Answer to Question #218957 in Calculus for anto

Question

Find the maclaurin series expansion for y=(8+e^x)^1/2 in ascending powers of x upto and including the term in x³ hence estimate (8+e^0.54)^1/2 to three decimal places

Answer 1

$f(x)=(8+e^x)^{1/2}$

$f(0)=(8+e^0)^{1/2}=3$

$f'(x)=\dfrac{1}{2}e^x(8+e^x)^{-1/2}$

$f'(0)=\dfrac{1}{6}$

$f''(x)=\dfrac{1}{2}e^x(8+e^x)^{-1/2}-\dfrac{1}{4}e^{2x}(8+e^x)^{-3/2}$

$f''(0)=\dfrac{1}{6}-\dfrac{1}{108}=\dfrac{17}{108}$

$f'''(x)=\dfrac{1}{2}e^x(8+e^x)^{-1/2}-\dfrac{1}{4}e^{2x}(8+e^x)^{-3/2}$

$-\dfrac{1}{2}e^{2x}(8+e^x)^{-3/2}+\dfrac{3}{8}e^{2x}(8+e^x)^{-5/2}$

$f'''(0)=\dfrac{1}{6}-\dfrac{1}{108}-\dfrac{1}{54}+\dfrac{1}{648}=\dfrac{91}{648}$

$f(x)=3+\dfrac{1}{6}x+\dfrac{17}{216}x^2+\dfrac{91}{3888}x^3+O(x^4)$

$(8+e^{0.54})^{1/2}\approx3+\dfrac{1}{6}(0.54)+\dfrac{17}{216}(0.54)^2$

$+\dfrac{91}{3888}(0.54)^3\approx3.117$