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12:09 AM
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.
 
It’s certainly not true without continuity.
Unless you know the limit function is bounded.
 
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.
 
Yeah, $f_n$ bounded and $f_n\to f$ uniformly implies $f$ bounded. My guess is that they assume continuity.
 
Apparently while I was gone, either my system rebooted, or it hard crashed and restarted because I was playing Sonic Mega Collection on Dolphin. Bruh.
 
the OP there states continuity as an assumption
 
12:31 AM
Hi, Thor
 
1:35 AM
Would someone be able to tell me if I'm right to be suspicious of the answer here:
1
Q: Let $n,m\in\mathbb{N}$. Show that if there is a bijection between $\{1,..,n\}$ and $\{1,..,m\}$ then $n=m$.

user295645My 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)
 
2:32 AM
@SoumikMukherjee The basis of this vector space is $1,x,x^2,...,x^n$ where $x\in F$
 
X4J
2:49 AM
@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
 
3:25 AM
This isn’t quite the usual “condensation test.” I don’t know offhand.
 
X4J
@TedShifrin Is there a book you'd recommend me to focus if I want to expand my knowledge in infinite sums at the level of real analysis 2?
 
You might look at Polya/Szegö. Some analysis books have more esoteric things, like Raabe’s test.
 
X4J
Yes, I've introduced to Raabe's test only couple of days ago and never heard about it before
I find the idea of infinite series surprisingly unintuitive considered to the rest of mathematics I studied during the semester
 
This is not particularly something I’ve ever dabbled in other than teaching Spivak’s Calculus and some other exercises.
 
@X4J I think if $\limsup\limits_{k\to\infty}\frac{a_{2^{k+1}}}{a_{2^k}}\lt\frac12$, the series converges.
 
3:36 AM
Personally, I prefer multivariable analysis and differential geometry.
Robjohn, assuming monotonic?
 
@TedShifrin yes
 
That is the harmonic series boundary…
 
X4J
@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
 
$$\sum_{j=2^{k+1}}^{2^{k+2}-1}a_j\le2\limsup\sum_{j=2^{k}}^{2^{k+1}-1}a_j$$
by monoticity and the condition. So if the $\limsup\lt\frac12$, it should converge
 
You need $\sum 2^k a_{2^k}$ to converge.
 
3:46 AM
Break the series into pieces $a_{2^k}+\dots+a_{2^{k+1}-1}$
My condition says the ratio of these pieces is geometric
 
Indeed. My comment was to X4J, not to you :)
 
Ok. I was just sort of justifying to whoever might be listening ;-)
 
So there’s an obvious generalization to $m^k$ ….
i wonder how useful these things really are.
 
X4J
@TedShifrin I don't think it actually is, I just ask it to check if I understand
@robjohn Thank you mate, that helped
 
4:08 AM
0
Q: Write the function $f(x)=e^x$ as a sum of infinite series using Taylor's Theorem with Lagrange's Form of Remainder.

Thomas FinleyWrite 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...

Need help with this issue.
 
4:19 AM
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
 
Like what’s wrong, Shy?
 
@TedShifrin Can you please take a look at my post (, if you have time) ?
 
@ThomasFinley Yes, so now you can answer your question when $X$ is finite or countable. For the uncountable case, see what Jakobian wrote earlier.
 
thanks for helping! It just looks like an old craigslist site, with just text, hyperlinks and no other formatting at all. It’s unusable.
 
4:36 AM
@SoumikMukherjee Thanks! Now, I can do it for finite sets. But can you please point out, the message of Jakobian which you want me to look at?
I think I understand which message ur talking about , but still some portions of it, goes above my brains...
Can you please help me with the infinite case, @SoumikMukherjee? I really like your explanations :)
 
I just received a letter from Stack overflow support at 9:23, so I will bow out for now. Many thanks for your attention and help.
 
11 hours ago, by Jakobian
@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
 
@SoumikMukherjee Yeah, but as I said, it all goes over my head :(
 
@ThomasFinley The countable case can be done by the set of all polynomials over $\Bbb{R}$
 
@SoumikMukherjee I agree and the basis is, $\{1,x,x^2,...\}$ , right?
 
4:51 AM
@ThomasFinley Okay, first make this observation. Let $S$ be a finite set with $n$ elements
@ThomasFinley yes
 
@SoumikMukherjee Ok
 
Now consider the set of all functions from $S$ to $\Bbb{R}$
 
@SoumikMukherjee Ok
 
On the other hand, consider the set $\Bbb{R}^n$
 
@SoumikMukherjee hmm..
 
4:53 AM
Can you make some identification between these two sets?
 
