(1 - \beta_{k}^{2} )^{2} + (2\zeta \beta_{k} )^{2} ]^{ - 1}$ is the dynamic amplification factor (acceleration data); $\beta_{k} = f/{\text{f}}_{k}$; ${\text{f}}_{k} = k/T_{d}$ is the FFT frequency; $T_{d}$ is the data duration (s); and the sum is over ${\text{f}}_{k}$ between $f(1 \pm \zeta \kappa )$. The strategy is to express it as a Riemann sum and then approximate by an integral. This is asymptotically correct when $N_{f} \to \infty$. A closed form expression can be obtained by further considering $\zeta \to 0$.

1.1 Long Data

The frequency ratios $\{ \beta_{k} \}$ are not evenly spaced, i.e., $\beta_{k + 1} - \beta_{k}$ is not a constant of $k$. In order to write as a Riemann sum we define and work with an evenly spaced (dimensionless) coordinate. Let $u_{k} = ({\text{f}}_{k} /f) - 1 = (1/\beta_{k} ) - 1$ so that $\{ u_{k} \}$ are evenly spaced at $\Delta u = u_{k + 1} - u_{k} = \Delta f/f = 1/N_{c}$; $\Delta f = T_{d}^{ - 1}$ is the FFT frequency interval and $N_{c} =$ data duration/natural period. Since ${\text{f}}_{k}$ ranges over $f(1 \pm \kappa \zeta )$, $u_{k}$ ranges over $\pm \kappa \zeta$. Substituting $\beta_{k} = 1/(1 + u_{k} )$ and rearranging gives

$$D_{k} = \frac{{(1 + u_{k} )^{4} }}{{u_{k}^{2} (2 + u_{k} )^{2} + 4\zeta^{2} (1 + u_{k} )^{2} }}$$

(16.54)

$$\sum\limits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} } = ( - 1)^{b} 2^{ - 2a} N_{c} \sum\limits_{k} {\frac{{(1 + u_{k} )^{4a - b - c} u_{k}^{b} }}{{[u_{k}^{2} (1 + u_{k} /2)^{2} + \zeta^{2} (1 + u_{k} )^{2} ]^{a} }}\Delta u}$$

(16.55)

The Riemann sum on the RHS tends to an integral, which is asymptotically correct when $\Delta u = N_{c}^{ - 1} \ll 2\kappa \zeta$, or equivalently, $N_{f} \gg 1$. Thus

$$\sum\limits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} } \sim ( - 1)^{b} 2^{ - 2a} N_{c} \int\limits_{ - \kappa \zeta }^{\kappa \zeta } {\frac{{(1 + u)^{4a - b - c} u^{b} }}{{[u^{2} (1 + u/2)^{2} + \zeta^{2} (1 + u)^{2} ]^{a} }}{{d}}u} \quad N_{f} \to \infty$$

(16.56)

Changing variable from $u$ to $u/\zeta$ removes $\zeta$ from the integration limit:

$$\sum\limits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} } \sim ( - 1)^{b} 2^{ - 2a} N_{c} \zeta^{b - 2a + 1} \int_{ - \kappa }^{\kappa } {\frac{{(1 + \zeta u)^{4a - b - c} u^{b} }}{{[u^{2} (1 + \zeta u/2)^{2} + (1 + \zeta u)^{2} ]^{a} }}du} \quad N_{f} \to \infty$$

(16.57)

1.2 Small Damping

For small $\zeta$ the absolute value of the integrand is almost symmetric about the origin. When $b$ is even, the integrals on the positive and negative side are both positive. When $b$ is odd they tend to cancel out. Their difference can be captured by a Taylor approximation w.r.t. $\zeta$. It can be neglected when $b$ is even but it becomes the leading order when $b$ is odd.

Separate the integral in (16.57) on two intervals, $[0,\kappa ]$ and $[ - \kappa ,0]$. For the latter, change integration variable from $u$ to $- u$. This gives

$$\begin{aligned} & \sum\limits_{k}^{} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} } = ( - 1)^{b} 2^{ - 2a} N_{c} \zeta^{b - 2a + 1} \times \\ & \quad \left[ {\int\limits_{0}^{\kappa } {\frac{{(1 + \zeta u)^{4a - b - c} u^{b} }}{{[u^{2} (1 + \zeta u/2)^{2} + (1 + \zeta u)^{2} ]^{a} }}} } + {\frac{{(1 - \zeta u)^{4a - b - c} u^{b} ( - 1)^{b} }}{{[u^{2} (1 - \zeta u/2)^{2} + (1 - \zeta u)^{2} ]^{a} }}{{d}}u} \right] \\ \end{aligned}$$

