What is a continuous extension? - Mathematics Stack Exchange
The continuous extension of f(x) f (x) at x = c x = c makes the function continuous at that point. Can you elaborate some more? I wasn't able to find very much on "continuous extension" …
probability theory - Why does a C.D.F need to be right-continuous ...
May 10, 2019 · This function is always right-continuous. That is, for each x ∈ Rk x ∈ R k we have lima↓xFX(a) =FX(x) lim a ↓ x F X (a) = F X (x). My question is: Why is this property important? …
Proof of Continuous compounding formula - Mathematics Stack …
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 …
What's the difference between continuous and piecewise …
Oct 15, 2016 · A continuous function is a function where the limit exists everywhere, and the function at those points is defined to be the same as the limit. I was looking at the image of a …
Difference between continuity and uniform continuity
Jan 27, 2014 · To understand the difference between continuity and uniform continuity, it is useful to think of a particular example of a function that's continuous on R R but not uniformly …
Continuity and Joint Continuity - Mathematics Stack Exchange
Jan 13, 2012 · the difference is in definitions, so you may want to find an example what the function is continuous in each argument but not jointly
The definition of continuously differentiable functions
Jan 24, 2015 · Note the ending "-ly", which makes it an adverb, not an adjective. So "continuously differentiable" means "differentiable in a continuous way".
Prove that $\\sqrt{x}$ is continuous on its domain $[0, \\infty).$
As you have it written now, you still have to show x−−√ x is continuous on [0, a) [0, a), but you are on the right track. As @user40615 alludes to above, showing the function is continuous at …
Is the set of non-differentiable points for a singular continuous ...
A function f: [0, 1] → R f: [0, 1] → R is called singular continuous, if it is nonconstant, nondecreasing, continuous and f′(t) = 0 f (t) = 0 whereever the derivative exists. Let f f be a …
is bounded linear operator necessarily continuous?
In general, is a bounded linear operator necessarily continuous (I guess the answer is no, but what would be a counter example?) Are things in Banach spaces always continuous?