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I have learned that 1 0 is infinity, why isnt it minus infinity? An infinite number? Kind of, because I can keep going around infinitely However, I never actually give away that sweet This is why people say that 1 0 "tends to" infinity - we can't really use infinity as a number, we can only imagine what we are getting closer to as we move in the direction of infinity
sequences and series - What is the sum of an infinite resistor ladder . . . Here's a proof by induction that the resistance of a finite version of this ladder with $\ n\ $ rungs is indeed homogeneous of degree $1$ in the variable $\ R\ $ Taking the limit as $\ n\rightarrow\infty\ $ (assuming it exists $\left \right ^\color {red} {\dagger}$) then shows that the resistance of the infinite ladder depicted in Figure $2
Additivity of the matrix exponential of infinite matrices Definition 1 We shall use $\mathbf {M}$ to denote the class of infinite dimensional, real valued matrices as described in the original post Unless explicitly stated otherwise, We shall use the words matrix, matrices exclusively to denote members of $\mathbf {M}$
general topology - Is there any continuous transformation on a compact . . . This means that for infinite entropy we can't have differentiable self-maps of compact manifolds or Lipschitz self-maps of compact metric spaces of finite lower box dimension The simplest example that comes to mind is the shift acting on bi-infinite sequences taking values in an infinite compact space
lie groups - Infinite-dimensional representation theory - Mathematics . . . Thanks for replying My main source of confusion comes from p 112 of Varadarajan's 'Introduction to Harmonic Analysis on Semisimple Lie Groups' (you can find it on google books): he treats the representation theory of the Lie algebra on a purely algebraic level For example, 'irreducible' means no nontrivial invariant subspaces at all, whereas for the group one typically demands nonexistence
linear algebra - Is there a quick proof as to why the vector space of . . . Your further question in the comments, whether a vector space over $\mathbb {Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb {Q}$ that are countable as sets
reference request - Infinite dimensional Clifford algebras . . . Infinite dimensional clifford algebras are the setting of David Hestenes' so-called "universal geometric algebra" Hestenes uses this setting to embed vector manifolds---manifolds whose points are vectors in the UGA and thus admit a lot of niceties in terms of vector operations