@SoumikMukherjee the range of the functions is the set R^n precisely?
There is a 1-1 correspondance between those two sets u mentioned
This means they are the same sets?
 
@ThomasFinley Right, but what you wrote above is incorrect
 
@SoumikMukherjee Why?
Sorry, if I sound silly
 
Each function from $S$ to $\Bbb{R}$ can be identified with a point in $\Bbb{R}^n$
 
@SoumikMukherjee you mean if $f:S\to\Bbb R$ and $(a_1,a_2,...,a_n)\in\Bbb R$ then $f=a_k$ where $k\in \{1,...,n\}$ ?
 
4:59 AM
@ThomasFinley What you understood is correct, that there is 1-1 correspondence. But you expressed it incorrectly
 
X4J
@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)
 
@ThomasFinley No, $f$ can be identified with the whole $n$-tuple $(a_1,a_2,...,a_n)$
 
@SoumikMukherjee My understanding is: let $f:S\to\Bbb R$ then there exists $(a_1,a_2,...,a_n)\in\Bbb R^n$ such that $f\equiv (a_1,a_2,...,a_n)$.
 
@SoumikMukherjee ok
Now?
 
5:02 AM
@X4J I don't think that's true.
 
Now for finite case, you can identify the set of all functions from $n$ element set to $\Bbb{R}$ with $\Bbb{R}^n$
 
X4J
If we say the \leq
 
@SoumikMukherjee yes.
 
X4J
I mean, we want to deduce that $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} \leq \frac{1}{2}
 
So the set of all functions from $n$ element set to $\Bbb{R}$, is precisely the required Vector Space, right?
 
5:04 AM
Now the set of functions will be a vector space, so we should be able to add the functions, right?
 
@SoumikMukherjee yes
 
@X4J I don't think we can deduce that.
 
We will be adding the corresponding points in $\Bbb{R}^n$
 
@SoumikMukherjee yeah
@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 $\Bbb{R}$
 
This should not make any problem as if we add finite number of points, we still get a point in $\Bbb{R}^n$
@ThomasFinley yes
 
5:07 AM
@SoumikMukherjee yes.
 
X4J
@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
 
Now consider the infinite case
 
@SoumikMukherjee ok
The case when $S$ is infinite, right ?
 
@SoumikMukherjee ok, now?
 
5:12 AM
We consider $\bigoplus_{x \in S}\Bbb{R}$
 
@SoumikMukherjee I never encountered this symbol
What does this mean?
I only know bout the direct sum of two subspaces
of a vector space
That's the only place I have use the $\oplus$ symbol
 
This is defined as the set of tuples $(a_i)_{i \in S}$ with each $a_i \in \Bbb{R}$ such that $a_i=0$ for all but finitely many $i$
 
@X4J just because this holds in a nice example, does not mean that it holds in general. I am writing up an example.
 
@SoumikMukherjee should we consider tuples of all sizes?
 
X4J
@robjohn Of course, I was just contemplating it
Anyways I am certain that I fail to grasp something since it looks intuitively true to me, I will try to think of an example in the meantime
 
5:23 AM
@ThomasFinley I don't get it
 
@SoumikMukherjee I mean should we consider tuples of all possible length say, 1,2,3,...
Say, (12,3,0,0,0,...) and (1,5,0,0) and (0,0,0,0,1,3,3) and so on.?
 
no no
$i \in S$ denotes the coordinate
 
@SoumikMukherjee Oh! I thought, $\bigoplus_{x\in S}\Bbb R =\{(12,3,0,0,0,...),(1,5,0,0),...\}$
 
No, as we need same length to add things coordinate wise
 
@SoumikMukherjee Can you show me an example, just to clear up this messy little confusion?
 
5:30 AM
If the set is uncountable, then you can't write that in the form $(a_1,a_2,...)$, right?
 
@SoumikMukherjee umm...if you mean, $S$ is uncountable then how can we write $S=(a_1,a_2,...)$ ? Uncountable sets can't be enumerated
 
yes
19 mins ago, by Soumik Mukherjee
We consider $\bigoplus_{x \in S}\Bbb{R}$
So we use this type of notation
 
@SoumikMukherjee Oh! I get it...
@SoumikMukherjee But does this set have a single tuple or many tuples?
 
say our set $S$ is $[0,1]$, so for each $i \in [0,1]$ we consider a copy of $\Bbb{R}$
 
@SoumikMukherjee I think you mean, "we consider an element of $\Bbb R$" , right?
Also, $[0,1]$ and $\Bbb R$ have the same cardinality
 
5:35 AM
@ThomasFinley uncountable tuples, more precisely $\lvert S\rvert$ tuples
 
@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?
 
