Crazy Calculus

$\begin{aligned} y & = e^{\frac x{\sqrt 2}} \sin \bigg(\frac x{\sqrt 2}+2017\bigg) & \small \color{#3D99F6} \text{By Euler's formula } e^{i\theta} = \cos \theta + i \sin \theta \\ & = \Im \bigg[e^{\frac x{\sqrt 2}}e^{i\left(\frac x{\sqrt 2}+2017\right)} \bigg] & \small \color{#3D99F6} \text{where }\Im(z) \text{ is the imaginary part of }z. \\ & = \Im \bigg[e^{(1+i)\frac x{\sqrt 2}}e^{2017i}\bigg] \\ \implies y_n & = \Im \bigg[\bigg(\frac {1+i}{\sqrt 2}\bigg)^n e^{(1+i)\frac x{\sqrt 2}}e^{2017i}\bigg] \\ y_{2017} & = \Im \bigg[\bigg(\frac {1+i}{\sqrt 2}\bigg)^{2017} e^{(1+i)\frac x{\sqrt 2}}e^{2017i}\bigg] & \small \color{#3D99F6} \text{Note that }(1+i)^2 = 2i \\ & = \Im \bigg[i^{1008} \cdot \frac {1+i}{\sqrt 2} \cdot e^{(1+i)\frac x{\sqrt 2}}e^{2017i}\bigg] & \small \color{#3D99F6} \text{Note that } i^4 = 1 \\ & = \Im \bigg[\frac {1+i}{\sqrt 2} \cdot e^{\frac x{\sqrt 2}}\bigg(\cos \bigg(\frac x{\sqrt 2} + 2017\bigg) + i \sin \bigg(\frac x{\sqrt 2} + 2017\bigg) \bigg) \bigg] \\ & = \frac {e^{\frac x{\sqrt 2}}}{\sqrt 2} \bigg(\cos \bigg(\frac x{\sqrt 2} + 2017\bigg) + \sin \bigg(\frac x{\sqrt 2} + 2017\bigg) \bigg)\\ y_{2017} \bigg|_{x=0} & = \frac {\cos (2017)+ \sin(2017)}{\sqrt 2} \approx \boxed{0.773} \end{aligned}$

let $z=\frac { x }{ \sqrt { 2 } } +2017$ $\Rightarrow \qquad \frac { dz }{ dx } =\frac { 1 }{ \sqrt { 2 } } \qquad x=0\quad \rightarrow \quad z=2017$

$y={ e }^{ z-2017 }\sin { \left( z \right) } ={ e }^{ -2017 }{ e }^{ z }\sin { \left( z \right) }$

${ y }^{ \left( 1 \right) }={ e }^{ -2017 }{ e }^{ z }\left( \sin { \left( z \right) } +\cos { \left( z \right) } \right) \cdot \frac { dz }{ dx } =\frac { { e }^{ -2017 } }{ \sqrt { 2 } } \cdot { e }^{ z }\left( \sin { \left( z \right) } +\cos { \left( z \right) } \right)$

${ y }^{ \left( 2 \right) }=\frac { { e }^{ -2017 } }{ \sqrt { 2 } } \cdot { e }^{ z }\left( \sin { \left( z \right) } +\cos { \left( z \right) + } \cos { \left( z \right) -\sin { \left( z \right) } } \right) \cdot \frac { dz }{ dx } =\frac { { e }^{ -2017 } }{ \sqrt { 2 } } \cdot { e }^{ z }\cdot \left( 2\cos { \left( z \right) } \right) \cdot \frac { 1 }{ \sqrt { 2 } } ={ e }^{ -2017 }{ e }^{ z }\left( \cos { \left( z \right) } \right)$

${ y }^{ \left( 3 \right) }={ e }^{ -2017 }{ e }^{ z }\left( \cos { \left( z \right) -\sin { \left( z \right) } } \right) \cdot \frac { dz }{ dx } =\frac { { e }^{ -2017 } }{ \sqrt { 2 } } \cdot { e }^{ z }\left( \cos { \left( z \right) -\sin { \left( z \right) } } \right)$

${ y }^{ \left( 4 \right) }=\frac { { e }^{ -2017 } }{ \sqrt { 2 } } \cdot { e }^{ z }\left( \cos { \left( z \right) -\sin { \left( z \right) } - } \sin { \left( z \right) } -\cos { \left( z \right) } \right) \cdot \frac { dz }{ dx } =\frac { { e }^{ -2017 } }{ \sqrt { 2 } } \cdot { e }^{ z }\cdot \left( -2\sin { \left( z \right) } \right) \cdot \frac { 1 }{ \sqrt { 2 } } ={ -e }^{ -2017 }{ e }^{ z }\left( \sin { \left( z \right) } \right)$

${ y }^{ \left( 4 \right) }=-y\Rightarrow { \left( { y }^{ \left( 4 \right) } \right) }^{ 4 }={ \left( -y \right) }^{ 4 }\Rightarrow { y }^{ \left( 8 \right) }=-{ y }^{ \left( 4 \right) }=-\left( -y \right) =y$

Now ${ y }^{ \left( 2017 \right) }=\frac { d }{ dx } { \left( { y }^{ \left( 2016 \right) } \right) }={ y }^{ \left( 1 \right) }$

${ y }^{ \left( 1 \right) }{ | }_{ z=2017 }=\frac { { e }^{ -2017 } }{ \sqrt { 2 } } \cdot { e }^{ z }\left( \sin { \left( z \right) } +\cos { \left( z \right) } \right) { | }_{ z=2017 }$ $=\frac { { e }^{ -2017 }{ e }^{ 2017 }\left( \sin { \left( 2017 \right) } +\cos { \left( 2017 \right) } \right) }{ \sqrt { 2 } } \approx 0.77259$