Question

Give the first three non-zero terms of the Taylor series for f(x) = tan(x) about x 0· Use this to approximate tan(1) and give an upper bound on the error in this approximation

Answer 1

Answer:

$f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....$

Approximate error = 0.4426

Step-by-step explanation:

f(x)=tanx, a=0

Maclaurin series formula used is given below

$f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)x^{n}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^{2}+\dfrac{f'''(0)}{3!}x^{3}+....$

f(x)=tanx

f(0)=tan0=0

$f'(x)=sec^{2}x\\f'(0)=sec^{2}0=1\\f''(x)=2sec^{2}xtanx\\f''(0)=2sec^{2}0tan0=0\\f'''(x)=-4sec^{2}x+6sec^{4}x\\f'''(0)=-4sec^{2}0+6sec^{4}0=-4+6=2$

$f''''(x)=-8(2sec^{2}xtan^{2}x+sec^{4}x)+24(4sec^{4}xtan^{2}x)+sec^{6})\\f''''(0)=-8(0+1)+24(0+1)=-8+24=16$

$f(x)=0+x+0+\dfrac{2x^{3}}{3!}+\dfrac{16x^{4}}{4!}$

$f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}$

Hence, the Taylor series for f(x)=tanx is given by

$f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....$

Maclaurin series upper bound error formula used is given as

R_n(x)=|f(x)-T_n(x)|

R_3(x)=|tanx-T_3(x)|

$R_3(x)=|tanx-x-\dfrac{x^{3}}{3}-\dfrac{2x^{4}}{3}|$

Plugging this value x=1

$R_3(x)=|tan(1)-1-\dfrac{1}{3}-\dfrac{2}{3}|$

R_3(x)=|1.5574-1-0.333-0.666|

R_3(x)=|-0.4426|=0.4426

Hence, upper bound on the error approximation

tan(1)=0.4426