Consider this answer. The following claim is proved:
> Claim: If $g: \mathbb R \to \mathbb R$ is continuous and $f_n: [0,1] \to \mathbb R$ converges uniformly to $f$, then $g\circ f_n$ converges uniformly to $g\circ f$.
They write that since $f_n\to f$ uniformly on $[0,1]$, there exists an $M$ such that $|f_n(x)|\leq M$ for all $x\in[0,1]$. If a sequence of functions converges uniformly on a compact set, does that imply boundedness? Or are they also assuming that $(f_n)$ is a continuous sequence of functions? If the latter, then that makes sense. In case the former, then that is something I have not found in either Rudin or Spivak.
Bro I love how stable all software and systems are such that when software crashes, the whole system is capable of crashing when this should really be impossible.
My try for the question:
Let $X,Y$ be subsets of $\mathbb{N}$.
Let $f: X\rightarrow Y$ be a bijection function with $\operatorname{Card}(X)=n,\operatorname{Card}(Y)=m$ both in $\mathbb{N}$. I think that If we show Card$(X)$=Card$(Y)$ then we are done. So, how? Can you give a hint?
As I remark in a comment to the accepted answer, "How does showing 𝐶𝑎𝑟𝑑({1,...,𝑚})=𝐶𝑎𝑟𝑑({1,...,𝑛}) suffice, when we are (at least as far as I can tell) using this very theorem to show that 𝐶𝑎𝑟𝑑 is well-defined?"
Ostensibly, the argument proves $\Num{\{1,...,m\}} = \Num{\{1,...,n\}}$ but there are two problems with that: (1) We are using this theorem/lemma to prove that $\Num X$ is well-defined in the first place! Thus we cannot make mention of the notion which is not, as yet, well-defined. (2) Even if (1) were not a problem, $\Num{\{1,...,m\}} = \Num{\{1,...,n\}}$ just means, by definition, that both have a bijection to some $\{1,...,k\}$, which is to say (by composition of bijections) there is a bijection between $\{1,...,m\}$ and $\{1,...,n\}$. But that was our hypothesis in the first place, and …
(I ask here because the user does not appear to be active)
@TedShifrin If $(a_n)$ is a monotonic sequence of reals s.t $\lim_{k \to \infty} \frac{a_{2^{k+1}}}{a_{2^k}} = 0$, does it imply that the sum of $\sum a_n$ is finite? I wondered this when I observed sequences $(t_n)$ where the sequence $nt_n$ converges
@TedShifrin I initially asked this where the lim < \frac{1}{2} too, by considering various examples. However I thought it would be easier to consider the case where the limit equals zero
Write the function $f(x)=e^x$ as a sum of infinite series using Taylor's Theorem with Lagrange's Form of Remainder.
My solution goes like this:
Taylor's Theorem with Lagrange's form of remainder states that, if $f$ is a function defined on an interval $[a,a+h]$ where $h>0$ and if $f$ satisfies th...
Sorry to bother you, but almost none of the stack overflow sites have been displaying properly in any of my browsers on my Mac, including Mathematics for the last two days. They are working fine on my phone, however. Is there anything I can do to fix this? Many thanks
@ThomasFinley I'm sorry, I didn't read "basis" in your question. If $F$ is a given field, then direct sum $\bigoplus_{x\in X} F$ works perfectly well for an $F$-vector space with basis of size $|X|$. You can construct this vector space as the space of functions $f:X\to F$ with $f(x) = 0$ for all $x\in X$ except for finite amount. The functions $f_x$ defined as $f_x(y) = 1$ iff $x = y$ and $0$ otherwise, form a basis of this space
@robjohn I suppose the inverse of the claim is true also, isnt it? (namely we can deduce the limit if we assume the convergence of the series and the monotonicity)
@SoumikMukherjee yeah when we are looking in the vector space R^n, and we will be adding functions in the set of all functions from $n$ element set to $\Bbb{R}$
@robjohn My intuition was that taking $(a_n)$ to be $(\frac {1}{n})$ shows that the ratio equals $\frac{1}{2}$ exactly and maybe we can learn from the divergence of the harmonic series
@SoumikMukherjee K, I get it...and this is precisely why we can't really denote the elements of $\bigoplus _{x\in S}\Bbb R$ as (a1,a2,...,an,...) i.e in the way we denote finite tuples, right?
I mean that's precisely the reason why that standard notation of finite tuples makes no sense at all. Did I get it?
@SoumikMukherjee I think by coordinates you mean the elements of S, right? I think you are talking about the fact, that each elememt in R has an uncountable number of pre-images in S, under consider all the functions?
@SoumikMukherjee Btw, can you take a look at this? I need some validation, if my proof is correct? It's a Taylor series expansion of $e^x.$ Strangely, this question seemed to stump most of the people , and I dont know why...
