How can we expand the following by the binomial expansion, upto the term including $x^3$? That'll be 4 terms.
This the expression to be expanded: $\sqrt{2+x\over1-x}$
I understand how to do the numerator and denominator individually. Now this is what I'm doing – having expanded the denominator (using the standard expansion formula for $(1+x)^{-1}$), do I now simply need to multiply this expansion once with the numerator $(2+x)$? I'm not getting the correct answer, but is this the correct method?
Best Answer
We will use the expansion $\sqrt{1+x} = 1+x/2+x^2(1/2)(-1/2)/2 + x^3(1/2)(-1/2)(-3/2)/6 + ... = 1+x/2-x^2/8+x^3/8+... $ where "..." means "terms of higher order than $x^3$" both in this expansion and in the math below.
Note: I am doing the following math off the top of my head as I am entering it, so the chances for error are decidedly nonzero. However, the form of the result should be correct. Because we are only interested in terms up to order $x^3$, whenever a term of higher order occurs, it is dropped and subsumed into the "..." part. For those who know the "big-$O$" notation, the "+..." could also be written as $+O(x^4)$.
$\begin{align} \sqrt{2+x\over1-x} &=\sqrt{(2+x)(1+x+x^2+x^3+...)}\\ &=\sqrt{2+2x+2x^2+2x^3+...+x+x^2+x^3+...}\\ &=\sqrt{2+3x+3x^2+3x^3+...}\\ &=\sqrt{2}\sqrt{1+3x/2+3x^2/2+3x^3/2+...}\\ &=\sqrt{2}(1+(3x/2+3x^2/2+3x^3/2)/2 -(3x/2+3x^2/2)^2/8 +(3x/2)^3/8+...)\\ &=\sqrt{2}(1+3x/4+3x^2/4+3x^3/4 -(3x/2)^2(1+x)^2/8 +27x^3/64+...)\\ &=\sqrt{2}(1+3x/4+3x^2/4+3x^3/4 -(9x^2/32)(1+2x+x^2) +27x^3/64+...)\\ &=\sqrt{2}(1+3x/4+3x^2/4+3x^3/4 -9x^2/32-9x^3/16 +27x^3/64+...)\\ &=\sqrt{2}(1+3x/4+x^2(3/4-9/32)+x^3(3/4-9/16+27/64))+...\\ &=\sqrt{2}(1+3x/4+x^2((3*8-9)/32)+x^3((3*16-9*4+27)/64))+...\\ &=\sqrt{2}(1+3x/4+15x^2/32+39x^3/64)+...\\ \end{align} $
This is why computer algebra systems came to be.