What is the sigma notation for Maclaurin series of $\sqrt{1+x}$

Question

I have been able to expand the function $\sqrt{1+x}$ into binomial series:

The series is:

$\sqrt{1+x}=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3-\dfrac{5}{128}x^4+\dfrac{105}{3840}x^5...$

But I don't know how to form the sigma notation for this function. Can you help me just a little bit?

Answer 1

---

If you're unfamiliar with the generalized binomial coefficients, here's another (unfortunately rather longer) approach.

First, focus on the string of derivatives:

$$ \begin{align} f(x) &= (x+1)^{\frac{1}{2}}\\[1ex] f'(x) &= \frac{1}{2} (x+1)^{-\frac{1}{2}}\\[1ex] f''(x) &= \left(\frac{-1}{2}\right) \left(\frac{1}{2}\right) (x+1)^{-\frac{3}{2}}\\[1ex] f'''(x) &= \left(\frac{-3}{2}\right)\left(\frac{-1}{2}\right)\left(\frac{1}{2}\right) (x+1)^{-\frac{5}{2}}\\[1ex] f^{(4)}(x) &= \left(\frac{-5}{2}\right)\left(\frac{-3}{2}\right)\left(\frac{-1}{2}\right)\left(\frac{1}{2}\right) (x+1)^{-\frac{7}{2}}\\[1ex] & \vdots\\[1ex] f^{(n)}(x) &= \left(\frac{-(2n-3)}{2}\right) \left(\frac{-(2n-5)}{2}\right) \left(\frac{-(2n-7)}{2}\right) \cdot \cdot \left(\frac{-3}{2}\right)\left(\frac{-1}{2}\right)\left(\frac{1}{2}\right) (x+1)^{-\frac{2n-1}{2}}\\[2ex] &= \frac{(-1)^{n-1}}{2^n}(2n-3)(2n-5)(2n-7) \cdot \cdot \; (3)(1)(1) (x+1)^{-\frac{2n-1}{2}}\\[1ex] \color{white}{text}\\ \end{align} $$

To handle the string $\,(2n-3)(2n-5)(2n-7) \cdot \cdot \;(3)(1),\,$ we can do the following manipulation:

$$ \begin{align} (2n-3)(2n-5)(2n-7) \cdot \cdot \;(3)(1) &= \frac{(2n-3)(2n-4)(2n-5)(2n-6)(2n-7) \cdot \cdot \;(3)(2)(1)}{(2n-4)(2n-6)\cdot \cdot \;(2)}\\[1ex] &= \frac{(2n-3)(2n-4)(2n-5)(2n-6)(2n-7) \cdot \cdot \;(3)(2)(1)}{2(n-2)\; 2(n-3)\;\cdot \cdot \; 2(1)}\\[1ex] &= \frac{(2n-3)!}{2^{n-2}(n-2)!}\\[1ex] \end{align} $$

Then, combining with the above, we'll get:

$$ \begin{align} f^{(n)}(x) &= \frac{(-1)^{n-1}}{2^n}(2n-3)(2n-5)(2n-7) \cdot \cdot (3)(1)(1) (x+1)^{-\frac{2n-1}{2}}\\[1ex] &= \frac{(-1)^{n-1}}{2^n} \cdot \frac{(2n-3)!}{2^{n-2}(n-2)!} (x+1)^{-\frac{2n-1}{2}}\\[1ex] &= \frac{(-1)^{n-1}(2n-3)!}{2^{2n-2}(n-2)!} (x+1)^{-\frac{2n-1}{2}}\\[1ex] &= \frac{(-1)^{n-1}(2n-3)!}{4^{n-1}(n-2)!} (x+1)^{-\frac{2n-1}{2}}\\[1ex] &= \frac{(-1)^{n-1}(2n)(2n-1)(2n-2)(2n-3)!}{4^{n-1}(2n)(2n-1)(2n-2)(n-2)!} (x+1)^{-\frac{2n-1}{2}}\\[1ex] &= \frac{(-1)^{n-1}(2n)!}{4^{n-1}\;2(n)(2n-1)\;2(n-1)(n-2)!} (x+1)^{-\frac{2n-1}{2}}\\[1ex] &= \frac{(-1)^{n-1}(2n)!}{4^{n-1}\;4(2n-1)(n)(n-1)(n-2)!} (x+1)^{-\frac{2n-1}{2}}\\[1ex] &= \frac{(-1)^{n-1}(2n)!}{4^n (2n-1) \; n!} (x+1)^{-\frac{2n-1}{2}}\\[1ex] \end{align} $$

Finally, inserting this into the general term for a MacLaurin series:

$$ \begin{align} \frac{f^{(n)}(0)}{n!}x^n &= \frac{\frac{(-1)^{n-1}(2n)!}{4^n (2n-1) \; n!}}{n!} \; x^n\\[1ex] &= \frac{(-1)^{n-1}(2n)!}{4^n (2n-1) \; n!\;n!} \; x^n\\[1ex] &= \boxed {\binom{2n}{n} \; \frac{(-1)^{n-1}}{4^n (2n-1)} \; x^n \;} \end{align} $$

Answer 2

Recall that the generalized binomial coefficients are: $$\binom{\alpha}{k}=\cfrac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-k+1)}{k}$$ with $\alpha$ arbitrary and $k$ a non-negative integer. Thus, putting $\alpha=\tfrac12$ yields $$ (1+x)^{1/2}=\sum_{k=0}^\infty\binom{\tfrac12}{k}x^k $$

Answer 3

This is known as the binomial series: $$(1+x)^\alpha=\sum_{k=0}^\infty{\alpha \choose k}x^k=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\dots $$ which is convergent for $|x|<1$. The convergence at $|x|=1$ depends on $\alpha$. In your case, $\alpha=\frac12$: $$\sqrt{1+x}=1+\frac12x+\frac{\frac12\cdot\frac{-1}{2}}{2}x^2+\dots$$