I asked this question earlier and got a very useful hint, but I still can't get the full solution. Here is the original question:  I'm trying to show that for $b&gt;1$, $x&gt;0$ and $x$ irrational,  that $$\sup \{b^p:p\in Q,\;0&lt;p&lt;x\} = \inf \{b^q:q\in Q,\;x&lt;q\}$$  I know this follows immediately if we define $b^x=\exp (x\log(b))$  and use the fact that the exponential function is strictly increasing and  continuous over $\mathbb R$, but I'm not allowed to do that. In fact, I only have  that $b^{m/n}=(b^{1/n})^m$, where $b^{1/n}$ means the positive $n^{th}$ root of $b$;  $b^x$ is not yet defined for irrational $x$.  I know that the supremum and infimum both exist because if $b&gt;1$ and $p&lt;q$, then $b^p&lt;b^q$.  Also, by assuming that the inf $&lt;$ sup, I got a contradiction. The part I'm stuck on is deriving  a contradiction from assuming that sup $&lt;$ inf. Does anybody have any ideas?  The hint I got, courtesy of Greg Martin:  Try this one: let $S$ be the sup and $I$ be the inf. If $S&lt;I$, then set $\epsilon =I−S$. Choose $p&lt;x$ rational such that $b^p &gt; S-\frac\epsilon{100}$ (say) and $q&gt;x$ rational such that $b^q&lt;I+ \frac \epsilon{100}$. Then with $r=\frac12(p+q)$, use some inequalities to show that $b^r&gt;S$ and $b^r&lt;I$, contradiction.  This was my reply:  I have the inequality $S-\frac\epsilon{100}&lt;b^p&lt;b^r&lt;b^q&lt;I+\frac\epsilon{100}$, and I'm trying to show that for all $n$, $S-\frac\epsilon{n}&lt;b^r&lt;I+\frac\epsilon{n}$. This would show that $S&lt;b^r&lt;I$. But $p$ and $q$ depend on $n$, and $r$, being their midpoint, depends on $p$ and $q$, and hence $n$. How do I proceed from here? That is, how do I show that $r$ doesn't change when $n$ changes?  Can anybody help? Thank you.  The link to the question:
I'm trying to show that for $b&gt;1$, $x&gt;0$ and $x$ irrational, that $$\sup \{b^p:p\in Q,\;0&lt;p&lt;x\} = \inf\{b^q:q\in Q,\;x&lt;q\}$$ I know this follows immediately if we define $b^x=\exp(x\log(b))$ and use the fact that the exponential function is strictly increasing and continuous over $\mathbb R$, but I'm not allowed to do that. In fact, I only have that $b^{m/n}=(b^{1/n})^m$, where $b^{1/n}$ means the positive $n^{th}$ root of $b$; $b^x$ is not yet defined for irrational $x$.  I know that the supremum and infimum both exist because if $b&gt;1$ and $p&lt;q$, then $b^p&lt;b^q$. Also, by assuming that the inf $&lt;$ sup, I got a contradiction. The part I'm stuck on is deriving a contradiction from assuming that sup $&lt;$ inf. Does anybody have any ideas?  Thank you.
