mm-411 === Subject: Re: integral limits > Hi everybody!!! > I have some difficulties with such problems: > 1 > 1). Find limit ( x^k integral f(t)/t^(k+1) dt ) > x->0 x > where k>0 and f-is continious function on [0,1]. We can write this as an integral from x to d plus an integral from d to 1 where x << d << 1 and d is otherwise arbitrary; the result of the d to 1 integral is a function F(d) of d. This is independent of x, so x^k*F(d) vanishes when x -> 0 for all possible d. For the 0 to d integral, approximate f(t) by f(0) + O(t). integrand ~ f(0)/t^(k+1) + O(t^-k) integral ~ f(0)/k*x^k + O(x^-(k-1)) [the contribution from the upper limit d is O(1) and therefore doesn't contribute after multiplying by x^k and taking the limit.] so x^k * integral ~ f(0)/k + O(x) -> f(0)/k as x -> 0 > x > 2). limit (integral sqrt(1+t^4) dt )/x^3. > x->infty 0 Again, split the integral at t = R, 1 << R << x; the integral from 0 to R is bounded, and so doesn't contribute to the limit; for the other integral, rewrite the integrand as t^2 * sqrt(1 + 1/t^4), use the binomial expansion (sqrt(1 + 1/t^4) ~ 1 + 1/(2t^4) + O(1/t^8)) and integrate. Again, the contribution from the lower limit t = R is a constant, and therefore can be ignored. > 3). Function f is continious on [0;+infty] and f(x) tends to A when > x tends to +infty. > x > Find limit ( 1/x ) integral f(t)dt. > x->+infty 0 Split at R, 0 << R << x, and write f(t) ~ A + O(1/t) in the R-x integral. As usual, the 0-R integral and the lower limit of the R-x integral are bounded and so do not contribute to the limit. -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. === Subject: Re: integral limits >1). Find limit ( x^k integral f(t)/t^(k+1) dt ) > x->0 x >where k>0 and f-is continious function on [0,1]. > We can write this as an integral from x to d plus an integral from > d to 1 where x << d << 1 and d is otherwise arbitrary; the result of the d > to 1 integral is a function F(d) of d. This is independent of x, so > x^k*F(d) vanishes when x -> 0 for all possible d. > For the 0 to d integral, approximate f(t) by f(0) + O(t). This started well but f(t) need not be f(0) + O(t); example: sqrt(t). But observe that if |f(t)| < c, then the expression we're looking at is also < c. This is enough to make the argument go. >2). limit (integral sqrt(1+t^4) dt )/x^3. > x->infty 0 > Again, split the integral at t = R, 1 << R << x; the integral from 0 to R > is bounded, and so doesn't contribute to the limit; for the other > integral, rewrite the integrand as t^2 * sqrt(1 + 1/t^4), use the > binomial expansion (sqrt(1 + 1/t^4) ~ 1 + 1/(2t^4) + O(1/t^8)) and > integrate. Again, the contribution from the lower limit t = R is a > constant, and therefore can be ignored. Or let t = xs and the expression becomes integral_[0,1] sqrt{(1/x^4) + s^2} ds. >3). Function f is continious on [0;+infty] and f(x) tends to A when >x tends to +infty. > x >Find limit ( 1/x ) integral f(t)dt. > x->+infty 0 > Split at R, 0 << R << x, and write f(t) ~ A + O(1/t) in the R-x integral. > As usual, the 0-R integral and the lower limit of the R-x integral > are bounded and so do not contribute to the limit. Well, I think that one's too easy to even give a hint on after the first two. === Subject: Re: Scrambling Problem > Nobody can help?.. :| break it into cases. no repeating letters, 1 pair and 2 different, 1 triple and another, etc. -- === Subject: One simple question Let G=(-00,a). Then, I wonder wheather we can say -00 is in G or not. If we can't say -00 is in G, Can we say -00 not in G ? === Subject: Re: One simple question > Let G=(-00,a). Regardless of whether infinity is real or otherwise, this is an open interval - and by definition, neither a, nor -00 are in it. Similarly, (-3,3], then -3 is not in this interval, but 3 is. Also, you can never have [-00,00] - it is has