# kernel functions

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## 1—10 of 29 matching pages

##### 1: 10.62 Graphs
###### §10.62 Graphs
For the modulus functions $M\left(x\right)$ and $N\left(x\right)$ see §10.68(i) with $\nu=0$. … Figure 10.62.2: ker ⁡ x , kei ⁡ x , ker ′ ⁡ x , kei ′ ⁡ x , 0 ≤ x ≤ 8 . Magnify Figure 10.62.4: e x / 2 ⁢ ker ⁡ x , e x / 2 ⁢ kei ⁡ x , e x / 2 ⁢ N ⁡ ( x ) , 0 ≤ x ≤ 8 . Magnify
##### 2: 10.70 Zeros
###### §10.70 Zeros
10.70.1 $\frac{\mu-1}{16t}+\frac{\mu-1}{32t^{2}}+\frac{(\mu-1)(5\mu+19)}{1536t^{3}}+% \frac{3(\mu-1)^{2}}{512t^{4}}+\dotsi.$
$\mbox{zeros of \operatorname{ker}_{\nu}x}\sim\sqrt{2}(t+f(-t)),$ $t=(m-\tfrac{1}{2}\nu-\tfrac{5}{8})\pi$,
In the case $\nu=0$, numerical tabulations (Abramowitz and Stegun (1964, Table 9.12)) indicate that each of (10.70.2) corresponds to the $m$th zero of the function on the left-hand side. …
##### 3: 31.10 Integral Equations and Representations
###### KernelFunctions
31.10.9 $\mathcal{K}(\theta,\phi)=P\begin{Bmatrix}0&1&\infty&\\[1.0pt] 0&\frac{1}{2}-\delta-\sigma&\alpha&{\cos}^{2}\theta\\[1.0pt] 1-\gamma&\frac{1}{2}-\epsilon+\sigma&\beta&\end{Bmatrix}\*P\begin{Bmatrix}0&1&% \infty&\\[1.0pt] 0&0&-\frac{1}{2}+\delta+\sigma&{\cos}^{2}\phi\\[1.0pt] 1-\epsilon&1-\delta&-\frac{1}{2}+\epsilon-\sigma&\end{Bmatrix},$
Fuchs–Frobenius solutions $W_{m}(z)=\tilde{\kappa}_{m}z^{-\alpha}\mathit{H\!\ell}\left(1/a,q_{m};\alpha,% \alpha-\gamma+1,\alpha-\beta+1,\delta;1/z\right)$ are represented in terms of Heun functions $w_{m}(z)=(0,1)\mathit{Hf}_{m}\left(a,q_{m};\alpha,\beta,\gamma,\delta;z\right)$ by (31.10.1) with $W(z)=W_{m}(z)$, $w(z)=w_{m}(z)$, and with kernel chosen from …
###### KernelFunctions
31.10.19 $\mathcal{K}(u,v,w)=u^{1-\gamma}v^{1-\delta}w^{1-\epsilon}\mathscr{C}_{1-\gamma% }\left(u\sqrt{\sigma_{1}}\right)\*\mathscr{C}_{1-\delta}\left(v\sqrt{\sigma_{2% }}\right)\mathscr{C}_{1-\epsilon}\left(\mathrm{i}w\sqrt{\sigma_{1}+\sigma_{2}}% \right),$
##### 4: 10.67 Asymptotic Expansions for Large Argument
###### §10.67(i) $\operatorname{ber}_{\nu}x,\operatorname{bei}_{\nu}x,\operatorname{ker}_{\nu}x,% \operatorname{kei}_{\nu}x$, and Derivatives
10.67.13 ${\operatorname{ker}}^{2}x+{\operatorname{kei}}^{2}x\sim\frac{\pi}{2x}e^{-x% \sqrt{2}}\left(1-\frac{1}{4\sqrt{2}}\frac{1}{x}+\frac{1}{64}\frac{1}{x^{2}}+% \frac{33}{256\sqrt{2}}\frac{1}{x^{3}}-\frac{1797}{8192}\frac{1}{x^{4}}+\dotsb% \right),$
10.67.14 $\operatorname{ker}x\operatorname{kei}'x-\operatorname{ker}'x\operatorname{kei}% x\sim-\frac{\pi}{2x}e^{-x\sqrt{2}}\left(\frac{1}{\sqrt{2}}-\frac{1}{8}\frac{1}% {x}+\frac{9}{64\sqrt{2}}\frac{1}{x^{2}}-\frac{39}{512}\frac{1}{x^{3}}+\frac{75% }{8192\sqrt{2}}\frac{1}{x^{4}}+\dotsb\right),$
10.67.15 $\operatorname{ker}x\operatorname{ker}'x+\operatorname{kei}x\operatorname{kei}'% x\sim-\frac{\pi}{2x}e^{-x\sqrt{2}}\left(\frac{1}{\sqrt{2}}+\frac{3}{8}\frac{1}% {x}-\frac{15}{64\sqrt{2}}\frac{1}{x^{2}}+\frac{45}{512}\frac{1}{x^{3}}+\frac{3% 15}{8192\sqrt{2}}\frac{1}{x^{4}}+\dotsb\right),$
10.67.16 $\left(\operatorname{ker}'x\right)^{2}+\left(\operatorname{kei}'x\right)^{2}% \sim\frac{\pi}{2x}e^{-x\sqrt{2}}\left(1+\frac{3}{4\sqrt{2}}\frac{1}{x}+\frac{9% }{64}\frac{1}{x^{2}}-\frac{75}{256\sqrt{2}}\frac{1}{x^{3}}+\frac{2475}{8192}% \frac{1}{x^{4}}+\dotsi\right).$
