http://www.bing.com:80/search?q=Continuous+Line+Drawing+ClockSearch resultshttp://www.bing.com:80/s/a/rsslogo.gifhttp://www.bing.com:80/search?q=Continuous+Line+Drawing+ClockCopyright © 2026 Microsoft. All rights reserved. These XML results may not be used, reproduced or transmitted in any manner or for any purpose other than rendering Bing results within an RSS aggregator for your personal, non-commercial use. Any other use of these results requires express written permission from Microsoft Corporation. By accessing this web page or using these results in any manner whatsoever, you agree to be bound by the foregoing restrictions.https://math.stackexchange.com/questions/5114829/continuous-vs-discrete-variablesBoth discrete and continuous variables generally do have changing values—and a discrete variable can vary continuously with time. I am quite aware that discrete variables are those values that you can count while continuous variables are those that you can measure such as weight or height.Sat, 28 Mar 2026 22:48:00 GMThttps://math.stackexchange.com/questions/7923/are-continuous-functions-always-differentiableAn interesting fact is that most (i.e. a co-meager set of) continuous functions are nowhere differentiable. The proof is a consequence of the Baire Category theorem and can be found (as an exercise) in Kechris' Classical Descriptive Set Theory or Royden's Real Analysis.Fri, 10 Apr 2026 14:33:00 GMThttps://math.stackexchange.com/questions/75491/how-does-the-existence-of-a-limit-imply-that-a-function-is-uniformly-continuousThen the theorem that says that any continuous function on a compact set is uniformly continuous can be applied. The arguments above are a workaround this.Sat, 11 Apr 2026 22:03:00 GMThttps://math.stackexchange.com/questions/556137/eigenvalues-are-continuousThese functions aren't even defined, I don't see how they could be continuous. What is true is that the set of eigenvalues is continuous (for the right topology on the power set).Fri, 10 Apr 2026 02:59:00 GMThttps://math.stackexchange.com/questions/653100/difference-between-continuity-and-uniform-continuityTo understand the difference between continuity and uniform continuity, it is useful to think of a particular example of a function that's continuous on $\mathbb R$ but not uniformly continuous on $\mathbb R$.Sat, 11 Apr 2026 06:39:00 GMThttps://math.stackexchange.com/questions/775045/how-do-i-show-that-all-continuous-periodic-functions-are-bounded-and-uniform-conShow that every continuous periodic function is bounded and uniformly continuous. For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval $ [x_0,x_0+P]$. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded.Tue, 07 Apr 2026 12:06:00 GMThttps://math.stackexchange.com/questions/667939/is-the-set-of-non-differentiable-points-for-a-singular-continuous-function-nowheIn view of the correspondence of nondecreasing functions with positive measures, singular continuous functions correspond to singular continuous measures, i.e. an atomless positive Borel measures concentrated on a set of Lebesgue measure zero.Sat, 11 Apr 2026 14:17:00 GMThttps://math.stackexchange.com/questions/294683/showing-that-arctan-is-continuousAs such, $\arctan$ is continuous. If you define $\arctan$ by integrals or power series the result is immediate (the first by the Lipshitz continuity of the indefinite integral and the second from the uniform convergence of power series in compact sets)Tue, 07 Apr 2026 01:01:00 GMThttps://math.stackexchange.com/questions/4951094/continuous-surjection-mathbb-rm-to-mathbb-rn-that-is-not-a-quotient-mapIn fact, it turns out that every continuous function from a path connected space to $\mathbb R$ is a quotient map Note that the closed map lemma cannot be generalised, for example $ (0,1)\to [0,1]$ is not closed.Fri, 10 Apr 2026 23:51:00 GMThttps://math.stackexchange.com/questions/1277341/prove-that-ax-is-continuousIt can be shown that for any 2 functions f and g, if f is continuous on R and g is a linear function with nonzero slope, f ∘ g is continuous so for any positive real number a, if exp (x) is continuous on R, then exp (x ln a) is continuous on R but a x = exp (x ln a) so a x is continuous on R if exp (x) is continuous on R. Exp (x) is defined ...Mon, 13 Apr 2026 04:14:00 GMT