(16.58)

The two integrands mainly differ by terms involving $\pm \zeta$. For small $\zeta$,

$$\frac{{(1 \pm \zeta u)^{4a - b - c} u^{b} }}{{[u^{2} (1 \pm \zeta u/2)^{2} + (1 \pm \zeta u)^{2} ]^{a} }} = \frac{{u^{b} }}{{(1 + u^{2} )^{a} }}\left\{ {1 \pm \frac{\zeta u}{{1 + u^{2} }}\left[ {(2a - b - c) + (3a - b - c)u^{2} } \right]} \right\} + O(\zeta^{2} )$$

(16.59)

When $b$ is even the $O(\zeta )$ terms in (16.58) cancel out, giving

$$\sum\limits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} } \sim 2^{ - 2a + 1} N_{c} \zeta^{b - 2a + 1} \int \limits_{0}^{\kappa } {\frac{{u^{b} }}{{(1 + u^{2} )^{a} }}{{d}}u} \qquad \begin{array}{*{20}c} {b\,{\text{even,}}} \\ {N_{f} \to \infty ,} \\ {\zeta \to 0} \\ \end{array} $$

(16.60)

When $b$ is odd, the zeroth order terms cancel out, leaving the $O(\zeta )$ terms as the leading order:

$$ \begin{aligned} & \begin{array}{*{20}l}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,{\sum\limits_{k}{D_{k}^{a}(\beta_{k} - 1)^{b} \beta_{k}^{c} } } \\ \quad { \sim -2^{ - 2a + 1}N_{c} \zeta^{b - 2a + 2} \int\limits_{0}^{\kappa}{\displaystyle\frac{{u^{b + 1} }}{{(1 + u^{2} )^{a + 1} }}\left[(2a- b - c) + (3a- b - c)u^{2} \right]du} } \end{array}\\ & {{b\,{\text{odd}}, {N_{f}\to \infty , \zeta \to 0}}}\end{aligned} $$

(16.61)

The integrals in (16.60) and (16.61) can be evaluated analytically. Table 16.6 shows some results for $c = 0$ that are frequently used.

Table 16.6 Asymptotic expression for $\sum\nolimits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} }$ for $N_{f} \to \infty$ and $\zeta \to 0$

1.3 Caution with Shortcuts

When analyzing the asymptotic behavior of $\sum\nolimits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} }$ it is tempting to take $\beta_{k} \sim 1$ or $\beta_{k} - 1\sim 0$ but this need not be correct. When $b$ is even it is correct to take $\beta_{k}^{c} \sim 1$, i.e.,

$$\sum\limits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} \beta_{k}^{c} } \sim \sum\nolimits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} } \quad b\,{\text{even}}$$

(16.62)

because (16.60) does not depend on $c$. This rule can be generalized to

$$\sum\limits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} P(\beta_{k} )} \sim \sum\limits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} P(1)} \quad b\,{\text{even}}$$

(16.63)

for any polynomial $P(\beta_{k} )$ with $P(1) \ne 0$. When $b$ is odd, this rule does not apply and (16.61) should be used. On the other hand, terms of $(\beta_{k} - 1)$ should not be approximated by zero, since then their asymptotic order is grossed out. The sum $\sum\nolimits_{k} {D_{k}^{a} (\beta_{k} - 1)^{b} }$ should be evaluated using (16.60) or (16.61).

Appendix B Derivation of Small Damping Asymptotics (Zeroth Order)

This appendix derives the small damping (in addition to long data) asymptotic expressions in Table 16.2.

