Hanlin Posted June 30, 2004 Share Posted June 30, 2004 2004-06-30 Hello. My name is Hans Lindroth. I have a question about continuity. How do you verify/proove that, for instance, arcsine(x) is continuous? I mean, when I apply the definition of continuity on it I get: 0<=|arcsine(x) - arcsine(a)|<e (e>0) (Here "<=" means "smaller than or equal to"). I'm stuck here however since I don't know of any method to convert "arcsine(x) - arcsine(a)" into something I can modify. Can anybody help me out? If you can, please give me a hint. Thanks Hans Lindroth. Link to comment Share on other sites More sharing options...

JoeSF35 Posted July 4, 2004 Share Posted July 4, 2004 I believe you can use their derivatives. d/dx(arcsin(x)) = 1/sqrt(1-x^2) d/dx(arctan(x)) = 1/1+x^2 There sould be a table in your book for the others. From this you can verify that arcsin is only defined and continuous for -1=<x<=1, (testing limits). I am pretty sure this is along the line of being correct. Link to comment Share on other sites More sharing options...

Dave Posted July 4, 2004 Share Posted July 4, 2004 Or just use the inverse function theorem. Link to comment Share on other sites More sharing options...

bloodhound Posted July 7, 2004 Share Posted July 7, 2004 dave, did a google search on inverse function theorem, and it seems a lil bit hard.an you explain how you would use it to verify continuity for the given functions Link to comment Share on other sites More sharing options...

Dave Posted July 7, 2004 Share Posted July 7, 2004 It's not that hard, we did it in analysis II Can't remember it off the top of my head, but I'm pretty sure that it says something about the inverse of a continuous function being continous if a couple of requirements are met. Link to comment Share on other sites More sharing options...

e(ho0n3 Posted July 20, 2004 Share Posted July 20, 2004 I think the inverse function theorem is overkill. To prove that a function of a single variable x is contiunous on at a if: 1. f(a) is defined, 2. [math]\lim_{x \to a} f(x)[/math] exists, and 3. [math]\lim_{x \to a} f(x) = f(a) [/math]. 1 is necessarily trivial. You can prove 2 and 3 using a delta-epsilon argument which is what I think is being sought for. Link to comment Share on other sites More sharing options...

bloodhound Posted July 20, 2004 Share Posted July 20, 2004 why bother with that when u can reduce the problem to which is already solved Link to comment Share on other sites More sharing options...

e(ho0n3 Posted July 20, 2004 Share Posted July 20, 2004 I can't seem to formulate any proper delta-epsilon argument to prove the continuity of the arcsin function. Maybe using the inverse function theorem is the only viable path. Link to comment Share on other sites More sharing options...

Dave Posted July 20, 2004 Share Posted July 20, 2004 It would appear to be that way. I hate arcsin Link to comment Share on other sites More sharing options...

bloodhound Posted July 20, 2004 Share Posted July 20, 2004 i hate any inverse trig or hyperbolic functions Link to comment Share on other sites More sharing options...

Dave Posted July 20, 2004 Share Posted July 20, 2004 In case I didn't imply it, so do I. Link to comment Share on other sites More sharing options...

Gauss Posted September 8, 2004 Share Posted September 8, 2004 Several steps are involved in this and I will keep it as brief as possible. Hanlin The 'Limit formula' you used is applied for only a point. Since continuity at a point c is defined in terms of a limit, there is an [math]\in ,\mbox{ }\delta [/math]version of the definition. A direct trnslation of [math]\lim_{x \to c} f(x) = f©[/math] One method of proving that arcsin(x) is continuous is as follows: Step 1 Let (a, b) be an open interval. Then a function [math]f[/math] is continuous on (a, b) if it is continuous at each point c [math]\in [/math] (a, b). Step 2 Sin(x) has a polynominal expansion derived by using a taylor series. [math]P(x) = Sin(x)[/math] [math]P(x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} + ...[/math] Step 3 [math]\lim_{x \to c} P(x) = P©[/math] Proof omitted. In words, this result says that the limit of a polynomial P(x) as x approaches c not only exists, but is actaully the vlaue of P at c. Therefore polynomial functions are continuous at x = c for each number c. Step 4 [math]\lim_{x \to c} Sin(x) = Sin©[/math] Proof Take any number c. We can write [math]\lim_{x \to c} Sin(x) [/math] as [math]\lim_{h \to 0} Sin(c + h) [/math] [math]\lim_{h \to 0} Sin(c + h) = (Sin©) (\lim_{h \to 0}Cos(h)) + (Cos©)(\lim_{h \to 0}Sin(h))[/math] [math]\(Sin©)(1) + (Cos©)(0) = Sin© [/math] Step 5 Let f be a one to one function defined on an interval I. If [math]f[/math] is continuous, then its inverse [math]f^{-1}[/math] is also continuous. Proof omitted. [math]f(x) = Sin(x) [/math] The domain of [math]f[/math] is [math][-\pi/2, \pi/2][/math] [math]f'(x) = Cos(x) > 0[/math] for the domain of [math]f[/math]. Therefore [math]f(x) = Sin(x) [/math] has an inverse and therefore arcsin(x) is continous on the domain. Link to comment Share on other sites More sharing options...

