The Taylor series expansion, in general, is written as:
#sum_(n=0)^(oo) f^n(a)/(n!)(x-a)^n#
So, we will have to take #n# derivatives of #1/x^2#. #n = 3# is the bare minimum in my opinion if you want to see a significant chunk of a pattern, but let's just stop at #n = 4#; this derivative isn't too bad, I guess. You just get pretty large numbers past #n = 4#.
#f^0(x) = color(blue)(f(x)) = x^-2 color(blue)(= 1/x^2)#
#color(blue)(f'(x)) = -2x^-3 color(blue)(= -2/x^3)#
#color(blue)(f''(x)) = 6x^-4 color(blue)(= 6/x^4)#
#color(blue)(f'''(x)) = -24x^-5 color(blue)(= -24/x^5)#
#color(blue)(f''''(x)) = 120x^-6 = color(blue)(120/x^6)#
Now, let's plug them in. Just remember:
- #x -> a#, except for the term #(x-a)^n#, which remains as #(x-a)^n#.
- #n# does vary from #0# to your choice of #n#. We chose #n_(end) = 4#.
- #a# is a constant we choose. We chose #a = 1# for simplicity, and because it's close to #x = 0# (if we let #x->a# as #a -> 0#, then #x->0#, and our Taylor series becomes more accurate, usually). #a# does not vary.
Generally, the sum is written out to be:
#= f^0(1)/(0!)(x-1)^0 + (f'(1))/(1!)(x-1)^1 + (f''(1))/(2!)(x-1)^2 + (f'''(1))/(3!)(x-1)^3 + (f''''(1))/(4!)(x-1)^4 + ...#
Now plug in your newly-acquired derivatives:
#= (1/(1^2))/(1!)(1) + (-2/1^3)/(1!)(x-1) + (6/1^4)/(2!)(x-1)^2 + (-24/1^5)/(3!)(x-1)^3 + (120/1^6)/(4!)(x-1)^4 + ...#
Simplify:
#= 1 + (-2)(x-1) + (6)/(2)(x-1)^2 + (-24)/(6)(x-1)^3 + (120)/(24)(x-1)^4 + ...#
And simplify some more:
#= color(blue)(1 - 2(x-1) + 3(x-1)^2 - 4(x-1)^3 + 5(x-1)^4 + ...)#
