Jtdylktuy

$$f(x) = sqrt{x-1}$$ about $a=5$

So I found the polynomial, but getting it into a series is where I'm having trouble. The first two terms are positive, and then it alternates.
The $C_n$ term is $1$ divided by some power of $2$. That power goes $0, 2, 6, 9, 14, dots $

Any suggestions?

I can't post images apparently, but it's

$$2 + frac{1}{4}(x-5) - frac{1}{64}(x-5)^2 + frac{1}{512}(x-5)^3 - frac{5}{16384}(x-5)^4 + cdots$$

What I have so far:

$$sum_{n=0}^infty frac{(-1)^n(x-5)^n}{n!}$$

Let us forget the value of $5$ and be as lazy as I am.

Let $x=y+a$ to make
$$sqrt{x-1}=sqrt{y+(a-1)}=sqrt{(a-1)},sqrt{1+frac{y}{a-1}}$$ Now, let $z=frac{y}{a-1}$ and use the binomial expansion for $sqrt{1+z}=sum_{k=0}^inftybinom{1/2}{k}z^k$. Then, replace. So, around $x=a$,
$$sqrt{x-1}=sqrt{(a-1)},sum_{k=0}^inftybinom{frac 12}{k}left(frac{y}{a-1}right)^k=sum_{k=0}^inftyfrac{binom{frac 12}{k}}{(a-1)^{k-frac 12} }(x-a)^k $$

If you really want the specific closed form for the series, you need to express the $n$th derivative of $f$ evaluated at $x=5$ for all $n$.

Since $f(x)=(x-1)^{frac{1}{2}}$, we know that $f'(x)=frac{1}{2}(x-1)^{-frac{1}{2}}$. Note the change: the power on $(x-1)$ drops by 1, and you multiply on the current power.

This means the $n$th derivative of $f$, for $n>1$ is
$$begin{align*}
f^{(n)}(x)&=frac{1}{2}left(-frac{1}{2}right)dotscleft(frac{3}{2}-nright)(x-1)^{frac{1}{2}-n}\
&=frac{1}{2^n}left(1cdot(-1)cdot(-3)cdotdotsccdot(-(2n-3))right)(x-1)^{frac{1}{2}-n}\
&=frac{(-1)^{n-1}}{2^n}(1cdot3cdotdotsccdot(2n-3))(x-1)^{frac{1}{2}-n}\
&=-left(-frac{1}{2}right)^n(2n-3)!!(x-1)^{frac{1}{2}-n}
end{align*}$$
where $!!$ is the double factorial.

Now substitute $x=5$ to get $f(5)=4$, $f'(5)=1/4$, and
$$begin{align*}
f^{(n)}(5)&=-left(-frac{1}{2}right)^n(2n-3)!!cdot4^{frac{1}{2}-n}\
&=-2left(-frac{1}{8}right)^n(2n-3)!!
end{align*}$$

Putting that into the Taylor expansion gives
$$f(x)=4+frac{1}{4}(x-5)-2sum_{n=2}^{infty}frac{(-1)^n(2n-3)!!}{8^nn!}(x-5)^n$$

Note: if you really dislike the double factorial, rewrite it as
$$begin{align*}
(2n-1)!!&=1cdot3cdotdotsccdot(2n-1)\
&=frac{1cdot2cdot3cdotdotsccdot2n}{2cdot4cdot6cdotdotsccdot2n}\
&=frac{(2n)!}{2^nn!}
end{align*}$$

We know that taylor expansion is given as
begin{equation}
f(x) =
f(a)+frac {f'(a)}{1!} (x-a)+ frac{f''(a)}{2!} (x-a)^2+frac{f'''(a)}{3!}(x-a)^3+ +frac{f''''(a)}{4!}(x-a)^4+ cdots
end{equation}
In your case $a=5$, so
begin{equation}
f(x) =
f(5)+frac {f'(5)}{1!} (x-5)+ frac{f''(5)}{2!} (x-5)^2+frac{f'''(5)}{3!}(x-5)^3+ frac{f''''(5)}{4!}(x-5)^4+ cdots
end{equation}
Let's get all 4 derivatives

begin{align}
f'(x) &= dfrac{1}{2sqrt{x-1}}\
f''(x) &= -dfrac{1}{4left(x-1right)^frac{3}{2}}\
f'''(x) &= dfrac{3}{8left(x-1right)^frac{5}{2}}\
f''''(x) &= -dfrac{15}{16left(x-1right)^frac{7}{2}}
end{align}
Replacing for $x = a = 5$, we get
begin{align}
f(5) &= 2\
f'(5) &= frac{1}{4}\
f''(5) &= -frac{1}{32}\
f'''(5) &= frac{3}{256}\
f''''(5) &= -frac{15}{2048}
end{align}
Replacing we get
begin{equation}
f(x) =
2+ frac{1}{4} (x-5) -frac{1}{32(2)}(x-5)^2+frac{3}{3!(256)}(x-5)^3-frac{15}{4!(2048)}(x-5)^4+ cdots
end{equation}
that is
begin{equation}
f(x) =
2+ frac{1}{4} (x-5) -frac{1}{64}(x-5)^2+frac{1}{512}(x-5)^3-frac{15}{4!(2048)}(x-5)^4+ cdots
end{equation}

In General

We notice that for $n > 2$
begin{equation}
f^{(n)}(x)
=
(-1)^{n+1}
frac{prod_{k=1}^{n-2} (2k+1)}{2^{n}(x-1)^{frac{2n-1}{2}}}
end{equation}
so for $n>2$
begin{equation}
f^{(n)}(5)
=
(-1)^{n+1}
frac{prod_{k=1}^{n-2} (2k+1)}{2^{n}(5-1)^{frac{2n-1}{2}}}
=
(-1)^{n+1}
frac{prod_{k=1}^{n-2} (2k+1)}{2^{n}(4)^{frac{2n-1}{2}}}
=
(-1)^{n+1}
frac{prod_{k=1}^{n-2} (2k+1)}{2^{n}(2)^{2n-1}}
end{equation}
Replacing we get
begin{equation}
f(x)
=
2+
frac{1}{4}(x-5)
-
frac{1}{64}(x-5)^2
+
sum_{n=3}^{infty}
(-1)^{n+1}frac{prod_{k=1}^{n-2} (2k+1)}{underbrace{2^{n}(2)^{2n-1}}_{2^{3n-1}}n!}(x-5)^n
end{equation}