TITLE: Do there exist measurable transformations which are recurrent over spaces of infinite measure?
QUESTION [1 upvotes]: I am trying to find a measurable space $(X, \mathcal{B})$, with measure $\mu$ such that $\mu(X)=\infty$, and with a corresponding measure preserving transformation $T:X\rightarrow X$ such that $T$ is recurrent over $X$.  That is, $T$ is such that for all $B\in \mathcal{B}$, there exists infinitely many $n>0$ such that $T^{n}(B) \cap B \not= \emptyset$.
Does such a dynamical system exist?  If so, can you provide example or reference? If no such system exists, is there a particular reason (ie. it is not possible, or one has not been found)?

REPLY [1 votes]: Your question is related to, what is called, infinite ergodic theory. There are many interesting examples. An excellent reference to learn about this is the book of Jon Aaronson "An introduction to infinite ergodic theory"


TITLE: How would you approach this question on definite integral?
QUESTION [2 upvotes]: $$\int_{\frac \pi 4}^{\frac \pi 2}\left(2\cos\left(\frac x 2\right) - e^{-x}\right)^2dx$$

I have tried expanding it and then do the integral but it leads to $4 cos(\frac{x}{2})e^{-x}$ in the middle term of the expansion.
Am i missing something? Is there a "trick" to this? If so how would you approach the question? What are the things that i should keep in mind solving this kind of question? PS. the answer to this is $1.41$

REPLY [2 votes]: $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(2\cos\left(\frac{x}{2}\right)-e^{-x}\right)^{2}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(4\cos^{2}\left(\frac{x}{2}\right)+e^{-2x}-4\cos\left(\frac{x}{2}\right)e^{-x}\right)dx$$$$=4\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos^{2}\left(\frac{x}{2}\right)dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-2x}dx-4\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos\left(\frac{x}{2}\right)e^{-x}dx$$$$=2\color{green}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}1+\cos\left(x\right)dx}+\color{blue}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-2x}dx}-4\color{red}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx}$$
For the red part (where you have a problem with) using integration by parts we have:
$$\text{I}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx$$$$=-e^{-x}\cos \left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}-\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\sin\left(\frac{x}{2}\right)dx$$$$=-e^{-x}\cos\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}-\frac{1}{2}\left[-e^{-x}\sin\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}+\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx\right]$$$$=-e^{-x}\cos\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}+\frac{1}{2}e^{-x}\sin\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}-\frac{1}{4}\underbrace{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-x}\cos\left(\frac{x}{2}\right)dx}_\textrm{I}$$
Hence
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos\left(\frac{x}{2}\right)e^{-x}dx=\frac{4}{5}\left(-e^{-x}\cos\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}+\frac{1}{2}e^{-x}\sin\left(\frac{x}{2}\right)\Big|_\frac{\pi}{4}^\frac{\pi}{2}\right)\simeq \color{red}{{0.208396}
}$$
For the blue part setting $-2x \mapsto u$ follows:
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{-2x}dx=\frac{1}{2}\int_{-\pi}^{-\frac{\pi}{2}}e^{u}du=\frac{1}{2}\left(e^{\left(-\frac{\pi}{2}\right)}-e^{\left(-\pi\right)}\right)\simeq \color{blue}{0.0823328290435}$$
For the green part using some elementary integrals follows:
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}1+\cos\left(x\right)dx=\frac{\pi}{2}-\frac{\pi}{4}+\sin\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{4}\right)\simeq \color{green}{1.07829138221}$$
Summing the results gives the final answer:
$$2\left(1.07829138221\right)+\left(0.0823328290435\right)-4\left({0.208396}
\right)=1.4053315934635
$$
Which is the the answer you mentioned.

REPLY [0 votes]: Hint:
Integrate $$\int \cos\left(\frac x2\right)\exp(-x)\,\mathrm dx$$ by parts twice. The first time, let $f = \cos\left(\dfrac x2\right)$ and $g' = \exp(-x)$. The second time, let $f = -\sin\left(\dfrac x2\right)$ and $g' = \dfrac{\exp(-x)}4$.


