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" throughout the web. …
probability theory - Why does a C.D.F need to be right-continuous ...
2019年5月10日 · 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? Is there …
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 rate (as a
What's the difference between continuous and piecewise continuous ...
2016年10月15日 · 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
2014年1月27日 · 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 continuous on R R.
Continuity and Joint Continuity - Mathematics Stack Exchange
2012年1月13日 · 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
2015年1月24日 · 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 each point in the …
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 singular continuous …
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?