Very interesting productivity app that resembles a game. Do check it out!
A few months ago, I noticed someone in one of my Facebook groups posting about an interesting app called Habitica. It’s one of a host of time-management and productivity-increasing applications, both web- and mobile-based. What sets it apart, however, is that it turns your efforts at organizing your life into a game. Specifically, it turns your life into something reminiscent of a classic, pixelated, 8-bit RPG.
So no, the title of this post is not a metaphor. You literally get XP (and gold) for doing your tasks in real life.
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(L’Hopital’s Rule Proof)
” is incorrect.
instead, and this implies 
.
.
where 
.
that is finite and Galois, there is a one-to-one correspondence between its intermediate fields and subgroups of its Galois group.
where 
and 
is the corresponding fixed field (the set of those elements in 
which are fixed by every automorphism in 
).
where 
is an intermediate field of 
is the set of those automorphisms in 
which fix every element of 
holds iff the inclusion of fields 
holds.![|H|=[E:E^H]](https://s0.wp.com/latex.php?latex=%7CH%7C%3D%5BE%3AE%5EH%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
and ![|\text{Gal}(E/F)/H|=[E^H:F]](https://s0.wp.com/latex.php?latex=%7C%5Ctext%7BGal%7D%28E%2FF%29%2FH%7C%3D%5BE%5EH%3AF%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
.
(or equivalently, Galois extension, since any subextension of a separable extension is separable) iff 
and 
exist.
.
then 
, where 
. Consider 
, 
.
, so 
.
as 
. Hence 
converges to 
everywhere.
. The below Laurent Series formula may not be the most practical way to compute the coefficients, usually we will use known formulas, as the example below shows.
about a point 
is given by: 
where 
is anticlockwise around a closed, rectifiable path containing no self-intersections, enclosing 
in which 
. This function is holomorphic everywhere except at 
. Using the Taylor series of the exponential function 
we get
, can we make 
the subject, i.e. write 
? The implicit function theorem allows us to answer such questions, though like most Pure Math theorems, it only guarantees existence, the theorem does not explicitly tell us how to write out such a function.
be a continuously differentiable function, and let 
have coordinates 
. Fix a point 
with 
, where 
. If the matrix ![\displaystyle [(\partial f_i/\partial y_j)(\mathbf{a},\mathbf{b})]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5B%28%5Cpartial+f_i%2F%5Cpartial+y_j%29%28%5Cmathbf%7Ba%7D%2C%5Cmathbf%7Bb%7D%29%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
is invertible, then there exists an open set 
containing 
, an open set 
containing 
, and a unique continuously differentiable function 
such that 
to 
, the Jacobian matrix is
is the matrix of partial derivatives in the variables 
and 
is the matrix of partial derivatives in the variables 
.
as desired.
and 
.
is invertible iff 
. By the implicit function theorem, we see that we can locally write the circle in the form 
for all points where 
.
-dimensional (differentiable) manifold 
is a Hausdorff topological space with a countable (topological) basis, together with a maximal differentiable atlas.
where the domains of the charts, 
, form an open cover of 
are open in 
, the charts (local coordinates) 
are homeomorphisms, and every change of coordinates 
is differentiable on its domain of definition 
.
. It is one of the rarer books from the geometric approach, as opposed to the algebraic approach.
which is also a group such that the group multiplication 
(and the map sending 
) is a differentiable map.
, 
consists of those points that belong to infinitely many 
, and 
consists of those points that belong to all 
on (i.e. belong to all but finitely many 
.
![[a,b]](https://s0.wp.com/latex.php?latex=%5Ba%2Cb%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
. Let 
, and 
for all 
, 
be a differentiable curve given by the vector function 
, 
.
is continuous on 
such that 
for all 
is constant.
by Stewart) or too advanced. Many analysis books skip multivariable calculus altogether and just focus on measure and integration.
be a 
. Then there exists 
such that 
and 
.
, 
)
.
where 
.
, and let 
also be a topological space. If both 
and 
are continuous, then 
is continuous.
is closed in 
, which is continuous. Hence 
for some set 
is closed (in 
is also closed (in ![F[x]](https://s0.wp.com/latex.php?latex=F%5Bx%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
with at least one root in ![|\text{Aut}(E/F)|=[E:F]](https://s0.wp.com/latex.php?latex=%7C%5Ctext%7BAut%7D%28E%2FF%29%7C%3D%5BE%3AF%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
, that is, the number of automorphisms equals the degree of the extension.
.
, the trivial subgroup of 
.
be a PID. Let 
be a nonzero prime ideal. Suppose 
. Then 
for some 
.
implies 
or 
. Since 
, we have 
for some 
. Then, 
which implies 
, thus 
.
“. He is a very good author of Christian books.
is the center of 
are all the conjugacy classes in 
be an element in 
for each 
. Then we have: ![\displaystyle |G|=|Z(G)|+\sum_{i=1}^r[G:C_G(g_i)].](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%7CG%7C%3D%7CZ%28G%29%7C%2B%5Csum_%7Bi%3D1%7D%5Er%5BG%3AC_G%28g_i%29%5D.&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
.
. Then 
for all 
. Hence 
consists of a single element 
. Then
![\displaystyle |\text{Orb}(g_i)|=[G:\text{Stab}(g_i)]=[G:C_G(g_i)].](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%7C%5Ctext%7BOrb%7D%28g_i%29%7C%3D%5BG%3A%5Ctext%7BStab%7D%28g_i%29%5D%3D%5BG%3AC_G%28g_i%29%5D.&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
and 
be real or complex sequences.
converges to 
converges to 
.
has radius of convergence 
, and another power series 
has radius of convergence 
, then their Cauchy product converges to 
and has radius of convergence at least the minimum of 
.
is a continuous map from a closed subset 
, then there exists a continuous map 
with 
for all 
in 
, i.e., if 
, there is a continuous function 
and a closed set 
for which 
. Let 
be a sequence of measurable functions on 
is a function that is holomorphic (analytic) and injective, then 
for all 
in 
for some 
. Using Taylor series, 
for all 
, with 
, 
and 
where 
, we write 
where 
.
on a small circle centered at 
has at least two zeroes there.
but sufficiently close to 
, 
be the two roots of 
, 
. If 
, then 
for some analytic function 
. This means 
, which implies that 
Mathtuition88 