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The Maclaurin series for f(x) = \frac{x}{3 Squareroot {27 + 5x^2}}| is sigma^infinity_{n = 0} c_n...

Question:

The Maclaurin series for {eq}f(x) = \frac{x}{\sqrt[3] {27 + 5x^2}}|{/eq} is {eq}\sum^\infty_{n = 0} c_n x^n.|{/eq}

Find the first few coefficients.

{eq}C_0 =|{/eq}

{eq}c_1 =|{/eq}

{eq}c_2 =|{/eq}

{eq}c_3 =|{/eq}

{eq}c_4 =|{/eq}

{eq}c_5 =|{/eq}

The radius of convergence for the series is

Binomial Series and its Convergence:

{eq}\\ {/eq}

If {eq}k {/eq} is any number and {eq}\left| x \right| < 1 {/eq} then Binomial series expansion is given as:

{eq}{(1 + x)^k} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}} k \\ n \\ \end{array}} \right){x^n}} = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + ..... {/eq}

The above series representation is valid only when: {eq}\; \; \Longrightarrow |x| < 1 {/eq}

Answer and Explanation:

{eq}\\ {/eq}

{eq}\displaystyle f(x) = \dfrac {x}{\sqrt [3] {27 + 5x^{2}}} {/eq}

{eq}\displaystyle f(x) = \dfrac {\dfrac {x}{3}}{\sqrt [3] {1 + \dfrac {5x^{2}}{27}}} {/eq}

We know the 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}

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

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

{eq}\displaystyle \dfrac {1}{\sqrt [3] {1 + a}} = 1 - \dfrac {a}{3} + \dfrac {2a^{2}}{9} - \dfrac {14a^{3}}{81} + \cdots {/eq}

Now substitute the value of {eq}\; a = \dfrac {5x^{2}}{27} \; {/eq} in the above expression:

{eq}\displaystyle \dfrac {1}{\sqrt [3] {1 + \dfrac {5x^{2}}{27}}} = 1 - \dfrac {1}{3} \; \biggr( \dfrac {5x^{2}}{27} \biggr) + \dfrac {2}{9} \; \biggr( \dfrac {5x^{2}}{27} \biggr)^{2} - \dfrac {14}{81} \; \biggr( \dfrac {5x^{2}}{27} \biggr)^{3} + \cdots {/eq}

{eq}\displaystyle \Longrightarrow \boxed {f(x) = \dfrac {\dfrac {x}{3}}{\sqrt [3] {1 + \dfrac {5x^{2}}{27}}} = \dfrac {x}{3} \; \Biggr[ 1 - \dfrac {1}{3} \; \biggr( \dfrac {5x^{2}}{27} \biggr) + \dfrac {2}{9} \; \biggr( \dfrac {5x^{2}}{27} \biggr)^{2} - \dfrac {14}{81} \; \biggr( \dfrac {5x^{2}}{27} \biggr)^{3} + \cdots \Biggr] } {/eq}

The interval of convergence is given as: {eq}\; \; |\dfrac {5x^{2}}{27}| < 1 \; \; \; \Longrightarrow |x| < \sqrt {\dfrac {27}{5}} {/eq}

{eq}\displaystyle \Longrightarrow c_{0} = 0 \\ c_{1} = \dfrac {1}{3} \\ c_{2} = 0 \\ c_{3} = - \; \dfrac {5}{243} \\ c_{4} = 0 \\ c_{5} = \dfrac {50}{19683} {/eq}


Learn more about this topic:

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How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
9.8K

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