Jacobi–Anger Chebyshev truncation

The Chebyshev recurrence and $\cos ((k+1)\theta )+\cos ((k-1)\theta )=2\cos \theta \cos (k\theta )$ show by induction that $T_k(\cos \theta )=\cos (k\theta )$ for all $k\ge 0$. Every $x\in [-1,1]$ is $\cos \theta$ for some real $\theta$, so $\mathrm{\mid }{T}_k(x)\mathrm{\mid }\le 1$.

If $\tau \mathrm{\ne }0$, the ratio of consecutive absolute terms in the defining series of $J_k(\tau )$ is $\mathrm{\mid }\tau {\mathrm{\mid }}^2\mathrm{/}[4(j+1)(j+k+1)]$, which tends to zero, so the series converges absolutely; for $\tau =0$ the series is immediate. For the integral $I_{k,j}={\int }_0^1{u}^{j-1\mathrm{/}2}(1-u{)}^{k-1\mathrm{/}2}du$, splitting $1=u+(1-u)$ and integrating the derivative of $u^{j+1\mathrm{/}2}(1-u{)}^{k+1\mathrm{/}2}$ gives $I_{k,j+1}\mathrm{/}{I}_{k,j}=(j+1\mathrm{/}2)\mathrm{/}(j+k+1)$. Iterating yields $I_{k,j}\mathrm{/}{I}_{k,0}=(2j)!k!\mathrm{/}[4^{j}j!(j+k)!]$. Expanding $e^{i\tau t}$ uniformly on $[-1,1]$, substituting $u=t^2$, and using this moment ratio gives $J_k(\tau )=\frac{(\tau \mathrm{/}2{)}^k}{k!I_{k,0}}{\int }_{-1}^1{e}^{i\tau t}(1-t^2{)}^{k-1\mathrm{/}2}dt\mathrm{.}$ The weight is nonnegative and its normalized integral is one; [F4] therefore gives $\mathrm{\mid }{J}_k(\tau )\mathrm{\mid }\le (\mathrm{\mid }\tau \mathrm{\mid }\mathrm{/}2{)}^k\mathrm{/}k!$.

Put $L=\ln (1\mathrm{/}ϵ)>0$, ${\tau }_0=\mathrm{\mid }\tau \mathrm{\mid }>ϵ$, and $S={\tau }_0+L\mathrm{/}\ln (e+L\mathrm{/}{\tau }_0)$. If ${\tau }_0\ge 1\mathrm{/}4$, then $S\ge 1\mathrm{/}4$. If ${\tau }_0<1\mathrm{/}4$, put $u=1\mathrm{/}{\tau }_0>4$, so $L>\ln u$. The function $g(y)=y\mathrm{/}\ln (e+uy)$ is increasing for $y>0$, since its derivative has numerator $\ln (e+uy)-uy\mathrm{/}(e+uy)>0$; the first term exceeds $1$ and the second is below $1$. Also $\ln u\le u\mathrm{/}2$ for $u\ge 4$ (the difference has positive derivative and is positive at $4$), and $e\le u^2\mathrm{/}2$, so $e+u\ln u\le u^2$. Therefore $S\ge g(L)>g(\ln u)\ge 1\mathrm{/}2$. In all cases $S\ge 1\mathrm{/}4$. Set $m=\lceil 64S\rceil$; then $m\ge 16$, $m\ge 64{\tau }_0$, and $d=m-1\ge \max (1,\lceil {\tau }_0\rceil )$, so this is an eligible candidate in the OP4 loop starting rule [F7]. The real parameter $\tau$ is defined in [F1].

For $x=\cos \theta$, [F6] writes $e^{-i\tau x}$ as the product of the absolutely convergent exponential series for $e^{-i\tau e^{i\theta }\mathrm{/}2}$ and $e^{-i\tau e^{-i\theta }\mathrm{/}2}$. Absolute convergence permits grouping terms by the exponent difference $k$; for each $k\ge 0$ the coefficient of $e^{ik\theta }$ is $(-i{)}^k{J}_k(\tau )$, and the coefficient of $e^{-ik\theta }$ is the same. Pairing these terms and using [F1] gives $e^{-i\tau x}=J_0(\tau )+2{\sum }_{k\ge 1}(-i{)}^k{J}_k(\tau )T_k(x)\mathrm{.}$ By [F1], [F3], [F4], and [F6], the defining product is absolutely and uniformly convergent in $\theta$ on bounded intervals, so the regrouping is valid.

