Find the first four terms of the bionimial series of the given functions \left ( 1+\frac{x}{10}...

Question:

Find the first four terms of the bionimial series of the given functions

{eq}\left ( 1+\frac{x}{10} \right )^{-2} {/eq}

{eq}\left ( 1-\frac{x}{7} \right )^{1/3} {/eq}

{eq}\left ( 1+\frac{x}{7} \right )^{3} {/eq}

{eq}\left ( 1-\frac{x}{3} \right )^{-2/3} {/eq}

Standard Function of Binomial Rule:

There are different types of functions in this particular problem such as {eq}\displaystyle (1+\alpha )^{-m},\ (1-\alpha)^m,\ (1+\alpha)^{m} {/eq} and {eq}(1-\alpha)^{-m} {/eq}. For all these functions or first four terms of these function, we'll use the binomial formula of standard function {eq}(1+x)^m {/eq} by comparing the values of {eq}x {/eq} and {eq}m {/eq} with the given functions.

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

Answer and Explanation:

The given algebraic function is:

{eq}\displaystyle \left ( 1+\frac{x}{10} \right )^{-2} {/eq}

According to the standard function {eq}(1+x)^m {/eq}, we have:

{eq}x=\displaystyle \frac{x}{10}\\ m=-2 {/eq}


Substituting the values of varaible {eq}x {/eq} and exponent {eq}m {/eq} in the formula of binomial expansion, we get:

{eq}\begin{align*} \displaystyle (1+\frac{x}{10} )^{-2}&=1+(-2)\frac{x}{10}+\frac{-2(-2-1)}{2!}\left (\frac{x}{10} \right )^2+\frac{-2(-2-1)(-2-2)}{3!}\left (\frac{x}{10} \right )^3+\dots\\ &=\displaystyle 1-\frac{x}{5}+\frac{-2(-3)}{2(1)}\left (\frac{x^2}{100}\right)+\frac{-2(-3)(-4)}{3(2)(1)}\left (\frac{x^3}{1000} \right )+\dots&\because \left ( \frac{p}{q} \right )^n=\frac{p^n}{q^n}\\ &=\displaystyle 1-\frac{x}{5}+\frac{3}{100}x^2-\frac{1}{250}x^3+\dots&\because n!=n(n-1)(n-2)\dots\\ \end{align*} {/eq}


The algebraic function with positive fractional exponent is:

{eq}\displaystyle \left ( 1-\frac{x}{7} \right )^{1/3} {/eq}

Comparing the above function with standard form of binomial rule, we get:

{eq}\displaystyle x=-\frac{x}{7}\\ \displaystyle m=\frac{1}{3} {/eq}


To get the first four non-zero terms of the series expansion of the function {eq}\displaystyle \left ( 1-\frac{x}{7} \right )^{1/3} {/eq}, we'll substitute the above values in the binomial formula.

{eq}\begin{align*} \displaystyle (1-\frac{x}{7} )^{\frac{1}{3}}&=1+\left ( \frac{1}{3} \right )\left (-\frac{x}{7} \right )+\frac{\frac{1}{3}\left (\frac{1}{3}-1\right)}{2!}\left (-\frac{x}{7} \right )^2+\frac{\frac{1}{3}\left (\frac{1}{3}-1\right)\left (\frac{1}{3}-2\right)}{3!}\left (-\frac{x}{7} \right )^3+\dots\\ &=1-\frac{x}{21} +\frac{\frac{1}{3}\left (-\frac{2}{3}\right)}{2(1)}\left (\frac{x^2}{49} \right )+\frac{\frac{1}{3}\left (-\frac{2}{3}\right )\left (-\frac{5}{3}\right)}{3(2)(1)}\left (-\frac{x^3}{343} \right )+\dots&\because \left ( \frac{p}{q} \right )^n=\frac{p^n}{q^n}\\ &=1-\frac{x}{21} -\frac{1}{441}x^2+\frac{5}{81}\left (-\frac{x^3}{343} \right )+\dots\\ &=1-\frac{x}{21} -\frac{1}{441}x^2-\frac{5}{27783}x^3+\dots \end{align*} {/eq}


The given function with positive exponent is:

{eq}\displaystyle \left ( 1+\frac{x}{7} \right )^{3} {/eq}

Here, we have:

{eq}x=\displaystyle \frac{x}{7} \\ m=3 {/eq}


To get the first four non-zero terms of the series expansion of the function {eq}\displaystyle \left ( 1+\frac{x}{7} \right )^{3} {/eq}, we'll substitute the above values in the binomial formula.

{eq}\begin{align*} \displaystyle (1+\frac{x}{7} )^{3}&=1+\left ( 3 \right )\left (\frac{x}{7} \right )+\frac{3\left (3-1\right)}{2!}\left (\frac{x}{7} \right )^2+\frac{3\left (3-1\right)\left (3-2\right)}{3!}\left (\frac{x}{7} \right )^3+\dots\\ &=1+\frac{3}{7}x+\frac{3\left (2\right)}{2}\left (\frac{x^2}{49} \right )+\frac{3\left (2\right)\left (1\right)}{3(2)(1)}\left (\frac{x^3}{7^3} \right )+\dots\\ &=1+\frac{3}{7}x+\frac{3}{49}x^2+\frac{1}{343}x^3+\dots\\ \end{align*} {/eq}


Given:

{eq}\displaystyle \left ( 1-\frac{x}{3} \right )^{-2/3} {/eq}

Comparing the above function with the standard form, we get:

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


Substituting the above values after comparison in the binomial formula, we get:

{eq}\begin{align*} \displaystyle (1-\frac{x}{3} )^{-\frac{2}{3}}&=1+\left ( -\frac{2}{3} \right )\left (-\frac{x}{3} \right )+\frac{-\frac{2}{3}\left (-\frac{2}{3}-1\right)}{2!}\left (-\frac{x}{3} \right )^2+\frac{-\frac{2}{3}\left (-\frac{2}{3}-1\right)\left (-\frac{2}{3}-2\right)}{3!}\left (-\frac{x}{3}\right )^3+\dots\\ &=1+\frac{2}{9}x+\frac{-\frac{2}{3}\left (-\frac{5}{3}\right)}{2}\left (\frac{x^2}{9} \right )+\frac{-\frac{2}{3}\left (-\frac{5}{3}\right)\left (-\frac{8}{3}\right)}{3(2)(1)}\left (-\frac{x^3}{27}\right )+\dots\\ &=1+\frac{2}{9}x+\frac{5}{9}\left (\frac{x^2}{9} \right )+\frac{-40}{81}\left (-\frac{x^3}{27}\right )+\dots\\ &=1+\frac{2}{9}x+\frac{5}{81}x^2+\frac{40}{2187}x^3+\dots\\ \end{align*} {/eq}


Learn more about this topic:

Loading...
The Binomial Theorem: Defining Expressions

from Algebra II: High School

Chapter 12 / Lesson 7
11K

Related to this Question

Explore our homework questions and answers library