# Set up Taylor expansion \bb\color(red)\text(formula) for sqrt(x) around a=2?

## Check my work please? $f ' = \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}}$ $f ' ' = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) {x}^{- \frac{3}{2}}$ $f ' ' ' = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) {x}^{- \frac{5}{2}}$ ${f}^{4} = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) {x}^{- \frac{7}{2}}$ ${f}^{5} = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) \left(- \frac{7}{2}\right) {x}^{- \frac{9}{2}}$ $f ' \left(2\right) = \left(\frac{1}{2}\right) \cdot {2}^{- \frac{1}{2}}$ $f ' ' \left(2\right) = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \cdot {2}^{- \frac{3}{2}}$ $f ' ' \left(2\right) ' = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \cdot {2}^{- \frac{5}{2}}$ ${f}^{4} \left(2\right) = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) \cdot {2}^{- \frac{7}{2}}$ ${f}^{5} \left(2\right) = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) \left(- \frac{7}{2}\right) \cdot {2}^{- \frac{9}{2}}$ Okay, I don't know where to go from here. but I DO know that the exponents increase in this order... $\left(- \frac{1}{2}\right) + {\left(- 1\right)}^{\setminus} \beta$ (not sure if $\setminus \beta = n$ yet)

May 6, 2018

sqrtx=sqrt2+sum_(n=1)^oo(-1)^(n+1)(x-2)^n(2n-1)!!2^(1/2-2n)/(n!)

#### Explanation:

For this function, finding the $n t h$ derivative pattern will be a bit tougher, but it's possible.

We will not evaluate the derivatives at $a = 2$ just yet. We'll focus on finding ${f}^{\left(n\right)} \left(x\right) :$

$f \left(x\right) = {x}^{\frac{1}{2}} , f \left(2\right) = \sqrt{2}$

$f ' \left(x\right) = \frac{1}{2} {x}^{- \frac{1}{2}}$

$f ' ' \left(x\right) = - \frac{1}{2 \cdot 2} {x}^{- \frac{3}{2}}$

$f ' ' ' \left(x\right) = \frac{- 1 \cdot - 3}{{2}^{3}} {x}^{- \frac{5}{2}}$

${f}^{\left(4\right)} \left(x\right) = \frac{- 1 \cdot - 3 \cdot - 5}{{2}^{4}} {x}^{- \frac{7}{2}}$

If we look only at the derivatives and disregard $f \left(x\right) = {x}^{\frac{1}{2}} ,$ we have a pattern going.

That is, we start with $n = 1.$

The first derivative has no negative sign, the second derivative is negative, and this pattern continues for the following derivatives. Since we start at $n = 1 ,$ we can represent this with ${\left(- 1\right)}^{n + 1}$.

The exponent on the $x$ starts with $- \frac{1}{2} ,$ and decreases by $1$ for every following derivative. We can represent this with ${x}^{\frac{1}{2} - n} , n \ge 1.$

In the denominator, we ${2}^{n} .$

Now, for the numerator, we see we have the pattern $1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot \left(2 n - 1\right)$, that is, we multiply only odd numbers, starting at $1.$

We can represent this with the double factorial, (2n-1)!!, which indicates multiplying by integers that have the same parity (oddness or evenness) as $n$

Thus,

f^((n))(x)=(-1)^(n+1)(2n+1)!!x^(1/2-n)/(2^n)

f^((n))(2)=(-1)^(n+1)(2n+1)!!2^(1/2-n)/(2^n)

${2}^{\frac{1}{2} - n} / \left({2}^{n}\right) = {2}^{\frac{1}{2} - n - n} = {2}^{\frac{1}{2} - 2 n}$

f^((n))(2)=(-1)^(n+1)(2n+1)!!2^(1/2-2n)

Now, we use the general form for a Taylor series centered at $a$

f(x)=sum_(n=0)^oof^((n))(a)(x-a)^n/(n!)

to see that

sqrtx=sqrt2+sum_(n=1)^oo(-1)^(n+1)(x-2)^n(2n-1)!!2^(1/2-2n)/(n!)

Note that we did not include $f \left(2\right) = \sqrt{2} ,$ the $0 t h$ term, in our series, as we couldn't get it to originally fit in with our pattern.

We can't simplify the factorials any further.