Let $K \subset\mathbb R^{n}$ a compact set and $f : K \rightarrow\mathbb R^{m}$ a continuous and one-to-one function. Show that the function $f^{-1} : f(K) \rightarrow K$ it is continuous. Hint: By hip $K$ is compact and f is injective, so $f(K)$ is compact, then by the Heine-Borel theorem is closed and bounded.
This problem is bugging me for some time.     $f:(X,\mathcal{T}) \to (Y,\mathcal{T}')$ is a continuous bijection where $(X,\mathcal{T})$ is compact and $(Y,\mathcal{T}')$ is $T_2$ (i.e. Hausdorff and $T_1$). Then show that $f$ must be a homeomorphism.   So far I have failed to understand any relation between the compactness of domain and $T_2$-ness of the range. I tried in vain to prove that $f(U)$ is open in $(Y,\mathcal{T}')$ whenever $U$ is so in $(X,\mathcal{T})$. I know $f$ makes $X$ and $Y$ respectively $T_2$ and compact and hence makes both of them $T_4$ spaces. But that is not coming to any use. I would be glad if someone helps me with a solution. Thank you.
found a 1994 uncashed profit sharing retirement plan check I found an old un-cashed check dated 1994 for $487 from my Xerox Retirement plan with no expiration date. Can I still cash it? How to convert image of map into vector format? Normalizing point cloud data How I can get Digital Height Model (only object heights, DSM - DTM = DHM)?  Many applications can do this, but they convert it to GRID or TIN format. I was wondering if it was possible to keep the data as point clouds. How to view monitor refresh rate in macOS? I just purchased a Dell P2175Q monitor for my 15" MBP w/Touchbar. I'm connecting with a USB-C to DisplayPort cable, but when I go to Apple > System Preference > Displays, there is nothing about the Refresh Rate. Pressing the Option key does not show any additional options.  I've seen countless screenshots where this information should appear. Is there another way to check the refresh rate?  I'm on macOS 10.13.4  Thanks. Sometimes we assume that regressors are fixed, i.e. they are non-stochastic. I think that means all our predictors, parameter estimates etc. are unconditional then, right? Might I even go so far that they are no longer random variables?  If on the other hand we accept that most regressors in economics say are stochastic because no outside force determined them with some experiment in mind. Econometricians then condition on these stochastic regressors.  How is this different from treating them as fixed?  I understand what conditioning is. Mathematically, it means we make all observations and inference conditional on that particular set of regressors and have no ambitions to say that inferences, parameter estimates, variance estimates etc. would have been the same had we seen a different realization of our regressors (such is the crux in time series, where each time series is only ever seen once).  However, to really grasp the difference between fixed regressors vs. conditioning on stochastic regressors, I am wondering if anyone here knows of an example of an estimation or inference procedure that is valid for say fixed regressors but breaks down when they are stochastic (and will be conditioned on).  I am looking forward to seeing those examples! Efficiently replace all accented characters in a string? For a poor man's implementation of near-collation-correct sorting on the client side I need a JavaScript function that does efficient single character replacement in a string. Here is what I mean (note that this applies to German text, other languages sort differently):  native sorting gets it wrong: a b c o u z ä ö ü collation-correct would be:   a ä b c o ö u ü z  Basically, I need all occurrences of &quot;ä&quot; of a given string replaced with &quot;a&quot; (and so on). This way the result of native sorting would be very close to what a user would expect (or what a database would return). Other languages have facilities to do just that: , in , , . But what about JavaScript? Here is what I have right now. // s would be a rather short string (something like  // 200 characters at max, most of the time much less) function makeSortString(s) {   var translate = {     &quot;ä&quot;: &quot;a&quot;, &quot;ö&quot;: &quot;o&quot;, &quot;ü&quot;: &quot;u&quot;,     &quot;Ä&quot;: &quot;A&quot;, &quot;Ö&quot;: &quot;O&quot;, &quot;Ü&quot;: &quot;U&quot;   // probably more to come   };   var translate_re = /[öäüÖÄÜ]/g;   return ( s.replace(translate_re, function(match) {      return translate[match];    }) ); }  For starters, I don't like the fact that the regex is rebuilt every time I call the function. I guess a closure can help in this regard, but I don't seem to get the hang of it for some reason. Can someone think of something more efficient?  Answers below fall in two categories:  String replacement functions of varying degrees of completeness and efficiency (what I was originally asking about) A  of  , which is now  among JS engines (not so much at the time of the question) and could solve this category of problem much more elegantly. Is "TABLESAMPLE BERNOULLI(1)" not very random at all? I found this question in a book.   How many terms of the Maclaurin expansion of $(1+x)^{82}$ are needed to guarantee finding a value of $1.1^{82}$ to an accuracy of $10^{-6}$?  This is how I tried to do it:  We want the Lagrange error to be less than $10^{-6}$:  $$R_{n}(0.1)&lt;10^{-6}$$  Now to find $R_{n}(0.1)$:  $$R_{n}(0.1)=\frac{f^{(n+1)}(0.1)\cdot0.1^{n+1}}{(n+1)!}$$  After writing the values of ${f^{(n+1)}}(x)$, I find a pattern and say that  $${f^{(n+1)}}(x)=\frac{82!}{(82-n)!}(1+x)^{82-n}$$  $${f^{(n+1)}}(0.1)=\frac{82!}{(82-n)!}(1.1)^{82-n}=\frac{82!(1.1)^{82-n}}{(82-n)!}$$  Now just substitute all of this back to $R_{n}(0.1)&lt;10^{-6}$:  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}&lt;10^{-6}$$  Great! Now I have an inequality that I can solve. I can't graph this so I just use trial and error (which seems like quite an intensive way to do it, so if anyone has a better way I'd love to know).  