Write the function $f(x)=e^x$ as a sum of infinite series using Taylor's Theorem with Lagrange's Form of Remainder.
My solution goes like this:
Taylor's Theorem with Lagrange's form of remainder states that, if $f$ is a function defined on an interval $[a,a+h]$ where $h>0$ and if $f$ satisfies th...
Does anyone know what the probability distribution for a best of N match would be, given the probability of winning each game is $p$? E.g. in a best of 3, you win 2 of the three and then you win the best of 3.
I thought it might be a binomial distribution, but if $p=1$, then the probability of winning two of them in a best of 3 is zero, since you win all 3. Summing over binomial distributions though sounds like it would include scenarios not present in a real best of 3.
@Jakobian the result by Moore I was looking at yesterday says (in modern language) that if you quotient the plane by an upper semicontinuous decomposition in nonseparating continua, then the quotient is homeomorphic to the plane. This is supposedly a famous theorem. Do you happen to know a reference with a proof in modern notation and language? The original paper is rather unreadable
@shintuku If you're talking about those removed messages, then here what it was: I was asking the user SoumikMukherjee to take a look at my post. That's all. But since he isn't online, I thought 'bout deleting it :/
@AlessandroCodenotti in modern terminology, this would mean that if $q:\mathbb{R}^2\to X$ is a monotone perfect quotient map, then $X\cong \mathbb{R}^2$
Let $x\in R$ for a Boolean ring $R$. We have
$$\begin{align}
x&=x^2\, \text{(Boolean)}\\
&=xx\, \text{(Indices)}\\
&=xx^2\, \text{(Boolean)}\\
&=x\color{red}{x}x\, \text{(Indices)}.
\end{align}$$
So just let $y=\color{red}{x}$.
The above got two downvotes recently. I have no idea why. Any suggestions?
Does anyone know in which kind of books I can find the definition of uniform convergence of a net of functions? Under what topic does this go? I have looked at some introductory topology books and functional analysis books, but haven't found a plain definition of what it means. Can we also use the sup-norm to describe uniform convergence of a net of functions?
If you google uniform convergence of net of functions, you end up with a lot of specialized content, whereas I'm looking for a textbook-kind of definition.
@ThomasFinley You have written down Taylor's theorem for [a,a+h], h>0. You could prove a similar result for [a-h,a], h>0 as well using the proof for the former almost word to word.
@ThomasFinley You have written down Taylor's theorem for [a,a+h], h>0. You could prove a similar result for [a-h,a], h>0 as well using the proof for the former almost word to word.
Are the only palindrome numbers of the form $2^a+3^b$ with non-negative integers $a,b$ : $$[2, 3, 4, 5, 7, 9, 11, 33, 131, 515, 737, 19691]$$ There are no more for $a,b\le 3\ 000$ , so the list should be complete. But can we prove it ?
@Koro that means normally, we can have, in [a,a+h] taylor's theorem as, $f(a+h)=f(a)+hf'(a)+....+\frac{h^{n-1}f^{n-1}(a)}{(n-1)!}+R_n$ but now, I have to prove, $f(a)=f(a+h)-hf'(a+h)+....+\frac{(-1)^{n-1}h^{n-1}f^{n-1}(a+h)}{(n-1)!}+R_n$ , right?
Ok, but what's the Taylor series of $f$. If I got u correctly then you want me to show Taylor series of f converges to f ? But if I knew the taylor series of f, I would be done?
@Koro I think I need to show g(-x) converges to $f(x)=e^{-x}$. Isn't it?
I am assuming that $g$ is the Taylor series expansion of $e^x$ when $x\gt 0$ ...
@ThomasFinley what do you mean by "g(-x) converges to f(x)". This makes no sense at all. 1) g(-x)= f(x)., 2) Taylor series of f about 0 is: f(0)+xf'(0)+...= 1-x+x^2/2-... and it converges to f(x) because its R_n(x) goes to 0. 3) So by 2) & 1), it converges to g(-x) because g(-x)=f(x).
4) Therefore, your problem of considering [a-h,a] is solved.
My try for the question:
Let $X,Y$ be subsets of $\mathbb{N}$.
Let $f: X\rightarrow Y$ be a bijection function with $\operatorname{Card}(X)=n,\operatorname{Card}(Y)=m$ both in $\mathbb{N}$. I think that If we show Card$(X)$=Card$(Y)$ then we are done. So, how? Can you give a hint?
Let $x\in R$ for a Boolean ring $R$. We have
$$\begin{align}
x&=x^2\, \text{(Boolean)}\\
&=xx\, \text{(Indices)}\\
&=xx^2\, \text{(Boolean)}\\
&=x\color{red}{x}x\, \text{(Indices)}.
\end{align}$$
So just let $y=\color{red}{x}$.