Gauss Posted September 8, 2004 Share Posted September 8, 2004 e(ho0n3 Your conditions only apply for continuity at a point. Link to comment Share on other sites More sharing options...

Gauss Posted September 8, 2004 Share Posted September 8, 2004 Hanlin The other inverse fucntions are treated in a similar fashion. I hope this helps you. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted September 8, 2004 Share Posted September 8, 2004 I think the easiest way is to use the inverse function theorem like says dave. Show that the functions sine etc. satisfy the requirements and conclude that the functions arcsine etc are also continuous. Joe => It is impossible to use the derivatives of the arcsine functions to show continuity since they can only exists if your functions is continuous, making your arguementation circular. Gauss => A taylor series is not a polynomial ! showing that polynomials are continuous is easily done using epsilon-delta definition. But your step 3 involves changing two limits ! (that of the partial sums and that of x tending to c) which can be rather tricky.... Mandrake Link to comment Share on other sites More sharing options...

e(ho0n3 Posted September 8, 2004 Share Posted September 8, 2004 Sin(x) has a polynominal expansion derived by using a taylor series. [math]P(x) = Sin(x)[/math] [math]P(x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} + ...[/math] This is a no no. You can't use the Taylor series expansion because Taylor series are constructed using derivatives and derivatives automatically imply continuity. [math]\lim_{x \to c} P(x) = P©[/math] Proof omitted. You destroy the purpose of this exercise by 'omitting' the proof. Let f be a one to one function defined on an interval I. If [math]f[/math] is continuous' date=' then its inverse [math']f^{-1}[/math] is also continuous. Proof omitted. You're doing it again. e(ho0n3 Your conditions only apply for continuity at a point. Naturally, but if the point is an arbitrary point in the domain of the function in question, then it follows that the function in question is continuous in its domain. Link to comment Share on other sites More sharing options...

Gauss Posted September 8, 2004 Share Posted September 8, 2004 MandrakeRoot We can write Taylor polynominals: [math]P_n(x) = f^{0}(0) + f^{1}(0)x + \frac {f^{2}(0)}{2!}x^2 + . . . + \frac {f^{n}(0)}{2!}x^n[/math] in [math]\Sigma [/math] notation. If f is infinitely differentiable on an open interval I containing 0 then we have [math]P_n(x)= \sum\limits_{k = 0}^n {\frac{{f^{(k)}}}{{k!}}} x^k [/math] Hence we say that [math]f(x)= \sum\limits_{k = 0}^n {\frac{{f^{(k)}}}{{k!}}} x^k + R_{n+1}(x) [/math] and taking limits we say that f(x) can be expanded as a Taylor series in x and write [math]f(x)= \sum\limits_{k = 0}^{\infty} {\frac{{f^{(k)}}}{{k!}}} x^k[/math] Therefore Taylor polynomials can be expressed as Taylor Series. You said Quote "But your step 3 involves changing two limits ! (that of the partial sums and that of x tending to c) which can be rather tricky...." Sorry no it does not. [math]\lim_{x \to \infty}P(x) =P©[/math] Is simply basic limit theory for polynomials. Link to comment Share on other sites More sharing options...

