Consider the function 
. Determine the first five coefficients (
) in the Taylor series expansion of 
at 
.
First, we take some derivatives,
![\begin{align*} && f(x) &= (2+x^2)^{\frac{5}{2}} \\[9pt] \implies && f'(x) &= 5x(2+x^2)^{\frac{3}{2}} \\[9pt] \implies && f''(x) &= 15x^2 (2+x^2)^{\frac{1}{2}} + 5 (2+x^2)^{\frac{3}{2}} \\[9pt] \implies && f'''(x) &= \frac{15x^3}{(2+x^2)^{\frac{1}{2}}} + 45x (2+x^2)^{\frac{1}{2}} \\[9pt] \implies && f^{(4)} (x) &= -\frac{15x^4}{(2+x^2)^{\frac{3}{2}}} + \frac{90x^2}{(2+x^2)^{\frac{1}{2}}} + 45 (2+x^2)^{\frac{1}{2}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2b0c22a5496a721d80cf8c8daa25b1c0_l3.png "Rendered by QuickLaTeX.com")
Then, since the coefficients in the Taylor series at are of the form 
we can compute as follows:
![\begin{align*} a_0 &= \frac{f(0)}{0!} = 2^{\frac{5}{2}} = 4 \sqrt{2} \\[9pt] a_1 &= \frac{f'(0)}{1!} = \frac{0}{1} = 0 \\[9pt] a_2 &= \frac{f''(0)}{2!} = \frac{5 \cdot 2^{\frac{3}{2}}}{2} = 5 \sqrt{2} \\[9pt] a_3 &= \frac{f'''(0)}{6} = 0 \\[9pt] a_4 &= \frac{f^{(4)}(0)}{4!} = \frac{45 \sqrt{2}}{24} = \frac{15}{8} \sqrt{2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e66765b310d59feeda93be2b6172b255_l3.png "Rendered by QuickLaTeX.com")
Consider the function . Determine the first five coefficients () in the Taylor series expansion of at .
First, we take some derivatives,
Then, since the coefficients in the Taylor series at are of the form we can compute as follows: