Question

By using Maclaurin’s theorem expand log sec x up to the term containing x⁶.

Answer 1

By chain rule, we have

log(sec(x))' = \(\frac 1{\sec(x)}\)·sec(x)tan(x) = tan(x)

log(sec(x))⁽²⁾ = tan'(x) = sec²(x)

log(sec(x))⁽³⁾ (sec²(x))' = 2sec(x) (sec(x))' = 2sec²(x)tan(x)

log(sec(x))⁽⁴⁾ = (2sec²(x) tan(x))' = 2(sec⁴(x) + 2sec²(x) tan²(x))

log(sec(x))⁽⁵⁾ = (2(sec⁴(x) + 2sec²(x)tan²(x)) = 8sec²(x) tan(x) (2sec²(x) + tan²(x))

log(sec(x))⁽⁶⁾ = 8(2sec⁶(x) + 11sec⁴(x)tan²(x) + 2sec²(x)tan⁴(x))

We need to evaluate the derivatives in zero in order to get the maclaurin series.

Once sec(0) = 1 and tan(0) = 0, we have

log (sec(0)) = log(1) = 0

log(sec(0))' = tan(0) = 0

log(sec(0))⁽²⁾ = sec²(0) = 1

log(sec(0))⁽³⁾ = 2sec²(0) tan(0) = 0

log(sec(0))⁽⁴⁾ = 2(sec⁴(0) + 2sec²(0) tan²(0)) = 2

log (sec(0))⁽⁵⁾ = 8sec²(0) tan (0) (2sec²(0) + tan²(0)) = 0

log(sec(0))⁽⁶⁾ = 8(2sec⁶(0) + 11sec⁴(0) tan²(0) + 2sec²(0) tan⁴(0)) = 16

Hence, the maclaurin series is given by

\(M(x) = f(0) + f'(0)x + \frac{f^{(2)}(0)}2 x^2 + \frac{f^{(3)}(0)}6 x^3 + \frac{f^{(4)}(0)}{24} x^4 + \frac{f^{(5)}(0)}{120} x^5 + \frac{f^{(6)}(0)}{720} x^6\)

⇒ \(M(x) = \frac 12x^2 + \frac 1{12}x^4 + \frac 1{45}x^6\)

Answer 2

Let y (x) = logsecx,

y (0) = logsec0 = 0