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).

  • $\begingroup$ 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

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)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.