Show that between any two real solutions of $e^x \sin x = 1 $, there is at least one real solution of $e^x \cos x = -1 $ Try: Let $f(x) = e^x \sin x - 1 $. Consider $I = [0, \pi/2]$, Then, $f( 0) = -1 $ and $f( \pi/2) = e^{\pi/2} - 1 &gt;0 $. Thus, one can find a solution $c_1 \in I $ such that $f(c_1) = 0$. Similarly, if $J= [ \pi/2, \pi ] $, one can find $c_2 \in J$ such that $f(c_2) = 0 $. Consider the interval $[c_1,c_2]$. Let $F(x) = e^x \cos x + 1 $. We have $$F(c_1 ) = e^{c_1} \cos (c_1) + 1 &lt; e^{c_1} \cos 0 +1 =  e^{c_1} + 1 &gt;0 $$ and $$ F(c_2) = e^{c_2} \cos (c_2) - 1 &lt; e^{c_2} \cos \pi - 1 = - e^{c_2} - 1 &lt; 0 $$ so, $\mathbf{there \; exists}$ some $\alpha \in [c_1,c_2]$ such that $F(\alpha) = 0$ as we wanted to prove. Is this a correct proof?
The following is a homework problem:  Prove that between two roots of the function $f(x) = \exp(x) \sin(x) -1$ there must be at least one root of the function $\exp(x)\cos(x) +1$.  I think it is possible to prove it by simply computing the signs of the functions at $\frac{\pi}{2}k, \pi k, \frac{3\pi}{2}k, 2\pi k$. However, I'm looking for a more elegant way. It is intuitively clear that the roots of $f$ are ˋˋconverging'' to the roots of $sin$ and the ones of $g$ to the ones of $cos$. Do you have any hint how to construct a nice argument based on this observation?
Will wolves attack Ocelots? In my opinion the accepted answer should permanently stick as only the original question submitter has the ability to accept. I put a video in my pdf with Acrobat and there I can watch my video. But if I open the pdf on my mobile, I can't watch the video, I only see an image. Is there a way to watch videos of a pdf in a mobile? I have a samsung J5. Ackermann Function primitive recursive I am reading the wikipedia page on ackermann's function,   And I am having trouble understanding WHY ackermann's function is an example of a function which is not primitive recursive.  I understand that the ackermann function terminates, and thus is a total function. So why is it not primitive recursive? The only information that i can find on the wikipedia page is      [Ackermann's function] grows faster than any primitive recursive function and is therefore   not primitive recursive   Which isn't a good example of why it is not primitive recursive.  Could anyone explain to me how exactly ackermann's function is NOT primitive recursive please? Determine the following limit as x approaches 0: $\frac{\ln(1+x)}x$ $$\lim_{x\to 0} \frac{\ln(1+x)}x$$  The process I  want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule)   $$\lim_{h \to 0} \frac{f(x+h) - f(x)}h$$  $$\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h$$  $$\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}$$  At this point I get confused because I know the answer is $1$, but I am not getting this answer through simplification of my formula. I am trying to draw the below given figure. Please help me to plot it in Tikz. I am only able to plot concentric circles. . I would like to have different colors in between different circles. Also, the text should be along the curve. The arrow in between the two concentric circles does not have to overlap with the circle on the right hand side.  I would be grateful if you could help me in plotting this in Tikz. Please let me know if you need more information. Thanks!  \documentclass[conference]{IEEEtran}  \ifCLASSINFOpdf  \else  \fi  \usepackage[usenames,dvipsnames]{color} \usepackage[T1]{fontenc} \usepackage{tikz} \usetikzlibrary{positioning} \usetikzlibrary{arrows} \usetikzlibrary{shapes.multipart} \usepackage[caption=false]{subfig} \usepackage{tabularx,tikz} \usepackage{graphicx} \usepackage{color} \usepackage{pdfpages} \begin{document}  \title{Practice Tex}   \maketitle  \section{Introduction}  \begin{figure} \begin{center} \begin{tikzpicture}     \draw(0,0) circle (0.4cm)   node[draw=none,sloped,above]{cxvzxcv};         \draw(0,0) circle (0.75cm) node[draw=none,sloped,above]{ratio};    \draw(0,0) circle (1.25cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,2.2) {22};    \draw(0,0) circle (2cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,3.8) {33333};    \draw(0,-9) circle (0.4cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,-9) {fsgsdgf};      \draw(0,-9) circle (0.75cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,-7.8) {fgdsfg};    \draw(0,-9) circle (1.25cm) node[midway,sloped] at (0,-9) {dfsdfad};    \draw(0,-9) circle (2.8cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,-6.4) {dfsdfsd};    \draw(0,-9) circle (3.2cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,-6) {dfsdfsd};    \draw(0,-9) circle (3.6cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,-5.6) {sfdsfd};    \draw(0,-9) circle (4.4cm);   \node[draw=none,align=center, font=\scriptsize,text width = 1.25cm] at (0,-4.8) {sdfsf};     \pgfsetlinewidth{3pt}    \draw [-implies][-&gt;] (0,-4) -- (0,-4.6);   \end{tikzpicture}  \end{center} \end{figure} \end{document} Showing that $f:(0,2\pi]\to S^1$ is not a homeomorphism I want to show that $f:(0,2\pi]\to S^1$ defined by $t\to (\sin t, \cos t)$ is not a homeomorphism. I will do this by showing that its inverse is not continuous.    