Prove by induction that all coins greater than or equal to 8 pence  can be made using 3 pence and 5 pence coins .  Here is my thought : I looked at Z+ greater than 8 ... I considered multiples of 3(equivalent to using 3p coins with repetitions) and considered multiples of 5. Rest of them were mostly primes where I had to combine both 3 and 5 .  And I noticed that 3 and 5 are in the form of 4k+1 and 4k-1.  I would appreciate any information . Thanks
How would I prove that for all elements of a set {n | n ∈ ℕ ∧ n &gt; 7} (all natural numbers greater than 7), there are multiples of 3 and 5 which, when added up, are equal to that element of the set?  A := {n | n ∈ ℕ ∧ n &gt; 7} ∀ n ∈ A: n = x*3 + y*5   I do understand why this is true:   8 = 1*3 + 1*5 9 = 3*3 + 0*5 ...  15 = 0*3 + 3*5   and this pattern repeats once 16 is reached, as 8 can be replaced with 1*3 + 1*5, so I'd basically have to add 1 to x and y for every time 8 fits in n.  I am however not sure how I can formulate this mathematically correct.  Thanks in advance, CrushedPixel
Does text in brackets invoke the "an" vs. "a" rule? I would like to place a lower limit under two consecutive \sums, somewhere in between. So I have \sum_{j=1}^{\infty}\sum_{k=1}^{\infty} and would like to place j&gt;k in the lower limit between the two sums. I believe the multiple integral operator has this kind of functionality. How do I convert Word smart quotes and em dashes in a string? I'm having a difficult time understanding this statement. Can someone please explain with a concrete example? How to Uninstall Blender Software I installed Blender for a project working on and it's way more than I need. I'd like to uninstall it, but can't figure out how to do so. It's not showing up in Programs (using Windows 9) and I can't find instructions on how to remove it in the Blender documentation. Any assistance would be greatly appreciated. Thanks. Evaluating $\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$ The top bar is showing (again) the suggested edits count and 10K flags count, like it was before.  On Stack Overflow it looks like this: (that's the original color, same as in old top bar)        And on this very Meta site:        Same top bar. But different colors! Consistency broken.  I've just noticed the suggested edits count on Meta:        Black on black, I can see just a floating number - really confusing and out of any context.  In the name of Consistency, can we please have the same color on all sites? I have the problem, that PHP replaces all spaces with underscores in POST and GET variables. For example if I have the URL: http://localhost/proxy.php?user name=Max the browser will convert it to http://localhost/proxy.php?user%20name=Max. But if I give the $_GET parameters out, the key is not user name but user_name (note the underscore)! Is there any possibility to change this behaviour? How to find all questions that were duplicates of another question? I am interested to see a list of all questions that were closed because being a dup of another question.  I strongly want to research a particular question and reading all the answers inside the dups will help me a lot.  Is there such feature? if not, this is a feature request Calculate $E(X^{2n})$ where $X$ is normal (0,1) I need help proving the following:  Let $X$ be normally distributed with parameters $\sigma=0$ and $\mu=1$. Let $n$ be a positive integer. Show that:  $$E(X^{2n})=\frac{(2n)!}{2^nn!}=:(2n-1)!!$$  I've tried the change of variables $Y=X^2$ and then calculating the integral, but got nowhere. Any hints?  Thanks in advance! `lmodern` package reduces the size of automatically sized delimiters in math mode Using pgfplots, why do I get "undefined control sequence" when trying to use a foreach variable as a parameter to addplot? Since I have multiple plots with error bars, I would like to shift the error bars slightly to make the overall plot more readable. However, when I try to do this using the variable of a \foreach as the amount to shift, I get an "undefined control sequence" error.  \documentclass{article}  \usepackage{tikz} \usepackage{pgfplots}  \begin{document}  \begin{tikzpicture} \begin{axis} \foreach \x/\y in {a/-1cm, b/0cm, c/1cm} {     \addplot+ [         %every error bar/.append style={xshift=\y},         %every node/.style={xshift=\y},         error bars/y dir=both,         error bars/y explicit,     ] coordinates {         (0, 0) +- (0, 1)         (1, 0) +- (0, 1)         (2, 0) +- (0, 1)     }; } \end{axis} \end{tikzpicture}  \end{document}   (The document does not compile when I uncomment either one of the two comments.)  What's wrong here?  EDIT: Added another variable to the \foreach. In my actual use-case it would contain the filename to plot data from or the label to use for the curve. If $n\geq m$ then $(x^m+y^m)^{1/m} \ge (x^n+y^n)^{1/n}$ What laptop specs are important for intensive image editing work? How important is the following in purchasing a laptop that will be used for image editing:   processor speed amount of RAM integrated vs. dedicated graphics card HD speed (SSD; or 5400/7200/10000 rpm) The book says:     Concentration. Some spells require you to maintain concentration in order to keep   their magic active. If you lose concentration, such a spell ends.   So what happens if, for example, you are already concentrating on a spell, and cast Witch Bolt without concentrating. Would you get the initial arc of blue energy, or would it not even go off?  I am asking if you can cast a spell that says concentration without concentrating for an immediate effect. This would be useful to avoid interrupting your existing concentration. In essence, if you are concentrating on thing A (doesn't have to be a spell since other things need concentration), can you cast a concentration spell getting an immediate effect without losing concentration on thing A?  Since the definition of concentration is to keep the spell active (meaning it is already active) I would think that you would get the initial effect (if it had one) without the need to concentrate on it. Give a combinatorial proof to show the following for all integers $n \geq 2$. $$2^{n-2} n (n -1) = \sum\limits_{k=2}^n k (k - 1) \binom{n}{k}.$$  I'm completely stumped. I just have no idea how to do this. What I've tried so far has been simplifying the right hand side slightly to $\sum\limits_{k=2}^n \binom{n}{k - 2}$, and then proving that this somehow is equal to some manipulation of the left hand side. I can't seem to find any breakthroughs that way though.  If anyone could help out or point me in the right direction that would be great.  EDIT: I've actually just realized my simplification of the right hand side is wrong. Now I'm even more stuck. What is a word/phrase for using a term for a popular special case instead of a generic term? I have a set of coordinate points. The points are distributed in such a way that some points are clustered around a specific area. There is an option to find out mean coordinates(s) of the whole point layer. I would like to know if there is any way in QGIS to find out the mean coordinates of the points from a specific area from a point layer. Acount suspension reason is only shown on the main site profile not meta How do we know an $ \aleph_1 $ exists at all? I have two questions, actually. The first is as the title says: how do we know there exists an infinite cardinal such that there exists no other cardinals between it and $ \aleph_0 $? (We would have to assume or derive the existence of such an object before we label it something like $ \aleph_1 $.)  My second question is, can we say for certain if there's any limit to the number of cardinals existing between $ \aleph_0 $ and continuum (i.e. $ 2^{\aleph_0} $)? I mean, how could we know that there&#39;s not an infinite number of cardinals between the two - perhaps even more than $ \aleph_0 $?