# a) Use the binomial series formula to find the first four terms of the maclaurin series for 1 ?...

## Question:

a) Use the binomial series formula to find the first four terms of the maclaurin series for {eq}\frac{1}{\sqrt{1-x^2}} {/eq} .

(Hint: First find a series for {eq}\frac{1}{\sqrt{1+x}} {/eq} using the binomial series formula, and then use an appropriate substitution to get a series for the given function.) Show all work.

b) Use the series obtained above to determine the first four terms of the series for {eq}arcsin(x) {/eq}.

(Hint: First, think of a relationship between the function given above and the {eq}arcsin(x) {/eq} function.) Show all work.

## Binomial Series Representation:

{eq}\\ {/eq}

The definition of Binomial series representation {eq}\; \Biggr[ (1 + a)^{k} = 1 + ka + \dfrac {k (k -1)}{2!} \; a^{2} + \dfrac {k (k - 1) (k - 2)}{3!} \; a^{3} + \cdots \Biggr] \; {/eq} and some basic tools such as calculus and the method of substitution will be used in order to get the final expression of power series for the function {eq}\; \dfrac {1}{\sqrt {1 - x^{2}}} \; {/eq} and {eq}\; \arcsin(x) \; {/eq}. First of all, we will determine the power series for the function {eq}\; \dfrac {1}{\sqrt {1 + a}} \; {/eq} then we will go for the main function.

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

{eq}\\ {/eq}

Part (a)

{eq}\displaystyle F(x) = \dfrac {1}{\sqrt {1 + a}} {/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 (1 + a)^{-\dfrac {1}{2}} = 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 \dfrac {1}{\sqrt {1 + a}} = 1 - \dfrac {a}{2} + \dfrac {3a^{2}}{8} - \dfrac {5a^{3}}{16} + \cdots {/eq}

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

{eq}\displaystyle \Longrightarrow \boxed {\dfrac {1}{\sqrt {1 - x^{2}}} = 1 + \dfrac {x^{2}}{2} + \dfrac {3x^{4}}{8} + \dfrac {5x^{6}}{16} + \cdots } {/eq}

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

Part (b)

{eq}\displaystyle \dfrac {d \biggr( \arcsin(x) \biggr)}{dx} = \dfrac {1}{\sqrt {1 - x^{2}}} {/eq}

{eq}\displaystyle \int_{}^{} \; \dfrac {dx}{\sqrt {1 - x^{2}}} \; dx = \int_{}^{} \; \; \Biggr[ 1 + \dfrac {x^{2}}{2} + \dfrac {3x^{4}}{8} + \dfrac {5x^{6}}{16} + \cdots \Biggr] \; dx {/eq}

{eq}\displaystyle \arcsin(x) + \text {C} = x + \dfrac {x^{3}}{6} + \dfrac {3x^{5}}{40} + \dfrac {5x^{7}}{112} + \cdots {/eq}

Now put the value of {eq}\; x = 0 \; {/eq} in the above expression:

{eq}\displaystyle \arcsin(0) + \text {C} = 0 \; \; \; \; \Longrightarrow \text {C} = 0 {/eq}

{eq}\displaystyle \Longrightarrow \boxed {\arcsin(x) = x + \dfrac {x^{3}}{6} + \dfrac {3x^{5}}{40} + \dfrac {5x^{7}}{112} + \cdots } {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \boxed { -1 < x < 1} {/eq}