I'm using latex editor to insert a picture. When I view the output in pdf file the "Diagram.jpg" appear above the picture. How I can remove it?   the code written for insertion is:  \begin{figure}[h]     \centering         \includegraphics[width=1.00\textwidth]{Blcok Diagram.jpg}     \caption{Block Diagram}     \label{fig:Block Diagram} \end{figure}   and the output is like:
I have inserted an image in my LaTeX document and the image always displays with the file path of the image directly above it.  How can I prevent this file path from displaying?  I am inserting the image like this:  \begin{figure} \centering     \includegraphics{C:/Users/Name/Pictures/image1.png} \caption{a caption} \label{fig:reference} \end{figure}   Here is a working example of document that produces this behaviour:  \documentclass{article}  \usepackage{graphicx} \begin{document} \begin{figure} \includegraphics{C:/Users/Name/Pictures/image 1.png}  \end{figure}  \end{document}   The image was inserted using the TeXnicCenter Insert>Picture function.  When I generate the pdf document, directly above any image (I have inserted multiple images using this technique) the file path displays above the image, in the case above it would show: "file\path\of\image", how can I prevent this from displaying?
"Oriented" vs. "orientated" Joda Time: How to get dates of weekdays on some date interval? After I factory reset my Nokia 6, this icon appeared. I've never seen it before. What is this icon in the highlighted part? Landau's derivation of a free particle's kinetic energy- expansion of a function? I was reading a bit of Landau and Lifshitz's Mechanics the other day and ran into the following part, where the authors are about to derive the kinetic energy of a free particle. They use the fact that the Lagrangian of this particle must be the same (or at most, differ by the total time derivative of a function of co-ordinates and time) in different inertial frames.     We have $L'=L(v'^2)=L(v^2+2\mathbf{v}\cdot\boldsymbol{\epsilon}+\boldsymbol{\epsilon}^2)$. Expanding this expression in powers of $\boldsymbol{\epsilon}$ and neglecting terms above the first order, we obtain    $$L(v'^2)=L(v^2)+\frac{\partial L(v^2)}{\partial (v^2)}2\mathbf{v}\cdot\boldsymbol{\epsilon}.$$   I think I'm ok with all the physics in this section. What I don't get is just the part I quoted above (so maybe this post is better suited for the math site, but since this is book is so physics-y I thought I'd post it here). My math is pretty rusty, so I'm not really sure- how do the authors expand the function to arrive at the above expression? It reminds me a bit of a Taylor expansion, but not very much. What's the process used to arrive at it? Let $R$ be a ring (unital, not necessarily commutative), $M$ a finitely presented left $R$ module. Suppose $m_1,\ldots,m_n\in M$ generate $M$. This determines a surjection $f:R^n\to M$.     Must the kernel of $f$ be finitely generated?   This is true by definition if the generating set $m_1,\ldots,m_n$ is the one coming from the finite presentation of $M$. What if the hard disk is partitioned into more than 26 sections in Windows? The proof of shrinking coefficients using ridge regression through "spectral decomposition" I have understood how ridge regression shrinks coefficients towards zero geometrically. Moreover I know how to prove that in the special "Orthonormal Case," but I am confused how that works in the general case via "Spectral decomposition." How to prove and interpret $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$? What is so special about $\alpha=-1$ in the integral of $x^\alpha$? Of course, it is easy to see, that the integral (or the antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for $\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is $x^{\alpha+1}/(\alpha+1)$.  I was wondering if there is an intuitive (probably geometric) explanation why the case $\alpha=-1$ is so different and why the logarithm appears?  Some answers which I thought of but which are not convincing:   Taking the limit $\alpha=-1$ either from above or below lead to diverging functions. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity. I navigate with Ctrl-n,p,f,b (readline/ keys), so that I don't have to move my hands too much and get  again. Is there a way to disable Ctrl-b being hotkey for bold?  Note that this is not just an Emacs thing. Standard web browsers, like Firefox and Chrome, use things like  Ctrl + K and Ctrl + L to navigate to the Google search bar and address bar respectively. One divided by Infinity? Characterization convex function. Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function such that for all $a,b\in [0,1]$ with $a&lt;b$ $$f\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_a^b f(x)\,dx.$$ How to prove that $f$ is a convex function? Thank you for your help. Determinant of a square matrix in a field \begin{array}{rrrrr|r}     b &amp; a &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; b &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; a &amp; b &amp; \cdot \cdot \cdot &amp; a \\     \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ a &amp; a &amp; a &amp; \cdot \cdot  \cdot &amp; b   \end{array}  I have the above matrix $A\in M_{n\times n}(F)$ where $F$ is a field and $n\geq1$, $a,b\in F$.  I'm trying to find out how to use row operations to make it into an upper triangular matrix in order to figure out the determinant. But I'm not sure how I would approach it. Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$ This question is from [Number Theory George E. Andrews 1-1 #3].     Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$   This problem is driving me crazy. $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$  $(x^n-y^n)/(x-y) =$ the sum for the first $n$ numbers and then I added $(xy^{(n+1)-2}+y^{(n+1)-1})$ which should equal $(x^{n+1}-y^{n+1})/(x-y)$ but I can't figure it out    This is a similar problem in the book and I tried this method but it wasn't working out      $\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$    $\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$     $\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$.  $\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$ \eqalign{ \sum^{(k+1)-1}_ {j=0}x^j &amp; = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\ &amp;= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\ &amp;= \dfrac{x^{k+1}-1}{x-1}. }$$   Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.    $\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m&gt;1$, then $n&lt;m^n.$ Triggering actions when a variable is set I'm currently going through notes on a physics course and I'm having trouble understanding the difference between a tensor, a vector, and a matrix. I know that a vector is a kind of tensor and that a matrix is also a form of tensor, but what I'm having trouble understanding is how they differ, and how a vector is related to a matrix. I live in a town where fresh or whole milk is not readily available. I was wondering if I could make cheese from UHT Milk. If so, what kind of cheese and what method I should use? Or rather, Should I? How can I show that the set of rational numbers with denominator a power of two form a dense subset of the reals? How can I show that the set of rational numbers with denominator a power of two form a dense subset of the reals? File sync error (AUTH_FAILED) Is there an Ubuntu theme available to make it look like Windows 7? I want to know is there any theme available make it Ubuntu look like Windows 7 ?