Auto-Derivatives $\hat{L}^{(ff)}$, $\hat{L}^{(\zeta \zeta )}$, $\hat{L}^{(SS)}$, $\hat{L}^{{(S_{e} S_{e} )}}$ and $\hat{L}^{{({\varvec{\upvarphi \upvarphi }})}}$

From Table 16.1, $\hat{L}^{(ff)} \sim \sum\nolimits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)2} (1 + e_{k} )^{ - 2} } \sim \sum\nolimits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)2} }$ since $e_{k} \ll 1$. Substituting $(D_{k}^{ - 1} )^{(f)} = 4f^{ - 1} \beta_{k}^{2} (\beta_{k}^{2} - 1 + 2\zeta^{2} )$ from (16.23), writing $\beta_{k}^{2} - 1 = (\beta_{k} - 1)(\beta_{k} + 1)$ and expanding the square,

$$\hat{L}^{(ff)} \sim 16f^{ - 2} \left\{ {\sum\limits_{k} {D_{k}^{2} (\beta_{k} - 1)^{2} (\beta_{k} + 1)^{2} \beta_{k}^{4} } + 4\zeta^{2} \sum\limits_{k} {D_{k}^{2} (\beta_{k} - 1)(\beta_{k} + 1)\beta_{k}^{4} } + 4\zeta^{4} \sum\limits_{k} {D_{k}^{2} \beta_{k}^{4} } } \right\}$$

(16.64)

Since $(\beta_{k} + 1)$ and $\beta_{k}$ are $O(1)$, the first sum $\sum\nolimits_{k} {D_{k}^{2} (\beta_{k} - 1)^{2} (\beta_{k} + 1)^{2} \beta_{k}^{4} }$ has the same order as $\sum\nolimits_{k} {D_{k}^{2} (\beta_{k} - 1)^{2} }$, which is $O(N_{c} \zeta^{ - 1} )$ from Table 16.6. Similarly, the second sum $\sum\nolimits_{k} {D_{k}^{2} (\beta_{k} - 1)(\beta_{k} + 1)\beta_{k}^{4} }$ has the same order as $\sum\nolimits_{k} {D_{k}^{2} (\beta_{k} - 1)} = O(N_{c} \zeta^{ - 1} )$ and so the second term is $O(N_{c} \zeta )$. The third sum $\sum\nolimits_{k} {D_{k}^{2} \beta_{k}^{4} }$ has the same order as $\sum\nolimits_{k} {D_{k}^{2} } = O(N_{c} \zeta^{ - 3} )$ and so the third term is $O(N_{c} \zeta )$. Thus, the first term in the brace dominates, giving

$$\hat{L}^{(ff)} \sim 16f^{ - 2} \sum\limits_{k} {D_{k}^{2} (\beta_{k} - 1)^{2} (\beta_{k} + 1)^{2} \beta_{k}^{4} }$$

(16.65)

Since $(\beta_{k} - 1)^{2}$ has even power, using (16.63) we can take $\beta_{k} + 1\sim 2$ and $\beta_{k} \sim 1$, and so $\hat{L}^{(ff)} \sim 64f^{ - 2} \sum\nolimits_{k} {D_{k}^{2} (\beta_{k} - 1)^{2} }$. Using Table 16.6 to evaluate the sum ($a = 2$,$b = 2$) gives the expression in Table 16.2.

For $\hat{L}^{(\zeta \zeta )}$, from Table 16.1 and using $(1 + e_{k} )^{ - 2} \sim 1$,

$$\hat{L}^{(\zeta \zeta )} \sim \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(\zeta )2} (1 + e_{k} )^{ - 2} } \sim \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(\zeta )2} }$$

(16.66)

Substituting $(D_{k}^{ - 1} )^{(\zeta )} = 8\zeta \beta_{k}^{2}$ from (16.23), $\hat{L}^{(\zeta \zeta )} \sim 64\zeta^{2} \sum\nolimits_{k} {D_{k}^{2} \beta_{k}^{4} } \sim 64\zeta^{2} \sum\nolimits_{k} {D_{k}^{2} }$. Using Table 16.6 to evaluate the sum gives the expression in Table 16.2.

For $\hat{L}^{(SS)}$, from Table 16.1,

$$\hat{L}^{(SS)} \sim S^{ - 2} \sum\limits_{k} {(1 + e_{k} )^{ - 2} } \sim S^{ - 2} \sum\limits_{k} 1 = S^{ - 2} N_{f}$$

(16.67)

as in Table 16.2.

For $\hat{L}^{{(S_{e} S_{e} )}}$, from Table 16.1, $\hat{L}^{{(S_{e} S_{e} )}} \sim S_{e}^{ - 2} [(n - 1)N_{f} + \sum\nolimits_{k} {e_{k}^{2} (1 + e_{k} )^{ - 2} } ]$. The second term in the bracket is dominated by the first and so

$$\hat{L}^{{(S_{e} S_{e} )}} \sim S_{e}^{ - 2} (n - 1)N_{f}$$

(16.68)

as in Table 16.2.