Subtracting the degree-$d$ partial sum in step 2.1, applying the triangle inequality, and using $\mathrm{\mid }{T}_k(x)\mathrm{\mid }\le 1$ from step 1.1 gives error at most $2{\sum }_{k>d}\mathrm{\mid }{J}_k(\tau )\mathrm{\mid }$. With $m=d+1$, the ratio of consecutive factorial majorants $(\mathrm{\mid }\tau \mathrm{\mid }\mathrm{/}2{)}^k\mathrm{/}k!$ for $k\ge m$ is at most $u=\mathrm{\mid }\tau \mathrm{\mid }\mathrm{/}(2(m+1))<1$. The geometric-series bound from step 1.2 now yields the stated factorial tail.

By [F7], each of the $d+1$ OP4 coefficient errors is at most $\rho \mathrm{/}(32(d+1))$. Since a nonconstant Chebyshev term has coefficient multiplier $2$, their total uniform contribution is at most $(1+2d)\rho \mathrm{/}(32(d+1))<\rho \mathrm{/}16$; adding the tail $\rho \mathrm{/}8$ from step 3.1 gives a combined error $<\rho$. The coefficients and multiplier $2$ are those of the expansion in [F1].

Let $r=L\mathrm{/}{\tau }_0$. If $r\ge 1$, then $m\ge 64{\tau }_0(1+r\mathrm{/}\ln (e+r))$. Since $\ln y\le \sqrt{y}$ for $y\ge 1$ and $r\mathrm{/}(e+r)\ge 1\mathrm{/}(e+1)$, we have $r\mathrm{/}\ln (e+r)\ge \sqrt{e+r}\mathrm{/}(e+1)$. Thus $2m\mathrm{/}(e{\tau }_0)\ge [128\mathrm{/}(e(e+1))]\sqrt{e+r}>\sqrt{e+r}$, because $e<3$ gives $128\mathrm{/}(e(e+1))>128\mathrm{/}12>1$; hence $\ln (2m\mathrm{/}(e{\tau }_0))\ge \frac{1}{2}\ln (e+r)$. If $r<1$, $m\ge 64{\tau }_0$ gives $2m\mathrm{/}(e{\tau }_0)\ge 128\mathrm{/}e$, hence $\ln (2m\mathrm{/}(e{\tau }_0))>1$, since $e<3$ implies $128\mathrm{/}e>42>e$. For $r\ge 1$, these bounds give $m\ln (2m\mathrm{/}(e{\tau }_0))\ge 32{\tau }_0(\ln (e+r)+r)\ge 32(L+{\tau }_0)$, because $\ln (e+r)>1$. For $r<1$, they give $m\ln (2m\mathrm{/}(e{\tau }_0))>64{\tau }_0>32(L+{\tau }_0)$, because $L<{\tau }_0$. Also $u_m={\tau }_0\mathrm{/}[2(m+1)]\le 1\mathrm{/}128$, so $2\mathrm{/}(1-u_m)\le 256\mathrm{/}127<4$, and the factorial tail expression is bounded by the estimate of step 3.1. The real $\tau$ is defined in [F1]. $\ln (m!)={\sum }_{j=1}^m\ln j\ge {\int }_1^m\ln xdx\ge m\ln m-m,$ so $m!\ge (m\mathrm{/}e{)}^m$. The factorial-tail test at $d=m-1$ is consequently at most $\frac{2({\tau }_0\mathrm{/}2{)}^m}{m!(1-u_m)}\le 4\exp \text{\hspace{0.17em}⁣}(-m\ln \frac{2m}{e{\tau }_0})\le 4e^{-32(L+{\tau }_0)}\le e^{-2L}\mathrm{/}80000=\rho \mathrm{/}8.$ The last inequality holds because $L+{\tau }_0>1$: for $0<ϵ<1$, the function $\ln (1\mathrm{/}ϵ)-(1-ϵ)$ is positive (its derivative is $(ϵ-1)\mathrm{/}ϵ<0$ as $ϵ$ increases to $1$, and its value at $1$ is zero), and ${\tau }_0>ϵ$. Hence $30(L+{\tau }_0)>30>\ln (320000)$; the numerical comparison follows from $e>2$, which gives $e^{30}>2^{30}>320000$. Since $m\le 64S+1\le 68S$ using $S\ge 1\mathrm{/}4$, the passing degree $d=m-1$ is $O(S)$. The factorial tail is the estimate of step 3.1 and the OP4 threshold is [F7].

If ${\tau }_0\le ϵ$, the derivative fact [F5] and $\mathrm{\mid }{e}^{-isx}\mathrm{\mid }=1$ from [F4] give $e^{-i\tau x}-1={\int }_0^{\tau }-ix{e}^{-isx}ds,\mathrm{\mid }{e}^{-i\tau x}-1\mathrm{\mid }\le \mathrm{\mid }\tau x\mathrm{\mid }\le ϵ$ for every $x\in [-1,1]$. Thus the wrapper [F8] has the stated small-time error, including $\tau =0$.