Given that y = sec x y = \sec x y = sec x
(a) show that
d 3 y d x 3 = sec x tan x ( p sec 2 x + q ) \begin{align*}
\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} =\,& \sec x \tan x(p\sec^2 x + q)\\[2mm]
\end{align*} d x 3 d 3 y = sec x tan x ( p sec 2 x + q ) where p p p q q q
(4)
(b) Hence determine the Taylor series expansion about π 3 \frac{\pi}{3} 3 π sec x \sec x sec x ( x − π 3 ) \left( x - \frac{\pi}{3} \right) ( x − 3 π ) ( x − π 3 ) 3 \left( x - \frac{\pi}{3} \right)^3 ( x − 3 π ) 3
(3)
(c) Use the answer to part (b) to determine, to four significant figures, an approximate value of sec ( 7 π 24 ) \sec \left(\frac{7\pi}{24}\right) sec ( 24 7 π )
(2)
解答
(a)
Given y = sec x y = \sec x y = sec x x x x
d y d x = sec x tan x \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
= &\,\sec x \tan x
\end{align*} d x d y = sec x tan x Differentiate again using the product rule:
d 2 y d x 2 = sec x ⋅ sec 2 x + ( sec x tan x ) ⋅ tan x = sec 3 x + sec x tan 2 x \begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
= &\,\sec x \cdot \sec^2 x + (\sec x \tan x) \cdot \tan x\\[4mm]
= &\,\sec^3 x + \sec x \tan^2 x
\end{align*} d x 2 d 2 y = = sec x ⋅ sec 2 x + ( sec x tan x ) ⋅ tan x sec 3 x + sec x tan 2 x Using the identity tan 2 x = sec 2 x − 1 \tan^2 x = \sec^2 x - 1 tan 2 x = sec 2 x − 1
d 2 y d x 2 = sec 3 x + sec x ( sec 2 x − 1 ) = 2 sec 3 x − sec x \begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
= &\,\sec^3 x + \sec x (\sec^2 x - 1)\\[4mm]
= &\,2\sec^3 x - \sec x
\end{align*} d x 2 d 2 y = = sec 3 x + sec x ( sec 2 x − 1 ) 2 sec 3 x − sec x Differentiate a third time:
d 3 y d x 3 = 6 sec 2 x ( sec x tan x ) − sec x tan x = 6 sec 3 x tan x − sec x tan x = sec x tan x ( 6 sec 2 x − 1 ) \begin{align*}
\frac{\mathrm{d}^3y}{\mathrm{d}x^3}
= &\,6\sec^2 x (\sec x \tan x) - \sec x \tan x\\[4mm]
= &\,6\sec^3 x \tan x - \sec x \tan x\\[4mm]
= &\,\sec x \tan x (6\sec^2 x - 1)
\end{align*} d x 3 d 3 y = = = 6 sec 2 x ( sec x tan x ) − sec x tan x 6 sec 3 x tan x − sec x tan x sec x tan x ( 6 sec 2 x − 1 ) This is in the required form, with integers p = 6 p = 6 p = 6 q = − 1 q = -1 q = − 1
(b)
To find the Taylor series about x = π 3 x = \frac{\pi}{3} x = 3 π y y y π 3 \frac{\pi}{3} 3 π
y ( π 3 ) = sec ( π 3 ) = 2 y ′ ( π 3 ) = sec ( π 3 ) tan ( π 3 ) = 2 ( 3 ) = 2 3 y ′ ′ ( π 3 ) = 2 ( 2 ) 3 − 2 = 16 − 2 = 14 y ′ ′ ′ ( π 3 ) = 2 ( 3 ) ( 6 ( 2 ) 2 − 1 ) = 2 3 ( 24 − 1 ) = 2 3 ( 23 ) = 46 3 \begin{align*}
y\left(\frac{\pi}{3}\right)
= &\,\sec\left(\frac{\pi}{3}\right) = 2\\[4mm]
y'\left(\frac{\pi}{3}\right)