@ThomasFinley Yes, as we have uncountable tuples
@ThomasFinley yes
That's why we write this notation
 
@SoumikMukherjee and uncountable elements in each tuple, right?
 
yes, but we have another restriction
 
@SoumikMukherjee what is that ?
 
5:39 AM
say you pick an arbitrary element, it has uncountably many coordinates, right?
now our condition is that only finitely many of them can be nonzero
 
@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?
 
@ThomasFinley Each element of $S$ denote a coordinate, right?
 
@SoumikMukherjee yes
 
lets think this way, say $S=\{1,2,3\}$ and consider the set of all functions from $S$ to $\Bbb{R}$
 
@SoumikMukherjee ok
 
5:44 AM
we can view this set as $\Bbb{R}^3$ right?
 
@SoumikMukherjee yes
 
for each element of $S$, we have an coordinate of $\Bbb{R}^3$, right
 
@SoumikMukherjee yes
 
for 1, we have the first coordinate, for 2 the middle and for 3 the last
 
@SoumikMukherjee yup
 
5:48 AM
Now let's go back to the uncountable case, for each $i \in S$ we have a coordinate of $\Bbb{R}$
 
@SoumikMukherjee ok
 
now an element of $\Bbb{R}^3$ looks like $(a_1,a_2,a_3)$
 
@SoumikMukherjee yeah
 
Similarly we write an element of $\bigoplus_{i \in S}\Bbb{R}$ as $(a_i)_{i \in S}$
 
@SoumikMukherjee perfect
All's clear up until here...
 
5:53 AM
okay, now we use the condition that finitely many $a_i$ should be non zero
 
@SoumikMukherjee ok
 
Why this condition is used will be clear later
 
@SoumikMukherjee fine
 
Now lets get back to $\Bbb{R}^3$, how to construction the standard basis of it?
 
@SoumikMukherjee (1,0,0),(0,1,0) and (0,0,1)
 
5:56 AM
Yes
 
@SoumikMukherjee fine... all's clear up until this point...
 
so we take $1$ in a coordinate and $0$ in the rest
 
@SoumikMukherjee Ok
 
@ThomasFinley So what is the cardinality of this set? $3$ as there are $3$ coordinates
 
@SoumikMukherjee and the set of all such tuples have the same cardinality as $S$ right?
 
5:58 AM
:64453897 yes, you mean $S$ right?
yes
 
So we found a basis having same cardinality as $S$ , right? Also, we constructed such a vector space
That solves the problem, i guess
 
Yes
yes, this is what Jackobian wrote earlier, can you now understand that answer?
 
@SoumikMukherjee Thanks a ton! Good gracious! The problem's finally solved and we are done, right? I can't believe it myself as well...
So gimme a cofirmation, please :?)
 
@ThomasFinley A little part is left, why the finiteness condition
 
@SoumikMukherjee where ?
 
6:02 AM
@ThomasFinley That finitely many coordinates can be non zero
 
@SoumikMukherjee so that we can make a one-one correspondance of the tuples in basis to the elements of S?
 
this is because, we need to write each element as finite linear combination of basis elements
 
@SoumikMukherjee oh!
@SoumikMukherjee got it! That's the reason why we took that assumption.
So, finally all complete, correct?
:)
 
yes
@ThomasFinley otherwise we had to extend the size of our basis
 
@SoumikMukherjee Thanks a lot! I got the whole thing now...
@SoumikMukherjee yeah...
 
6:06 AM
You are welcome
 
@SoumikMukherjee Adieu, for now!
Again, thanks a thousand times!
 
no problem
 
What a relief! Good night....
 
Good night
 
@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...
0
Q: Write the function $f(x)=e^x$ as a sum of infinite series using Taylor's Theorem with Lagrange's Form of Remainder.

Thomas FinleyWrite 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...

 
X4J
6:28 AM
@robjohn You got a counterexample?
 
Bye all, 💤
 
X4J
night night
 
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.
 
6:45 AM
@ThomasFinley Not a problem :) I will take a look at it afterwhile
 
 
2 hours later…
8:26 AM
@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
 
 
2 hours later…
10:21 AM
Ok, no one?
 
the curse of the self-learner
 
10:49 AM
yeah that's right, you have no right insulting me
 
@shintuku umm...not sure whom ur talkin about?
I don't ever remember insulting you :/
 
pff ok whatever
 
@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 :/
 
i've no time for this
 
@shintuku me too.
 
