$\newcommand{\N}{\mathbb N}$Exercise 3.5.12.  This exercise will establish a rigorous version of Proposition 2.1.16. Let $f : \N×\N → \N$ be a function, and let $c$ be a natural number. Show that there exists a function $a : \N → \N$ such that $a(0) = c$ and $a(n+1) = f(n,a(n))$ for all $n ∈ \N$, and furthermore that this function is unique.  (Hint: first show inductively, by a modification of the proof of Lemma 3.5.12, that for every natural number $N \in \N$, there exists a unique function $a_N \colon \{n ∈ \N : n ≤ N\}\to\N$ such that $a_N(0) = c$ and $a_N(n+1) = f(n,a_N(n))$ for all $n \in \N$ such that $n&lt;N$.)    Lemma 3.5.12 (Finite choice). Let $n \ge 1$ be a natural number, and for each natural number $1 \le i \le n$, let $X_i$ be a non-empty set. Then there exists an $n$-tuple $(x_i)_{1\le i\le n}$ such that $x_i \in X_i$ for all $1 \le i \le n$. In other words, if each $X_i$ is non-empty, then the set $\prod_{i\le i\le n} X_i$ is also non-empty.   Proof  We induct on $n$. Starting with the base case $n = 1$; the claim     follows from Lemma 3.1.6 (why?) now suppose inductively that the claim     has been proven for some $n$, we will prove it for $n+1$.  Let $X_1,\dots,X_{n+1}$ be a collection of non-empty sets.     By the inductive hypothesis, we can find an $n$-tuple $(x_i)_{1\le i\le n}$     such that $x_i \in X_i$ for all $1 \le i\le n$. Since $X_{n+1}$ is     nonempty, by Lemma 3.1.6 we may find an object $a$ such that $a \in  X_{n+1}$. If we thus define the $(n+1)$-tuple $(y_i)_{1\le i \le n+1}$     by setting $y_i := x_i$ when $1 \le i \le n$ and     $y_{n+1} := a$, it is clear that $y_i \in X_i$ for all $i$     with $1 \le i \le n+1$; this proves the claim.    My attempted solution was to consider the function $a_N$ as an $N$-tuple and use mathematical induction on $N$:  Base case: $a_0(0) = c$ by definition.  Inductive hypothesis: $a_N(0)=c$ and $a_N(n+1) = f(n,a(n))$ for all $n \le N$. It follows from the order of $\N$ that there exists an $n$ such that $a_N(N) = f(n,a(n))$.  Then I show by the same method of the lemma that $a_{N+1}(n) = a_N(n)$ for $n \le N$ and that $a_{N+1} = f(N,a(N))$ from the inductive hypothesis. Is my attempt true. If not I would consider a proof that follows the hint.  Note The question was asked here      but answers doesn't follow the hint.
Exercise 3.5.12. Let $ f: \mathbb N \times \mathbb N \to \mathbb N$ be a function, and let c be a natural number. Show that there exists a function $a:\mathbb N\to \mathbb N$ such that $$a(0)=c$$ and $$a(n++)=f(n,a(n))\;for\;all\;n\in \mathbb N,$$ and furthermore that this function is unique. Tao gives the hint: first show inductively, by a modification of the proof of Lemma 3.5.12, that for every natural number $N\in \mathbb N$, there exists a unique function $a_N:\{ n\in \mathbb N:n\lt N\}\to \mathbb N $ such that $a_N(0)=c$ and $ a_N(n++)=f(n,a(n))$ for all $n\in \mathbb N$ such that $n\lt N$.  In my opinion, we could solve it by proving that $\forall n\in \mathbb N,a(n)\in \mathbb N$ according to mathematcial induction.  The property P(n) is:$$\forall n\in \mathbb N, a(n)\;is\;unique$$ P(0) is true. And if P(n) holds, P(n++) is true because different natural numbers must have different successors. $$a(n++)=f(n,a(n))$$ $$a(m++)=f(m,a(m))$$ $$n++=m++\Rightarrow n=m\Rightarrow a(n++)=a(m++)$$ It's similar to the proof of  Proposition 2.1.16 (Recursive definitions): Suppose for each natural number $n$, we have some function $f_{n} : \mathbb{N} \rightarrow \mathbb{N}$ from the natural numbers to the natural numbers. Let $c$ be a natural number. Then we can assign a unique natural number $a_{n}$ to each natural number $n$, such that $a_{0} = c$ and $a_{n{++}} = f_{n} (a_{n})$ for each natural number $n$.  This proof indicates both the existence and uniqueness of the function $a$. I don't know why Tao says it's informal.  But regarding the hint, how do we get the function f even if we've got the conclusion from the hint. Also it seems like the function a(n) can be diffentent considering different N. I don't know if the hint is correct.
