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The Maclaurin series for f(x) = x / cubic root 27 + 4x^2 is sigma_{n = 0}^infinity c_n x^n. Find...

Question:

The Maclaurin series for {eq}\displaystyle f(x) = \frac{x} {\sqrt [3] {27 + 4x^2}} {/eq} is {eq}\displaystyle \sum_{n = 0}^\infty c_n x^n {/eq}.

Find the first few coefficients.

(a) {eq}c_0 = \_\_\_\_\_\_\_\_ {/eq}

(b) {eq}c_1 = \_\_\_\_\_\_\_\_ {/eq}

(c) {eq}c_2 = \_\_\_\_\_\_\_\_ {/eq}

(d) {eq}c_3 = \_\_\_\_\_\_\_\_ {/eq}

(e) {eq}c_4 = \_\_\_\_\_\_\_\_ {/eq}

(f) {eq}c_5 = \_\_\_\_\_\_\_\_\_ {/eq}

The radius of convergence for the series is {eq}\_\_\_\_\_\_\_\_ {/eq}.

Binomial Series Expansion:

To form the power series representation of the function about the point {eq}\;x = 0 \; {/eq}, we will use the method of the Binomial theorem for the fractional negative powers. First of all, we will determine the power series for the standard function then we will go for the whole function.

{eq}\displaystyle (1 + b)^{n} = 1 + nb + \dfrac {n(n -1)}{2!} \; b^{2} + \dfrac {n(n -1)(n -2)}{3!} \; b^{3} + \cdots {/eq}

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

Answer and Explanation:

{eq}\displaystyle f(x) = \dfrac {x}{\sqrt [3] {27 + 4x^{2}}} {/eq}

{eq}\displaystyle f(x) = x \; \biggr(27 + 4x^{2} \biggr)^{-\dfrac {1}{3}} = \biggr( \dfrac {x}{3} \biggr) \; \biggr(1 + \dfrac {4x^{2}}{27} \biggr)^{-\dfrac {1}{3}} {/eq}

We know the standard Binomial series expansion for the fractional negative powers:

{eq}(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 (1 + a)^{-\dfrac {1}{3}} = 1 - \dfrac {a}{3} + \dfrac {2a^{2}}{9} - \dfrac {14a^{2}}{81} + \dfrac {35}{243} \; a^{3} \cdots {/eq}

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

Now replace the value of {eq}\; a = \biggr( \dfrac {4x^{2}}{27} \biggr) \; {/eq} in the above expression:

{eq}\displaystyle \biggr(1 + \dfrac {4x^{2}}{27} \biggr)^{-\dfrac {1}{3}} = 1 - \dfrac {1}{3} \; \biggr( \dfrac {4x^{2}}{27} \biggr) + \dfrac {2}{9} \; \biggr( \dfrac {4x^{2}}{27} \biggr)^{2} - \dfrac {14}{81} \; \biggr( \dfrac {4x^{2}}{27} \biggr)^{3} + \dfrac {35}{243} \; \biggr( \dfrac {4x^{2}}{27} \biggr)^{4} + \cdots {/eq}

The above series representation is valid only when: {eq}\; \; |\biggr( \dfrac {4x^{2}}{27} \biggr)| < 1 \; \; \; \Longrightarrow |x| < \dfrac {2}{3 \sqrt {3}} {/eq}

Finally, the power series representation of the function is given as:

{eq}\displaystyle \Longrightarrow \boxed {f(x) = \dfrac {x}{\sqrt [3] {27 + 4x^{2}}} = \dfrac {x}{3} \; \Biggr[ 1 - \dfrac {1}{3} \; \biggr( \dfrac {4x^{2}}{27} \biggr) + \dfrac {2}{9} \; \biggr( \dfrac {4x^{2}}{27} \biggr)^{2} - \dfrac {14}{81} \; \biggr( \dfrac {4x^{2}}{27} \biggr)^{3} + \dfrac {35}{243} \; \biggr( \dfrac {4x^{2}}{27} \biggr)^{4} + \cdots\Biggr]} {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow |x| < \dfrac {2}{3\sqrt {3}} {/eq}

{eq}\Longrightarrow \boxed {c_{0} = 0, \; \; c_{1} = \dfrac {1}{3}, \; \; c_{2} = 0, \; \; c_{3} = - \; \dfrac {4}{243}, \; \; \text {and} \; \; c_{5} = \dfrac {32}{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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