11:30 AM
@AlessandroCodenotti in modern terminology, this would mean that if $q:\mathbb{R}^2\to X$ is a monotone perfect quotient map, then $X\cong \mathbb{R}^2$
right?
maybe I went too far with calling it perfect
 
Hmm I'm not sure, you need the nonseparating part
Upper semicontinuous means that the quotient map is closed
 
that's true
did you try looking here?
I don't know this theorem, and it's so specific it's hard for me to figure out what to look for
there's this expository paper, that theorem should be equivalent I think
 
11:51 AM
That looks great! Thanks!
 
12:11 PM
@ThomasFinley I don't really understand what's happening in chat yet, I came here because of Alessandro's ping
 
12:25 PM
-1
A: Why is every Boolean ring regular?

ShaunLet $x\in R$ 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?
And I know I can't know for sure . . .
 
isn't this trivial
 
It's not if you overthink it.
 
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.
 
12:56 PM
@ThomasFinley Please stop spamming the chat with a link to your question.
Once is enough.
 
Suppose
 
16 messages deleted
 
that (X, F, mu) is a measure space. And it is given that f:X-->[0,oo) is F measurable.
For every E\subset F, where E is a sigma field, there exists a g:X--->[0,oo) such that $\int_A g= \int_A f$ for every A in E.
I don't see why the following gives a counterexample: take E={\emptyset, X}
assume mu to be sigma finite.
 
@XanderHenderson Ok, I am sorry, I was desparate. Can you please help me with it?
Also, @Koro you once helped me with a calculus problem. Can you please take a look at this: math.stackexchange.com/questions/4773440/…
? I am trying to solve it for 8 hours but I failed :(
 
1:13 PM
@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.
so in your case e^x, x>0 or x<0 doesn't matter.
 
@Koro Wait,
The thing is
Can we expand it in [x,0]
?
 
yes, you can about 0. There is nothing special about left end or right end.
prove this:
2 mins ago, by Koro
@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.
using the proof for [a,a+h], h>0.
 
I proved it
already
But then?
 
you expand about 0 in [x,0], x<0
 
@Koro by taking a=x and h=0, right?
 
1:18 PM
a=0, h=-x
 
@Koro oh! Ok, just a sec
Wait, then we get an expansion like, $f(0)=f(x)-xf'(x)+...$, right?
@Koro expanding it in that way, I got, $1=e^x(1-x+x^2/2+...+(-1)^nx^ne^{-\theta x}/n!)$
But then?
Ok, then, $e^{-x}=1-x+x^2/2+...+(-1)^nx^ne^{-\theta x}/n!$
Now, I need to prove, $\lim_{n\to\infty}\frac{(-1)^nx^ne^{-\theta x}}{n!}=0$
 
1:36 PM
@ThomasFinley no, you expand about 0. That is, f(x)=f(0)+(0-x)f'(0)+... etc.
(atleast, that's what I meant.)
36 mins ago, by Koro
I don't see why the following gives a counterexample: take E={\emptyset, X}
it does.
 
@Koro Wait, isn't the Taylor's Theorem in this case be, $f(a)=f(a-h)+hf'(a-h)+h^2\frac{f''(a-h)}{2!}+\cdots +R_n.$ ?
R_n is the Lagrange's Remainder
And you said, h=-x and a=0 ?
 
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 ?
 
1:53 PM
Wow, now I am trying this for 9 hours! It's so good!
 
@ThomasFinley gg
 
@SineoftheTime I think I'll go for a stroll now...
 
good idea
 
@SoumikMukherjee Did you take a look at it, buddy?
 
Not yet, I am stuck with something else
 
1:59 PM
My analysis isn't strong enough to comment Thomas, sorry :(
 
@EE18 Np
@SoumikMukherjee Fine :)
 
@ThomasFinley no. it will be f(a-h)= f(a)+ (0-h) f'(a)+ (0-h)^2 f''(a)/2+...
expanding about a means that you have f(a), f'(a), f''(a) etc. on right.
 
@Koro oh! So you are considering the interval $[a,a-h]$ right?
 
no, I am considering [a-h, a].
48 mins ago, by Koro
prove this:
if you don't like this, then consider Taylor series for $f(x)= e^{-x}$ for x>0.
show that the Taylor series of f converges to f with the theorem you wrote in your post.
 
@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?
@Koro from here I couldn't understand...
 
2:10 PM
@ThomasFinley no, prove that on [a-h,a], h>0, f(a-h)= f(a)+ (0-h) f'(a)+ (0-h)^2 f''(a)/2+...
 
