问道03的题

zhunshangUID: 2556536

我觉得这题只有第二个选项是正确的。
Let  fn(x)=2xn^{2}exp{-n^{2}x^{2}} defined on [0,1], then fn(x)->f(x)=0 for all x in [0,1], as n->\infinity.
Then F(x)=0,  Fn(x)=(1-exp{-n^{2}x^{2}})->1 as n->\infinity, thus, Fn!->F.

cobyssUID: 2638623

Hi! I don't know whether it's just your typing error, but if not, the first statement's absolutely false, since you can always add any constant to the infinite integrals F_n and F then the finite integral of them will vary largely and the equality can never be told for sure.

Again, I guess you mean F'_n goes to f instead of being equal to each other as you type, in the second statement. Right, it's true. A function can determine it's derivative --- none or unique --- but has a family of arbitrary infinite integrals.

As for the final one, based on the same notation, you can find a specific example to illustrate it (as we usually love to do when in high school). Let f_n = x ^ (-n), write down the expression for integral(0,x,f_n) then you'll find it does not even converge at x=1.

cobyssUID: 2638623

Sorry I forgot to give the conclusion just now:  Only the second statement is true.

cobyssUID: 2638623

Hi! I don't know whether it's just your typing error, but if not, the first statement's absolutely false, since you can always add any constant to the infinite integrals F_n and F then the finite integral of them will vary largely and the equality can never be told for sure.

Again, I guess you mean F'_n goes to f instead of being equal to each other as you type, in the second statement. Right, it's true. A function can determine it's derivative --- none or unique --- but has a family of arbitrary infinite integrals.

As for the final one, based on the same notation, you can find a specific example to illustrate it (as we usually love to do when in high school). Let f_n = x ^ (-n), write down the expression for integral(0,x,f_n) then you'll find it does not even converge at x=1.

fishingaloneUID: 2664682

6# zhunshang 请仔细看下一致收敛是什么意思：sup|fn(x)-f(x)|->0，你那个函数fn(x)在x=1/n点->Inf,它的极限函数在x=0根本不可能等于0，反例不成立

fishingaloneUID: 2664682

9# cobyss 你提的问题是对的，毕竟看的是回忆题，没那么精确。比如我们都知道sinx原函数是cosx，不用加什么常数。
至于后面那个问题，x^(-n)在[0,1]上根本就没有极限函数，x=0那点没法定义。

zhunshangUID: 2556536

10# fishingalone 请楼主重新看一下我给出的反例函数， 还有题目里说的是fn点点收敛到f, 好像跟一致收敛没关系吧