- Proof of Continuous compounding formula - Mathematics Stack Exchange
Following is the formula to calculate continuous compounding A = P e^(RT) Continuous Compound Interest Formula where, P = principal amount (initial investment) r = annual interest rate (as a
- is bounded linear operator necessarily continuous?
3 This property is unrelated to the completeness of the domain or range, but instead only to the linear nature of the operator Yes, a linear operator (between normed spaces) is bounded if and only if it is continuous
- Topological Continuous Functions - Mathematics Stack Exchange
The pasting lemma for finitely many closed sets now says that h h is continuous on X X (a) would follow from the following lemma: If Y Y is an ordered topological space, L = {(y,y′) ∈Y2: y ≤y′} L = {(y, y) ∈ Y 2: y ≤ y} is closed in Y2 Y 2 Assuming this lemma, (a) follows from standard facts on the product topology:
- $f$ is a homeomorphism iff $f$ is bijective, continuous and open
2 Homeomorphism means a continuous bijection whose inverse is continuos too Now use the fact that f is continuous iff for every open set U U of Y , f−1(U) f 1 (U) is open in X The bijection is needed for the other direction, when you have to prove f is homeomorphism f−1 f 1 exists since it is a bijection and continuos as f is an open map
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