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 Nov 16 comment Formula for nth term of sequence? @Enigma99 - Don't let the terminology get in the way. Just put $x_n = r^n$ into the recurrence, reduce the result to a quadratic equation, and use the quadratic formula to solve for the two roots, $r_1,r_2$. Then any linear combination $\alpha r_1^n+\beta r_2^n$ must also solve the same recurrence, but imposing the given initial conditions will force particular values of $\alpha, \beta$. Nov 11 revised Why can a probability measure be defined over power set of countable sample space? added 33 characters in body Nov 11 answered Why can a probability measure be defined over power set of countable sample space? Nov 8 comment Can any Real number be typed in a computer? @RobertIsrael -- I,too, think this answer is wrong, because of its claim about what can be defined in "two ASCII-pages or so". Many reals in $[0,10]$ can of course be defined by explicitly listing their binary digits, and a trillion (say) independent random bits is practically certain to define a number that cannot be expressed in "two ASCII-pages or so". (This supposes that after generating these bits, I would say that I have "thought of" the number.) Nov 8 comment Probability of flipping an infinite number of heads "$P\{\omega\}=0\text{ a.s.}$" is a category error, since "$P\{\omega\}=0$" is not an event -- "almost sure" applies standardly to an event whose probability equals $1$. The opposite extreme might be called "almost impossible", i.e., an event with probability equal to $0$. Thus, in your example, $P\{\omega\}=0$ (exactly, not "almost surely"), so each atomic event $\{\omega\}$ is almost impossible and its complement $\Omega\backslash\{\omega\}$ is almost sure. Oct 18 answered Are hyperoperators primitive recursive? Oct 18 comment Are hyperoperators primitive recursive? It's worth mentioning that although each hyperoperation $H_n:\mathbb{N}^2 \to \mathbb{N}$ is primitive recursive, the function $H: \mathbb{N}^3 \to \mathbb{N}: (n,x,y) \mapsto H_n(x,y)$ is not primitive recursive. A simple proof of this is skecthed in Introduction to Computability Theory by Zucker & Pretorius. Oct 15 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? fix typo Oct 15 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? fix typo Oct 15 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? major re-write to show an extremely improved bound on the number of iterations Oct 13 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? restate in terms a more-general result Oct 12 awarded Good Question Oct 12 comment Confused about Continuous Random variable @Quality - Be sure to distinguish between the random variable $X$ and its possible values $x$. The "bounding" notation $3 \le x \lt 4$ is just identifying one of the intervals of possible values in which the probability $P(X\le x)$ is a constant (specific to that interval). (Also, see my comment to you your question.) Oct 12 comment Confused about Continuous Random variable The random variable is discrete (not continuous), so $F(x)=P(X\le x)$ changes (increases) only at values of $x\in\{1,2,3,4,5,6 \}$, which are the only points with nonzero probability. In the intervals between these points, $P(X\le x)$ is constant (a different constant for each interval). Oct 11 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? fix typo Oct 11 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? fix typo Oct 11 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? improve phrasing Oct 11 revised With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? use correct tags; correct the title and value of googolplex Oct 11 answered With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? Oct 9 comment With $f(n) = n!$, what is the least $k$ such that $f^k(\text{googolplex}) > \text{Graham's number}$? Although it won't make any difference to the answer, you should use the correct definitions: $\tt googol = 10^{100}, googolplex = 10^{googol}=10^{(10^{100})}.$