How do you know if #f(s) = 4s^(3/2)# is an even or odd function?

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Answer 1 · 2016-01-03T11:06:13

Find that #f(s)# satisfies neither the condition for an even function nor the condition for an odd function, so it is neither.

An even function satisfies #f(-x) = f(x)# for any #x# in the domain.

An odd function satisfies #f(-x) = -f(x)# for any #x# in the domain.

If #f(s)# is considered as a Real valued function, then it's not defined for #s < 0#.

If #f(s)# is considered as a Complex valued function, then we have to think about how #s^(3/2)# is defined for negative values of #s#.

I have changed my mind on this one. I thought it was not well defined, but I think that if #s < 0# then #s^(3/2) = -i (-s)^(3/2)#.

You can arrive at this definition by considering the polar representation of negative numbers #s = (-s, pi)#.

Hence #s^(3/2) = ((-s)^(3/2), (3 pi)/2) = -i(-s)^(3/2)#

Hence we find: #f(-s) = -i f(s)#

For example, #f(-1) = 4 (-1)^(3/2) = -4i 1^(3/2) = -4i = -i f(1)#

so #f(s)# is neither even nor odd.