@Koro for the time being I am assiming this true
Then?
 
put a=0, h=-x, x<0
 
@Koro ok, I got f(x)=f(0)+(0-x)f'(0)+...
 
now, remainder term goes to 0.
(as it did in case of x>0)
 
Wait, what's the expression of R_n in here?
$\frac{f''(a-\theta h)(-h)^n}{n!}$ is the term in case of f(a-h)= f(a)+ (0-h) f'(a)+ (0-h)^2 f''(a)/2+... , right , @Koro ?
where $\theta\in (0,1)$ ?
 
2:15 PM
it is what you would expect: $(-h)^nf^{(n)}(z_x)/n!,$ where $z_x$ is some point in (a-h,a).
@ThomasFinley yes.
 
@Koro then it will be $+\theta h$ isn't it ?
 
why? a-\theta h is in (a-h,a)
 
No, no what I wrote previously is correct...
@Koro yes
Also, $f^{(n)}(-\theta x)$ is bounded
Is it?
So, $\lim R_n=0$ , right?
 
or you could use the ratio test.
 
I mean the remainder term goes to zero as $f^{(n)}(-\theta)$ is bounded and $\frac{(-1)^nx^n}{n!}=0$ ?
Isn't it a valid way?
@Koro yeah, sure, but I think this way is valid as well, right?
 
2:22 PM
$f^{(n)} (z)$ is bounded on [a-h,a].
but rest is okay.
 
@Koro Fine, and we are done, aren't we ?
 
I think so?
 
The only thing remaining is to prove, on [a-h,a], h>0, $f(a-h)= f(a)+ (0-h) f'(a)+ (0-h)^2 f''(a)/2+...+(-h)^nf^{(n)}(z_x)/n!,$ ?
 
yes.
 
Is this a corollary of Taylor's theorem?
 
2:24 PM
and I already told you to repeat (almost word to word) the proof for [a,a+h] to prove for [a-h,a].
 
Ok, let me try then...
 
@Koro Is Luffy beating Goku now?
 
@Koro would you advice me to remember this form as well?
 
note that all this can be avoided if
18 mins ago, by Koro
if you don't like this, then consider Taylor series for $f(x)= e^{-x}$ for x>0.
 
@Koro how I don't get that at all?!
 
2:26 PM
ok so write Taylor series for f(x) using your theorem (in your post).
 
@Koro would you advice me to remember this form as well? Also in an exam , should I do all this?
@Koro ok
Done
You mean for $f(x)=e^x$ or in general any $f(x)$ @Koro?
 
suppose g(x)= e^x. You know the series for g(x) if x>0. You want to know the series of e^x if x<0 and this is given by f(x) when x>0.
@冥王Hades not a fan of Goku anymore. DBS is probably cancelled now.
 
@Koro Ok, then I assume the series for $e^{-x}$ is represented by $f(x)$
 
yes
 
@Koro fine
 
2:29 PM
Arcs after universal tournament were boring @冥王Hades
 
@Koro Then?
 
@冥王Hades Kizaru is super strong too. :-)
@ThomasFinley take your time. See the comments again.
 
@Koro you told to prove something bout convergence which I didn't get :/
 
answers are already there in the comments.
 
24 mins ago, by Koro
show that the Taylor series of f converges to f with the theorem you wrote in your post.
@Koro You talking about this?
 
2:33 PM
yes, with f(x)= e^{-x},x>0
 
But I dont get this thing?!
 
what don't you get?
 
Do you mean that $f$ converges to $g(-x)$ ?
 
f(x) =g(-x).
 
@Koro meaning of " $f$ converges to $f$ "
 
2:35 PM
@ThomasFinley I didn't say that. I said 'Taylor series of f converges to f'.
 
@Koro yes, but I should prove that $f(x)=g(-x)$ but how?
@Koro I dont know what $f$ is ?
 
2 mins ago, by Koro
yes, with f(x)= e^{-x},x>0
 
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 $x\gt 0$ ...
 
if i have two matrices $A,B$ with $A$ constant, is there a compact way to write the gradient of their frobenius product?
 
@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.
 
2:44 PM
@Koro how did you write (2) ? You essentially ised the analogous taylor seies in [a-h,a]?
Wait, is $x\gt 0$ ?
If so, then the normal Maclaurin's Formula for $[0,x]$ yields $(2)$ .
 
yes, that's what I have been trying to tell you
 
@Koro My goodness! Then we can wholly do the thing by just assuming $f(x)=e^{-x}$ and $x\gt 0$
Then, as $f$ satisfies the criterial of Taylor's Theorem, we expand it in $[a,a+h]$
Then, we take, $a=0,h=x$
Finally, we get, $f(x)=f(0)+xf'(0)+...+R_n$ and since, $R_n\to 0$
as $n\to\infty$ so, we can say, $1-x+x^2/2-...=f(x)=e^{-x}$ and we are done, aren't we @Koro?
 
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