Question

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

Answer 1

`(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 = 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 (1/2-j)` in terms of factorials and powers of `2`. Let us see if we can find one...

Let:

`a_k = prod_(j=0)^(k-1) (1/2-j)`

Then:

`a_0 = 1`

`a_1 = 1/2`

`a_2 = (1/2)(-1/2) = -1/4`

`a_3 = (1/2)(-1/2)(-3/2) = 3/8`

`a_4 = (1/2)(-1/2)(-3/2)(-5/2) = -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 `1/2` and alternating signs into account:

Let us propose the following formula for `k >= 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`