Question

use maclaurine series find lnsecx

Answer 1

Look at this http://mathworld.wolfram.com/MaclaurinSeries.html,
by the definition,
\[y=\ln (\sec x)\]
\[y'=(secxtanx)\frac{ 1 }{ secx}=tanx\]
y'(0)=0
\[y''=\sec ^{2}x\]
y''(0)=1
\[y'''=(2secx)secxtanx=2sec ^{2}xtanx\],
y'''(0)=0
\[y''''=2(tanx 2secx secxtanx+\sec^{2}x \sec^{2}x)\]
y''''(0)=2
And y(0)=0, so the final equation is
y=\[\frac{ 1 }{ 2 }x^2+\frac{ 1 }{ 12 }x^4+......\]