Question

expand log(secx) upto the term containing x6 using mactaurin's series
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Answer 1

Answer:

Step-by-step explanation:

By chain rule, we have

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

$log(sec(x))^{(2)} =  tan'(x) = sec^2(x)$

$log(sec(x))^{(3)} = (sec^2(x))' = 2sec(x)(sec(x))' = 2sec^2(x)tan(x)$

$log(sec(x))^{(4)} = (2sec^2(x)tan(x))' = 2(sec^4(x)+2sec^2(x)tan^2(x))$

$log(sec(x))^{(5)} = (2(sec^4(x)+2sec^2(x)tan^2(x))' = 8 sec^2(x) tan(x) (2 sec^2(x) + tan^2(x))$

$log(sec(x))^{(6)} = 8(2 sec^6(x) + 11 sec^4(x) tan^2(x) + 2 sec^2(x) tan^4(x))$

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

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

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

$log(sec(0))^{(2)} = sec^2(0) = 1$

$log(sec(0))^{(3)} = 2sec^2(0)tan(0) = 0$

$log(sec(0))^{(4)} = 2(sec^4(0)+2sec^2(0)tan^2(0)) = 2$

$log(sec(0))^{(5)} = 8 sec^2(0) tan(0) (2 sec^2(0) + tan^2(0)) = 0$

$log(sec(0))^{(6)} = 8(2 sec^6(0) + 11 sec^4(0) tan^2(0) + 2 sec^2(0) tan^4(0)) = 16$

Hence, the maclaurin series is given by

$M(x) = f(0) + f'(0)x+\frac{f^{(2)}(0)}{2}x^2+\frac{f^{(3)}(0)}{6}x^3+\frac{f^{(4)}(0)}{24}x^4+\frac{f^{(5)}(0)}{120}x^5+\frac{f^{(6)}(0)}{720}x^6\\\Rightarrow M(x) = \frac{1}{2}x^2+\frac{1}{12}x^4+\frac{1}{45}x^6$