# Changes between Version 5 and Version 6 of research/trigTweet

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Timestamp:
Oct 13, 2011, 4:42:46 PM (9 years ago)
Comment:

 v5 {{{ #!latex $\big\vert\sin(x) - P(x)\big\vert \le E \qquad \forall x \in \bigg[-\frac{\pi}{2}, \frac{\pi}{2}\bigg]$ $\max_{x \in [-\pi/2, \pi/2]}{\big\vert\sin(x) - P(x)\big\vert} = E$ }}} {{{ #!latex $\big\vert\sin(x) - xQ(x^2)\big\vert \le E \qquad \forall x \in \bigg[-\frac{\pi}{2}, \frac{\pi}{2}\bigg]$ $\max_{x \in [-\pi/2, \pi/2]}{\big\vert\sin(x) - xQ(x^2)\big\vert} = E$ }}} {{{ #!latex $\big\lvert\sin(\sqrt{y}) - \sqrt{y}Q(y)\big\rvert \le E \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\big\lvert\sin(\sqrt{y}) - \sqrt{y}Q(y)\big\rvert} = E$ }}} {{{ #!latex $\bigg\lvert\frac{\sin(\sqrt{y})}{\sqrt{y}} - Q(y)\bigg\rvert \le \frac{E}{|\sqrt{y}|} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})}{\sqrt{y}} - Q(y)\bigg\rvert}{\dfrac{1}{|\sqrt{y}|}}} = E$ }}} If we want to force the asymptotic behaviour in x=0, we substitute Q(y) with 1+yR(y): If we want to force the asymptotic behaviour at x=0, we substitute Q(y) with 1+yR(y): {{{ #!latex $\bigg\lvert\frac{\sin(\sqrt{y})}{\sqrt{y}} - 1 - yR(y)\bigg\rvert \le \frac{E}{|\sqrt{y}|} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})}{\sqrt{y}} - 1 - yR(y)\bigg\rvert}{\dfrac{1}{|\sqrt{y}|}}} = E$ }}} {{{ #!latex $\bigg\lvert\frac{\sin(\sqrt{y})-\sqrt{y}}{y\sqrt{y}} - R(y)\bigg\rvert \le \frac{E}{|y\sqrt{y}|} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})-\sqrt{y}}{y\sqrt{y}} - R(y)\bigg\rvert}{\dfrac{1}{|y\sqrt{y}|}}} = E$ }}}