= &\,\sec\left(\frac{\pi}{3}\right)\tan\left(\frac{\pi}{3}\right) = 2(\sqrt{3}) = 2\sqrt{3}\\[4mm]
y''\left(\frac{\pi}{3}\right)
= &\,2(2)^3 - 2 = 16 - 2 = 14\\[4mm]
y'''\left(\frac{\pi}{3}\right)
= &\,2(\sqrt{3})(6(2)^2 - 1) = 2\sqrt{3}(24 - 1) = 2\sqrt{3}(23) = 46\sqrt{3}
\end{align*} y ( 3 π ) = y ′ ( 3 π ) = y ′′ ( 3 π ) = y ′′′ ( 3 π ) = sec ( 3 π ) = 2 sec ( 3 π ) tan ( 3 π ) = 2 ( 3 ) = 2 3 2 ( 2 ) 3 − 2 = 16 − 2 = 14 2 ( 3 ) ( 6 ( 2 ) 2 − 1 ) = 2 3 ( 24 − 1 ) = 2 3 ( 23 ) = 46 3 Using the Taylor series expansion formula:
y = y ( π 3 ) + y ′ ( π 3 ) ( x − π 3 ) + y ′ ′ ( π 3 ) 2 ! ( x − π 3 ) 2 + y ′ ′ ′ ( π 3 ) 3 ! ( x − π 3 ) 3 + … = 2 + 2 3 ( x − π 3 ) + 14 2 ( x − π 3 ) 2 + 46 3 6 ( x − π 3 ) 3 = 2 + 2 3 ( x − π 3 ) + 7 ( x − π 3 ) 2 + 23 3 3 ( x − π 3 ) 3 \begin{align*}
y
= &\,y\left(\frac{\pi}{3}\right) + y'\left(\frac{\pi}{3}\right)\left(x-\frac{\pi}{3}\right) + \frac{y''\left(\frac{\pi}{3}\right)}{2!}\left(x-\frac{\pi}{3}\right)^2 + \frac{y'''\left(\frac{\pi}{3}\right)}{3!}\left(x-\frac{\pi}{3}\right)^3 + \dots\\[4mm]
= &\,2 + 2\sqrt{3}\left(x-\frac{\pi}{3}\right) + \frac{14}{2}\left(x-\frac{\pi}{3}\right)^2 + \frac{46\sqrt{3}}{6}\left(x-\frac{\pi}{3}\right)^3\\[4mm]
= &\,2 + 2\sqrt{3}\left(x-\frac{\pi}{3}\right) + 7\left(x-\frac{\pi}{3}\right)^2 + \frac{23\sqrt{3}}{3}\left(x-\frac{\pi}{3}\right)^3
\end{align*} y = = = y ( 3 π ) + y ′ ( 3 π ) ( x − 3 π ) + 2 ! y ′′ ( 3 π ) ( x − 3 π ) 2 + 3 ! y ′′′ ( 3 π ) ( x − 3 π ) 3 + … 2 + 2 3 ( x − 3 π ) + 2 14 ( x − 3 π ) 2 + 6 46 3 ( x − 3 π ) 3 2 + 2 3 ( x − 3 π ) + 7 ( x − 3 π ) 2 + 3 23 3 ( x − 3 π ) 3
(c)
We need an approximation for sec ( 7 π 24 ) \sec\left(\frac{7\pi}{24}\right) sec ( 24 7 π ) x = 7 π 24 x = \frac{7\pi}{24} x = 24 7 π ( x − π 3 ) \left(x-\frac{\pi}{3}\right) ( x − 3 π )
7 π 24 − 8 π 24 = − π 24 \begin{align*}
\frac{7\pi}{24} - \frac{8\pi}{24}
= &\,-\frac{\pi}{24}
\end{align*} 24 7 π − 24 8 π = − 24 π Substitute this into our Taylor series expansion:
sec ( 7 π 24 ) ≈ 2 + 2 3 ( − π 24 ) + 7 ( − π 24 ) 2 + 23 3 3 ( − π 24 ) 3 ≈ 2 − 0.4534498 + 0.119946 − 0.029809 ≈ 1.63668... ≈ 1.637 (to 4 sig figs) \begin{align*}
\sec\left(\frac{7\pi}{24}\right)
\approx &\,2 + 2\sqrt{3}\left(-\frac{\pi}{24}\right) + 7\left(-\frac{\pi}{24}\right)^2 + \frac{23\sqrt{3}}{3}\left(-\frac{\pi}{24}\right)^3\\[4mm]
\approx &\,2 - 0.4534498 + 0.119946 - 0.029809\\[4mm]
\approx &\,1.63668...\\[4mm]
\approx &\,1.637 \quad \text{(to 4 sig figs)}
\end{align*} sec ( 24 7 π ) ≈ ≈ ≈ ≈ 2 + 2 3 ( − 24 π ) + 7 ( − 24 π ) 2 + 3 23 3 ( − 24 π ) 3 2 − 0.4534498 + 0.119946 − 0.029809 1.63668... 1.637 (to 4 sig figs) 