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

1 Answer
Dec 7, 2016

Answer:

#(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#