Do questions about specific StackOverflow implementation details belong on SO or Meta? has several discussion comments going back and forth about whether the question belongs on SO or Meta.  The question is asking about how a specific feature of the site is implemented from a coding perspective (so that the OP might duplicate it in their own implementation), which can make it a legitimate programming question.  According to the Meta FAQ: "If your "question is about ... Stack Overflow / The core Stack Exchange engine", it should be asked here.  Here are some similar example questions on Meta:         Where do these questions belong? Do I need a transit vsa in Kuwait; what about checked baggage? I have a client that has a Links page in SharePoint 2013. On the page is a column called URL that is set as a hyperlink column, but they want the column to accept pw://example.com/... instead of just ...  Is there a way to change the hyperlink column to allow for this behavior? 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Edit:  The reason that I'm looking for a one-line expression here is not that I like incredibly terse code, but because I'm having to write a lot of code like this:  x = get_first_list() if x:     # do something with x[0]     # inevitably forget the [0] part, and have a bug to fix y = get_second_list() if y:     # do something with y[0]     # inevitably forget the [0] part AGAIN, and have another bug to fix   What I'd like to be doing can certainly be accomplished with a function (and probably will be):  def first_item(list_or_none):     if list_or_none: return list_or_none[0]  x = first_item(get_first_list()) if x:     # do something with x y = first_item(get_second_list()) if y:     # do something with y   I posted the question because I'm frequently surprised by what simple expressions in Python can do, and I thought that writing a function was a silly thing to do if there was a simple expression could do the trick.  But seeing these answers, it seems like a function is the simple solution. Prevent Windows 10 from installing/updating a specific driver? It seems that the recent Synaptics driver update for Synaptics SMBus TouchPad on Windows 10 (build 100049) somehow doesn't see my middle-button. Uninstalling, and installing an older driver for Windows 8.1 from the HP site (my machine is HP EliteBook 8570w) restores the middle button functionality.    Few days later, the driver updates itself again, and I lose the middle-button.  I realize I'm running unsupported software, but is it possible to prevent updating this particular driver, without either disabling updates altogether or preventing other drivers from updating? Clarification for non-answers One of my flags for "non-answer" was declined and I was wondering if I misunderstood what "non-answer" means. 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Thanks in advance!  ftp -v -n $SERVER  &gt;&gt; $LOG_FILE &lt;&lt;-EOC             user $USERNAME $PWD             binary             cd $DIR1             mkdir $dir_lock             get $FILE             bye EOC Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$ I've been looking at  $$\int\limits_0^\infty  {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$  It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:  $$\displaystyle \int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$  $$\int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$  $$\int\limits_0^\infty  {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$  So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $\dfrac{1}{4}$, $\dfrac{1}{3}$ and $\dfrac{2}{5}$. Is there any way of finding it? I'll work on it and update with any ideas.    UPDATE:  The integral reduces to finding   $$\int\limits_{ - \infty }^\infty  {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt} $$  With $a =\dfrac{n+1}{m}$ which converges only if   $$0 &lt; a &lt; 1$$  Using series I find the solution is     $$\sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} $$ How can I change the color of the lines (the frame) in a tabular environment? In the following MWE:  \begin{tabular}{c|c|c} a&amp;b&amp;c \\ \hline d&amp;e&amp;f \\ \hline g&amp;h&amp;i \end{tabular}   I would like the horizontal and vertical lines of the table to have, say a blue color.