##### 5: 10.61 Definitions and Basic Properties
10.61.2 $\operatorname{ker}_{\nu}x+i\operatorname{kei}_{\nu}x=e^{-\nu\pi i/2}K_{\nu}% \left(xe^{\pi i/4}\right)=\tfrac{1}{2}\pi i{H^{(1)}_{\nu}}\left(xe^{3\pi i/4}% \right)=-\tfrac{1}{2}\pi ie^{-\nu\pi i}{H^{(2)}_{\nu}}\left(xe^{-\pi i/4}% \right).$
10.61.3 $x^{2}\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}x}^{2}}+x\frac{\mathrm{d}w}{\mathrm{d% }x}-(ix^{2}+\nu^{2})w=0,$ $w=\begin{array}[t]{cc}\operatorname{ber}_{\nu}x+i\operatorname{bei}_{\nu}x,&% \operatorname{ber}_{-\nu}x+i\operatorname{bei}_{-\nu}x\\ \operatorname{ker}_{\nu}x+i\operatorname{kei}_{\nu}x,&\operatorname{ker}_{-\nu% }x+i\operatorname{kei}_{-\nu}x.\end{array}$
10.61.4 $x^{4}\frac{{\mathrm{d}}^{4}w}{{\mathrm{d}x}^{4}}+2x^{3}\frac{{\mathrm{d}}^{3}w% }{{\mathrm{d}x}^{3}}-(1+2\nu^{2})\left(x^{2}\frac{{\mathrm{d}}^{2}w}{{\mathrm{% d}x}^{2}}-x\frac{\mathrm{d}w}{\mathrm{d}x}\right)+(\nu^{4}-4\nu^{2}+x^{4})w=0,$ $w=\operatorname{ber}_{\pm\nu}x,\operatorname{bei}_{\pm\nu}x,\operatorname{ker}% _{\pm\nu}x,\operatorname{kei}_{\pm\nu}x$.
10.61.11 $\operatorname{ker}_{\frac{1}{2}}\left(x\sqrt{2}\right)=\operatorname{kei}_{-% \frac{1}{2}}\left(x\sqrt{2}\right)=-2^{-\frac{3}{4}}\sqrt{\frac{\pi}{x}}e^{-x}% \sin\left(x-\frac{\pi}{8}\right),$
10.61.12 $\operatorname{kei}_{\frac{1}{2}}\left(x\sqrt{2}\right)=-\operatorname{ker}_{-% \frac{1}{2}}\left(x\sqrt{2}\right)=-2^{-\frac{3}{4}}\sqrt{\frac{\pi}{x}}e^{-x}% \cos\left(x-\frac{\pi}{8}\right).$
##### 6: 10.63 Recurrence Relations and Derivatives
###### §10.63(i) $\operatorname{ber}_{\nu}x$, $\operatorname{bei}_{\nu}x$, $\operatorname{ker}_{\nu}x$, $\operatorname{kei}_{\nu}x$
$\sqrt{2}\operatorname{kei}'x=-\operatorname{ker}_{1}x+\operatorname{kei}_{1}x.$
##### 7: 10.69 Uniform Asymptotic Expansions for Large Order
10.69.3 $\operatorname{ker}_{\nu}\left(\nu x\right)+i\operatorname{kei}_{\nu}\left(\nu x% \right)\sim e^{-\nu\xi}\left(\frac{\pi}{2\nu\xi}\right)^{\ifrac{1}{2}}\left(% \frac{xe^{3\pi i/4}}{1+\xi}\right)^{-\nu}\sum_{k=0}^{\infty}(-1)^{k}\frac{U_{k% }(\xi^{-1})}{\nu^{k}},$
10.69.5 $\operatorname{ker}_{\nu}'\left(\nu x\right)+i\operatorname{kei}_{\nu}'\left(% \nu x\right)\sim-\frac{e^{-\nu\xi}}{x}\left(\frac{\pi\xi}{2\nu}\right)^{\ifrac% {1}{2}}\left(\frac{xe^{3\pi i/4}}{1+\xi}\right)^{-\nu}\*\sum_{k=0}^{\infty}(-1% )^{k}\frac{V_{k}(\xi^{-1})}{\nu^{k}},$
##### 9: 10.68 Modulus and Phase Functions
10.68.2 $N_{\nu}\left(x\right)e^{i\phi_{\nu}\left(x\right)}=\operatorname{ker}_{\nu}x+i% \operatorname{kei}_{\nu}x,$
##### 10: 10.65 Power Series
10.65.3 $\operatorname{ker}_{n}x=\tfrac{1}{2}(\tfrac{1}{2}x)^{-n}\sum_{k=0}^{n-1}\frac{% (n-k-1)!}{k!}\cos\left(\tfrac{3}{4}n\pi+\tfrac{1}{2}k\pi\right)(\tfrac{1}{4}x^% {2})^{k}-\ln\left(\tfrac{1}{2}x\right)\operatorname{ber}_{n}x+\tfrac{1}{4}\pi% \operatorname{bei}_{n}x+\tfrac{1}{2}(\tfrac{1}{2}x)^{n}\sum_{k=0}^{\infty}% \frac{\psi\left(k+1\right)+\psi\left(n+k+1\right)}{k!(n+k)!}\cos\left(\tfrac{3% }{4}n\pi+\tfrac{1}{2}k\pi\right)(\tfrac{1}{4}x^{2})^{k},$