TITLE: When is $[0,1]^K$ submetrizable or even metrizable?
QUESTION [1 upvotes]: Let $I=[0,1]$ and $K$ is a compact space. Then could the function space $I^K$ be submetrizable, even metrizable? In other words, in general, if $I^A$ can be submetrizable (metrizable) for some space $A$, what's condition that $A$ should satisfying?

REPLY [4 votes]: If $A$ is compact, $I^A$ is metrizable with the metric being the uniform norm.  That is, $d(f,g):=\sup_{a\in A} d(f(a),g(a))$.


TITLE: Number of increasing functions
QUESTION [4 upvotes]: Let $a > 0$, and $0 < n < m$ be given. How many functions $f:\{0, \ldots, n\} \rightarrow \{0, \ldots, m \}$ are there which satisfy

$f$ is non-decreasing
For all $i \in \{0, \ldots, n \}$ we have $f(i) \leq a + i$

I am interested in the case where $a, m = \Theta(n)$. In this case, as $n \rightarrow \infty$, I expected that the number of such functions is exponential in $n$. What is this rate of exponential increase?

REPLY [2 votes]: If $m$-sequence is increasing then all terms of it are distinct, from n disponible objects we choose m objects in $\binom{n}{m}$ ways that can be arranged in increasing sequences in an unique way.So number of sequences with desired properties is
$$\binom{n}{m}=\frac{n!}{m!(n-m)!}$$
in the second case we have sequences
$$(a,a+1,a+2,...,a+m)$$
there $n-m\leq a+m\leq n$ or $a=1,2,...,n-m$ that mean ther exists $$n-m$$ sequences with second propperties


TITLE: Which integers are representable by $x^2+2y^2+7z^2$?
QUESTION [3 upvotes]: I have been trying to characterize integers representable by several ternary forms and reached a roadblock with this particular form:
$$x^2+2y^2+7z^2$$
Ideally, I am looking for a characterization of square-classes represented. Computer calculations were unable to find any class number 1 ternary sublattices or an entire ternary genus. Any help with this problem or direction towards useful sources would be very much appreciated.

REPLY [2 votes]: other genus:
=====Discriminant  52  ==Genus Size==   2
   Discriminant   52
  Spinor genus misses     no exceptions
        52:    1     2          7      2    0    0 vs. s.g.   5  10  14  21  46
        52:    2     3          3      2    0    2 vs. s.g.   1  6  26  34  741
--------------------------size 2
The 150 smallest numbers represented by full genus
     1     2     3     4     5     6     7     8     9    10
    11    12    14    15    16    17    18    19    20    21
    22    23    24    25    26    27    28    29    30    31
    32    33    34    35    36    37    38    40    41    42
    43    44    45    46    47    48    49    50    51    53
    54    55    56    57    58    59    60    61    62    63
    64    65    66    67    68    69    70    71    72    73
    74    75    76    77    78    79    80    81    82    83
    84    85    86    87    88    89    90    91    92    93
    94    95    96    97    98    99   100   101   102   103
   104   105   106   107   108   109   110   111   112   113
   114   115   116   118   119   120   121   122   123   124
   125   126   127   128   129   131   132   133   134   135
   136   137   138   139   140   141   142   143   144   145
   146   147   148   149   150   151   152   153   154   155

The 150 smallest numbers NOT represented by full genus
    13    39    52   117   130   156   182   208   221   286
   299   325   351   377   390   455   468   494   520   546
   559   624   637   663   689   715   728   793   806   832
   858   884   897   962   975
Disc: 52
==================================

Comparison up to 1,000,000, ignore multiples of 169:
exceptions for first form   (   1   2   7   2   0   0) are

          5         10         14         21         46         61
         65         78        141        174        181        210
        229        265        394        429        474        481
        598        874        894        949       1209       1261
       1270       1417       1794       2266       2301       2470
       2561       3406       4030       4966       5446       5486
       5941       6006       6149       6565       7189       8034
      10894      11089      11401      12246      12610      15405
      15574      19409      19669      20254      26169      37609
      44421      51181      51649      68809      85930     109486
     125086     140374

There were 62  exceptions for first form

exceptions for second form  (   2   3   3   2   0   2) are

   1:     1          6         26         34        741        754
       2509       3289

There were 8  exceptions for second form

       1 1,000,000
do you want to do another one?  type y or n: n


TITLE: What is the millionth decimal digit of the $ 10^{10^{10^{10}}} $-th prime?
QUESTION [31 upvotes]: What is the millionth decimal digit of the $10^{10^{10^{10}}}$th prime?

