# Taylor Series Expansion of Sqrt(x) at a=4

• Apr 10th 2013, 10:44 PM
EliteAndoy
Taylor Series Expansion of Sqrt(x) at a=4
Hi Everyone! Today we are asked to find a Taylor Series for $\displaystyle \sqrt{x}$ at about its center, a=4. I've been working on this one problem for hours now, and I feel so dumb not being able to find a Taylor series for it. So far I've been working on $\displaystyle f^n(x)$formulas. So far, the nth derivatives goes like this:$\displaystyle \frac{1}{4},\frac{-1}{32},\frac{3}{256},\frac{-15}{2048},\frac{105}{16384}...$. The easy part is that the denominator is a base 2 power function and can be given by $\displaystyle 2^{3n-1}$. The alternating signs can be given by $\displaystyle (-1)^{n+1}$. But the numerator is the real problem. It can be given by 1,1,1*3,1*3*5,1*3*5*7... which is somewhat equal to simply $\displaystyle \frac{(2n-1)!}{2^{n-1}(n-1)!}$. But the problem is, instead of the numerator going: 1,3 ,15,105... it instead goes: 1,1,3,15,105...
That first numerator is giving me problems. Anyone got any ideas on how I should proceed with this or at least, how to give find the Taylor series for this function? Thanks everyone in advance.(Nod)
• Apr 11th 2013, 12:09 AM
Gusbob
Re: Taylor Series Expansion of Sqrt(x) at a=4
Write out explicitly the first term, then write a general equation for the rest. To see what I mean, consider

$\displaystyle e^x = \displaystyle{\sum_{n=0}^{\infty}\frac{x^n}{n!}}=1 + \displaystyle{\sum_{n=1}^{\infty}\frac{x^n}{n!}}$

or even

$\displaystyle 5+e^x = 5+ \displaystyle{\sum_{n=0}^{\infty}\frac{x^n}{n!}}=6 + \displaystyle{\sum_{n=1}^{\infty}\frac{x^n}{n!}}$
• Apr 11th 2013, 12:19 AM
EliteAndoy
Re: Taylor Series Expansion of Sqrt(x) at a=4
If you were referring for the final equation then yes, I've already figured that out. But my main problem, like I said, is not being able to represent the formula for nth derivative since the first derivative doesn't correspond with the formula that I came up with. If for some reason I took your idea wrong, please enlighten me more by elaborating on your solution. :D
• Apr 11th 2013, 01:24 AM
EliteAndoy
Re: Taylor Series Expansion of Sqrt(x) at a=4
Just figured it out right about now. The first two 1's in the numerator are from a phenomenon called double factorial. $\displaystyle \frac{(2n)!}{2^{n}n!(2n-1)}$ worked perfectly to define the behavior of the numerators. I'd explain everything tomorrow as I have to finish this work and go to sleep. Thank you guys for all your efforts. :D
• Apr 12th 2013, 04:42 AM
EliteAndoy
Re: Taylor Series Expansion of Sqrt(x) at a=4
Just as promised, here's my derivation of the for the formula:

$\displaystyle a_n=1,1,3,5,15,105...$

1) To start this problem, let's first take a look at the definition of $\displaystyle n!!$:
$\displaystyle n!!=\begin{cases} & \n(n-2)\cdot5\cdot3\cdot1 ,n>0(odd)\\ & \n(n-2)\cdot6\cdot4\cdot2, n>0(even)\\ & \1,n=-1,0 \end{cases}$
From that definition, we can say that if n is odd, n should be included in the product of non negative odd numbers that is an element of natural numbers.

2) For now, let's disregards the fact that we have to satisfy the other terms of the sequence and focus on the first two numbers. By the definition of $\displaystyle n!!$, then it should be true that $\displaystyle (2n-3)!!$ should satisfy the sequence $\displaystyle a_n=1,1,3,5,15,105...$. Bu the problem with$\displaystyle (2n-3)!!$ is that it is not defined when n=3. So what we do is try to find the nearest odd number alternative for $\displaystyle (2n-3)!!$ that will still satisfy the first two terms of the sequence. My first guess is $\displaystyle (2n-1)!!$. But how does that relate to $\displaystyle (2n-3)!!$?

$\displaystyle (2n-3)!!=(2n-3)\cdot(2n-5)\cdot(2n-7)\cdot(2n-9).....$

$\displaystyle (2n-1)!!=(2n-1)\cdot2n-3)\cdot(2n-5)\cdot(2n-7)\cdot(2n-9).....$

Notice that if we divide $\displaystyle (2n-1)!!$ by $\displaystyle 2n-1$, then it should be identical as $\displaystyle (2n-3)!!$. By that, then it should be true to say that:
$\displaystyle 2n-3)!!=\frac{(2n-1)!!}{(2n-1)}$

3) That's basically it, but how the hell are we supposed to relate that to factorials? Well there's actually a double factorial identity that relates double factorials to single factorials. That is usually expressed as:
$\displaystyle (2n-1)!!\cdot(2^n)(n!)=(2n)!$
$\displaystyle (2n-1)!!=\frac{(2n)!}{2^nn!}$
Now, recall that $\displaystyle (2n-3)!!=\frac{(2n-1)!!}{2n-1}$
So by substituting:
$\displaystyle (2n-3)!!=\frac{(2n)!}{2^nn!\cdot(2n-1)}$
And that's the formula for the sequence $\displaystyle a_n=1,1,3,5,15,105...$(Clapping)
So that:
$\displaystyle a_n=\frac{(2n)!}{2^nn!\cdot(2n-1)}$