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a) Use the binomial series to find the first three non-zero of the Maclaurin series of the...

Question:

a) Use the binomial series to find the first three non-zero of the Maclaurin series of the function {eq}y = \frac{1}{\sqrt{4 + t^{2}}} {/eq}.

b) Recall that

$$\frac{d}{dx} \ln(x + \sqrt{4 + x^{2}} = \frac{1}{\sqrt{4 + x^{2}}}. $$

Therefore,

$$\int_{0}^{x} \frac{1}{\sqrt{4 + t^{2}}} \, dt = \left( \ln(t + \sqrt{4 + t^{2}}) \right)_{t=0}^{x} = \ln \left( x + \sqrt{4 + x^{2}} \right) - \ln(2). $$

Find the first four non-zero terms of the Maclaurin series of the function {eq}y = \ln(x + \sqrt{4 + x^{2}}) {/eq}.

Binomial Series Representation:

{eq}\\ {/eq}

To get the series representation of the given functions, we will use the Binomial theorem for the fractional negative powers. First of all, we will convert the given function into a standard form then we will apply the Binomial theorem in order to get the series expansion and its interval of convergence.

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

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

{eq}\displaystyle \int_{}^{} \; t^{n} \; dt = \biggr( \dfrac {t^{n+1}}{n+1} \biggr) + \text {C} {/eq}

{eq}\text {C} \; \rightarrow \; \text {constant of integration} {/eq}

Answer and Explanation:

{eq}\\ {/eq}

Part (a)

{eq}\displaystyle f(t) = \dfrac {1}{\sqrt {4 + t^{2}}} \\ \displaystyle f(t) = \dfrac {\biggr( \dfrac {1}{2} \biggr)}{\sqrt {1 + \biggr( \dfrac {t}{2} \biggr)^{2}}} {/eq}

{eq}\dfrac {1}{\sqrt {1 + \biggr( \dfrac {t}{2} \biggr)^{2}}} = \Biggr[ 1 + \biggr( \dfrac {t}{2} \biggr)^{2} \Biggr]^{-\dfrac {1}{2}} {/eq}

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

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

The above series expansion holds only when: {eq}\; \; \Longrightarrow |a| < 1 {/eq}

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

Now replace the value of {eq}\; a = \biggr( \dfrac {t}{2} \biggr)^{2} \; {/eq} in the above series representation:

{eq}\displaystyle \Biggr[1 + \biggr( \dfrac {t}{2} \biggr)^{2} \Biggr]^{-\dfrac {1}{2}} = 1 - \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{2} + \biggr( \dfrac {3}{8} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{4} - \biggr( \dfrac {5}{16} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{6} + \cdots {/eq}

The above series representation holds properly when: {eq}\; \; |\biggr( \dfrac {t}{2} \biggr)^{2}| < 1 \; \; \; \Longrightarrow \; \; |t| < 2 {/eq}

Now finally, the power series representation of the main function is given as:

{eq}\displaystyle \Longrightarrow \boxed {f(t) = \dfrac {1}{\sqrt {4 + t^{2}}} = \biggr( \dfrac {1}{2} \biggr) \; \Biggr[ 1 - \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{2} + \biggr( \dfrac {3}{8} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{4} - \biggr( \dfrac {5}{16} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{6} + \cdots \Biggr] } {/eq}

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

Part (b)

{eq}\displaystyle F(x) = \int_{0}^{x} \; \dfrac {dt}{\sqrt {4 + t^{2}}} {/eq}

We know the power series representation of the function {eq}\; \biggr( \dfrac {1}{\sqrt {4 + t^{2}}} \biggr) \; {/eq}:

{eq}\displaystyle F(x) = \biggr( \dfrac {1}{2} \biggr) \; \int_{0}^{x} \; \Biggr[ 1 - \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{2} + \biggr( \dfrac {3}{8} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{4} - \biggr( \dfrac {5}{16} \biggr) \; \biggr( \dfrac {t}{2} \biggr)^{6} + \cdots \Biggr] \; dt {/eq}

{eq}\displaystyle F(x) = \biggr( \dfrac {1}{2} \biggr) \; \Biggr[ t - \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {t^{3}}{2^{2} \times 3} \biggr) + \biggr( \dfrac {3}{8} \biggr) \; \biggr( \dfrac {t^{5}}{2^{4} \times 5} \biggr) - \biggr( \dfrac {5}{16} \biggr) \; \biggr( \dfrac {t^{7}}{2^{6} \times 7} \biggr) \Biggr]_{t = 0}^{t = x} {/eq}

{eq}\displaystyle F(x)= \int_{0}^{x} \; \dfrac {dt}{\sqrt {4 + t^{2}}} = \biggr( \dfrac {1}{2} \biggr) \; \Biggr[ x - \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {x^{3}}{2^{2} \times 3} \biggr) + \biggr( \dfrac {3}{8} \biggr) \; \biggr( \dfrac {x^{5}}{2^{4} \times 5} \biggr) - \biggr( \dfrac {5}{16} \biggr) \; \biggr( \dfrac {x^{7}}{2^{6} \times 7} \biggr) \Biggr] {/eq}

{eq}\displaystyle \int_{0}^{x} \; \dfrac {dt}{\sqrt {4 + x^{2}}} = \ln \Biggr[x + \sqrt {4 + x^{2}} \Biggr] - \ln(2) {/eq}

{eq}\displaystyle \ln \Biggr[ x + \sqrt {4 + x^{2}} \Biggr] - \ln(2) = \biggr( \dfrac {1}{2} \biggr) \; \Biggr[ x - \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {x^{3}}{2^{2} \times 3} \biggr) + \biggr( \dfrac {3}{8} \biggr) \; \biggr( \dfrac {x^{5}}{2^{4} \times 5} \biggr) - \biggr( \dfrac {5}{16} \biggr) \; \biggr( \dfrac {x^{7}}{2^{6} \times 7} \biggr) \Biggr] {/eq}

{eq}\displaystyle \Longrightarrow \boxed {\ln \Biggr[x + \sqrt {4 + x^{2}} \Biggr] = \ln(2) + \biggr( \dfrac {1}{2} \biggr) \; \Biggr[ x - \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {x^{3}}{2^{2} \times 3} \biggr) + \biggr( \dfrac {3}{8} \biggr) \; \biggr( \dfrac {x^{5}}{2^{4} \times 5} \biggr) - \biggr( \dfrac {5}{16} \biggr) \; \biggr( \dfrac {x^{7}}{2^{6} \times 7} \biggr) \Biggr] } {/eq}

The above series expansion is valid only when: {eq}\; \; \Longrightarrow |x| < 2 {/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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