how to solve $x$ in $4^x = x^4$ My try: $$ 4^x = x^4$$ $$ x \ln4 = 4 \ln x $$ $$ 2x\ln2 = 4 \ln x $$ $$ \frac{1}{2} x \ln2 = \ln x $$ $$ e^{\frac{1}{2} x \ln2} = x ....(1)$$ $$$$ $$\text{Let} : u = \frac{1}{2} x \ln 2 ....(2)$$ $$\text{then},  x = \frac{2u}{\ln 2} ....(3)$$ sub (2) and (3)into (1): $$e^{u} = \frac{2u}{\ln 2}$$ And I have no idea of how to go on ...
What is the best approach for solving the equations of the form      $p^x = x^q$ ; where $p$ and $q$ are some constants.   Basically, I wanted to solve $2^x = x^2$ to analyze algorithms with respective growth functions.
UAE visit visa with expired passport number $\displaystyle \int\nolimits_0^8 \sqrt{x^4+4x^2}\,dx$.   Alright, so I thought I had this figured out. Here's what I did:   I factor out an $x^2$ to get $\sqrt{x^2(x^2+4)}$. I let $x = 2\tan(\theta)$, therefore the integrand is $\sqrt{4\tan^2(\theta)  (4\tan^2(\theta) + 4)}$. Factor out a 4 and it becomes $\sqrt{(16\tan^2(\theta)  (\tan^2(\theta) + 1))}$ Which equals $\sqrt{16\tan^2(\theta) \sec^2(\theta)}$ This is easy to take the sqrt of. The integrand becomes $4\tan(\theta)\sec(\theta)$. Now, the integral of this is $4\sec(\theta)$ And it's evaluated from $0$ to $\arctan(4)$ right? Because as $x$ goes to $0$, so does $\theta$, and as $x$ goes to $8$, $\theta$ goes to $\arctan(4)$... But the end result $(4 (\sec(\arctan(4)) - 1) )$ isn't the correct answer   I put it into WolframAlpha and I get $(8/3) (17\sqrt{17} - 1)$, which is the right answer. How did they get that? (there's no "show steps" option)  Any help is greatly appreciated!  PS, what's the syntax for doing sqrts and exponentials? Coffeescript existential operator sometimes checks against undefined, but not always Brace expansion not working in a script for i in {1..40} do     echo $i done   I got  {1..40}   and I would like to have something like  1 2 3 and so on   so I can use the variable i inside a command's parameter. What is the difference between == and === in php. I am unsure when to use both.  Updated note: So that it shows up in StackOverflow search, the difference between == and === is the same as the difference between != and !==. What do you recommend as an optimal point and orientation for mounting a rear-view mirror? I'm open to handlebar-, glasses-, or helmet-mounted.  Some of the considerations, as I understand it, are:   Ease of mounting. View provided. Vibration. Likelihood of breaking. Robustness for repeated off and on (for glasses-mounted).   I've got a hybrid bike with flat handlebars.  I'd also welcome specific product recommendations and recommendations about other aspects to look for in a mirror. I have 10 years of daily returns data for 28 different currencies. I wish to extract the first principal component, but rather than operate PCA on the whole 10 years, I want to rollapply a 2 year window, because the currencies' behaviours evolve and so I wish to reflect this. However I have a major problem, that is that both the princomp() and prcomp() functions will often jump from positive to negative loadings in adjacent PCA analyses (ie 1 day apart). Have a look at the loading chart for the EUR currency:    Clearly I can't use this because adjacent loadings will jump from positive to negative, so my series which uses them will be erroneous. Now take a look at the absolute value of the EUR currency loading:    The problem is of course that I still cannot use this because you can see from the top chart that the loading does go from negative to positive and back at times, a characteristic which I need to preserve.   Is there any way I can get around this problem? Can I force the eigenvector orientation to always be the same in adjacent PCAs?   By the way this problem also occurs with the FactoMineR PCA() function. The code for the rollapply is here:  rollapply(retmat, windowl, function(x) summary(princomp(x))$loadings[, 1], by.column = FALSE, align = "right") -&gt; princomproll How to prove the function is a constant Let $f\in C[0,1]$ be such  that $\forall \phi\in C[0,1]$ with $\int_0^1\phi \, dx=0$,we have $\int_0^1f\phi \, dx=0$.  Then $f$ is constant.  May be someone can help me proof this problem. Thanks! Application of Kolmogorov's Zero-One Law Most of the books I read, they only state some examples of tail events. One of them is  $[\sum_n X_n \ converges]$. My main problem is to show that this is indeed a tail event. Is Adzerk safe / legitimate? The number of (non-equal) forests on the vertex set V = {1, 2, ...,n} that contains exactly 2 connected components is given by The number of (non-equal) forests on the vertex set V = {1, 2, ...,n} that contains exactly 2 connected components is given by $\sum_{k=1}^{n-1} {n-1 \choose k-1} k^{k-2} (n-k)^{n-k-2}$.  I am unsure how to approach this question. Could someone guide me through the proof? If G is a spanning tree and it is disconnected then I know it is a spanning forest. Also, I know by Cayley's thm that there are $n^{n-2}$ nonequal tree on a vertex set V = {1, 2, ...,n} which is where i'm assuming the $k^{k-2}$ comes from but I do not understand where the rest of it is coming from. A question about long exact sequence of finite dim vector spaces Q) Prove that if $0\rightarrow V_1\rightarrow V_2\rightarrow...