For $\hat{L}^{{({\varvec{\upvarphi \upvarphi }})}}$, from Table 16.1, using $1 + e_{k} \sim 1$ and $e_{k}^{ - 1} = SD_{k} /S_{e}$,

$$\hat{L}^{{({\varvec{\upvarphi \upvarphi }})}} \sim [2\sum\limits_{k} {(1 + e_{k} )^{ - 1} e_{k}^{ - 1} } ]({\mathbf{I}}_{n} - {\bar{\varvec{\upvarphi }}}\bar{\varvec{\upvarphi} }^{T} )\sim (\frac{2S}{{S_{e} }}\sum\limits_{k} {D_{k} } )({\mathbf{I}}_{n} - {\bar{\varvec{\upvarphi }}}\bar{\varvec{\upvarphi}}^{T} )$$

(16.69)

Using Table 16.6 to evaluate the sum gives the expression in Table 16.2.

Cross Derivatives $\hat{L}^{(f\zeta )}$, $\hat{L}^{(fS)}$ and $\hat{L}^{(\zeta S)}$

From Table 16.1,

$$\hat{L}^{(f\zeta )} \sim \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)} (D_{k}^{ - 1} )^{(\zeta )} (1 + e_{k} )^{ - 2} } \sim \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)} (D_{k}^{ - 1} )^{(\zeta )} }$$

(16.70)

Substituting $(D_{k}^{ - 1} )^{(f)} = 4f^{ - 1} \beta_{k}^{2} (\beta_{k}^{2} - 1 + 2\zeta^{2} )$ and $(D_{k}^{ - 1} )^{(\zeta )} = 8\zeta \beta_{k}^{2}$, and expanding,

$$\hat{L}^{(f\zeta )} \sim 32f^{ - 1} \zeta \left[ {\sum\limits_{k} {D_{k}^{2} (\beta_{k} - 1)\beta_{k}^{5} } + \sum\limits_{k} {D_{k}^{2} (\beta_{k} - 1)\beta_{k}^{4} } + 2\zeta^{2} \sum\limits_{k} {D_{k}^{2} \beta_{k}^{4} } } \right]$$

(16.71)

Evaluating the first two sums using (16.61) and the third using Table 16.6 gives the expression in Table 16.2.

For $\hat{L}^{(fS)}$, from Table 16.1,

$$\hat{L}^{(fS)} \sim - S^{ - 1} \sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(f)} (1 + e_{k} )^{ - 2} } \sim - S^{ - 1} \sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(f)} }$$

(16.72)

Substituting $(D_{k}^{ - 1} )^{(f)} = 4f^{ - 1} \beta_{k}^{2} (\beta_{k}^{2} - 1 + 2\zeta^{2} )$ and expanding,

$$\hat{L}^{(fS)} \sim - 4f^{ - 1} S^{ - 1} \left[ {\sum\limits_{k} {D_{k} (\beta_{k} - 1)\beta_{k}^{3} } + \sum\limits_{k} {D_{k} (\beta_{k} - 1)\beta_{k}^{2} } + 2\zeta^{2} \sum\limits_{k} {D_{k} \beta_{k}^{2} } } \right]$$

(16.73)

Evaluating the first two sums using (16.61) and the third using Table 16.6 gives the expression in Table 16.2.

For $\hat{L}^{(\zeta S)}$, from Table 16.1,

$$\hat{L}^{(\zeta S)} \sim - S^{ - 1} \sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(\zeta )} (1 + e_{k} )^{ - 2} } \sim - S^{ - 1} \sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(\zeta )} }$$

(16.74)

Substituting $(D_{k}^{ - 1} )^{(\zeta )} = 8\zeta \beta_{k}^{2}$, $\hat{L}^{(\zeta S)} \sim - 8\zeta S^{ - 1} \sum\nolimits_{k} {D_{k} \beta_{k}^{2} } \sim - 8\zeta S^{ - 1} \sum\nolimits_{k} {D_{k} }$. Evaluating the sum using Table 16.6 gives the expression in Table 16.2.

