Would you consider buying a used canon 5d Mark II in mint condition with 120 000 shutter actuations for 500$? The seller is a nice lady, she is a professional phototographer. Here are some photos of the camera that she has sent me.
When looking at used camera bodies, a question that comes up a lot, is "how many shutter actuations, or clicks, does it have?"   What I am wondering is, how many is too many. In other words, is there a reference for knowing when a sensor is more likely to go? How should I judge a camera body with 20000 vs 5000? Is that a significant difference, or is that not even relevant.  Obviously this might something that is dependent on model, so if there is some information based on model that would be even more interesting.  One of the guys at Canon Tech Support said that this is completely irrelevant for the Rebel series cameras, but he is obviously slightly biased. :)
This question is a possibly harder version of: .  Question.  Let $f,g :\mathbb R\to\mathbb R$, such that \begin{align} f(x-y)=f(x)\, g(y)-f(y)\, g(x), \tag{1}\\ g(x-y)=g(x)\, g(y)+f(x)\, f(y), \tag{2} \end{align} for all $x,y \in \mathbb{R} $. If $g$ is continuous at $x=0$ and not identically zero, then there exists an $\alpha\in\mathbb R$, such  $$ f(x)=\sin \alpha x\quad\text{and}\quad g(x)=\cos \alpha x. $$ Is there a pair of discontinuous $f$ and $g$ satisfying $(1)$ and $(2)$?  Update. If $\ell :\mathbb R\to\mathbb R$ is a linear functional over $\mathbb Q$ (i.e., $\ell(qx+ry)=q\ell(x)+r\ell(y)$, for all $x,y\in\mathbb R$ and $q,r\in\mathbb Q$), then $$ \sin\big(\ell(x)\big), \quad \cos\big(\ell(x)\big), $$ satisfy $(1)$ and $(2)$. Discontinuous such functionals do exist, and they are obtainable using Zorn's Lemma (equivalently the Axiom of Choice.) This takes care of the second question. Proving $\prod \limits_{k=0}^{n}(1-a_k) \geq1- \sum\limits_{k=0}^{n}a_k$ Clean way of building simple taxonomy browser of arbitrary depth Prove the inequality $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$ Proof that the set of irrational numbers is dense in reals Is it possible to trade Pokemon from the 1st generation games to the most recent ones? How to edit 'etc/hosts' file in non-rooted phone? iPhone Passcode for Powering Off I am looking for an option to require a passcode to power off my iPhone. That way, if the phone is lost, the person who finds it cannot simply turn it off, rendering Find My iPhone useless. Is there a way to accomplish this? jQuery .delay does not delay How to show google.com in an iframe? My PC has both Ubuntu and Windows 7 installed. I have installed all my drivers in Windows like Nvidia graphics, motherboard etc. Shall I install them in Ubuntu too?  I looked on the web for my Nvidia drivers and I have found a .run file. How could I install it on my Ubuntu? PHP replaces spaces with underscores In a question related to introductory Galois theory, I was asked to given an example of a tower of fields $F \subset K \subset E$ such that $E$ is a splitting field for some polynomial $f(x) \in F[x]$, but that $K$ need not be a splitting field. The example that comes to mind for me is $\mathbb Q \subset \mathbb Q(\sqrt[3]{2}) \subset \mathbb Q(\omega_3, \sqrt[3]{2})$.  So I know that the minimal polynomial for $\mathbb Q(\omega_3, \sqrt[3]{2})$ over $\mathbb Q$ is $x^3-2$. This polynomial does not split in the intermediate field $\mathbb Q(\sqrt[3]{2})$.  I wanted to make sure that I'm justifying that this intermediate field is not a splitting field correctly. Now I may be confused about the books terminology. When we say something is not a splitting field, we always have to refer to a specific polynomial correct? I've read a term called a normal extension which I think refers to one in which every polynomial splits, but my book doesn't mention these.   If splitting field means with respect to the polynomial $x^3-2$, then I've already demonstrated that $\mathbb Q(\sqrt[3]{2})$ is not a splitting field. Also, it couldn't be a normal extension either because that polynomial doesn't split.  Does it sound like I'm understanding this correctly or are there some subtleties I might be missing? Tor browser not opening I just checked my server's /var/log/auth.log and found that I'm getting over 500 failed password/break-in attempt notifications per day! My site is small, and its URL is obscure. Is this normal? Should I be taking any measures? Finite Abelian groups with the same number of elements for all orders are isomorphic Let $A$ and $B$ be finite abelian groups. Suppose that for every natural number $m$, the number of elements of order $m$ in $A$ is equal to the number of elements of order $m$ in $B$. Prove that $A$ and $B$ are isomorphic.  Idea Given that these groups are finite, I think you have to use the primary decomposition theorem somehow. Is there a way to find self-answers with the advanced search? Preferably, not using the  and including results of all time and not just the recent ones with the new share your knowledge, Q&amp;A-style Decompose vector fields on product manifolds So, I know that tangent bundle of a product manifold $M \times N$ splits in a sum $$ T_{(x,y)}(M \times N) = T_xM \oplus T_yN, $$ so that is obvious that the sum $X \oplus Y$ of smooth vector fields $X \in \mathcal{T}(M)$ and $Y \in \mathcal{T}(N)$ is a smooth vector field of $M \times N$. I've been told that, though not every vector field in $\mathcal{T}(M \times N)$ is a sum, locally one can always find one such decomposition, which will in turn be unique due the fact the sum is a direct one.   How can I show that this decomposition exists locally? More than that, if $X = X_1 + X_2$ is the decomposition, is there a way the express the coordinate functions of $X_1$ and $X_2$ in terms of those of $X$?   First I thought about taking two frames that locally spam $TM$ and $TN$ and write down $X$ using them, but then the coordinate functions are of the form $X^i: M \times N \to \mathbb R$, and the vector field components in each subspace are not exactly fields of $M$ and $N$ because their coordinate functions don't have the right domains. Is there another better way to see this decomposition holds locally? If $p,q,r$ are all primes,and $p|qr-1$,$q|pr-1$ and $r|pq-1$,find all possible values of $pqr$. When is a gerund supposed to be preceded by a possessive adjective/determiner? I assume that the following sentences are grammatically correct:   He resents your being more popular than he is. Most of the members paid their dues without my asking them. They objected to the youngest girl's being given the command position. What do you think about his buying such an expensive car? We were all sorry about Jane's losing her parents like that.   I'm still getting used to this possessive gerund structure. It sounded so weird to me at first. Is the structure used in both formal and informal contexts? Are there any alternative structures that result in the same meaning and are more frequently used? (Examples taken from )