\rightarrow V_k\rightarrow 0$ is an exact sequence of finite dimensional vector spaces, then $\sum_{i=1}^{k}(-1)^idim (V_i) = 0$.  For $k=3$, I see that the statement is true. For $k&gt;3$, am I correct in understanding that $V_2 = V_1\oplus V_3 (\implies dim(V_2)=dim(V_1)+dim(V_3))$ and $V_3 = V_2\oplus V_4(\implies dim(V_3)=dim(V_2)+dim(V_4))$ and so on?   If it is true, then for $k=4$, $\sum_{i=1}^{4}(-1)^idim (V_i) = -m_1+m_1+m_3-m_2-m_4+m_4$ and am not sure why $m_2=m_3$ for the sum to be $0$? where $m_i = dim(V_i)$. Thanks. Can we please hold the hats? I really like the hats game. But I would love to keep the hats longer then only one winter. I want to leave them in my wardrobe during summer. Last year, I collected some cool hats, and I apparently lost them during summer.  Maybe we could make a system that allows hats to last for about 2 or 3 years, so we don't have to start over each winter. Or even specify per type of hat a number of years they last? Very very rare hats could last for ever.  I asked this, simply because I love hats! :D "Which" vs. "what" — what's the difference and when should you use one or the other? Connected graph with colored edges We have connected undirected graph with colored edges in two way (green or blue). And also each vertex have the same number of green and blue edges. How to prove that there are alternate colored (green-blue) path between any two nodes? TL:DR: Is it ever a good idea to train an ML model on all the data available before shipping it to production? Put another way, is it ever ok to train on all data available and not check if the model overfits, or get a final read of the expected performance of the model?    Say I have a family of models parametrized by $\alpha$. I can do a search (e.g. a grid search) on $\alpha$ by, for example, running k-fold cross-validation for each candidate.   The point of using cross-validation for choosing $\alpha$ is that I can check if a learned model $\beta_i$ for that particular $\alpha_i$ had e.g. overfit, by testing it on the "unseen data" in each CV iteration (a validation set). After iterating through all $\alpha_i$'s, I could then choose a model $\beta_{\alpha^*}$ learned for the parameters $\alpha^*$ that seemed to do best on the grid search, e.g. on average across all folds.  Now, say that after model selection I would like to use all the the data that I have available in an attempt to ship the best possible model in production.  For this, I could use the parameters $\alpha^*$ that I chose via grid search with cross-validation, and then, after training the model on the full ($F$) dataset, I would a get a single new learned model $\beta^{F}_{\alpha^*}$   The problem is that, if I use my entire dataset for training,  I can't reliably check if this new learned model $\beta^{F}_{\alpha^*}$ overfits or how it may perform on unseen data. So is this at all good practice? What is a good way to think about this problem? asp.net mvc put controllers into a separate project I'm just learning asp.net mvc and I'm trying to figure out how to move my controllers into a separate project.  Typically when I have designed asp.net web apps before, I created one project for my models, another for my logic, and then there was the web.    Now that I'm learning asp.net mvc I was hoping to follow a similar pattern and put the models and controllers each into their own separate projects, and just leave the views/scripts/css in the web.  The models part was easy, but what I don't understand is how to make my controllers in a separate project be "found".  Also, I would like to know if this is advisable.  Thanks! Is $\frac{1}{\sqrt{2}}$ a fraction or not?   I just came across a Grade 8 student's class work where the teacher has given the above as an example to differentiate fractions from rational numbers. I think this is wrong fundamentally, as $\sqrt2$ is irrational while all fractions would be rational. Please advise. Thanks! There seems to be something wrong with the sorting of Stack Exchange sites by name:    elementary OS and Русский язык appear in the wrong place at the end of the list. How does the secondary effect of the Heat Metal spell work? The  spell description states:  If a creature is holding or wearing the object and takes the damage from it, the creature must succeed on a Constitution saving throw or drop the object if it can. If it doesn't drop the object, it has disadvantage on attack rolls and ability checks until the start of your next turn...  I am unsure how this disadvantage mechanic functions as a whole. When do you suffer from it and in what ways (if any) can you end it? For example: What happens if you fail the save but cannot drop the object? Does dropping the object later, on your turn, do anything? What happens if you succeed on the save?  Does dropping the object later, on your turn, do anything? I would like it to be assumed that ; however, heat metal can still force you to drop the weapon. I would also like it to be assumed that .