# 1. Use the binomial series to find the Maclaurin series of the following. f ( x ) = \frac { 1 } {...

## Question:

1. Use the binomial series to find the Maclaurin series of the following. {eq}f ( x ) = \frac { 1 } { \sqrt { 1 + x ^ { 3 } } } {/eq}

2. Use part (a) to evaluate {eq}f^{(9)} (0). {/eq}

## Higher-Order Derivative at Zero:

According to the binomial formula of {eq}(1+x)^{-m} {/eq} and the given quotient function, we'll expand the quotient expression as the power series to get the coefficient of the term {eq}x^9 {/eq}.

• {eq}\displaystyle (1+x)^{-m}=1-mx+\frac{m(m+1)}{2!}x^2-\frac{m(m+1)(m+2)}{3!}x^3+\frac{m(m+1)(m+2)(m+3)}{4!}x^4-\dots {/eq}

After that, we'll compare the obtained coefficient with the coefficient of the term {eq}x^9 {/eq} of the formula of the Maclaurin series to get the value of the given derivative function at zero.

• {eq}f(x)=\displaystyle f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f''''(0)}{4!}x^4+\frac{f'''''(0)}{5!}x^5+\frac{f''''''(0)}{6!}x^6+\frac{f'''''''(0)}{7!}x^7+\frac{f''''''''(0)}{8!}x^8+\frac{f'''''''''(0)}{9!}x^9+\dots {/eq}

1.

The given quotient function is:

{eq}\begin{align*} \displaystyle f ( x )& = \frac { 1 } { \sqrt { 1 + x ^ { 3 } } }\\ & = \frac { 1 } { (1 + x ^ { 3 })^{\frac{1}{2}} }\\ & =(1 + x ^ { 3 })^{-\frac{1}{2}}&\because \frac{1}{t^n}=t^{-n} \\ \end{align*} {/eq}

According to the standard form of function {eq}(1+x)^{-m} {/eq} in the binomial formula, we have:

{eq}x=x^3\\ m=\displaystyle \frac{1}{2} {/eq}

Substitute {eq}x=x^3 {/eq} and {eq}m=\displaystyle \frac{1}{2} {/eq} in the formula of binomial expansion and simplify each term.

{eq}\begin{align*} \displaystyle (1+x^3)^{- \frac{1}{2}}&=1- \frac{1}{2}x^3+\frac{ \frac{1}{2}( \frac{1}{2}+1)}{2!}(x^3)^2-\frac{ \frac{1}{2}( \frac{1}{2}+1)( \frac{1}{2}+2)}{3!}(x^3)^3+\frac{ \frac{1}{2}( \frac{1}{2}+1)( \frac{1}{2}+2)( \frac{1}{2}+3)}{4!}(x^3)^4-\dots\\ &=\displaystyle 1- \frac{1}{2}x^3+\frac{ \frac{1}{2}( \frac{3}{2})}{2}x^{3(2)}-\frac{ \frac{1}{2}( \frac{3}{2})( \frac{5}{2})}{3(2)(1)}x^{3(3)}+\frac{ \frac{1}{2}( \frac{3}{2})( \frac{5}{2})( \frac{7}{2})}{4(3)(2)(1)}x^{3(4)}-\dots&\because (x^m)^n=x^{mn}\\ &=\displaystyle 1- \frac{1}{2}x^3+\frac{3}{8}x^{6}-\frac{ 5}{16}x^{9}+\frac{ 35}{128}x^{12}-\dots \end{align*} {/eq}

2.

The nth derivative of a function is:

{eq}f^{(9)} (0)=? {/eq}

From part (a), we have:

{eq}\displaystyle (1+x^3)^{- \frac{1}{2}}=\displaystyle 1- \frac{1}{2}x^3+\frac{3}{8}x^{6}-\frac{ 5}{16}x^{9}+\frac{ 35}{128}x^{12}-\dots {/eq}

Comparing the coefficient of the term {eq}x^9 {/eq} of the above power series with the coefficient of the same term of the Maclaurin series, we get:

{eq}\begin{align*} \displaystyle \frac{f'''''''''(0)}{9!}&=-\frac{ 5}{16}\\ f^9(0)&=-\frac{ 5}{16}(9!)\\ &=-\frac{ 5}{16}(9(8)(7)(6)(5)(4)(3)(2)(1))&\because n!=n(n-1)(n-2)(n-3)(n-4)\dots\\ &=-5(9(4)(7)(6)(5)(3)(1))\\ &=-113400\\ \end{align*} {/eq}