1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Binomial_series and en.wikipedia.org/wiki/…. It seems you were told about the binomial series awhile ago, so you should already know the coefficients are $\binom{1/2}{n}$; are you having trouble simplifying $\binom{1/2}{n}$ to something involving factorials, or what's up? In fact, it looks like you simplified the terms $\binom{-1/2}{n}(-x^2)^n$ yourself in that previous question. so the same logic should help with this. $\endgroup$ – runway44 Nov 6 '19 at 5:07
  • $\begingroup$ It may help to know that, for $n\ge1$, $$\binom{\frac12}n=\frac{(-1)^{n-1}}{n\cdot2^{2n-1}}\cdot\binom{2n-2}{n-1}$$ $\endgroup$ – bof Nov 6 '19 at 6:10
2
$\begingroup$

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:

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

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $$ (1+x)^{1/2}=\sum_{k=0}^\infty\binom{\tfrac12}{k}x^k $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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, $\alpha=\frac12$: $$\sqrt{1+x}=1+\frac12x+\frac{\frac12\cdot\frac{-1}{2}}{2}x^2+\dots$$

| cite | improve this answer | |
$\endgroup$

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.