Restrictions on n for 1/(a^n) when a<0 ?

Question

Wondering what the answer to this would be: what are the restrictions on n for 1/(a^n) when a<0 ? Explain.
Thanks!

Answer 1

(The reiprocal -- "1/(.....)" does not make any restricion.)
If a is a positive integer, what is implied is the product of the relevant number -- n -- of a's; if negative, use (-n) a's, and take reciprocal; but how will you define a fractional power of a negative number, e.g., square root -- without going to complex numbers?
If n is a rational number with odd denominator, say p/q, then it can be done:
- ((-a) ^ (1/q)) ^ p is a well-defined real number which fits the requirement. But if n is not rational or has even denominator, then the expression is not defined.

Answer 2

In general, the function 1/x is defined for all x not equal to zero. So our only concern is if a^n=0. But a<0, so a^n can never be zero.

EDIT: Okay the answer above me caught what I missed. There are restrictions on n if you want to stay in the reals. I guess the question doesn't say that explicitly, but if you're not in a chapter about complex numbers, the real values are probably what they mean. It's a good point and something interesting to think about.

Answer 3

we are in a position to instruct with the aid of induction that the nth spinoff of e^(-a million/x^2) has the type g_n(x) * e^(-a million/x^2) the place g_n(x) is a rational function. Then we use the shrink definition of spinoff, and induction on n. If f^(n)(x) exists for all x, then f^(n+a million)(0) = lim x->0 f^(n)(x)/x = lim x->0 g_n(x)*e^(-a million/x^2)/x. This shrink is 0 via fact e^(-a million/x^2) strategies 0 quicker than the denominator of g_n(x). EDIT: Oops, John D published a miles better answer on an identical time as i grew to become into composing mine.