I am trying to find the Taylor series for $\sqrt{x}$ centered at $a=16$.
I begin by finding some derivatives so that i can work out a formula for the $n_{th}$ derivative as the Taylor series for a function is given by:
$$f(x) = \sum_{n=0}^\infty \frac{f^{n}(a)}{n!}(x-a)^n$$
I obtained the following for the first four derivatives:
$$f'(x)=\frac{1}{2}x^{-\frac{1}{2}}$$
$$f^2(x) = \frac{1}{2}.\frac{-1}{2}x^{-\frac{3}{2}}$$
$$f^3(x)=\frac{1}{2}.\frac{-1}{2}.\frac{-3}{2}x^{-\frac{5}{2}}$$
$$f^4(x)=\frac{1}{2}.\frac{-1}{2}.\frac{-3}{2}.\frac{-5}{2}x^{-\frac{7}{2}}$$
from this i have determined that successive terms alternate in signs beginning with a positive, so i might say that $f^n(a) = (-1)^{n+1}$ and checking for n=1 results in a positive, n=2 negative etc. Next, it is easily seen that success derivatives have the pattern $\frac{1}{2^n}$, the powers of x, $\text{}$ $x^{\frac{-2n+1}{2}}$
I am having difficulty obtaining a pattern for $1.1.3.5.7…$
I notice that sometimes in these problems you can only get a pattern for the derivatives from say n=2, or n=3. My instinct says knock off the first derivative and obtain a pattern for n=2 onwards then we have $1.3.5.7 $
I've tried a couple of things using $Tn=a+(n-1)d$ to get the pattern but i'm not having much luck. Could someone provide me some tips on how to determine this pattern and also how to know when you have to 'knock off a term', say only take a pattern from n=2?
edit
In order to determine the pattern you can do the following. given a sequence that starts at n=2 and is $3.7.11…$ for example. Take the difference between the terms, so 4. then multiply by n: $4n$ now substitute values of n to see what this gives and compare to what is expected. so for $n=2$, $4(2)=8$ but we want $3$ so we need to subtract 5. Likewise for n=3, $4(3)=12$ but we want 7 so subtracting 5 will give 7. Hence the pattern is $3.7.11…(4n-5)$
Best Answer
Note that by the extended binomial theorem, for $|z|\leq 1$, $$\sqrt{1+z}=\sum_{n=0}^{\infty}\binom{1/2}{n}z^n$$ where $\binom{1/2}{0}=1$, $\binom{1/2}{1}=1/2$ and for $n>1$, $$\binom{1/2}{n}=\frac{1}{n!}\prod_{k=0}^{n-1}\left(\frac{1}{2}-k\right)= \frac{(-1)^{n-1}(1\cdot3\cdot5\cdot7\cdots(2n-3))}{n!2^n} =\frac{(-1)^{n-1}\binom{2n}{n}}{(2n-1)4^n}.$$ Hence for $|x-16|\leq 16$, $$\sqrt{x}=\sqrt{16+(x-16)}=4\sqrt{1-\frac{(x-16)}{16}}=4\sum_{n=0}^{\infty}\frac{\binom{1/2}{n}}{16^n}(x-16)^n\\ =\sum_{n=0}^{\infty}\frac{(-1)^{n-1}\binom{2n}{n}}{(2n-1)4^{3n-1}}(x-16)^n.$$