Cross Derivatives $\hat{L}^{{(fS_{e} )}}$, $\hat{L}^{{(\zeta S_{e} )}}$ and $\hat{L}^{{(SS_{e} )}}$

From Table 16.1,

$$\hat{L}^{{(fS_{e} )}} \sim - S^{ - 1} \sum\limits_{k} {(D_{k}^{ - 1} )^{(f)} (1 + e_{k} )^{ - 2} } \sim - S^{ - 1} \sum\limits_{k} {(D_{k}^{ - 1} )^{(f)} }$$

(16.75)

Substituting $(D_{k}^{ - 1} )^{(f)} = 4f^{ - 1} \beta_{k}^{2} (\beta_{k}^{2} - 1 + 2\zeta^{2} )$ and expanding gives

$$\hat{L}^{{(fS_{e} )}} \sim - 4f^{ - 1} S^{ - 1} \left[ {\sum\limits_{k} {(\beta_{k} - 1)\beta_{k}^{3} } + \sum\limits_{k} {(\beta_{k} - 1)\beta_{k}^{2} } + 2\zeta^{2} \sum\limits_{k} {\beta_{k}^{2} } } \right]$$

(16.76)

Evaluating the first two sums using (16.61) and noting $\sum\nolimits_{k} {\beta_{k}^{2} } \sim N_{f} = 2\zeta \kappa N_{c}$ gives the expression in Table 16.2.

For $\hat{L}^{{(\zeta S_{e} )}}$, from Table 16.1,

$$\hat{L}^{{(\zeta S_{e} )}} \sim - S^{ - 1} \sum\limits_{k} {(D_{k}^{ - 1} )^{(\zeta )} (1 + e_{k} )^{ - 2} } \sim - S^{ - 1} \sum\limits_{k} {(D_{k}^{ - 1} )^{(\zeta )} }$$

(16.77)

Substituting $(D_{k}^{ - 1} )^{(\zeta )} = 8\zeta \beta_{k}^{2}$ gives

$$\hat{L}^{{(\zeta S_{e} )}} \sim - 8\zeta S^{ - 1} \sum\limits_{k} {\beta_{k}^{2} } \sim - 8\zeta S^{ - 1} N_{f} = - \frac{{16N_{c} \zeta^{2} \kappa }}{S}$$

(16.78)

as in Table 16.2.

For $\hat{L}^{{(SS_{e} )}}$, from Table 16.1,

$$\hat{L}^{{(SS_{e} )}} \sim S^{ - 2} \sum\limits_{k} {D_{k}^{ - 1} (1 + e_{k} )^{ - 2} } \sim S^{ - 2} \sum\limits_{k} {D_{k}^{ - 1} }$$

(16.79)

Evaluating the sum using (16.60) gives the expression in Table 16.2.

Appendix C First Order of NLLF Derivatives w.r.t. $f,\zeta ,S$

In this appendix we derive expressions for $\hat{L}^{(ff)}$, $\hat{L}^{(\zeta \zeta )}$, $\hat{L}^{(SS)}$ and $\hat{L}^{(\zeta S)}$ up to the first order of the n/e ratio $\nu = S_{e} /S$. It leads to results in the form of (16.43).

Second Derivative $\hat{L}^{(ff)}$

From Table 16.1 and using $(1 + e_{k} )^{ - 2} \sim 1 - 2e_{k}$,

$$\hat{L}^{(ff)} = \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)2} (1 + e_{k} )^{ - 2} } \sim \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)2} (1 - 2e_{k} )} = \hat{L}_{0}^{(ff)} + \hat{L}_{1}^{(ff)}$$

(16.80)

where $\hat{L}_{0}^{(ff)} = \sum\nolimits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)2} }$ is the zeroth order whose expression is given in Table 16.2; and, using $D_{k} e_{k} = \nu$,

$$\hat{L}_{1}^{(ff)} = - 2\sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(f)2} e_{k} } = - 2\nu \sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(f)2} }$$

(16.81)

is the first order. Substituting $(D_{k}^{ - 1} )^{(f)} = 4f^{ - 1} \beta_{k}^{2} (\beta_{k}^{2} - 1 + 2\zeta^{2} )$ from (16.23), writing $\beta_{k}^{2} - 1 = (\beta_{k} - 1)(\beta_{k} + 1)$ and expanding the square,

$$\begin{aligned} \sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(f)2} } & = \frac{16}{{f^{2} }}\left[ {\sum\limits_{k} {D_{k} (\beta_{k} - 1)^{2} (\beta_{k} + 1)^{2} \beta_{k}^{4} } } \right. \\ & \quad + \left. {4\zeta^{2} \sum\limits_{k} {D_{k} (\beta_{k} - 1)(\beta_{k} + 1)\beta_{k}^{4} } + 4\zeta^{4} \sum\limits_{k} {D_{k} \beta_{k}^{4} } } \right] \\ \end{aligned}$$