The inverse is defined by $(\sin t, \cos t)\to t$. So at $(0,1)$ we want to have $\lvert 2(1-\cos t)\rvert &lt;\delta \Rightarrow \lvert t-2\pi\rvert&lt;\epsilon$ for any $\epsilon$. If we take $\epsilon=\pi/2$, no matter how close $\cos t$ and $1$ are there will always be $t$'s close to $0$. So $f^{-1}$ is not continuous.  What do you think? Are there any mistakes? Cost Distance Analyst In QGIS? Are there any tools, plugins or methods in QGIS which allows for an analysis of the best path between two points using least cost path method? As shown below ArcGIS has several distance tools. Unless I missed something key during Descent 1/2, how is it that the Borg are once again a collective? How to model non-negative zero-inflated continuous data? I'm currently trying to apply a linear model (family = gaussian) to an indicator of biodiversity that cannot take values lower than zero, is zero-inflated and is continuous. Values range from 0 to a little over 0.25. As a consequence, there is quite an obvious pattern in the residuals of the model that I haven't managed to get rid of:   Does anyone have any ideas on how to solve this? Prove that every terms of a sequence defined by a recurrence relation is a perfect square. Seeking Python script for creating .mxd files? Interview question: three arrays and O(N*N) I want to install the  on my iPhone. On the phone I have iOS 7. When I try to install the App, it tells me it needs at least version 8 of iOS.  I do not want to upgrade my iOS version, especially for the sole purpose of installing an App. However, I do want to use it, it's kind of important.  I searched for a while about how to do this, but I could only find solutions that would work if you installed the App before and it shows up on the "Purchases" page in iTunes or something like that. And this doesn't apply to this case.    How can I install an older version of the App (one that requires "only" iOS 7) without having to upgrade the operating system on my iPhone ? Why does signing on to one site not sign me on to all of them? I've seen this question asked 3 times, but in none of those threads did I find a satisfactory answer.  I'm not asking about reputation: it makes sense that different sites should have different reputations.  However, consider the StackExchange drop-down menu at the top left of the screen.  When looking through the "Hot questions", the links are often to sister websites of StackExchange (programmers.stackexchange i.e.).  It doesn't make sense to me that if I go to one of those sister sites, I'm no longer logged in.  I feel like it's counterintuitive to click a link to go answer a question, only to find that you can't answer that question until you log in.  What is the reason for this?  Is there an option to do as I'm suggesting? "If $1/a + 1/b = 1 /c$ where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer" If $1/a  +  1/b  =  1 /c$   where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer.  I found this question in RMO 1992 paper !  Can anyone help me to prove this ?  May be it's too easy and simple . But , I'm just novice in number theory ! Beam size and material for 15 feet long swing I need to create a 15 feet span beam for kid swing (300 lbs load / 150 lbs and 2 swings assumption). One end will bolt to the tree (just for convenience).  What "treated lumber" lumber size and material I should use? Or is it better / cheaper to use metal pipe (from the plumbing department)?  I guess 4x8x16 or two (2) 2x8x16 would do it according to this website . But I do like a clean look of metal. My wife would concern about big piece of lumber running on top of the kid (that looks heavy visually). How can the speed of sound increase with an increase in temperature? How do I install a .tar.gz (or .tar.bz2) file? This is Exercise 26 from Chapter V Algebraic Extensions in Lang's Algebra.  $F$ is finite normal extension over $k$ and $f(x)$ is irreducible in $k[x]$. If $f(x)=g(x)h(x)k(x) \in F[x]$ where $g(x),h(x)$ are monic irreducible factors in $F[x]$, I need to show that there exists an automorphism of $F$ which takes $g(x)$ to $h(x)$.  Any hints would be appreciated.  Sending one root to another is okay but how to do that for a collection of roots?