Hassan H. answered • 08/03/13
Math Tutor (All Levels)
Hello Sun,
To complete this problem, you simply have to recall the appropriate definitions, as well as a result from calculus. Let's generalize a bit, and then proceed.
For the equation
(*) f(x)y'' + g(x)y' + h(x)y = 0,
define the related functions
P(x) = g(x)/f(x), and
Q(x) = h(x)/f(x).
Equation (*) is now equivalent to
(**) y'' + P(x)y' + Q(x)y = 0,
provided we account for places where f(x) = 0.
Now, we define a singular point of (**) as any point x = x0 where P(x) and Q(x) are not both analytic. A function is analytic at a point x0 if it can be represented by a power series about x0 with positive radius of convergence (this is where your memory of calculus comes into play!).
Your equation has P(x) = (1/2)/(ln x) and Q(x) = 1/(ln x). So, in order first to show that x=1 is a singular point of your equation, you must be able to convince someone either P(x) or Q(x) (or both) is (are) not analytic there. But this is apparent, since the functions both become unbounded near x=1.
To classify singular points further, we say that a singular point x = x0 is regular if both (x-x0)P(x) and (x-x0)2Q(x) are analytic at x=x0; otherwise, we call the singular point irregular.
So, your task will be to show that
(x-1)P(x) = (1/2)(x-1)/(ln x)
and
(x-1)2Q(x) = (x-1)2/(ln x)
are analytic at x=1. You may protest that neither of these functions is actually defined at x=1; this is a valid complaint, but remember that the discontinuity at x=1 is removable, which is to say that
limx→1 (x-1)P(x) and limx→1 (x-1)2Q(x)
are both finite, so we can redefine the functions to be equal to their limits there. Now, just show that the functions are both smooth near x=1, either by explicitly exhibiting their power series expansions about x=1, or by referring to some known and accepted result, such as "away from any of its zeros, the reciprocal of an analytic function is analytic".
I hope this clears up the definitions and concepts at least.
Regards,
Hassan H.
