So i took my senior pictures and it said it was at 72 dpi. My school said they needed pictures to at least 300 dpi. I researched and found out that 300 dpi is the best printing quality. I changed the dpi from 72 to 300 but i did not resample the image so the length and width of the picture remained the same. Is this something good or bad? I'm scared that I did something wrong and that my picture will be blurry in the yearbook.  Help please!
I have two cameras, a 6mpix Panasonic FZ8, and 14mpix Canon A2200.  When I compare shots from two cameras, one of differences I notice is dpi number that camera reports in JPG info:   FZ8 reports dpi=72 A2200 reports dpi=180   What physical meaning can these numbers have? I am at loss for guesses.    I am fairly familiar with notion of dpi in scanning and printing. I can calculate density of pixels on the sensor of the camera. But then, the linear density of pixels on the sensor will be hundreds times larger than number above. So what, if anything, does it mean?
What is the best way to draw figures with canvas and set their angle? Why is the inclusion of the tensor product of the duals into the dual of the tensor product not an isomorphism? Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$? How to differentiate formally? How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n} $ I have put his on wolfram and obtained answer as follows:  $\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$  And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$  However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper? $m$ is the Lebesgue measure. I was thinking that: $\Bbb Q$ is dense and $m(\Bbb Q)=0&lt;1$ but it fails to be open, but maybe I could construct an open set from this fact, or also using that $(0,1)$ is a dense and open subset but $m(0,1)=1$.   Any hints or ideas will be very appreciated. Thanks. How to simulate traceroute using ping? I am learning computer networks and looking for a way to simulate traceroute using the ping command. I think it has something to do with ttl, but I don't know how to develop that further. Find all polynomials $P(x)$ such that $P(x^2)=P(x)^2$ Find all polynomials $P:\mathbb{C}\rightarrow\mathbb{C}$ such that $$P(x^2)=P(x)^2 .$$   Here is what I tried:  First, it is easy to see the constant solutions, namely $P\equiv 0,P\equiv 1$.  Let $r$ be a root of $P$ (i.e. $P(r)=0$). It follows that all terms in the infinite sequence $r,r^2,r^4,\dots,r^{2n},\dots$ are roots of $P$. To avoid a polynomial with infinite roots, we have to have the sequence be periodic. So it turns out that either $r=0$, or the roots of $P$ are roots of unity of degree $2n$ for some $n\in\mathbb{N}$.  In the first case, we get $P(x)=xQ(x)$. Plugging in, we get $$x^2Q(x^2)=x^2Q(x)^2\implies Q(x^2)=Q(x)^2 ,$$ i.e. that $Q$ satisfies the same condition as $P$.  In the second case, we have $P(x)=(x^{2n}-1)Q(x)$. Plugging in, we get $(x^{2n}-1)Q(x^2)=(x^{2n}-2x^n+1)Q(x)^2$. I was not sure how to proceed from here. What intuitive notion is formalized by "natural transformation" in category theory? Space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable iff $(\Sigma, \rho_\Delta)$ is separable Word for killing off or attempting to kill off an entire bloodline? PDOException “could not find driver” How many squares actually ARE in this picture? Is this a trick question with no right answer? This is one of those popular pictures on sites like Facebook. I always see a huge variation of answers such as $8, 9, 16, 17, 24, 28, 30, 40, 41, 52,$ etc., yet I've never seen a definitive answer on any website.    I counted $16$ whole squares ($4$ going down on left side, $4$ going down on right side, $8$ small squares in the centre). With this method the answer could also be $17$ if you want to count the entire shape as one big square.  I can also see the answer being $24$ if you count the smaller squares separately, as if they were floating above the main grid instead of intersecting with it.  Is there a right answer, or is it possible that there is no 'right' answer and it just depends on your counting method? How can I import a .c4d static model (Cinema 4d) on Blender? Is there any plugin could help me? I didn't notice anything on Internet. Short story: rocket fuel which oxidizes iron destroys life on earth I'm looking for the title of a short science fiction story, probably published in a Hugo winner collection 30 years ago. In the story, scientists have developed an extremely efficient rocket fuel that has one dangerous side-effect: it oxidizes iron. A man takes off in a rocket powered by this fuel but returns to find the earth completely devoid of life. His subsequent investigation reveals that a single drop of the fuel began a domino effect, as the substance oxidized all iron it came into contact with. In the end, the man is dying.  He drags himself to the ocean, believing that if he can get his living cells into the water they will restart the primordial soup. Do people sometimes get blocked by a webservice for things someone else has done with their dynamic IP-Address It is common practice for operators of web services (e.g. web sites) to block IP addresses when there was malicious traffic originating from that address.  It is also common practice that internet users don't have a static IP address but get one assigned by their ISP via DHCP.  Now I would expect that it sometimes happens that a specific IP address gets banned from a service and later that address gets assigned to another user, leaving the new user (temporarily) banned from the service for no apparent reason. However, I never had this problem or heard of someone having that problem.  Does this sometimes happen? If not, what prevents this from happening? Image with axis How to convert a column number (e.g. 127) into an Excel column (e.g. AA) How do you convert a numerical number to an Excel column name in C# without using automation getting the value directly from Excel.  Excel 2007 has a possible range of 1 to 16384, which is the number of columns that it supports. The resulting values should be in the form of excel column names, e.g. A, AA, AAA etc. What kind of power management circuit do I need to embedded a rechargeable battery in my electronics project? I am doing a project where my electronics runs on 5V which I wanted to power from a 7.2v lipo regulated down to that voltage. I'd really like to include the opportunity to charge the lipo without removing it from the project. What kind of circuit/components do I need to do it properly?  Say, would putting a battery charging ic at the battery's terminals in parallel to my system be enough to achieve what I need? In what occasion could the word “precious” be taken disapprovingly, or sarcastically? I understand that the word “precious” in the following quote of Maureen Dowd’s article “Heart of Darkness” (NYT, March 20) literally means “2. valuable or important and not to be wasted” as defined in OALD:     Congressman Jones read an e-mail from a former boss of General   Allen’s, giving the congressman this unvarnished assessment:   “Attempting to find a true military and political answer to the   problems in Afghanistan would take decades. Would drain our nation of   precious resources, with the most precious being our sons and daughters. Simply put, the United States cannot solve the Afghan   problem, no matter how brave and determined our troops are.   But I didn’t know “precious” has another meaning as defined in OALD: 5. (disapproving, people and their behavior) very formal, exaggerated and not natural, and Readers English Dictionary: 2.(colloquial) terrible, almost worthless, until I was told by my respected forum mate a few days ago.   Being encouraged by his suggestion for me to “ask another question if I’m not sure of what he means by describing his input as “precious” is actually rather “precious” phrasing, I would like to ask:   What is an example of “precious” being used as “disapproving” or sarcastic way (as neither OALD nor CALD provides examples)? Is there an easy way to discern the instance “precious” being used as “disapproving” from the case being used in the sense of genuine “great value” as used in the above quote other than judging from the context? Explicit bijection between the successor of $\alpha$ and $\alpha$ (infinite ordinal). I would like to construct an explicit bijection from $\alpha^*$ to $\alpha$, where $\alpha$ is an infinite ordinal. I have been told a possible bijection would be:  \begin{equation} f(\beta) = \begin{cases} 0 &amp; \text{if } \beta = \alpha \\ \beta^* &amp; \text{if } \beta \text{ is finite} \\ \beta &amp; \text{otherwise} \end{cases} \end{equation}  I don't understand how this is constructed, or why it is a bijection.