Please can you check my solution so far to see if I am on the right track. Thanks
\(\displaystyle \L{\rm Use Maclaurin's Theorum to expand: }2\sin ^2 2x - 1{\rm (first 4 non - zero terms) hence evaluate cos(0.4)}\)
\(\displaystyle \L2\sin ^2 2x - 1 = - \cos 4x\)
\(\displaystyle \L
\begin{array}{l}
f(x) = - \cos 4x{\rm }f(0) = - 1 \\
f'(x) = - 4\sin 4x{\rm } f'(0) = 0 \\
f''(x) = - 16\cos 4x{\rm } f''(0) = - 16 \\
f'''(x) = 64\sin 4x{\rm }f'''(0) = 0 \\
f^{iv} (x) = 256\cos 4x{\rm }f^{iv} (0) = 256 \\
f^v (x) = - 1024\sin 4x{\rm }f^v (0) = 0 \\
f^{vi} (x) = - 4096\cos 4x{\rm } f^{vi} (0) = - 4096 \\
\end{array}\)
\(\displaystyle \L
\begin{array}{l}
{\rm Maclaurins Theorum:} \\
f(x) = f(0) + xf'(0) + \frac{{x^2 }}{2}f''(0) + \frac{{x^3 }}{3}f'''(0) + \frac{{x^4 }}{4}f^{iv} (0) + \frac{{x^5 }}{5}f^v (0) + \frac{{x^4 }}{4}f^{vi} (0)... \\
\end{array}\)
\(\displaystyle \L
{\rm make x = 0}{\rm .1}\)
\(\displaystyle \L f(x) =( - \cos 4x) = - 1 - \frac{{2^4 (0.1)^2 }}{{2!}} + \frac{{2^8 (0.1)^4 }}{{4!}} - \frac{{2^{12} (0.1)^6 }}{{6!}}\)
\(\displaystyle \L = - 1 - \frac{2}{{25}} + \frac{2}{{1875}} - \frac{4}{{70312}}\)
Thank you
\(\displaystyle \L{\rm Use Maclaurin's Theorum to expand: }2\sin ^2 2x - 1{\rm (first 4 non - zero terms) hence evaluate cos(0.4)}\)
\(\displaystyle \L2\sin ^2 2x - 1 = - \cos 4x\)
\(\displaystyle \L
\begin{array}{l}
f(x) = - \cos 4x{\rm }f(0) = - 1 \\
f'(x) = - 4\sin 4x{\rm } f'(0) = 0 \\
f''(x) = - 16\cos 4x{\rm } f''(0) = - 16 \\
f'''(x) = 64\sin 4x{\rm }f'''(0) = 0 \\
f^{iv} (x) = 256\cos 4x{\rm }f^{iv} (0) = 256 \\
f^v (x) = - 1024\sin 4x{\rm }f^v (0) = 0 \\
f^{vi} (x) = - 4096\cos 4x{\rm } f^{vi} (0) = - 4096 \\
\end{array}\)
\(\displaystyle \L
\begin{array}{l}
{\rm Maclaurins Theorum:} \\
f(x) = f(0) + xf'(0) + \frac{{x^2 }}{2}f''(0) + \frac{{x^3 }}{3}f'''(0) + \frac{{x^4 }}{4}f^{iv} (0) + \frac{{x^5 }}{5}f^v (0) + \frac{{x^4 }}{4}f^{vi} (0)... \\
\end{array}\)
\(\displaystyle \L
{\rm make x = 0}{\rm .1}\)
\(\displaystyle \L f(x) =( - \cos 4x) = - 1 - \frac{{2^4 (0.1)^2 }}{{2!}} + \frac{{2^8 (0.1)^4 }}{{4!}} - \frac{{2^{12} (0.1)^6 }}{{6!}}\)
\(\displaystyle \L = - 1 - \frac{2}{{25}} + \frac{2}{{1875}} - \frac{4}{{70312}}\)
Thank you