In Lubuntu, there's no 'Add New Program...' button in Desktop Session Settings. Is there an easy way to add new autostart programs in Lubuntu? Regex for parsing directory and filename I'm trying to write a regex that will parse out the directory and filename of a fully qualified path using matching groups.  so...  /var/log/xyz/10032008.log   would recognize group 1 to be "/var/log/xyz" and group 2 to be "10032008.log"  Seems simple but I can't get the matching groups to work for the life of me.  NOTE:  As pointed out by some of the respondents this is probably not a good use of regular expressions.  Generally I'd prefer to use the file API of the language I was using.  What I'm actually trying to do is a little more complicated than this but would have been much more difficult to explain, so I chose a domain that everyone would be familiar with in order to most succinctly describe the root problem. What do you call a person who does/thinks the opposite of the crowd for just that reason? Notepad++ incrementally replace Is the “sum of all natural numbers” unique? A while ago, there was a great hype about the “identity” $$\sum_{n=1}^{\infty} n = -\frac{1}{12}.$$  Apart from some series manipulations where the validity seems to be at least questionable, the derivation of this always goes through the zeta function:  Where the series converges, the zeta function is defined by $$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}$$ and outside that range by analytic continuation. And it turns out that inserting $s=-1$ formally results in $$\zeta(-1) = -\frac{1}{12} = \sum_{n=1}^\infty n$$  However looking at the series in isolation, there is no indication that the zeta function should be chosen.  An obvious way to get an analytic function that at one point gives the sum of all natural numbers is $$f(x) = \sum_{n=1}^{\infty} nx^n$$ at $x=1$, however (not surprisingly) that function diverges at $1$.  Therefore my question:     Is it possible to get another finite value for the series by analytic continuation of another series?   Concretely, do there exist continuous functions $f_1, f_2, f_3, \ldots$ such that   On some non-empty open subset $S$ of $\mathbb C$, $f(x)=\sum_{n=1}^\infty f_n(x)$ converges to an analytic function. At some point $x_0$, $f_n(x_0) = n$ for all positive integers $n$. The analytic continuation of $f$ is well defined and finite at $x_0$. $f(x_0) \ne -1/12$   What if we demand the functions $f_n$ to be analytic rather than just continuous? Double integral with transformation; possible error in limits I'm asked to show that the following integral is true with the transformation $u = x+y$ and $y=uv$:  $$\int_0^1 \int_0^{1-x} e^{y/(x+y)} dx dy = \frac{e-1}2 $$  I found the determinant of the jacobian of the transformation is $-u$, and the exponent is just $v$ so the integral becomes $$\iint_R |-u|e^v du dv = \frac {u^2}2 e^v|_R$$ $$=\frac{(x+y)^2}2e^{y/(x+y)}|_{x=0}^{1-x}\;|_{y=0}^1$$ And plugging it all in doesn't work (AFAICT). For one, the x's don't all cancel.  I've looked at it 100 times. It's definitely written dxdy not dydx even though there is definitely an x in the integral. Maybe this is a typo and why my calculation isn't working out? I tried reversing the order to dydx and it didn't work. I tried changing the limit of integration to $1-y$ and it didn't work out.  I'm not even sure how to figure out the domain of integration in the uv-plane. Flipping things around yields y=uv and x=u-uv. But if my domain is bounded by x=0 and y=0 and y=1 (ignoring the last limit for now), then I have uv=o=u and uv=1. !?!  Am I just making an algebra mistake plugging everything in? Is there a way to handle a variable in it's own integral that I'm unaware of? How do I show for an ideal transformer $M^2=L_1L_2$? I've been stuck on this problem for about an hour.     In an ideal transformer, the same flux passes through all turns of the primary and of the secondary . Show that in this case $M^2=L_1L_2$, where $M$ is the mutual inductance of the coils, and $L_1$, $L_2$ are their individual self-inductances.   My first instinct was the following: If we isolate the response of each coil to the other one, we get the equations $N_1\Phi_1=MI_2$ and $N_2\Phi_2=MI_1$, where $\Phi$ is magnetic flux going through a single turn of the coil and $I$ is current. Likewise, if we isolate the response of each coil to itself, we get the equations $N_1\Phi_1=L_1I_1$ and $N_2\Phi_2=L_2I_2$. I would now naturally want to say "Now just multiply $N_1\Phi_1$ and $N_2\Phi_2$ both ways, and get the desired result!" i.e.,  $$(N_1\Phi_1)(N_2\Phi_2)=(MI_2)(MI_1)=(L_1I_1)(L_2I_2)\implies M^2=L_1L_2$$  But that's incorrect, because the response of the coils to themselves and to the other coil are different quantities (i.e. not necessarily equal).  