(16.82)

In the bracket, the first sum is of the same order as $\sum\nolimits_{k} {D_{k} (\beta_{k} - 1)^{2} } = O(N_{c} \zeta )$ from Table 16.6. Similarly, the second sum is of the same order as $\sum\nolimits_{k} {D_{k} (\beta_{k} - 1)} = O(N_{c} \zeta )$ and so the second term is $O(N_{c} \zeta^{3} )$. The last sum is of the same order as $\sum\nolimits_{k} {D_{k} } = O(N_{c} \zeta^{ - 1} )$ and so the last term is also $O(N_{c} \zeta^{3} )$. Thus, the first term in the bracket dominates:

$$\sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(f)2} } \sim 16f^{ - 2} \sum\limits_{k} {D_{k} (\beta_{k} - 1)^{2} (\beta_{k} + 1)^{2} \beta_{k}^{4} }$$

(16.83)

Since $(\beta_{k} - 1)^{2}$ has even power, using (16.63) we can take $\beta_{k} + 1\sim 2$ and $\beta_{k} \sim 1$ to get $\sum\nolimits_{k} {D_{k} (D_{k}^{ - 1} )^{(f)2} } \sim 64f^{ - 2} \sum\nolimits_{k} {D_{k} (\beta_{k} - 1)^{2} }$. Using Table 16.6 to evaluate the sum,

$$\sum\limits_{k} {D_{k} (D_{k}^{ - 1} )^{(f)2} } \sim \frac{{32N_{c} \zeta }}{{f^{2} }}(\kappa - \tan^{ - 1} \kappa )$$

(16.84)

Substituting into (16.81),

$$\hat{L}_{1}^{(ff)} = - \frac{{64N_{c} \nu \zeta }}{{f^{2} }}(\kappa - \tan^{ - 1} \kappa )\qquad \frac{{\hat{L}_{1}^{(ff)} }}{{\hat{L}_{0}^{(ff)} }} = - \underbrace {{\left[ {\frac{{4(\kappa - \tan^{ - 1} \kappa )}}{{\tan^{ - 1} \kappa - \kappa /(\kappa^{2} + 1)}}} \right]}}_{\displaystyle {c_{ff} }}\underbrace {{(4\nu \zeta^{2} )}}_{\displaystyle {\gamma^{ - 1} }}$$

(16.85)

We can then write

$$\hat{L}^{(ff)} \sim \hat{L}_{0}^{(ff)} (1 - c_{ff} \gamma^{ - 1} )\qquad c_{ff} = \frac{{4(\kappa - \tan^{ - 1} \kappa )}}{{\tan^{ - 1} \kappa - \kappa /(\kappa^{2} + 1)}}$$

(16.86)

Second Derivative $\hat{L}^{(\zeta \zeta )}$

From Table 16.1 and using $(1 + e_{k} )^{ - 2} \sim 1 - 2e_{k}$,

$$\hat{L}^{(\zeta \zeta )} = \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(\zeta )2} (1 + e_{k} )^{ - 2} } \sim \sum\limits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(\zeta )2} (1 - 2e_{k} )} = \hat{L}_{0}^{(\zeta \zeta )} + \hat{L}_{1}^{(\zeta \zeta )}$$

(16.87)

where $\hat{L}_{0}^{(\zeta \zeta )} = \sum\nolimits_{k} {D_{k}^{2} (D_{k}^{ - 1} )^{(\zeta )2} }$ is the zeroth order in Table 16.2; and $\hat{L}_{1}^{(\zeta \zeta )} = - 2\nu \sum\nolimits_{k} {D_{k} (D_{k}^{ - 1} )^{(\zeta )2} }$ is the first order. Substituting $(D_{k}^{ - 1} )^{(\zeta )} = 8\zeta \beta_{k}^{2}$ from (16.23) gives $\hat{L}_{1}^{(\zeta \zeta )} = - 128\nu \zeta^{2} \sum\nolimits_{k} {D_{k} \beta_{k}^{4} } \sim - 128\nu \zeta^{2} \sum\nolimits_{k} {D_{k} }$. Using Table 16.6 to evaluate the sum,

$$\hat{L}_{1}^{(\zeta \zeta )} \sim - 64N_{c} \nu \zeta \tan^{ - 1} \kappa \qquad \frac{{\hat{L}_{1}^{(\zeta \zeta )} }}{{\hat{L}_{0}^{(\zeta \zeta )} }}\sim - \underbrace {{\left[ {\frac{{4\tan^{ - 1} \kappa }}{{\tan^{ - 1} \kappa + \kappa /(\kappa^{2} + 1)}}} \right]}}_{\displaystyle {c_{\zeta \zeta } }}\underbrace {{(4\nu \zeta^{2} )}}_{\displaystyle {\gamma^{ - 1} }}$$

(16.88)

We can then write

$$\hat{L}^{(\zeta \zeta )} \sim \hat{L}_{0}^{(\zeta \zeta )} (1 - c_{\zeta \zeta } \gamma^{ - 1} )\qquad c_{\zeta \zeta } = \frac{{4\tan^{ - 1} \kappa }}{{\tan^{ - 1} \kappa + \kappa /(\kappa^{2} + 1)}}$$

(16.89)

Second Derivative $\hat{L}^{(SS)}$

From Table 16.1 and using $(1 + e_{k} )^{ - 2} \sim 1 - 2e_{k} ,\hat{L}^{(SS)} \sim S^{ - 2} \sum\nolimits_{k} {(1 - 2e_{k} )} = \hat{L}_{0}^{(SS)} + \hat{L}_{1}^{(SS)}$ where $\hat{L}_{0}^{(SS)} = S^{ - 2} N_{f}$ is the zeroth order; and $\hat{L}_{1}^{(SS)} = - 2\nu S^{ - 2} \sum\nolimits_{k} {D_{k}^{ - 1} }$ is the first order. Using (16.60) to evaluate the sum gives

$$\hat{L}_{1}^{(SS)} = - 8S^{ - 2} N_{f} \nu \zeta^{2} \left( {1 + \frac{{\kappa^{2} }}{3}} \right)\qquad \frac{{\hat{L}_{1}^{(SS)} }}{{\hat{L}_{0}^{(SS)} }} = - \underbrace {{2\left( {1 + \frac{{\kappa^{2} }}{3}} \right)}}_{\displaystyle {c_{SS} }}\underbrace {{(4\nu \zeta^{2} )}}_{\displaystyle {\gamma^{ - 1} }}$$

(16.90)

We can then write

$$\hat{L}^{(SS)} \sim \hat{L}_{0}^{(SS)} (1 - c_{SS} \gamma^{ - 1} )\qquad c_{SS} = 2\left( {1 + \frac{{\kappa^{2} }}{3}} \right)$$

(16.91)

Cross Derivative $\hat{L}^{(\zeta S)}$

From Table 16.1, using $(1 + e_{k} )^{ - 2} \sim 1 - 2e_{k}$ and $(D_{k}^{ - 1} )^{(\zeta )} = 8\zeta \beta_{k}^{2}$,

$$\hat{L}^{(\zeta S)} \sim - S^{ - 1} \sum\limits_{k} {D_{k} (8\zeta \beta_{k}^{2} )(1 - 2e_{k} )} = \hat{L}_{0}^{(\zeta S)} + \hat{L}_{1}^{(\zeta S)}$$

(16.92)

where $\hat{L}_{0}^{(\zeta S)} = - 8S^{ - 1} \zeta \sum\nolimits_{k} {D_{k} \beta_{k}^{2} } \sim - 4S^{ - 1} N_{c} \tan^{ - 1} \kappa$ is the zeroth order and

$$\hat{L}_{1}^{(\zeta S)} = 16S^{ - 1} \nu \zeta \sum\limits_{k} {\beta_{k}^{2} } \sim 32S^{ - 1} N_{c} \nu \zeta^{2} \kappa$$

(16.93)

is the first order. Thus

$$\frac{{\hat{L}_{1}^{(\zeta S)} }}{{\hat{L}_{0}^{(\zeta S)} }} = - \underbrace {{\left( {\frac{2\kappa }{{\tan^{ - 1} \kappa }}} \right)}}_{\displaystyle {c_{\zeta S} }}\underbrace {{(4\nu \zeta^{2} )}}_{\displaystyle {\gamma^{ - 1} }}$$

(16.94)

and we can write

$$\hat{L}^{(\zeta S)} \sim \hat{L}_{0}^{(\zeta S)} (1 - c_{\zeta S} \gamma^{ - 1} )\qquad c_{\zeta S} = \frac{2\kappa }{{\tan^{ - 1} \kappa }}$$

(16.95)