to be (-00,00) Kev === Subject: Re: One simple question > Let G=(-00,a). > Then, I wonder wheather we can say -00 is in G or not. G = (-00, a) is shorthand for G is the set of all real numbers less than a. > If we can't say -00 is in G, > Can we say -00 not in G ? Minus infinity is not a real number, therefore it is not in G. You can say it's not in G in the same way you can say George Washington is not in G -- neither -00 nor George Washington are real numbers. -- If this message helped you, consider buying an item from my wish list: Let G=(-00,a). > Then, I wonder wheather we can say -00 is in G or not. > If we can't say -00 is in G, > Can we say -00 not in G ? What would you say about whether a is in G? === Subject: linear functional I got the following problem in a linear analysis class and was hoping someone could help me smooth out the rough spots: Let X be normed vector space. Let {x_k} be a sequence of elements in X and let {a_k} be a sequence of scalars. Show that a necessary and sufficient condition for the existence of an f in X' (the dual space -- i.e., the space of bounded linear functionals on X) satisfying f'(x_k) = a_k for all k, and ||f|| = M (*) is that | b_1*a_1 + ... + b_n*a_n | <= M || b_1*x_1 + ... + b_n*x_n || (**) holds for each n and scalars b_1,...,b_n. This is what I've done so far. (==>) Suppose there is an f in X' satisfying (*). Then | b_1*a_1 + ... + b_n*a_n | = |f(b_1*x_1 + ... + b_n*x_n) | < = C ||b_1*x_1 + ... + b_n*x_n || by the definition of the norm ||f|| = M. (<==) Suppose (**) holds. Then consider the subspace generated by the x_k; call this subspace Y. Then each x in Y can be written in the form x = b_1*x_1 + b_2*x_2 + .... Now, define a functional F on Y by F(x) = F(b_1*x_1 + b_2*x_2 + ...) = b_1*a_1 + b_2*x_2 + ... Claim 1: F is linear. Pf: Write y = c_1*x_1 + c_2*x_2 + ... . Then F(x+y) = (b_1+c_1)*a_1 + (b_2+c_2)*a_2 + ... = b_1*a_1 + b_2*a_2 + ... + c_1*a_1 + c_2*a_2 + ... = F(x)+F(y). Also we clearly have F(qx)=qF(x) for some scalar q. Claim 2: F is bounded. Pf: Take limits as n --> infinity on (**): |b_1*a_1 + b_2*a_2 + ... | = lim | b_1*a_1 + ... + b_n*a_n | <= lim M || b_1*x_1 + ... + b_n*x_n || = M ||b_1*x_1 + b_2*x_2 + ... || using the continuity of the norm function. Claim 3: F(x_k)=a_k. Pf: Obvious from the definition of F. CLAIM 4: ||F|| = M. This is the one I'm stuck on. It's clear that ||F|| <= M, but I don't know how to go about showing equality. By the way, ||F|| is defined as the smallest number M for which |F(x)| <= M ||x||, where the norm of x is taken on X. Once I get CLAIM 4 out of the way, I can invoke the Hahn-Banach Linear Functional Theorem and claim F extends to a linear functional f on all of X -- can someone help me resolve CLAIM 4? === Subject: planar area of computations,{area of ciricle} and {triangle} {rectangle} all seperately by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1BDmZw13120; I request to become familiar with areas of circle for calculas does a impecable implimentaion come into being. and a idea to develop java applets expediantly for calculas tutorials use. Devin === Subject: Question about abelian groups by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1BGIDM26385; Hi everyone, I`ve got a problem I can`t solve. Suppose G is a group and every element a in G has the property that a^3=e (the identity) Can G be a non-abelian group?? I can`t figure it out. It`s obvious that a^2=a* (where a* is the inverse of a) I`ve tried manipulating products, but it doesn`t give anything me new. === Subject: vector calculus clue I need a clue please: I've got a cone (z >= 0) z^2 = x^2 + y^2 and a cylinder: (x-a)^2 + y^2 = a^2 I've to find the area of the cone which is inside the cylinder and the clue given says to use the area of a circle. But - I've a bit lost! Any clues would be good... Kev