Given #f(x) = sqrt(1-x)#, how do you write the Taylor series about c = -3 with the first four terms?

Question

Emma Aldridge · Accepted Answer

The general formula is #f(c)+f'(c)(x-c)+(f''(c))/(2!)(x-c)^2+(f'''(c))/(3!)(x-c)^3+\cdots#. Since #f(x)=\sqrt{1-x}=(1-x)^{1/2}# we get #f'(x)=-\frac{1}{2}(1-x)^{-1/2}#, #f''(x)=-\frac{1}{4}(1-x)^{-3/2}#, and #f'''(x)=-\frac{3}{8}(1-x)^{-5/2}#. Since #c=-3#, we get #f(c)=\sqrt{4}=2#, #f'(c)=-\frac{1}{2\sqrt{4}}=-\frac{1}{4}#, #f''(c)=-\frac{1}{4\cdot 4^{3/2}}=\frac{1}{32}#, and #f'''(c)=-\frac{3}{8\cdot 4^{5/2}}=-\frac{3}{256}#. Therefore, #\sqrt{1-x}=2-\frac{1}{4}(x+3)-\frac{1}{64}(x+3)^{2}-\frac{1}{512}(x+3)^{3}+\cdots#.

Evelyn Agard · Answer

undefined The Taylor series expansion of $ f(x) = \sqrt{1-x} $ about $ c = -3 $ with the first four terms is:

$$ f(x) = f(c) + f'(c)(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 $$

Substitute $ c = -3 $ and find $ f(c) $, $ f'(c) $, $ f''(c) $, and $ f'''(c) $. Then plug these values into the formula to obtain the first four terms of the Taylor series.