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- ...intervals or, moreover, of $1$-''chains''. We can see this idea in the new form of the additivity property: ...s. The form on the left is what we call the ''exterior derivative'' of the form on the right.34 KB (5,619 words) - 16:00, 30 November 2015
- *[[cubical form|cubical form]] *[[differentiable form|differentiable form]]16 KB (1,773 words) - 00:41, 17 February 2016
- where $f:R \to R$ is continuous and $x:I \to R$ is differentiable on an open interval $I$. The equation has to be satisfied for all $t \in I$ ...ut $dx$ its [[exterior derivative]], a $1$-form, and $dt$ the constant $1$-form.9 KB (1,561 words) - 16:06, 27 August 2015
- *Solve the following system, expressing the solution in vector form: $$\begin{array}{lll}x_2-x_3=1,\\x_1+x_2+x_3=2.\end{array}$$ **c. If the partial derivatives exist then the function is differentiable.14 KB (2,538 words) - 18:35, 14 October 2017
- These two sequences of numbers form a sequence of intervals: Thus to compare two functions $f$ and $g$, at infinity or at a point, we form a fraction from them and evaluate the limit of the ratio:59 KB (10,063 words) - 04:59, 21 February 2019
- ...$f$ is [[continuous]]. But if $b > 1 + \frac{a}{2} \pi$, $f$ is nowhere [[differentiable]]. (For another example, consider the issue of [[lengths of curves]] in the | $0$-form10 KB (1,471 words) - 12:50, 12 August 2015
- ...rem ([[Mean Value Theorem]]).''' Let $[ a, b ] \subset D(h)$, where $h$ is differentiable on $( a, b )$ and $h$ is continuous on $[ a, b ]$. Then ...f which isn't true. This is what it does and does not say about a function differentiable at $X=C$:7 KB (1,171 words) - 20:28, 10 July 2018
- '''Definition:''' Given a $0$-form $\varphi$, the ''integral of the form $\varphi$ over oriented $0$-manifold'' $M=\{p\}$ is: *$0$-form $=$ number assigned to each vertex.12 KB (1,906 words) - 17:44, 31 December 2012
- Recall, a continuous $1$-form in ${\bf R}^3$ is a function ...$(dx,dy,dz)$. The function is [[linear operator|linear]] on $dx,dy,dz$, [[differentiable]] on $x,y,z$.6 KB (1,177 words) - 15:53, 5 November 2012
- *right then down: we acquire a $0$-form $g$ by sampling function $f$, and then we acquire $dg$ by taking the differ ...he derivative $f'$ of $f$, and then we find the exterior derivative (a $1$-form) $dg$ by integrating $f'$ on the segments:21 KB (3,664 words) - 02:02, 18 July 2018
- ...e domain and a new edge for each node. Together, these new nodes and edges form a new copy of the domain. [[image:form and its dual.png| center]]64 KB (11,521 words) - 19:48, 22 June 2017
- ...hat, continuous and discrete forms are very similar in the sense that they form vector spaces that behave similarly. ...on we need to be able to match two very different entities -- a continuous form and a discrete one -- one by one. Let's review what they are.9 KB (1,483 words) - 13:54, 13 April 2013
- provided $x=x(t),\ y=y(t)$ are differentiable at $t$ and $u=f(x,y),\ v=g(x,y)$ are continuous at $(x(t),y(t))$. ...ir of functions $x=x(t)$ and $y=y(t)$ (a parametric curve) with either one differentiable on an open interval $I$ such that for every $t$ in $I$ we have:63 KB (10,958 words) - 14:27, 24 November 2018
- ...(Topological property of exterior derivative).''' If $\varphi$ is a $C^2$-form in $\Omega ^k({\bf R}^n)$, then $d_{k+1}(d_k \varphi)=0$ (we can also write ...uct]] and $A=A(x^1,...,x^n)$ is a coefficient function, twice continuously differentiable. Then, using the definition of exterior derivative and its formula for $0$-9 KB (1,423 words) - 20:53, 13 March 2013
- The four coefficients of $x,y$ form the first table: Now form a matrix:46 KB (7,625 words) - 13:08, 26 February 2018
- ...al "generically". They are usually required to be [[continuous]] or even [[differentiable]]. Recall $y = f(x)$ is a $0$-form and $dy=f'(x)dx$ is a $1$-form.10 KB (1,588 words) - 17:11, 27 August 2015
- ...[[First derivative test]]).''' Suppose $a$ is a local extreme point of a [[differentiable function]] $f$. Then which form a line.9 KB (1,511 words) - 16:07, 17 August 2011
- : for any [[differentiable function]] $f\cdot g$ wrt $x$. : for differentiable $f$ and $g$.6 KB (1,004 words) - 16:00, 2 May 2011
- ...aled versions of the old ones. Let's recast this statement in the integral form. ...ched versions of the old ones. Let's recast this statement in the integral form:69 KB (11,727 words) - 03:34, 30 January 2019
- where $A,B$ are continuously differentiable functions and $dX,dY$ are basis elements of $ \Omega^k,\Omega^m$ respective '''Theorem.''' $d(d\varphi) = 0$ for any $C^2$-form $\varphi \in \Omega^k$.8 KB (1,539 words) - 18:17, 22 August 2015