Why Gov. Rick Perry represents the Mental Ward of the GOP, Part DCCXVI

He is apparently trying to maximize the number of uninsured Texans.

Consider the case of Texas, which with 25 percent uninsured, leads the nation in not providing for its residents. If the state pulls out of Medicaid, as Gov. Rick Perry (R) is suggesting, that would put it at 40 percent uninsured, as Medicaid covers 15 percent of the state. Texas might try some other form of coverage, but it will have lost hundreds of millions of dollars of federal funding. You can occasionally do less with more, but when you have a lot less, you generally just do less. Whatever the state tried next would cover fewer people with less-comprehensive insurance, and it’s a safe bet that the rate of uninsured would ultimately settle above 30 percent. Some legacy.

Conversely, if Perry does nothing, the federal government is going to come in and pick up most of the cost of a massive coverage expansion. Texas, in fact, will be one of the biggest winners from health-care reform, as its huge pool of uninsured residents means the state will get an uncommonly large amount of subsidies to bring that down to manageable levels. Texas “can expect to see Medicaid enrollment rise by 46 percent while state spending on Medicaid rises by about 3 percent.” Pretty good deal.

Find the function


Given real functions \(f\), \(g\), and \(h\) which satisfy the following:

\(f’=2 f^{2} g h +\displaystyle\frac{1}{g h}\),

\(g’=f g^{2} h +\displaystyle\frac{4}{f h}\),

\(h’=3 f g h^{2} +\displaystyle\frac{1}{f g}\),

with \(f(0)=1\), \(g(0)=1\), and \(h(0)=1\), find the function \(f(x)\).


Clearly, one can divide each equation above by the functions \(f\), \(g\), and \(h\), respectively.  Adding the equations together produces a single equation for the function \(u=f g h\):

\(\displaystyle\frac{u’}{u}=6 \left (u+\displaystyle\frac{1}{u}\right )\),

from which a solution for \(u\) is produced:

\(6 x+C=\displaystyle\int^{u}\displaystyle\frac{du’}{1+u’^{2}}=\tan^{-1}u\)

Because \(u(0)=1\), \(u(x)=\tan \left (6 x+\displaystyle\frac{\pi}{4}\right )\).

This result is plugged back into the first equation, which may be written as follows:

\(\displaystyle\frac{f’}{f}=2 u+\displaystyle\frac{1}{u}=2 \tan \left (6 x+\displaystyle\frac{\pi}{4}\right )+\cot \left (6 x+\displaystyle\frac{\pi}{4}\right )\).

Integrating both sides, we can solve for the unknown function \(f\):

\(f(x)=-\displaystyle\frac{2}{6}\log \left [\cos \left (6 x+\displaystyle\frac{\pi}{4}\right) \right ] + \frac{1}{6}\log \left [\sin\left (6 x+\displaystyle\frac{\pi}{4}\right) \right ] + C’\).

Using \(f(0)=1\), the solution takes the form:

\(f(x)=2^{-\frac{1}{12}} \left [ \displaystyle\frac{\sin\left (6 x+\displaystyle\frac{\pi}{4}\right)}{\cos^{2}\left (6 x+\displaystyle\frac{\pi}{4}\right)} \right ]^{\frac{1}{6}}\)

Here is a plot of the solution.

Is anti-Semitism no longer a problem in the US?

Not according to the FBI:

According to the findings, of the 1,575 victims of anti-religious hate-crime in 2009:

  • 71.9 percent were victims because of an offender’s anti-Jewish bias.
  • 8.4 percent were victims because of an anti-Islamic bias.
  • [A]ccording to the lowest possible population estimates, Jews are four times more likely than Muslims to be victims of hate crimes in the United States; according to the highest possible estimates, Jews are around ten times more likely to be attacked.

    Despite all this, we are still told by agitators on both sides of the Atlantic that ‘Islamophobia’ has replaced anti-Semitism, and that Muslims now face a situation similar to European Jews during the run up to the Holocaust.

    I do not make light of the bigotry shown Muslims lately, especially through some of the media. But still, and I don’t wish to propagate a permanent victim mentality, but I find the whole phenomenon of anti-Semitism insanely perplexing. I will have more to say after I finish The Finkler Question, with which I am almost done.

    What do Tom Finneran and Sarah Palin have in common?

