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I found the first four derivatives and have the first few terms of the series. I'm getting stuck with putting it into summation notation.

$1+\sum\limits_{k=1}^{\infty}\dfrac {(-1)^{k+1}(x-3)^{k}}{k!2^k}$

That is the closest I have come to how to write it.. it takes care of everything except for the numerator coefficients, I think. Can anyone help me out on how to add that in to the summation? The coefficients are 1, 1, 3, 15, 105, ... (I factored out the first term so I could write it as an alternating series).

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  • $\begingroup$ The coefficients should be $ 1, \frac{1}{2}, -\frac{1}{8}, \frac{1}{16}, -\frac{5}{128}, \frac{7}{256}, -\frac{21}{1024}, ...$ $\endgroup$ – Dando18 May 11 '17 at 0:45
  • $\begingroup$ Oops, I was talking about the numerator parts of the coefficients and I didn't reduce them because I'm looking for a pattern in the numerator since k!*2^k works for the denominator. $\endgroup$ – calmcalculus May 11 '17 at 0:50
  • $\begingroup$ For $2 < x < 4$ we have $$ \sqrt{x - 2} = \sum_{n=0}^\infty (x-3)^n \binom{1/2}{n} $$ $\endgroup$ – Dando18 May 11 '17 at 0:56
  • $\begingroup$ Is there a way to write it without using binomial series notation? $\endgroup$ – calmcalculus May 11 '17 at 1:16
  • $\begingroup$ $\binom{n}{m} = \frac{n!}{m!(n-m)!}$ and $(1/2)! = \sqrt{\pi}/2$ $\endgroup$ – Dando18 May 11 '17 at 1:21
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Consider Taylor's Theorem, that is for $c\in [a,x]$:

$$f(x)=\sum_{k=0}^{n-1}\frac {(x-a)^k}{k!}f^{k}(a)+\frac {(x-a)^n}{n!}f^{(n)}(c)$$

Note we call $R_n=\frac {(x-a)^n}{n!}f^{(n)}(c)$ the Lagrange remainder.

So let's consider the function $f(x)=\sqrt{x-2}$ at $x$ about $a=3$.

The derivatives of $f(x)$ evaluate as follows:

$f^{(0)}(x)=(x-2)^{1/2}$, $f^{(1)}(x)=\frac {1}{2(x-2)^{1/2}}$, $f^{(2)}(x)=\frac {-1}{4(x-2)^{3/2}}$, $f^{(3)}(x)=\frac {3}{8(x-2)^{5/2}}$, ...

Thus, we see:

$$f(x)=(x-2)^{1/2}=1+\frac 12(x-3)-\frac 18(x-3)^2+\frac {1}{16}(x-3)^3-\frac {5}{128}(x-3)^4+...$$

Which is equivalent to:

$$f(x)=\sum_{k=1}^{\infty}\bigg[{\frac 12 \choose k}(x-3)^k\bigg]+1$$

Noting ${\frac 12 \choose k}= \frac {(-1)^{k-1}}{2^{2k-1}k}{2k-2 \choose k-1}$, we have:

$$f(x)=\sum_{k=1}^{\infty}\bigg[\frac{(-1)^{k-1}}{2^{2k-1}k}{2k-2 \choose k-1}(x-3)^k\bigg]+1$$

Regarding ${\frac 12 \choose k}$, you may want to look at this link: Binomial coefficients (1/2, k)

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