# How do you use the binomial series to expand  (1+x)^.5?

##### 1 Answer
Dec 7, 2016

(1+x)^(1/2) = 1 + sum_(k=1)^oo (-1)^(k-1)((2k-2)!)/(2^(2k-1) (k-1)!k!) x^k

#### Explanation:

In general, the binomial theorem tells us that:

(a+b)^n = 1/(0!) a^n + n/(1!) a^(n-1) b + (n(n-1))/(2!) a^(n-2) b^2 + (n(n-1)(n-2))/(3!) a^(n-3) b^3 + ...

color(white)((a+b)^n) = sum_(k=0)^oo (prod_(j=0)^(k-1) (n-j))/(k!) a^(n-k) b^k

In our example, $a = 1$, $b = x$ and $n = \frac{1}{2}$.

Now $1$ raised to any power is $1$, so the formula simplifies to:

(1+x)^(1/2) = sum_(k=0)^oo (prod_(j=0)^k (1/2-j))/(k!) x^k

It would be nice to have a formula for ${\prod}_{j = 0}^{k} \left(\frac{1}{2} - j\right)$ in terms of factorials and powers of $2$. Let us see if we can find one...

Let:

${a}_{k} = {\prod}_{j = 0}^{k - 1} \left(\frac{1}{2} - j\right)$

Then:

${a}_{0} = 1$

${a}_{1} = \frac{1}{2}$

${a}_{2} = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) = - \frac{1}{4}$

${a}_{3} = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) = \frac{3}{8}$

${a}_{4} = \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right) \left(- \frac{5}{2}\right) = - \frac{15}{16}$

...

One thing that might help us is:

((2m)!)/(2^m m!) = overbrace((2m-1)(2m-3)...(3)(1))^"m terms"

and hence:

((2m)!)/(2^(2m) m!) = overbrace((1/2)(3/2)(5/2)...((2m-1)/2))^"m terms"

So putting $m = k - 1$, taking the extra factor of $\frac{1}{2}$ and alternating signs into account:

Let us propose the following formula for $k \ge 1$:

a_k = (-1)^(k-1)((2k-2)!)/(2^(2k-1) (k-1)!)

Hence we can write:

(1+x)^(1/2) = sum_(k=0)^oo (prod_(j=0)^k (1/2-j))/(k!) x^k

color(white)((1+x)^(1/2)) = 1 + sum_(k=1)^oo ((-1)^(k-1)((2k-2)!)/(2^(2k-1) (k-1)!))/(k!) x^k

color(white)((1+x)^(1/2)) = 1 + sum_(k=1)^oo (-1)^(k-1)((2k-2)!)/(2^(2k-1) (k-1)!k!) x^k