    “Speaker Finneran was there at the creation,’’ Ware said. “I had hoped to elicit the truth from Speaker Finneran, and instead I was stonewalled.’’

    But Finneran regained his voice the day after the release of Ware’s report. He defended himself on his morning radio show on WRKO-AM.

    Criminals like Finneran only talk when in their own cocoon.  No nasty questions about a system of patronage that costs Massachusetts residents so much more than just money.  Because in the end, as Billy Bulger’s son Chris stated, hey, this is the way things work on Beacon Hill.  Not to visit a father’s sins upon his son, but if anyone, he should know.  So shut up, get on with your miserable lives, and every other year, remember to vote in the “D” column no matter what.

    Speaking of cocoons: does an over-reliance on a cocoon imply that Sarah Palin is a criminal?  [Now I sound like an anchor from Fox news.]  Who knows.  But she surely has something to hide.  Even Jennifer Rubin, whom I have raked over the coals in these pages for her fangirl support of an obviously unqualified candidate, believes that Palin needs to get out of her cocoon.  But just remember one thing: a candidate that gets elected from within his or her own coocon will truly give us the government we deserve.

    Maybe he was brushing his teeth with Ben-Gay

    Galluccio claimed he had used a toothpaste that mistakenly triggered the chemical breath test machine installed in his home. Galluccio failed the test the first time he was required to take it. Nestor scornfully sentenced Galluccio to one year’s imprisonment.

    Perhaps I am too harsh on our State Reps.  But please allow me the smallest amount of schadenfraude for seeing such an idiot finally be put behind bars.  Maybe the next question we can ask ourselves is how many of our State Reps end up with a criminal record upon exiting Public Service?

    The reason why the public is so angry with the TSA

    In studying the media’s coverage, officials have come to conclude that a slow news week, combined with the president’s being overseas and Congress being out of session, created the perfect storm of bad coverage.

    Yeah, that’s the ticket.  It’s not the fact that this stuff catches only the stupid terrorists or that it does not work against cavity bombs.  And when someone tries to set off a cavity bomb, to what sort of collective punishment will we be subjected?

    An unintuitive limit

    Problem: Let \(f(n)\) = the number of zeros in the decimal representation of \(n\).  For example, \(f(1009)=2\).  For \(a>0\), define



    \(L=\displaystyle\lim_{N\to\infty}\frac{\log S(N)}{\log N}\).

    Solution: The key is to recognize that the sum \(S(N)\) is best evaluated according to how many digits are in \(N\).  As an example, suppose \(N=9\): none of the single-digit numbers between 1 and 9 have zeros, so the sum \(S(N)\) is equal to 9.  Within the two-digit numbers, note that there are 9 numbers with 1 zero [e.g., 10, 20,…,90] and the rest [90-9=81] with no zeros; in this case \(S(99)=9 a+81+S(9)=9 a+9^2+S(9)\).  For three-digit numbers, there are 3 possibilities: the number can have 2, 1, or no zeros.  There are 9 such numbers with two zeros.  Numbers with 1 zero include 101 and 110; note that there are 2 different numbers resulting from two non-zero digits and one zero digit.  Given that there are \(9^2\) combinations of digits, it is clear that there are \(2\times 9^2\) three-digit numbers having exactly one zero digit.  Finally, there are \(900-162-9=729=9^3\) three-digit numbers with no zeros.  Therefore,

    \(S(999)=S(10^3-1)=9 a^2+2\times 9^2 a+9^3+S(10^2-1)\).

    The pattern here is clear to the [mathematically inclined] observer:

    \(S(10^k-1)=9 a^{k-1}+\binom{k-1}{1}\times 9^2 a^{k-2}+\ldots+\binom{k-1}{k-2}\times 9^{k-1} a+9^{k}+S(10^{k-1}-1)\)


    \(S(10^k-1)=9 (a+9)^{k-1}+S(10^{k-1}-1)=9 \displaystyle\sum_{j=0}^{k-1} (a+9)^{j}\).

    Summing the series, we arrive at the following:

    \(S(10^k-1)=9 \frac{(a+9)^{k}-1}{a+8}\).

    In evaluating \(L\), observe that the limit \(N\to\infty\) is equivalent to \(k\to\infty\). In this limit, \(\log N \approx k\). We therefore get the final result:

    \(L=\log (a+9)\)