In Random Forest, cross-validation can be avoided using the out-of-bag sample?
Out of Bag Error makes CV unnecessary in Random Forests?
Convert string to Title Case with JavaScript Which Harry Potter works are considered canon? "I", "me" and "myself" Possible Duplicate:           What is correct? We are a family of four: my father, my mother, my brother and me. or We are a family of four: my father, my mother, my brother and I. or We are a family of four: my father, my mother, my brother and myself. Phone Call On TextView Click Setting chain length My chain broke and in a moment of silliness (I was cold wet and slightly annoyed)  I threw the broken one away before thinking about measuring the new one.  Is there a guide to getting the chain length/tension correct ?  It's an 8 speed road bike (Shimano Sora back, double ring front) How did half of Darth Maul survive? If I push or hit an object in space will it rotate or move along a straight line? Multigroup (CCK3) / FlexiField-like solution for compound fields? I've used  in the past to do multiply occurring series of field groups; now, working on a Drupal 7 project, I'm setting up a "Resume" content type and need a way to set up repetitive entries for education/work experience.   IE: for "Education", I need the ability to dynamically add new rows comprised of a combination of dates and places for each institution studied at.  What is the best way to accomplish CCK3 multigroup/FlexiField-like functionality in Drupal 7?  Thanks! Proof that $\mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $ Where $ f: V \rightarrow V  $ is a linear operator on an inner product space $V$, I'm trying to show that $$ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $$  i.e. the Kernel of the adjoint is equal to the orthogonal complement of the image of f.  I taking the inner product of $f(v)$, where $v \in V$  and an arbitary element of the kernel, $k \in \mathrm{ker}(f^*) $  $$\langle f(v), k\rangle = \langle v, f^*(k)\rangle = \langle v, 0\rangle = 0$$  And then taking the inner product of $f(v)$ with $w$, where $w \in  \mathrm{im}(f)^{\perp}$  By definition  of the orthogonal complement to the image of f,   $$ \langle f(v), w\rangle = 0 $$  My question is, now that we have   $$ \langle f(v), k\rangle = \langle f(v), w\rangle = 0 $$  What can we conclude? I would like to conclude that because $w$ and $k$ are arbitary elements of their respective subspaces, that these must be equal, i.e. $ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $  However, I am not sure this is mathematically sound, is that conclusion valid or are there any technical barriers or additional statements that would have to be made to make that watertight? Find an equation in spherical coordinates for the surface represented by the rectangular equation The rectangular equation is $$x^2+y^2-8z^2=0$$ $$x^2+y^2=8z^2$$ Know in the relationship between rectangular and spherical coords. we can manipulate our given to fit the form: $$x^2+y^2+z^2=9z^2$$ $$\rho=x^2+y^2+z^2, \space z=\rho\cos(\phi)$$ $$\rho^2=9\rho^2\cos^2(\phi)$$ $$1=9\cos(\phi)$$ $$\frac{1}{3}=\cos(\phi)$$ $$\arccos(\frac{1}{3})=1.23 \space rads$$ And so the equation in spherical coords. is $\phi=1.23$ I know my math is correct but I have the wrong answer so I'm not sure where I went wrong. Redefining "cited on" string (and others) in biblatex "Started to work" vs "Started working" When your 10-year old boy says “It’s meta,” what does it mean? In what situation and of what sort of object they use this phrase? I asked about  a few days ago, quoting Maureen Dowd’s review of the movie, “J. Edgar” in New York Times.   I received six answers. But I still don’t get a clear idea of what “It’s meta” means because I don't understand (or have a total inability to comprehend) the concept of “self-referential.”   An answerer : “Meta in this fairly recent, casual context is supposed to mean self-referential, or recursive in some way. This is the sense in which my teenagers would use this term.”  So let me resubmit the question on “meta” in simpler format. When your teenager boy says “It’s (or this is) meta,” what does it mean? In what situation and of what sort of object they use this phrase?  I’m sorry for many users who lent me kind answers to my previous question. But I would like to get it fully on the meaning and usage of “it’s meta,” as a colloquial expression, not the meaning of meta as a prefix. I have already proved Triangle Inequality $\lVert v+w\rVert \le \lVert v \rVert+\lVert w \rVert$ in vector space $\mathbb{R}^n$. However, I'm having difficulties proving the Reverse Triangle Inequality in the vector space.  I have started with   $$\lVert v \rVert^2=v \cdot v-w\cdot w+w\cdot w \le \lVert v-w \rVert+\lVert w \rVert^2$$ $$\lVert w \rVert^2=w \cdot w-v\cdot v+v\cdot v \le \lVert w-v \rVert+\lVert v \rVert^2$$  based on the Triangle Inequality but I don't know if it is correct. If this is right, I think I can go on with proving it the same way I would the normal Reverse Triangle Inequality. A Product function for matrix products What is a good word to describe someone who is empathetic, quick, and witty in conversation Given an integer $n$, we are asked to investigate about the existence of integer divisors of $n^2+1$ of the form $4k+3$. Can you provide some insights about it? Why do people go to the Wall in Game of Thrones? Is the wall a bad or a good place to be? Some people are sent there as punishment, others go because they want to (Jon Snow), others to get away from harm (Arya).  If I am not completely correct, it is because I have only started watching the series recently, and it is a bit confusing!  Please also correct me if I am wrong with any facts! Subtracting (erasing) polygons from polygon using ArcGIS for Desktop? I have two polygon layers, one is green that represents all non-public right-of-way land. I have another polygon that is just a circle with a specified radius.  I would like my output to be a polygon layer that represents all the white space (the gaps) in the private land polygon layer that is also within the black circle.    How does one do this (In ArcGIS 10.2)? For $p,q$ prime and $p \neq q$ show that $C_p \times C_q \cong C_{pq}$ Here are my thoughts so far:    $C_p = \langle x \mid x^p =e\rangle$ $C_q = \langle y \mid y^q =e\rangle$ $C_p \times C_q$ has elements of the form $(x^a,y^b)$ There are $p$ possible values for $x^a$ and $q$ possible values for $y^b$. So there are $pq$ possible elements in $C_p \times C_q$. $C_{pq} = \langle z\mid z^{pq} = e\rangle$ and there are $pq$ elements in this group. For an isomorphism from $C_p \times C_q$ to $C_{pq}$ we send $(e,e)$ to $e$ to ensure that the identities are mapped to each other. However, I'm not sure how to define an isomorphism for the other elements.