Find the first four nonzero terms of the Taylor series about 0 for the function f(x) = square...

Question:

Find the first four nonzero terms of the Taylor series about 0 for the function {eq}f(x) = \sqrt {1 - 2 x} {/eq}. Note that you may want to find these in a manner other than by direct differentiation of the function.

Binomial Series Representation:

The definition of Binomial series representation for the fractional powers is very useful in order to get an expression for the series representation of functions such as {eq}\; \sqrt {1 + g(x)} \; {/eq}. First of all, we will derive the expression for the general function such as {eq}\; \sqrt {1 + a} \; {/eq} then we will use the method of substitution in order to reach the final expression.

{eq}\displaystyle (1 + a)^{k} = 1 + ka + \dfrac {k (k-1)}{2!} \; a^{2} + \dfrac {k (k-1) (k-2)}{3!} \; a^{3} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; |a| < 1 {/eq}

{eq}\\ {/eq}

{eq}\displaystyle f(x) = \sqrt {1 - 2x} {/eq}

We know the standard Binomial series representation:

{eq}\displaystyle (1 + a)^{k} = 1 + ka + \dfrac {k (k-1)}{2!} \; a^{2} + \dfrac {k (k-1) (k-2)}{3!} \; a^{3} + \cdots {/eq}

{eq}\displaystyle \sqrt {1 + a} = 1 + \dfrac {a}{2} + \dfrac {\biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {1}{2} - 1 \biggr)}{2!} \; a^{2} + \dfrac {\biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {1}{2} - 1 \biggr) \; \biggr( \dfrac {1}{2} - 2 \biggr)}{3!} \; a^{3} + \cdots {/eq}

{eq}\displaystyle \sqrt {1 + a} = 1 + \dfrac {a}{2} - \dfrac {a^{2}}{8} + \dfrac {a^{3}}{16} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; |a| < 1 {/eq}

Now substitute the value of {eq}\; a= -2x \; {/eq} in the above expression:

{eq}\displaystyle \sqrt {1 - 2x} = 1 - \dfrac {(2x)}{2} - \dfrac {(2x)^{2}}{8} - \dfrac {(2x)^{3}}{16} + \cdots {/eq}

Finally, the Taylor series representation for the function {eq}\; \sqrt {1 -2x} \; {/eq} along with its interval of convergence is given as:

{eq}\displaystyle \Longrightarrow \boxed {\sqrt {1 - 2x} = 1 - x - \dfrac {x^{2}}{2} - \dfrac {x^{3}}{2} - \cdots } {/eq}

The interval of convergence is given as: {eq}\; \; |2x| < 1 \; \; \; \Longrightarrow \; |x| < \dfrac {1}{2} {/eq}