I have these two binomial expansions to be expanded up to the first 4 terms:
a) $(3+x)^{-2}$ for $|x| \lt 3$
b) $(3+x)^{-2}$ for $|x| \gt 3$
I don't understand how these will be any different? Or how the expansion will vary?
If anyone could give me a club that'd be great!
On
If you develop $(3+x)^{-2}$ in an infinite series, this series is a real number for $|x|<3,$ and definitely not a real number whenever $|x|>3.$ That is, this series does not give any definite real number in this other range.
If you swap the summands, we also have that $(x+3)^{-2}$ may be expanded in powers of $3/x,$ by factoring out the $x.$ This will converge to a real number whenever $|3/x|<1,$ or in other words for $|x|>3.$
The convenient form of the Binomial series is $$(1+z)^p=1+pz+p(p-1)z^2/2!+p(p-1)(p-2)z^3/3!+p(p-1)(p-2)(p-3)z^4/4!+....+ad-inf, ~if~ |z|<1~~~~(1)$$ If you want to expand $(3+x)^{-2}, |x|<3$ then write it as $\frac{1}{9}(1+\frac{x}{3})^{-2}.$ So $z=x/3$ and $p=-2$ in (1) will give you correct expansion in the powers of $\frac{x}{3}.$ There you may go upto the number of desired terms (five here).
But if you want to expand $((3+x)^{-2}$ when $|x|>3$, then write $(3+x)^{-2}= x^{-2} (1+\frac{3}{x})$ and take $z=\frac{3}{x}$ and $p=-2$ in (1) and go upto the desired number of terms.