Changes between Version 5 and Version 6 of research/trig
 Timestamp:
 Oct 13, 2011, 4:42:46 PM (9 years ago)
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research/trig
v5 v6 11 11 {{{ 12 12 #!latex 13 \[\ big\vert\sin(x)  P(x)\big\vert \le E \qquad \forall x \in \bigg[\frac{\pi}{2}, \frac{\pi}{2}\bigg]\]13 \[\max_{x \in [\pi/2, \pi/2]}{\big\vert\sin(x)  P(x)\big\vert} = E\] 14 14 }}} 15 15 … … 18 18 {{{ 19 19 #!latex 20 \[\ big\vert\sin(x)  xQ(x^2)\big\vert \le E \qquad \forall x \in \bigg[\frac{\pi}{2}, \frac{\pi}{2}\bigg]\]20 \[\max_{x \in [\pi/2, \pi/2]}{\big\vert\sin(x)  xQ(x^2)\big\vert} = E\] 21 21 }}} 22 22 … … 25 25 {{{ 26 26 #!latex 27 \[\ big\lvert\sin(\sqrt{y})  \sqrt{y}Q(y)\big\rvert \le E \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]\]27 \[\max_{x \in [0, \pi^2/4]}{\big\lvert\sin(\sqrt{y})  \sqrt{y}Q(y)\big\rvert} = E\] 28 28 }}} 29 29 … … 32 32 {{{ 33 33 #!latex 34 \[\ bigg\lvert\frac{\sin(\sqrt{y})}{\sqrt{y}}  Q(y)\bigg\rvert \le \frac{E}{\sqrt{y}} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]\]34 \[\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})}{\sqrt{y}}  Q(y)\bigg\rvert}{\dfrac{1}{\sqrt{y}}}} = E\] 35 35 }}} 36 36 37 If we want to force the asymptotic behaviour inx=0, we substitute Q(y) with 1+yR(y):37 If we want to force the asymptotic behaviour at x=0, we substitute Q(y) with 1+yR(y): 38 38 39 39 {{{ 40 40 #!latex 41 \[\ bigg\lvert\frac{\sin(\sqrt{y})}{\sqrt{y}}  1  yR(y)\bigg\rvert \le \frac{E}{\sqrt{y}} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]\]41 \[\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})}{\sqrt{y}}  1  yR(y)\bigg\rvert}{\dfrac{1}{\sqrt{y}}}} = E\] 42 42 }}} 43 43 … … 46 46 {{{ 47 47 #!latex 48 \[\ bigg\lvert\frac{\sin(\sqrt{y})\sqrt{y}}{y\sqrt{y}}  R(y)\bigg\rvert \le \frac{E}{y\sqrt{y}} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]\]48 \[\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})\sqrt{y}}{y\sqrt{y}}  R(y)\bigg\rvert}{\dfrac{1}{y\sqrt{y}}}} = E\] 49 49 }}} 50 50