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}-10^{-6}&lt;0$$  This seems to work first for $n=28$. That means the highest power of $x$ that I need is 28, so the number of terms in the expansion needs to be $n+1=29$. But the book's answer is       30   What am I doing wrong here? I'm still quite new to the power series expansions so if there's anything unclear in my working, please let me know (also, as a side note, the book itself may be wrong, though that isn't very likely I'd assume). The three files are read in this order?  .bash_profile .profile .bashrc   This is not happening when I first open a terminal.   I have trace statements in the files appending to the file init.log. Please take a look at the following. It begins after opening a terminal. I have placed a comment to show where the log picks up after the su command.  stephen@debian:~$ cat init.log reading .bashrc done reading .bashrc stephen@debian:~$ su - stephen Password:  stephen@debian:~$ cat init.log reading .bashrc done reading .bashrc # # after su # reading .bash_profile reading .profile reading .bashrc done reading .bashrc done reading .profile done reading .bash_profile stephen@debian:~$    So the su - login triggers the expected sequence however the initial login reads only the bashrc. This cannot be correct. Can someone explain under what conditions this would occur. I could modify the bashrc and profile files so that the initial read includes all expected files but I would rather get to the root of the problem and fix it there. Can I leave South Africa before my 90 days expire, and return after a few days to stay for another 90 days? How many times can an Australian leave and re-enter South Africa (90 days at a time without a visa) in a year, and do they need to be out of the country for a certain period of time before they re-enter?  I want to be in South Africa for as long as I can, however as an Australian without a visa I can only stay for 90 days at a time. What I would like to do is simply leave South Africa and go to another country in Africa for a few days, before returning with a new 90 day allowance.    Is this possible, or must I wait a certain period before re-entering South Africa/is there a limit to the amount of times I can do this in one year? How do I calculate the power consumed by a lightbulb? I'm studying a lightbulb and its variable resistance, given by the expression:  $R(T) = Ro[1 + α(T-T_0)]$, where $R_0$ is the resistance of the lamp at $T_0$.  In this case, $R$ is not given by Ohm's law ($V=Ri$). So, which expression can I use to calculate the power consumed by a lightbulb?   $P = R\cdot i^2$ $P = \frac{V^2}{R}$ $P = V\cdot i$ (I don't think I can use this one)   I have measured the current and the e.p.d. at the bulb using instruments. If $\int_{0}^{x}f(t)dt\rightarrow \infty$ as $|x|\rightarrow \infty\;,$ Then every line $y=mx$ Intersect $ y^2+\int_{0}^{x}f(t)dt=2$ If $f$ is a continuous function with $\displaystyle \int_{0}^{x}f(t)dt\rightarrow \infty$ as $|x|\rightarrow \infty\;,$ Then show that every line $y=mx$ Intersect the curve $\displaystyle y^2+\int_{0}^{x}f(t)dt=2$  $\bf{My\; Try::}$ Put $y=mx$ in $$\displaystyle y^2+\int_{0}^{x}f(t)dt=2\;,$$ We get $$\displaystyle m^2x^2+y^2+\int_{0}^{x}f(t)dt=2.$$ We can write it as $$\displaystyle 2\int_{0}^{x}2m^2tdt+\int_{0}^{x}f(t)dt=2$$ So $$\displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2 =0$$ Now Let $$g(x) = \displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2.\;,$$ When $x\rightarrow \infty\;,$ Then $g(x)\rightarrow \infty.$ (Where $g(x)$ is an Continuous function.) Now I did not Understand How can i solve that, Help me Thanks Black screen after suspend-after moving to the nvidia official drivers (not priority) Making images from single.php pointing to an attachment .php template Shell programming temp=$1 vs temp=1 What's the difference between temp=$1 and temp=1? Why do I need the dollar sign?  For example:   #!/bin/bash  temp=$1 cell=$((($temp-32)*5/9))  echo $cell limit find output AND avoid signal 13 I have a directory with ~1M files and need to search for particular patterns. I know how to do it for all the files:  find /path/ -exec grep -H -m 1 'pattern' \{\} \;   The full output is not desired (too slow). Several first hits are OK, so I tried to limit number of the lines:  find /path/ -exec grep -H -m 1 'pattern' \{\} \; | head -n 5   This results in 5 lines followed by  find: `grep' terminated by signal 13   and find continues to work. This is well explained . I tried quit action:  find /path/ -exec grep -H -m 1 'pattern' \{\} \; -quit   This outputs only the first match.  Is it possible to limit find output with specific number of results (like providing an argument to quit similar to head -n)? Let f(x) be a  polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$  $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ . And then there are some question related to this. In the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3. $f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$ Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?. What is Ambient Occlusion? What is Ambient Occlusion and what are some of it's uses? I see it is used in creating what might be called "shadow catchers" in the following node tree setup: Using sed with an integer variable I want to sed using a variable.  For example, sed -n '1,10p' file. So, instead of 10, I want to use $k where k is an integer. Why does triply periodic univariate function not exist? A univariate function $f$ is periodic with period $p_1,\ldots,p_k$ if   $$f(z) = f(z + \sum_{i=1}^k n_i \cdot p_i)$$   for all complex $z$ and integers $n_i$. Elliptic function is an example of doubly periodic function.  that a triply periodic univariate function cannot exist, but why? Can't we follow the construction of elliptic functions that uses Schwarz-Christoffel formula to map upper-plane onto a triangle, and use the reflection principle to create a lattice which corresponds to three periods?     Which step fails to hold when we mimic the construction of an elliptic function to create a triply period meromorphic univariate function?   Sorry if the question is stupid. I do not have an access to the original proof that a triply periodic function cannot exist, either. So references are welcome!