Gauss Posted September 8, 2004 Share Posted September 8, 2004 e(ho0n3 You state "This is a no no". I am sorry but you are wrong. I can and I did use the Taylor polynomial (series) with the explicit intention to show that sin(x) is continuous over its domain. Hence Sin(x) was shown to be continutous in Step 3 of the solution. If someone is just 'given' the Talyor polynomial then they can use basic limit theory for polynomials to show that Sin(x) is continuous without knowing about derivatives of functions which inturn automatically imply continuity. When you say "You destroy the purpose of this exercise by 'omitting' the proof". I totally disagree. The problem was to show that inverse fucntions, especially the trigonometric ones are continous. This was done in the solution I provided, the question did not ask for proofs of basic limit theory. I agree when you say "You're doing it again", a proof would have been nice here, however I did say "I will keep it as brief as possible". However, I did leave some work for Hanlin to do. The defintion for Continuity at a Point states: Let [math]f[/math] be a function defined at least on an open interval [math](c - p, c + p) [/math] with p > 0. Then [math]f[/math] is continuous at c iff [math]\lim_{x \to \infty}f(x) = f© [/math] Therefore according to this defintion a fucntion is continuous at a point c iff: [math]f[/math]is defined at c. [math]\lim_{x \to c}f(x)[/math]exists. [math]\lim_{x \to c}f(x) = f© [/math] The defintion for Continuity on an interval states: Let (a, b) be an open interval. Then a function [math]f[/math]is continuous on (a, b) if it is continuous at each point c [math]\in[/math](a, b). This means that continuity at point can only be applied at that particular point. To prove that the function is continuous over the requried interval then you need to prove it by either step 3 or step 4 as was shown in the solution. Therefore your statement "but if the point is an arbitrary point in the domain of the function in question, then it follows that the function in question is continuous in its domain" is not correct. Link to comment Share on other sites More sharing options...

matt grime Posted September 8, 2004 Share Posted September 8, 2004 No, Gauss, you are completely in the wrong, you may not use Taylor series to show a function is continuous since you are implying it is differentiable, and therefore already continuous (a circular argument if you will). (and the OP did not say that Taylor series of trig functions were to be assumed to exist.) Link to comment Share on other sites More sharing options...

Gauss Posted September 8, 2004 Share Posted September 8, 2004 Yes I agree with you that it is a "circular argument". Anybody can generate a polynomial function without knowing about Sin or the Taylor series (yes I know that everybody who gets to this stage should know about Sin). Therefore polynomial functions are continuous at x = c for each number c. So your statement "you are completely in the wrong" is wrong. The original problem also did not say that the Taylor series of trigonometric functions were not to be assumed to exist. Again a "circular argument". Link to comment Share on other sites More sharing options...

e(ho0n3 Posted September 8, 2004 Share Posted September 8, 2004 The original problem also did not say that the Taylor series of trigonometricfunctions were not to be assumed to exist. Again a "circular argument". So much for that...I still believe this problem is solvable using a delta-epsilon argument. Link to comment Share on other sites More sharing options...

Dave Posted September 8, 2004 Share Posted September 8, 2004 I'd just like to point out that a google search for "continuity of arcsin" reveals this thread as the top link I've not got any further with it using a delta-epsilon argument. The problem seems to manipulating arcsin (which is a bit of a git to get around any day of the week) to get what you want out of it. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted September 9, 2004 Share Posted September 9, 2004 MandrakeRoot We can write Taylor polynominals: [math]P_n(x) = f^{0}(0) + f^{1}(0)x + \frac {f^{2}(0)}{2!}x^2 + . . . + \frac {f^{n}(0)}{2!}x^n[/math] in [math]\Sigma [/math] notation. If f is infinitely differentiable on an open interval I containing 0 then we have [math]P_n(x)= \sum\limits_{k = 0}^n {\frac{{f^{(k)}}}{{k!}}} x^k [/math] Hence we say that [math]f(x)= \sum\limits_{k = 0}^n {\frac{{f^{(k)}}}{{k!}}} x^k + R_{n+1}(x) [/math] and taking limits we say that f(x) can be expanded as a Taylor series in x and write [math]f(x)= \sum\limits_{k = 0}^{\infty} {\frac{{f^{(k)}}}{{k!}}} x^k[/math] Therefore Taylor polynomials can be expressed as Taylor Series. You said Quote "But your step 3 involves changing two limits ! (that of the partial sums and that of x tending to c) which can be rather tricky...." Sorry no it does not. [math]\lim_{x \to \infty}P(x) =P©[/math] Is simply basic limit theory for polynomials. You havent defined P and an infinite sum is a limit of the partial sums by definition. Since for what i understood of your argument you were using the fact that polynomials are continuous, writing a function f as an infinite sum (a power series) and then taking the limit and wanting to apply your argumentation would involve interchanging two limits. Or the use of arguments of the type that you can get the tail uniformly small or whatever. Mandrake Link to comment Share on other sites More sharing options...

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