# The Taylor series for f(x) = sqrt(100+x); at; a=0 is sum n=0 infty cn(x)n. c_1 = 10 c_2 = 1/20...

## Question:

The Taylor series for {eq}f(x) = \sqrt(100+x)\; at\; a=0 {/eq} is {eq}\sum_{n=0}^{\infty} \; c_{n} x^{n}. \\ c_1 = 10\\ c_2 = 1/20\\ c_3 = -1/8000\\ c_4 = 1/600000 {/eq}.

Find the error in approximating {eq}\sqrt {101} = f(1) {/eq} using the third degree Taylor polynomial of f at a = 0. That is find the error of the approximation {eq}\sqrt {(101)}01 \approx 73(1) {/eq}.

## Taylor Series Representation:

{eq}\\ {/eq}

To get the required power series representation for the given function, we will use the Binomial series expansion for the fractional powers. First of all, we will convert the given function into the standard form then with the help of binomial theorem, we will determine the series representation 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 expansion holds only when: {eq}\; \; \Longrightarrow |a| < 1 {/eq}

{eq}\\ {/eq}

{eq}f(x) = \sqrt {x + 100} {/eq}

{eq}\Longrightarrow f(x) = 10 \; \Biggr[1 + \biggr( \dfrac {x}{100} \biggr) \Biggr]^{\dfrac {1}{2}} {/eq}

We know the Binomial series representation for the fractional powers:

{eq}(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 holds good only when: {eq}\; \; \Longrightarrow |a| < 1 {/eq}

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

Now replace the value of {eq}\; a = \biggr( \dfrac {x}{100} \biggr) \; {/eq} in the above expression:

{eq}\displaystyle \sqrt {1 + \biggr( \dfrac {x}{100} \biggr)} = 1 + \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {x}{100} \biggr) - \biggr( \dfrac {1}{8} \biggr) \; \biggr( \dfrac {x}{100} \biggr)^{2} + \biggr( \dfrac {1}{16} \biggr) \; \biggr( \dfrac {x}{100} \biggr)^{3} + \cdots {/eq}

{eq}\displaystyle \Longrightarrow \boxed {f(x) = \sqrt {x + 100} = 10 \; \Biggr[ 1 + \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {x}{100} \biggr) - \biggr( \dfrac {1}{8} \biggr) \; \biggr( \dfrac {x}{100} \biggr)^{2} + \biggr( \dfrac {1}{16} \biggr) \; \biggr( \dfrac {x}{100} \biggr)^{3} + \cdots \Biggr]} {/eq}

{eq}c_{0} = 10 \\ c_{1} = 10 \times \biggr( \dfrac {1}{2 \times 100} \biggr) = \dfrac {1}{20} \\ c_{2} = 10 \times \Biggr[ - \; \biggr( \dfrac {1}{8} \biggr) \times \biggr( \dfrac {1}{100} \biggr)^{2} \Biggr] = - \; \dfrac {1}{8000} \\ c_{3} = 10 \times \Biggr[ \biggr( \dfrac {1}{16} \biggr) \times \biggr( \dfrac {1}{100} \biggr)^{3} \Biggr] = \dfrac {1}{600000} {/eq}

The above series representation holds properly only when: {eq}\; \; |\biggr( \dfrac {x}{100} \biggr)| < 1 \; \; \; \Longrightarrow |x| < 100 {/eq}

{eq}\displaystyle \Longrightarrow \boxed {f(1) = \sqrt {100 + 1} = \sqrt {101} \approx 10 \; \Biggr[ 1 + \biggr( \dfrac {1}{2} \biggr) \; \biggr( \dfrac {1}{100} \biggr) - \biggr( \dfrac {1}{8} \biggr) \; \biggr( \dfrac {1}{100} \biggr)^{2} + \biggr( \dfrac {1}{16} \biggr) \; \biggr( \dfrac {1}{100} \biggr)^{3} \Biggr] \approx 10.04987} {/eq}

{eq}{/eq}