Realizing this, I write down the more general equations.  $$N_1\Phi_1=MI_2+L_1I_1$$ $$N_2\Phi_2=MI_1+L_2I_2$$  Now even with the hypothesis that $\Phi_1=\Phi_2=\Phi$, I still can't see how the identity $M^2=L_1L_2$ follows. Could you help clear up my confusion?  *Taken from D. Griffiths Introduction to Electrodynamics (4th ed.), Problem 7.58. Why is log-of-sum-of-exponentials $f(x)=\log\left(\sum_{i=1}^n e^ {x_i}\right)$ a convex function for $x \in\mathbb R^n$? How to prove $f(x)=\log\left(\displaystyle\sum_{i=1}^n e^{x_i}\right)$ is a convex function?  EDIT1: for above function $f(x)$, following inequalities hold:  $$\max\{x_1,x_2,\ldots,x_n\}\leqslant f(x)\leqslant\max\{x_1,x_2,\ldots,x_n\}+\log n$$  and I have tried proving its convexity via definition of a convex function with above inequalities, but that didn't work.  EDIT2: I have posted my answers below. Objects with transparent texture not always show transparency in 3dview For some reason objects don't show their transparency all the time.  In this image the circles in red are suposed to be transparent like the ones in green.    When I select the object in the back, all objects in front of it become opaque.    When I select the two objects that are opaque, then they become transparent as they sould be.    The display is in GLSL mode  These are the material and texture properties      I'm not worried about the final render, just the editing view. I would like to see the actual transparency at editing time only, because the scene is to be imported in a third party game engine.  Blender is v2.68   is the blend file Analysing Bounce Rate and Time on Page Invalid Topology [Topoengine error] from ArcGIS for Desktop? Using ArcGIS 10.1 for Desktop, I'm trying to run the Identity function on two large data sets but it fails every time. I don't think it is a lack of hardware as I can run other processes on the same data sets.   I've already updated software, and added the 64 bit geoprocessing add-in, checked and repaired geometry, and done the fix in  (set ARCTMPDIR env variable to a writable local folder).   Executing: Identity DudleyStamp OA2011 "D:\Case Studies\National2\National.gdb\DudleyStampOA11" ALL "1 Meters" NO_RELATIONSHIPS  Start Time: Wed May 08 21:11:43 2013 Reading Features... Processing Tiles... Assembling Tile Features... ERROR 999999: Error executing function. The table was not found. The table was not found. [DudleyStampOA11] The table was not found.   The table was not found. [DudleyStampOA11]   Invalid Topology [Topoengine error.] Failed to execute (Identity). Failed at Wed May 08 21:45:41 2013 (Elapsed Time: 33 minutes 58 seconds)   Based on comments so far I tried copying data to new file geodatabase, and running the tool from ArcCatalog still get exactly the same error.  I have ordered 16GB of the fastest RAM my computer can take (I current have 4GB). I also have a 6 Core AMD Phenom II X6 1090T, SSD, and AMD Radeon HD 6970 Graphics Card. So I'm not sure what more hardware I can add.  I've also tried increasing the size of the page file which was a suggestion I found on Google with no effect. I just read this very insightful post , where the author stated that the variance of $\hat\beta$ is:  $$\text{var}(\hat\beta) = \sigma^2(\textbf{X}^\prime \textbf{X})^{-1}.$$  I couldn't figure out why it is like this. Can anyone elaborate a bit? Get real client IP in a Servlet How are the best director award winners chosen? I do not understand how the best director awards are determined each year. It seems impossible to assess the method of a movie director unless you are part of the crew.  So, how do the nominators/judges assess the directors' performance each year? What counts as a 'large' raid 5 array? Is it possible to Pivot data using LINQ? jQuery doesn't submit a form Prove that if $n$ is not the square of a natural number, then $\sqrt{n}$ is irrational. Let $m$ be the largest real root of the equation $\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4$ find $m$ Let $m$ be the largest real root of the equation $$\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4.$$ Find $m$.   do we literally add all the fractions or do we do something else  I have no clue how to solve this Express partial derivatives of second order (and the Laplacian) in polar coordinates $z=f(x,y)$ where $x=rcosθ$ and $y=rsinθ$  Find $ \frac{\partial z}{\partial x}$ and $ \frac{\partial^2 z}{\partial x^2}$  I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have...  $ \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}$. So I'm guessing that $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question.   Full Question: