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  • finance - 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 decimal) t = number of years A = amount after time t The above is specific to continuous compounding
  • probability theory - Why does a C. D. F need to be right-continuous . . .
    This fact is useful to resolve this natural question: Let $\{X_i\}_{i=1}^{\infty}$ be i i d random variables uniform over $[-1,1]$
  • What is the difference between differentiable and continuous
    $\begingroup$ @user135626: What I wrote is correct You are misreading it I'm not saying the derivative is zero, I'm saying that if the derivative exists, the numerator of the difference quotient necessarily converges to zero (not that the difference quotient itself must)
  • is bounded linear operator necessarily continuous?
    Added @Dimitris's answer prompted me to mention, beyond the fact that the implication on normed spaces indeed is an equivalence, that it's the converse which holds in the wider context of topological vector spaces, while the proposition mentioned here fails: there are bounded discontinuous linear operators, yet every continuous operator remains
  • Difference between continuity and uniform continuity
    I understand the geometric differences between continuity and uniform continuity, but I don't quite see how the differences between those two are apparent from their definitions For example, my book
  • Proving the inverse of a continuous function is also continuous
    Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
  • Why Do We Care About Hölder Continuity?
    The regularity of Brownian paths is expressed quite precisely by it being $\alpha$-Holder continuous for all $\alpha< \frac 12$ but not $\alpha$-Holder continuous for $\alpha \geq \frac 12$ In fact, this generalizes to a lot of solutions of various SPDEs, where the solution ends up having Holder-regular sample paths $\endgroup$


















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