Answer to Question #108830 in Calculus for harry

Question

Find a third degree McLaurin expansion of f (x) =sec2x

Answer 1

Consider the function $f(x)=sec(2x)$

The McLaurin series of a function $f(x)$ is given by the expansion formula as follows:

$f(x)\approx f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...+...$

Now, find the first derivative of the function as,

$f'(x)=2sec(2x)tan(2x)$

Find the second derivative of the function as,

$f''(x)=2[2sec(2x)tan(2x)tan(2x)+sec(2x)(2sec^2(2x))]$

$f''(x)=4sec(2x)tan^2(2x)+4sec^3(2x)$

Find the third derivative of the function as,

$f'''(x)=4[sec(2x)\frac{d}{dx}tan^2(2x)+tan^2(2x)\frac{d}{dx}sec(2x)]+4[\frac{d}{dx}sec^3(2x)]$

$f'''(x)=4[2tan(2x)sec^3(2x)+2tan^3(2x)sec(2x)]+4[6sec^3(2x)tan(2x)]$

Find the fourth derivative of the function as,

$f^{iv}(x)=2[(-4\left(-2\sec \left(2x\right)+4\sec ^3\left(2x\right)\right)+24\left(6\sec ^3\left(2x\right)\tan ^2\left(2x\right)+2\sec ^5\left(2x\right)\right)]$

Next, find the function value and derivative value at $x=0$ as,

$f(0)=1$

$f'(0)=0$

$f''(0)=4$

$f'''(0)=0$

$f^{iv}(0)=80$

So, the third order McLaurin expansion is given by,

$f(x)\approx 1+\frac{0}{1}x+\frac{4}{2}x^2+\frac{0}{6}x^3+\frac{80}{24}x^4+...$

$f(x)\approx 1+2x^2+0\cdot x^3+...$

Therefore, the third degree McLaurin expansion of the function $f(x)=sec(2x)$ is $f(x)\approx 1+2x^2+0\cdot x^3$