eddo
Oct10-04, 05:19 PM
This is my first encounter with this function. In the question at hand it's defined as
A(x) = 1 + x^3/(2)(3) + x^6/(2)(3)(5)(6) + x^9/(2)(3)(5)(6)(8)(9) + ...
and on like that to infinity. The question is to find the domain of this function. Here's what I have so far. The domain would mean the values for which the sum converges. Since this function is smaller than the summation of x^(3n), from n=o to n=infinity, and that function is convergent for |x|<=1, this function must also be convergent for |x|<=1 by the comparison test (a sort of modified squeeze theorem). But this alone doesn't prove that values of |x|>= 1 won't work, it simply shows that values less than 1 will work. Are there values greater than 1 that still cause the function to converge due the the decreasing effect of the denominator? If so, or if not, how can it be proven? I suspect that values >=1 won't work but i don't know how to prove it. Can anyone help me out?
Sorry that I don't know how to make the equations look nice.
A(x) = 1 + x^3/(2)(3) + x^6/(2)(3)(5)(6) + x^9/(2)(3)(5)(6)(8)(9) + ...
and on like that to infinity. The question is to find the domain of this function. Here's what I have so far. The domain would mean the values for which the sum converges. Since this function is smaller than the summation of x^(3n), from n=o to n=infinity, and that function is convergent for |x|<=1, this function must also be convergent for |x|<=1 by the comparison test (a sort of modified squeeze theorem). But this alone doesn't prove that values of |x|>= 1 won't work, it simply shows that values less than 1 will work. Are there values greater than 1 that still cause the function to converge due the the decreasing effect of the denominator? If so, or if not, how can it be proven? I suspect that values >=1 won't work but i don't know how to prove it. Can anyone help me out?
Sorry that I don't know how to make the equations look nice.