(This prime, with more than $10^{10^{10}}$ decimal digits, is far larger than the largest "known" prime.) The answer should include a proof of correctness. I'm posting this question in the spirit of this advice, and will eventually post an answer (with proof of a more general result) if no one else does so.
NB: The notation $10^{10^{10^{10}}}$ means the same as $10^{(10^{(10^{10})})}$, and the "millionth digit" means the millionth digit from the left, as usually written (i.e., the most significant digit is leftmost and is called the 1st digit).
Afterthought: It might have been more impressive to have asked for the $10^{10}$th digit of, say, the $10^{10^{10^{10^{10}}}}$th prime (that digit being $8$), since perhaps no one has ever before computed the ten-billionth digit of a particular prime.

REPLY [13 votes]: A similar approach as in the accepted answer, but using a tighter bound from Dusart  ...
Theorem: If $m$ is a positive integer, then the first $k-1$ digits of the $10^{10^m}$th prime are just those of $\log 10$, where $k$ is the greatest integer such that $(k \le \lfloor m - \log_{10} m \rfloor \wedge d_k \lt 9)$, and $d_k$ is the $k$th digit of $\log  10$.
Proof: Rosser proved that $p_n \gt n \log n$, and Dusart proved that $p_n \le (n \log n)(1 + r_n)$ for all $n \ge 39017$, where $r_n = \frac{\log \log n - 0.9484}{\log n} \gt 0$. Therefore,
$p_n = n \log n  + (n \log n) \ \epsilon_n $ for all $n \ge 39017$, where $0 \lt \epsilon_n \le r_n$.
Now suppose $n = 10^{10^m}$, where $m$ is a positive integer ($m \ge 1$ ensures $n \ge 39017$, so Dusart's bound applies). Then
$n \log n = 10^{10^m + m} \log 10$
which, in base $10$, is just $\log 10$ with the decimal point shifted $10^m + m$ places to the right.
Now $\log \log 10 < 0.9484$, so
$r_n = 10^{-m} \ (m + \frac{\log \log 10 - 0.9484}{\log 10}) < 10^{-m} m \le 10^{-\lfloor m - \log_{10} m \rfloor}$,
and hence
$(n \log n) \ \epsilon_n <  (n \log n) \ 10^{-\lfloor m - \log_{10} m \rfloor}$
where the right-hand side is seen to be, in base $10$, just $n \log n$ with the decimal point shifted $\lfloor m - \log_{10} m \rfloor$ places to the left.
Thus, if $k$ is the greatest integer satisfying both (1) $k \le \lfloor m - \log_{10} m \rfloor$, and (2) the $k$th digit is less than 9 (ensuring that no carry from the right can affect the $(k-1)$th digit), then adding $(n \log n) \ \epsilon_n$ to $n \log n$ does not affect the first $k-1$ digits of the latter term. QED
NB: This result generalizes to the $b^{b^{m}}$th prime, for any integer base $b$ such that $2 \le b \le 13$, and for integer $m \ge \log_b \log_b 39017$.  The restrictions are to ensure that Dusart's bound can be applied.
Example: For any $m \ge 10^6 + 8$ (e.g., $m = 10^{10}$, as in the posted question), the first million digits of the $10^{10^m}$th prime are just those of $\log 10$. This is because for $m = 10^6 + 8$, $\lfloor m - \log_{10} m \rfloor = 10^6 + 1$, giving $k$ as the greatest integer such that $(k \le 10^6 + 1 \wedge d_k \lt 9)$. Direct computation (e.g., using Sage, which took less than a minute) shows that $k-1 = 10^6$, $d_{k-1} = 5$, $d_k = 0 \ (\lt 9)$.