What is the Maclaurin series for? \(: \sqrt{1-x}\)
Answer:
\(f(x)=1-\frac{1}{2} x-\frac{1}{8} x^{2}-\frac{1}{16} x^{3}-\frac{5}{128} x^{4}+\ldots\)
Explanation:
Let:
We seek a Taylor Series, as no pivot is supplied, it is assumed that an expansion about the pivot \(x = 0\) is required. This is known as a Maclaurin series and is given by
\(f(x)=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\ldots \frac{f^{(n)}(0)}{n !} x^{n}+\ldots\)Although we could use this method, it is actually quicker, in this case, to use a Binomial Series expansion.
The binomial series tell us that:
\((1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots\)And so for the given function, we have: