Good evening, I was wondering how to prove easily this inequality: \begin{equation} py^{p-1}(x-y)\leq x^p-y^p\leq px^{p-1}(x-y) \end{equation} for all $x,y\geq0$ and $p\in[1,+\infty)$. I tried to study both the functions $\frac{x^p-y^p}{y^{p-1}(x-y)}$ and $\frac{x^p-y^p}{x^{p-1}(x-y)}$, but I'm not getting anywhere.
I was trying to solve the following:  Given: $0 &lt; a &lt; b$ and $n&gt;1$, prove: $$na^{n-1}(b-a) &lt; b^n-a^n &lt; nb^{n-1}(b-a)$$  I managed to get this far using the mean value theorem: $$a^n(b-a)&lt;b^n -a^n&lt;b^n (b-a)$$ Any idea how to continue?
What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$ How can I find the formula for the following equation?  $$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$$  More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by $n+2$, but that's about as far as I have been able to get:  $\frac12 + \frac16 + \frac1{12} + \frac1{20} + \frac1{30}...$ the denominator increases by $4,6,8,10,12,\ldots$ etc.  So how should I approach finding the formula? Thanks! How to avoid duplicating assigned materials when duplicating a mesh? how can I prevent Blender 2.8 from duplicating materials every time (?) I duplicate a mesh. I am working on a media project that contains little clips from movies and I wonder if there any way to avoid downloading all of the data instead only the specified parts of the movies? Asking users to accept answer when they comment that it answers their question Is there some trick to burning a cd with the CD/DVD creator?    I'm installing k3b because that's a known entity.  What is this, and where is the "burn" button?  Help just tells me this is Nautilus 3.4.1, but I don't see a .  There's no option to burn in the Files menu, nor does clicking on the whitespace indicate an option to burn anything. Why Was It 'Essential' That Voldemort Kill Harry Potter? What is the difference between topological and metric spaces? What is the difference between a topological and a metric space? How to use aes-encryption in PostgreSQL? I tried aes-encryption by using following statement:  SELECT encrypt('test', 'key', 'aes');   which worked, but I am not able to decrypt the value. I inserted it in a field of datatype bytea but I'm not sure if that was the right way.   SELECT decrypt(pw, 'key', 'aes') FROM table WHERE ID = 1;   gives me the error     ERROR:  function decrypt(bytea, unknown, unknown) does not exist   LINE 1: SELECT decrypt(pw, 'key', 'aes') FROM tabelle WHERE ID = 7;                  ^   HINT:  No function matches the given name and argument types. You might need to add explicit type casts.   Does that really mean that encrypt() is an existing function, but not decrypt()? How else could I retreive aes-encrypted values? How do I exclude plugins from getting automatically updated? There is an opt-in filter that allows all plugins on my site to receive automatic updates:  add_filter( 'auto_update_plugin', '__return_true' );  I like this feature, but I don't want all my plugins to be updated automatically. How can I allow some plugins to be updated automatically, while excluding those I want to do manually? I met this equation frequently in Guass-Newton optimizations. But I dont understand why the left and right side of the equation can be equal.  Lets say the Jacobian is $2$ by $2$   and Hessian is $$\begin{bmatrix}\frac{\partial^2f_1}{\partial^2 x_1 } &amp; \frac{\partial^2f_1}{\partial^2 x_2  } \\ \frac{\partial^2f_2}{\partial^2 x_1 } &amp; \frac{\partial^2f_2}{\partial^2 x_2  }\end{bmatrix}$$   But the right hand side of the equation is  $$J^TJ =\begin{bmatrix} \left( \frac{\partial f_1}{\partial x_1} \right)^2 + \left( \frac{\partial f_2}{\partial x_1} \right)^2  &amp;  \frac{\partial f_1}{\partial x_1}\frac{\partial f_1}{\partial x_2} +\frac{\partial f_2}{\partial x_1}\frac{\partial f_2}{\partial x_1}\\ \frac{\partial f_1}{\partial x_1}\frac{\partial f_1}{\partial x_2} +\frac{\partial f_2}{\partial x_1}\frac{\partial f_2}{\partial x_1} &amp;\left( \frac{\partial f_1}{\partial x_2} \right)^2 + \left( \frac{\partial f_2}{\partial x_2} \right)^2\end{bmatrix} $$  Why can these two be equal as presented in papers? Reading and searching long man pages I finally got fed up when wanting to read about bash's read and it's -s option with man bash. I found the right spot eventually (around line 4500), but it was a frustrating as usual, since both /read and even /\s-s\s searches have way too many matches.  So, the question is: How can I read long man pages efficiently, or get same information in other ways, locally? As a specific example, how to reach the relevant documentation after seeing read -s pwd in a shell script? A good answer could be a shell script snippet, or hint about some tool and how it is used, or something else entirely, as long as it helps in finding the right spot to read.  Note: I'm not tagging with  because I want the question to be about man page reading in general, even though that quite possibly is the most commonly encountered humongous man page. Ubuntu 19.10 change desktop environment Is the function differentiable at $0$? Let $$f(x) = \begin{cases}\begin{align*}&amp;\cos{\dfrac{1}{x}}, &amp;x \neq0 \\ &amp;0, &amp;x=0. \end{align*}\end{cases}$$ Is the function $F(x) = \displaystyle \int_{0}^x f dx$ differentiable at $0$?  We can see that the function $f(x)$ is continuous everywhere except $x=0$. In order to show differentiability we will need to show the derivatives from the left and right are equal. So we need to show that $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h}.$$ The derivative from the right is $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^+} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}$$ and the derivative from the left is $$\lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}.$$ I am not sure how to proceed. Wikipedia lists the following pseudocode for HMAC:  function hmac (key, message)     if (length(key) &gt; blocksize) then         key = hash(key) // keys longer than blocksize are shortened     end if     if (length(key) &lt; blocksize) then         // keys shorter than blocksize are zero-padded ('∥' is concatenation)          key = key ∥ [0x00 * (blocksize - length(key))]      end if      // Where blocksize is that of the underlying hash function        o_key_pad = [0x5c * blocksize] ⊕ key      // Where ⊕ is exclusive or (XOR)     i_key_pad = [0x36 * blocksize] ⊕ key       // Where '∥' is concatenation        return hash(o_key_pad ∥ hash(i_key_pad ∥ message))  end function   Why are they there?  How were they chosen?  It seems to me that  $hash(key + hash(message))$ (where + is concatenation) would suffice.    Obviously I'm missing something.  What is it? How can one cite theorems of references? I have observed that in gedit if I edit a file, another file is created in the same directory (the one with the same filename and a tilde '~' suffix). The extra file remains even if I close gedit.  I understand the need for a temp file (eg. in case of a crash), but vim for example deletes the extra file it creates, when I close it.  Is there a way to do the same with gedit? Some configuration perhaps? Updated title of arxiv not recognised by Google Scholar? I submitted a paper to a journal and one of the referees suggested to slightly change the title of the paper. The paper was already posted on arxiv. I have updated the paper on arxiv after the revision, but this update is not recognised by Google Scholar. The main problem with this is that Google Scholar is not detecting the new citations to the arxiv paper since the new citations use the new title.  Any suggestions on how to solve this issue?  Update: I updated the arxiv version 2 months ago but Google Scholar does not detect it. It detects new papers that cite my manuscript, but not the updated arxiv.  Disclaimer. I am aware of the following questions, but they are related to different issues. Why $∀x(P(x)→Q)$ is equivalent to $∃xP(x)→Q$ Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.  I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question. I have added the png transparent image with the setting and its not showing transparent black background is seen. I have the base color aluminum and i want to add that logo of png on to it. The image is black letter with transparent background why its showing in white letters and black background. May i know where im going wrong any suggestion or help. What is the difference between encrypting something and hashing something?  in what situations would I want one or the other?